# Geometric Complexity Theory - Lie Algebraic Methods for Projective Limits of
Stable Points

Bharat Adsul Computer Science Department, Indian Institute of Technology,
Mumbai, India<EMAIL_ADDRESS>Milind Sohoni Computer Science Department,
Indian Institute of Technology, Mumbai, and Indian Institute of Technology,
Goa, India<EMAIL_ADDRESS>K. V. Subrahmanyam Chennai Mathematical
Institute, Chennai, India<EMAIL_ADDRESS>
(31st December, 2021)

## Abstract

Let $G$ be a connected reductive algebraic group over ${\mathbb{C}}$, with Lie
algebra ${\cal G}$, acting rationally on a complex vector space $V$, and the
corresponding projective space $\mathbb{P}V$. Let $x\in V$ and let $H\subseteq
G$ be its stabilizer and ${\cal H}\subseteq{\cal G}$, its Lie algebra. Our
primary objective is to understand the points $[y]$, and their stabilizers,
which occur in the vicinity of $[x]$ in $\mathbb{P}V$.

Towards this, we construct an explicit Lie algebra action of ${\cal G}$ on a
suitably parametrized neighbourhood of $x$. As a consequence, we show that the
Lie algebras of the stabilizers of points in the neighbourhood of $x$ are
parameterized by subspaces of ${\cal H}$. When ${\cal H}$ is reductive, our
results imply that these are in fact, Lie subalgebras of ${\cal H}$. If the
orbit of $x$ were closed this would also follow from a celebrated theorem of
Luna [Lun73].

To construct our Lie algebra action we proceed as follows. We identify the
tangent space to the orbit $O_{x}$ of $x$ with a complement of ${\cal H}$ in
${\cal G}$. Let $N$ be any linear subspace of $V$ that is complementary to
$TO_{x}$, the tangent space to the orbit $O_{x}$. We construct an explicit map
$\theta:V\times N\rightarrow V$ which captures the action of the nilpotent
part of ${\cal H}$ on $V$. We show that the map $\theta$ is intimately
connected with the local curvature form of the orbit at that point. We call
this data, of a neighbourhood of $x$ with the Lie algebra morphism from ${\cal
G}$ to vector fields in this neighbourhood, a local model at $x$.

The action of $G$ on $V$ extends to an action on ${\mathbb{P}}(V)$. We
illustrate the utility of the local model in understanding when
$[x]\in{\mathbb{P}}V$ is in the projective orbit closure of
$[y]\in{\mathbb{P}}V$ via two applications.

The first application is when $V=Sym^{d}(X)$, the space of forms in the
variables $X:=\\{x_{1},\ldots,x_{k}\\}$. We consider a form $f$ (as the “$y$”
above) and elements in its orbit given by $f(t)=A(t)\cdot f$, where $A(t)$ is
an invertible family in $GL(X)$, with $t\in{\mathbb{C}}$. We express $f(t)$ as
$f(t)=t^{a}g+t^{b}f_{b}+\mbox{ higher terms}$, with $a<b$. We call $g\in
Sym^{d}(X)$, the leading term as the limit point (the ”$x$” above) and $f_{b}$
as the direction of approach. Let ${\cal K}$ be the Lie algebra of the
stabilizer of $f$ and ${\cal H}$ the Lie algebra of the stabilizer of $g$. The
local analysis gives us a flattening ${\cal K}_{0}$ of ${\cal K}$ as a
subalgebra of ${\cal H}$, thereby connecting the two stabilizers. There is a
natural action of ${\cal H}$ on $\overline{N}=V/TO_{g}$. We show that ${\cal
K}_{0}$ also stabilizes $\overline{f_{b}}\in\overline{N}$. We show that there
is an $\epsilon$-extension Lie algebra ${\cal K}(\epsilon)$ of ${\cal K}_{0}$
whose structure constants are incrementally closer to ${\cal K}$. The
extendability of ${\cal K}_{0}$ to ${\cal K}(\epsilon)$ translates into
certain Lie algebra cohomology conditions.

We then specialize to the projective orbit closures $\overline{O(f)}$ of forms
$f$ whose $SL(X)$-orbits are affine, and where the form $g$ is such that
$\overline{O(g)}$ is of co-dimension $1$ in $\overline{O(f)}$. When $g$ is
obtained as the limit of a 1-PS $\lambda(t)$, we examine the tangent of exit,
i.e., the direction along which the path $\lambda(t)\cdot f$ leaves $f$. We
show that (i) either ${\cal H}$ has a very simple structure, or (ii)
conjugates of the elements of ${\cal K}$, the stabilizer of $f$ also stabilize
$g$ and the tangent of exit. This is the triple stabilizer condition.

The second application looks at the space $Mat_{n}(\mathbb{C})$ of $n\times
n$-matrices under conjugation by $GL_{n}$. We show that for a general
diagonalizable matrix $X$, the signature of nilpotent matrices in the
projective closure of its orbit are determined by the multiplicity data of the
spectrum of $X$.

Finally, we formulate the path problem of finding paths with specific
properties from $y$ to its limit points $x$. This connects the algebraic study
of the neighborhood of $y$, with the local differential geometry of the orbit
thereby allowing an optimization formulation.

The questions we study are motivated from Geometric complexity theory, an
approach to solve the fundamental lower bound problems in complexity theory.
This theory was proposed by the second author and Ketan Mulmuley in [MS01].

##  1 Introduction

Let $G$ be a connected reductive algebraic group over ${\mathbb{C}}$ and let
$\rho:G\rightarrow End(V)$ be a rational representation of $G$ on a complex
vector space $V$. We say that $G$ acts on $V$ via the representation $\rho$.
Let $x,y\in V$. Denote by $O(y)$ or $G\cdot y$ the orbit $O(y)=\\{\rho(g)\cdot
y|g\in G\\}$ of $y$ in $V$ under the given representation. In general the
orbit is not Zariski closed, it is only constructible. Clearly $G$ also acts
on projective space ${\mathbb{P}}(V)$. For a nonzero $v\in V$, let $[v]$
denote the line corresponding to $v$ in ${\mathbb{P}}(V)$.

A fundamental problem of invariant theory is that of obtaining a good
description of the $G$ orbit closures of points $y\in V$ and deciding whether
a point $x$ belongs to the orbit closure of $y$. These problems go back to
Hilbert and are of fundamental importance in the construction of moduli
spaces, see [MFK94]. The ubiquitous appearance of this problem in many areas
of mathematics is also surveyed in the introduction of [Pop09].

We recall some definitions from Geometric Invariant Theory which we will use
throughout the paper. Standard references for this are [MFK94] and [Dol03].
Unless stated otherwise, we work over $\mathbb{C}$, the field of complex
numbers.

We say a point $v\in V$ is unstable (or equivalently, is in the nullcone) for
the action of $G$ if $\overline{Gv}$ contains $0$. A point which is not
unstable is said to be semistable. The semistable locus is open in $V$. A
point is said to be polystable if its orbit is closed in the open set of
semistable points. If, furthermore, the stabilizer in $G$ of a polystable
point has dimension 0, we say it is stable. We use the same adjectives, viz.,
unstable, polystable and stable for the projective point $[v]$, based on the
status of a representative point $v\in[v]$.

The central motivation for our investigation arises from computational
complexity, which is the study of algorithms to compute functions, typically
polynomials and forms. In Appendix A we give a brief introduction to algebraic
complexity theory and the fundamental lower bounds problems therein. We also
discuss the geometric complexity theory (GCT) approach to these problems, as
proposed by Mulmuley and Sohoni [MS01],[MS08]. In this formulation, lower
bound problems from algebraic complexity translate to determining the
membership of special points $x$ within the projective orbit closure of $y$,
where $y$ and $x$ have distinctive stabilizers. In this setting one of the
central problems is to determine the orbit closure in
${\mathbb{P}}(Sym^{n}(\mathbb{C}^{n^{2}})^{*})$ of the determinant, as a
polynomial of degree $n$ in $n^{2}$ variables and to determine the stabilizers
which arise in this orbit closure. This problem, in turn, is closely connected
to determining strata in orbit closures, and their classification.

Determining projective closures is intimately connected to many classical
problems. For example, when $y$ is an unstable point in $V$, determining the
projective limits of $[y]$ requires an analysis of the stratification of the
nullcone as a projective variety, [Hes79]. This subsumes, e.g., Hilbert’s work
on classifying null forms, and the combinatorial structure of the variety of
nilpotent matrices.

Another situation where projective closures arise is in the analysis of
leading terms. Consider, for example, a finite dimensional $SL(n)$-module $V$,
and let $\lambda:\mathbb{C}^{*}\rightarrow SL(n)$ be a one parameter subgroup
(1-ps) of $SL(n)$. Let the action of $\lambda$ on a point $y\in V$ be given
by:

$\lambda(t)\cdot y=\sum_{j}t^{j}y_{j}.$

Let $x$ be the least degree term i.e., $x=y_{k}$ where $y_{k}\neq 0$,
$y_{i}=0$ for all $i<k$. We call $x$ the leading term of $y$ under $\lambda$.
Then it is easily seen that $[x]$ is in the orbit closure of $[y]$ in
projective space ${\mathbb{P}}(V)$. Whence, projective orbit closures contain
all such limit points which arise by the action of 1-ps on $[y]$. Thus, while
the orbit of $y$ may be closed in $V$, the orbit $[y]\in\mathbb{P}V$ is almost
always not closed, and picks up points from the nullcone which are intimately
connected to it.

We give a brief review of what is known (to the best of our knowledge) about
the affine $G$-orbit closures of points in $V$ and the projective $G$-orbit
closure of points in ${\mathbb{P}}(V)$.

The set of unstable points in $V$ is described by the Hilbert-Mumford theory,
see [MFK94, Kem78]. Every unstable point is driven to $0$ by a one parameter
subgroup. In [Kem78], Kempf showed that for every unstable point $y$, there is
a canonical parabolic subgroup and an optimal one parameter subgroup which
drives $y$ to zero. The corresponding computational question remains far from
solved, except in a few special cases, see [BGO+18], [GGOW19], [DM17],
[IQS17].

Hesselink [Hes79] gave a description of the the strata of the nullcone in
terms of Kempf’s optimal 1-PS. These strata are locally closed, irreducible
and $G$-invariant. Popov [Pop03] gave a combinatorial algorithm to determine
the strata.

When $x$ is polystable it follows from Matsushima [Mat60] that the stabilizer
$H$ of $x$ is reductive. In this case Luna’s étale slice theorem gives
information about stabilizers in a Zariski neighbourhoods of $x$. Luna’s
theorem asserts the existence of an $H$-stable affine variety $N_{x}\subset V$
passing through $x$ and a $G$-invariant open neighbourhood $U$ of the orbit
$Gx$ in $V$, having the following properties. Let $G\times^{H}N_{x}$ be the
fibre bundle over $G/H$ with fibre $N_{x}$(see Section 3.2 for definitions).
Luna’s theorem states that the map $\psi:G\times^{H}N_{x}\rightarrow U$,
sending $[g,v]$ to $gv$ is strongly étale. It follows from Luna’s theorem that
the every point in a Zariski neighbourhood of $x$ in $V$ has a stabilizer
which is contained in a conjugate of $H$. So, if $H$ does not contain a
conjugate of the stabilizer of $y$, then $x$ cannot belong to the affine orbit
closure of $y$.

Of particular interest in GCT is the study of forms $f\in Sym^{n}(W^{*})$ with
reductive stabilizers and their orbit closures. In this connection, we have
Matsushima’s result [Mat60] that when $f$ has a reductive stabilizer, the
$SL(W)$-orbit of $f\in V$ is an affine variety. Whence, the closure in
projective space $\overline{O([y])}\backslash O([y])$ is either zero or is
pure of codimension one. Thus, there are forms $g_{1},\ldots,g_{k}$ whose
orbits constitute the closure.

The action of $GL(W)$ on $Sym^{n}(W^{*})$ can be extended to a natural action
of $GL(W)\otimes\mathbb{C}((t))$ on $Sym^{n}(W^{*})$, where $\mathbb{C}((t))$
denotes the Laurent series in $t$. Hilbert [Hil93] showed that for any $[g]$
in the orbit closure of $[f]$ there is a $\sigma\in
GL(W)\otimes\mathbb{C}((t))$ such that $P:=[f]\circ\sigma$ satisfies
$(P)_{t=0}=[g]$. In [LL89] (see also [BLMW11, Section 9.4], [HL16, Section
7.1]), this was reproved. They show the existence of a $\tilde{\sigma}\in
End(W)\otimes\mathbb{C}[t]$ such that
$f\circ\tilde{\sigma}=t^{K}g\>mod\>(t^{K+1})$. Thus, leading terms under
1-parameter substitutions do lead to all points in the closure, and therefore
also the $g_{i}$ above.

Although the above results tell us how a form $g$ arises in the closure of a
form $f$, how is one to determine these forms? In GCT, however one starts with
forms $f$ which are characterized by their stabilizers and one expects the
problem to be tractable. Even so very little is known. For example,
determining the codimension one components of the boundary of the determinant
form is an important open problem.

The orbit closure in projective space of a form $f$ as above can in principle
be determined by resolving the indeterminacy locus of a certain rational map.
This is outlined in [BLMW11] and makes clear the connections to Geometric
invariant theory. In [Hüt17], the author works this out in the case of the
determinant form of a 3 x 3 matrix and shows that there are precisely two
codimension one components, thereby proving a conjecture of Landsberg[Lan15].

In Popov [Pop09] the author gives a constructive algorithm to determine if $x$
is in the orbit closure of $y$. He does this in both, the affine as well as
the projective setting. This establishes the decidability of the orbit closure
membership problem in the sense of computability theory.

###  1.1 Our results and the organization of the paper.

In the spirit of Luna [Lun73], in this paper we develop a model for a
neighbourhood of $x$, when $x$ has a non trivial stabilizer. We aim to study
the set of such points $y$ for which $x$ may appear as a limit point.
Moreover, in the sense of [MS01], we consider points $x$ and $y$ which have
distinctive stabilizers. While the stabilizer $K\subseteq G$ of $y$ is
reductive, the stabilizer $H$ of $x$ is typically not reductive.

To describe our approach it will be useful to think of $G$ as a complex Lie
group with Lie algebra ${\cal G}$, see [Lee01]. Let $H=\\{g|\rho(g)\cdot
x=x\\}$ be the stabilizer of $x$ in $G$ and ${\cal H}$ its Lie algebra. Our
main technical construction, Theorem 2.7, is an explicit Lie algebra
homomorphism from ${\cal G}$ to vector fields at points in a neighbourhood of
$x$. As a consequence we show that the Lie algebras of the stabilizers of
points in this neighbourhood are parameterized by subspaces of ${\cal H}$.
When $x$ has a reductive stabilizer, our results imply that the Lie algebras
of stabilizers of points in the neighbourhood are Lie subalgebras of ${\cal
H}$. Such a conclusion would also follow from Luna’s theorem for polystable
$x$.

To describe the ${\cal G}$-action, we start with $N$, a linear subspace of $V$
that is complementary to $TO_{x}$, the tangent space to the orbit $O_{x}$. Let
$M$ be any locally closed submanifold of $G$ passing through $e$ that is
complementary to $H$. Then $M\times N$ is a locally closed submanifold of
$G\times V$ containing $(e,0)$. Next, the map $G\times V\to G\times^{H}V$ is
${\cal G}$-equivariant, and the locally closed embedding of $M\times N\to
G\times V$ gives us a quotient ${\cal G}$ action on $M\times N$. The map from
$G\times^{H}V$ to $V$ sending $[g,v]\rightarrow g(x+v)$ when restricted to
$M\times N$ gives us a ${\cal G}$-equivariant diffeomorphim of a neighbourhood
$W$ of $(e,0)\in M\times N$ to a neighbourhood of $x$ in $V$. An important
step in this paper is an explicit description of this action of ${\cal G}$ on
$W$ in terms of the projections $W\to M$ and $W\to N$ coming from the product
$M\times N$. An essential component of this construction, described in Section
2.2, is a map $\theta:V\times N\rightarrow V$ which captures the action of the
nilpotent part of ${\cal H}$. In Remark 2.11 we show that this map is
intimately related to the second fundamental form at $x$. We call this data,
of a neighbourhood of $x$ with an explicitly defined ${\cal G}$-action, a
local model at $x$. We develop the local model in Section 2.

We illustrate the utility of the local model in two applications. In Section 3
we apply this to the situation $V=Sym^{d}(X)$, the space of degree $d$ forms
in the variables $X$. Let $f$ and $g$ be two forms with stabilizer (Lie
algebras) ${\cal K}$ and ${\cal H}$ respectively. We consider a form $f$ (as
the “$y$” above) and elements in its orbit, of form $f(t)=A(t)\cdot f$, where
$A(t)$ is an invertible family in $GL(X)$. We express $f(t)$ as
$f(t)=t^{a}g+t^{b}f_{b}+\mbox{ higher terms}$, with $a<b$. We call $g\in
Sym^{d}(X)$, the leading term as the limit point (and “$x$” above) and $f_{b}$
as the tangent of approach. We construct a basis
$\\{\mathfrak{k}_{i}(t)=\mathfrak{h}_{i}(t)+\mathfrak{s}_{i}(t)\\}_{i=1}^{dim({\cal
K})}$ of $f(t)$ for generic $t$, with the coefficients of
$\mathfrak{h}_{i}(t)$ belonging to ${\cal H}$. In Theorem 3.13 we relate the
stabilizer ${\cal H}$ of $g$ with ${\cal K}$, the stabilizer of $f$ as
follows. The above data gives us an action of ${\cal H}$ on
$\overline{N}=V/TO_{g}$. We show that there is a Lie subalegbra ${\cal
K}_{0}\subset{\cal H}$ of the same dimension as ${\cal K}$ such that ${\cal
K}_{0}$ stabilizes $\overline{f_{b}}\in\overline{N}$. Furthermore ${\cal
K}_{0}$ is the Lie subalgebra spanned by $\\{{\mathfrak{h}}_{i}(0)\\}$.

It turns out that there is an $\epsilon$-extension Lie algebra ${\cal
K}(\epsilon)$ of ${\cal K}_{0}$ whose structure constants are incrementally
closer to ${\cal K}$. The extendability of ${\cal K}_{0}$ to ${\cal
K}(\epsilon)$ translates into certain cohomology conditions. We discuss this
connection to Lie algebra cohomology in Section 6.

In Section 4 we continue our discussion of forms and consider the case when
$g$ is the limit of a 2 block 1-parameter subgroup acting on $f$ (see Section
for definition). The grading we have on the underlying space by assuming so
allows us to prove stronger results. In Proposition 4.10 we show that either
there is an element which stabilizes $f,g$ and the tangent of approach (i.e, a
triple stabilizer), or the limiting Lie algebra ${\cal K}_{0}$ is nilpotent.
In Proposition 4.13 we extend this to the case when $g$ is the limit of a
general $1$-ps acting on $f$.

In Section 5 we illustrate the results of Section 4.1 when $f$ has a reductive
stabilizer and the orbit of $g$ is of codimension $1$ in the projective
closure of $f$. In particular we give details when $f$ is the determinant of a
$3\times 3$ matrix.

In Section 7 we apply the local model to $Mat_{n}(\mathbb{C})$, the vector
space of $n\times n$-matrices under conjugation by $GL_{n}$. We show in
Theorem 7.22 that for a general diagonalizable matrix $X$, the signature of
nilpotent matrices in the projective closure of its orbit are determined by
the multiplicity data of the spectrum of $X$.

In Section 8 we begin a study of differential geometry at the point $y$ in the
local model. We believe that differential geometric techniques may be useful
to further relate properties of the point $y$, the tangent vector of exit from
$y$ and the ultimate limit $x$. This is illustrated by an example of the
cyclic shift matrix in Section 8.3, which requires us to compute the
Reimannian curvature tensor. We also make connections to Kempf’s theorem in
Proposition 8.6.

Acknowledgements: The second author would like to thank Prof Ketan Mulmuley,
Computer Science Department, University of Chicago, for many discussions in
the period 1999-2002 during which the $\theta$ map was discovered. The third
author was supported by a MATRICS grant from the Department of Science and
Technology, India and by a grant from the Infosys foundation.

##  2 The local model

Let $G$ be a complex algebraic, connected, reductive group, with Lie algebra
${\cal G}$, acting on a complex vector space $V$, by a rational representation
$\rho:G\rightarrow GL(V)$. Our interest is in a local model for the action of
${\cal G}$, the Lie algebra of $G$, on a neighbourhood of a point $x\in V$,
with stabilizer $H$, and Lie algbra ${\cal H}$. Note that $\rho:G\rightarrow
GL(V)$ induces a Lie algebra map $\rho:{\cal G}\rightarrow End(V)$.

An elementary description of a neighborhood of $x$ is as follows.

  * •

The orbit $O(x)=G\cdot x$ is a smooth manifold of dimension $dim(G)-dim(H)$.

  * •

Let ${\cal S}\subseteq{\cal G}$ be a complement to ${\cal H}$ within ${\cal
G}$, i.e., ${\cal G}={\cal S}\oplus{\cal H}$. Then

${\cal S}\cdot x=\\{\rho(\mathfrak{s})\cdot x|\mathfrak{s}\in{\cal S}\\}$

is the tangent space $T_{x}O(x)$ to the orbit $O(x)$.

  * •

Let $N\subseteq T_{x}V$ be a complement to $T_{x}O\subseteq T_{x}V$, such that
$T_{x}V=T_{x}O\oplus N={\cal S}\cdot x\oplus N$. In any chosen neighborhood
$U\subseteq V$ of $x$, we identify $T_{u}V$ with $V$, for all $u\in U$. We may
choose a neighborhood $U$ of $x$ such that for all $u\in U$, $T_{u}V={\cal
S}\cdot u\oplus N$.

We would like to describe the action of any $\mathfrak{g}\in{\cal G}$ in this
neighborhood $U$ in terms of the above division of $T_{u}V={\cal S}\cdot
u\oplus N$. In particular, we will choose $u=x+n$, i.e., the section $x+N$, as
the domain of interest. The specification of $\rho(\mathfrak{g})$ on such a
section may be suitably extended to sections at other points $x^{\prime}\in
O(x)$ in the vicinity of $x$ and therefore to a neighborhood of $x$. We show
that the action of $\mathfrak{g}$ on this section is better understood by (i)
using the decomposition ${\cal G}={\cal S}\oplus{\cal H}$ to express
$\mathfrak{g}\in{\cal G}$ as $\mathfrak{g}=\mathfrak{s}+\mathfrak{h}$ with
$\mathfrak{s}\in{\cal S}$ and $\mathfrak{h}\in{\cal H}$, and (ii) obtaining
the descriptions of $\mathfrak{s}$ and $\mathfrak{h}$ separately, on the
section $x+N$. This description is the local model and it relies on the
following basic definition and theorem on the formation of quotients.

###  2.1 The quotient model

###### Definition 2.1

Let ${\cal X}$ be a smooth manifold and let $Vec({\cal X})$ be the Lie algebra
of smooth vector fields on ${\cal X}$. Let ${\cal G}$ be a Lie algebra. We say
that ${\cal X}$ is a ${\cal G}$-manifold via $\gamma$ if there is a Lie
algebra homomorphism $\gamma:{\cal G}\rightarrow Vec({\cal X})$. We denote
this by the data $({\cal G},\gamma,{\cal X})$.

Next, let $({\cal G},\gamma,{\cal X})$ and $({\cal G},\rho,{\cal Y})$ be two
${\cal G}$-manifolds. A map $\mu:{\cal X}\rightarrow{\cal Y}$ is called ${\cal
G}$-equivariant iff for all $x\in{\cal X}$ and $\mathfrak{g}\in{\cal G}$, we
have $\mu^{*}_{x}(\gamma(\mathfrak{g})(x))=\rho(\mathfrak{g})(\mu(x))$, where
$\mu^{*}_{x}:T_{x}{\cal X}\rightarrow T_{\mu(x)}{\cal Y}$ is the tangent map
corresponding to the map $\mu$.

We define ${\cal D}_{\mu}(x)\subseteq T_{x}{\cal X}$ as those elements $\Delta
x\in T_{x}{\cal X}$ such that $\mu^{*}_{x}(\Delta x)=0$. Thus ${\cal D}_{\mu}$
records the kernel of the map $\mu^{*}$ at every $x\in{\cal X}$.

###### Theorem 2.2

Let $({\cal G},\gamma,{\cal X})$ and $({\cal G},\rho,{\cal Y})$ be two ${\cal
G}$-manifolds of dimensions $m$ and $n$, respectively. Let $\mu:{\cal
X}\rightarrow{\cal Y}$ be ${\cal G}$-equivariant. Suppose that at the point
$x\in{\cal X}$, we have $\mu^{*}_{x}(T_{x}{\cal X})=T_{\mu(x)}{\cal Y}$.
Moreover, suppose that ${\cal M}$ is a locally closed submanifold of ${\cal
X}$ of dimension $m-n$ such that $x\in{\cal M}$ and the tangent space
$T_{x}(M)$ is transverse to ${\cal D}_{\mu}$ at $x$. Then:

  1. 1.

There is an open submanifold ${\cal M}^{\prime}\subseteq{\cal M}$ containing
$x$ such that for all $m\in{\cal M}^{\prime}$, the tangent space $T_{x}{\cal
M}^{\prime}$ is complementary to ${\cal D}_{\mu}(m)$.

  2. 2.

There is a corresponding ${\cal G}$-action $\gamma_{{\cal M}^{\prime}}$ on
${\cal M}^{\prime}$ such that the map $\mu:{\cal M}^{\prime}\rightarrow{\cal
Y}$ is a ${\cal G}$-equivariant diffeomorphism in a neighborhood of $x$.

  3. 3.

Moreover, for $m\in{\cal M}^{\prime}$, since we have $T_{m}{\cal X}=T_{m}{\cal
M}^{\prime}\oplus{\cal D}_{\mu}(m)$, let $\pi_{\cal M^{\prime}}:T_{m}{\cal
X}\rightarrow T_{m}{\cal M}^{\prime}$ be the corresponding projection. Then we
have:

$\gamma_{{\cal M}^{\prime}}(\mathfrak{g})(m)=\pi_{{\cal
M}^{\prime}}(\gamma(\mathfrak{g})(m))\mbox{ for all $\mathfrak{g}\in{\cal G}$
and $m\in{\cal M}^{\prime}$}$

For the definition of a ${\cal G}$-manifold see [AM95], [Mic08]. The proof of
the theorem is straightforward. It allows the construction of a local quotient
model ${\cal M}^{\prime}$ within ${\cal X}$ of the space ${\cal Y}$.

###### Definition 2.3

Let $G$ act on $V$ via $\rho$ as before. The $G$-space $G\times V$ is the
collection of tuples $(g,v)$ with $g\in G$ and $v\in V$. The action $\gamma$
of $G$ on $G\times V$ is given by $\gamma(g^{\prime})(g,v)=(g^{\prime}g,v)$.
Let $x\in V$ be a point with stabilizer $H\subseteq G$, as above. The map
$\mu:G\times V\rightarrow V$ is given by $\mu(g,v)=\rho(g)(x+v)$ (abbreviated
as $g\cdot(x+v)$).

###### Lemma 2.4

The following assertions hold:

  1. (i)

The map $\mu$ is $G$-equivariant.

  2. (ii)

Let $v\in V$, then the tangent space $T_{(e,v)}(G\times V)$ is
$(\mathfrak{g},\Delta v)$, where $\mathfrak{g}\in{\cal G}$ and $\Delta v\in
T_{v}V\cong V$.

  3. (iii)

The tangent map is given as follows: $\mu^{*}_{(e,v)}(\mathfrak{g},\Delta
v)=\mathfrak{g}\cdot(x+v)+\Delta v$. Whence, the distribution ${\cal
D}_{\mu}(e,v)=\\{(\mathfrak{g},\Delta v)|\mathfrak{g}\cdot(x+v)+\Delta
v=0\\}$.

  4. (iv)

$\mu^{*}_{(e,0)}=\mathfrak{g}\cdot x+\Delta v$. Thus $\mu^{*}$ is surjective
at $(e,0)$.

The proof is straightforward.

###### Proposition 2.5

Let $N$ be a vector space complement to $TO(x)_{x}$ within $V$. Let ${\cal S}$
be a complement to ${\cal H}$ within ${\cal G}$. Let $M$ be a locally closed
submanifold of $G$ of dimension $dim(G)-dim(H)$ such that $e\in M$ and
$T_{e}M={\cal S}$. Then there is an open submanifold $M^{\prime}\subseteq M$
containing $e$, such that ${\cal G}$ acts on $M^{\prime}\times N\subseteq
G\times V$ via $\gamma^{\prime}$. This action is derived from the action
$\gamma$ of ${\cal G}$ on $G\times V$ modulo the distribution ${\cal
D}_{\mu}$. Moreover, for this neighborhood of $M^{\prime}\times N$ containing
$(e,0)$, we have:

  * (i)

The map $\mu:M^{\prime}\times N\rightarrow V$ is a ${\cal G}$-equivariant
diffeomorphism from the space $({\cal G},\gamma^{\prime},M^{\prime}\times N)$
to $({\cal G},\rho,V)$. It maps a neighborhood of $[e,0]\in M^{\prime}\times
N$ to a neighborhood of $x\in V$.

  * (ii)

For any element $\mathfrak{g}=\mathfrak{s}+\mathfrak{h}$, with
$\mathfrak{s}\in{\cal S}$ and $\mathfrak{h}\in{\cal H}$, the action of
$\gamma^{\prime}(\mathfrak{g})$ at the point $(e,n)\in M^{\prime}\times N$ is
given by

$\gamma^{\prime}(\mathfrak{g})(e,n)=(\mathfrak{s}+\mathfrak{s}^{\prime},n^{\prime})\in
T_{(e,n)}(M^{\prime}\times N)$

where $\mathfrak{s}^{\prime}\in{\cal S}$ and $n^{\prime}\in N$ are unique and
satisfy the equation

$\mathfrak{s}^{\prime}\cdot(x+n)+n^{\prime}=\mathfrak{h}\cdot n.$

Proof: Let us verify that the conditions required by Theorem 2.2 hold for the
map $\mu$. From lemma 2.4, (i) and (iv), $\mu$ is $G$-equivariant and
$\mu^{*}$ is surjective at $(e,0)$. By the choice of $M$ and $N$, we see that
$\mu^{*}(e,0)(M\times N)={\cal S}\cdot x+N=V$. Thus $T_{(e,0)}M\times N$ is
transverse to ${\cal D}_{\mu}$. Thus Theorem 2.2 applies at the point $(e,0)$
and a suitable action $\gamma^{\prime}$ on $M^{\prime}\times N$ is available
for which $\mu$ is an equivariant diffeomorphism. Thus (i) is proved. We also
have that $M^{\prime}\times N$ is transverse to the distribution $D_{\mu}$. To
prove (ii), we must make an explicit computation of $\gamma^{\prime}$ at the
points $(e,n)\in M^{\prime}\times N\subseteq G\times V$ modulo the
distribution $D_{\mu}$. For this, we must evaluate
$\pi_{M^{\prime}}:T_{(e,n)}(G\times V)\rightarrow T_{(e,n)}(M\times N)$. We do
this in two steps:

$T_{(e,n)}G\times V\rightarrow T_{(e,n)}M\times V\rightarrow T_{(e,n)}M\times
N$

From Lemma 2.4(ii), for a typical element
$\mathfrak{g}=\mathfrak{s}+\mathfrak{h}$, we have
$\gamma(\mathfrak{g})(e,n)=(\mathfrak{s}+\mathfrak{h},0)\in T_{(e,n)}(G\times
V)$. We observe that for any $\mathfrak{h}\in{\cal H}$, we have

$\mu^{*}_{(e,n)}(\mathfrak{h},0)=\mathfrak{h}\cdot(x+n)=\mathfrak{h}\cdot
n=\mu^{*}_{(e,n)}(0,\mathfrak{h}\cdot n)$

Thus, we have:

$\gamma_{M}(\mathfrak{s}+\mathfrak{h})(e,n)=(\mathfrak{s}+\mathfrak{h},0)=(\mathfrak{s},\mathfrak{h}\cdot
n)\in T_{(e,n)}M\times V\mbox{ (modulo ${\cal D}_{\mu}$)}$

Thus, we have reduced the first component from ${\cal G}$ to ${\cal
S}=T_{e}M$.

Using Lemma 2.4(iii) it follows that

$\mu^{*}_{(e,n)}(\mathfrak{s},\mathfrak{h}\cdot
n)=\mathfrak{s}\cdot(x+n)+\mathfrak{h}\cdot n\in T_{(x+n)}V$

Consider the vector $\mathfrak{h}\cdot n\in T_{(x+n)}V$. Since $\mu$ is a
local diffeomorphism between $(e,0)\in M\times N$ and $x\in V$, we may assume
that $\mu$ is regular at $x+n$. Whence, there must be a unique
$\mathfrak{s^{\prime}}\in{\cal S}$ and $n^{\prime}\in N$ such that for
$(\mathfrak{s^{\prime}},n^{\prime})\in T_{(e,n)}M\times N$, we have
$\mu^{*}_{(e,n)}(\mathfrak{s}^{\prime},n^{\prime})=\mathfrak{h}\cdot n$, or in
other words:

$\mathfrak{s}^{\prime}\cdot(x+n)+n^{\prime}=\mathfrak{h}\cdot n.$

Hence

$\mu_{*(e,n)}(\mathfrak{g}\cdot[e,n])=\mu^{*}_{(e,n)}(\mathfrak{s},\mathfrak{h}\cdot
n)=\mathfrak{s}\cdot(x+n)+\mathfrak{h}\cdot
n=(\mathfrak{s}+\mathfrak{s}^{\prime})\cdot(x+n)+n^{\prime}=\mu^{*}_{(e,n)}(\mathfrak{s}+\mathfrak{s^{\prime}},n^{\prime})$

Since $(\mathfrak{s}+\mathfrak{s^{\prime}},n^{\prime})\in T(M\times
N)_{(e,n)}$, it must be the projection of $(\mathfrak{s},\mathfrak{h}\cdot n)$
to $T_{(e,n)}(M\times N)$, and thus from Theorem 2.2(3) it describes the
action of $\mathfrak{g}$ on $M\times N$ at the point $(e,n)\in M\times N$ .
This proves (ii). $\Box$

###  2.2 The $\theta$ map.

It is now clear that the computation modulo the distribution $D_{\mu}$ at
$(e,n)$ essentially revolves around the solution to the following equation in
$V$, where ${\Delta v}\in V\cong T_{(x+n)}V$ is a given and
${\mathfrak{s}}\in{\cal S}$, ${\mathfrak{n}}\in N$ are the unknowns:

${\mathfrak{s}}\cdot(x+n)+{\mathfrak{n}}=\Delta v.$

Since $V={\cal S}\cdot(x+n)\oplus N$, we may define projection $\lambda_{\cal
S}(n):V\rightarrow{\cal S}$ and $\lambda_{N}(n):V\rightarrow N$ so that the
above equation may be written as:

$\Delta v=(\lambda_{\cal S}(n)(\Delta v))(x+n)+(\lambda_{N}(n)(\Delta v))$

We now solve the above equation, with $n$ as a parameter. We note that for
$n=0$, the equation is easily solved by our choice of $N$ and the isomorphism
$V=T_{x}V=T_{x}O(x)\oplus N$ with $T_{x}O(x)={\cal S}\cdot x$. We may write
the projections $\lambda_{\cal S}(0):V\rightarrow{\cal S}$ and
$\lambda_{N}(0):V\rightarrow N$, or simply $\lambda_{\cal S},\lambda_{N}$,
such that

$\lambda_{\cal S}(\Delta v)\cdot x+\lambda_{N}(\Delta v)=\Delta v.$

By adding an “error term” $\lambda_{\cal S}(\Delta v)\cdot n$ on both sides,
we get

$\lambda_{\cal S}(\Delta v)\cdot(x+n)+\lambda_{N}(\Delta v)=\Delta
v+\lambda_{\cal S}(\Delta v)\cdot n.$

For a fixed $n\in N$, we may now define $\theta(n):V\rightarrow V$, as
follows: for $\Delta v\in V$, we define $\theta(n)(\Delta v)=\lambda_{\cal
S}(\Delta v)\cdot n$. We see that $\theta(n):V\rightarrow V$ is a linear
operator $V$ which depends linearly on $n$. Thus, the above equation may be
rewritten as:

$\lambda_{\cal S}(\Delta v)\cdot(x+n)+\lambda_{N}(\Delta v)=\Delta
v+\theta(n)(\Delta v)$

Substituting $\theta^{i}(\Delta v)$ for $\Delta v$ in the above equaltion
gives us the $i$-th approximation term as:

$\lambda_{\cal S}(\theta^{i}(n)(\Delta
v))\cdot(x+n)+\lambda_{N}(\theta^{i}(n)(\Delta v))=\theta^{i}(n)(\Delta
v)+\theta^{i+1}(\Delta v).$

Since $\theta(n)$ depends linearly on $n$, we may assume that $n$ is small
enough and hence all the eigen-values of $\theta(n)$ may be assumed to be less
than $1$ in magnitude. Thus the system,

${\mathfrak{s}}\cdot(x+n)+{\mathfrak{n}}=\Delta v,$

has the solution:

$[\lambda_{\cal S}\circ(1-\theta(n)+\theta^{2}(n)-\cdots)](\Delta
v)\cdot(x+n)+[\lambda_{N}\circ(1-\theta(n)+\theta^{2}(n)-\cdots)](\Delta
v)=\Delta v.$

Thus, we have proved the following basic lemma:

###### Lemma 2.6

For any element $\Delta v\in T_{(x+n)}V$, there are unique
$\mathfrak{s}\in{\cal S}$ and $n^{\prime}\in N$ such that:

$\mathfrak{s}\cdot(x+n)+n^{\prime}=\Delta v$

Moreover $\mathfrak{s}=\lambda_{\cal S}\circ((1+\theta(n))^{-1}(\Delta v))$
and $n^{\prime}=\lambda_{N}\circ(1+\theta(n))^{-1}(\Delta v)$. Thus, if ${\cal
I}(n):V\rightarrow V$ is the map $(1+\theta(n))^{-1}$, then $\lambda_{\cal
S}(n)=\lambda_{\cal S}\circ{\cal I}(n)$ and
$\lambda_{N}(n)=\lambda_{N}\circ{\cal I}(n)$.

We will use the above calculations and Proposition 2.5 to prove our main
theorem.

Recall that $H$ stabilizes the point $x$. We may choose a Levi decomposition
$H=RQ$, where $Q$ is the unipotent radical of $H$ and $R$ is a reductive
complement. Let ${\cal R}$ denote the Lie algebra of $R$ and ${\cal Q}$ denote
the Lie algebra of $Q$.

We may choose $N$ in the above proposition to be an $R$-invariant complement
to the space $O_{x}$ inside $V$. We may also choose ${\cal S}$ to be an
$R$-invariant complement to ${\cal H}$ with ${\cal G}$. Thus ${\cal S}$ now
acquires an $R$-module structure.

###### Theorem 2.7

Let $\rho$ be a representation of $G$ on a vector space $V$. Let $x\in V$ be a
point with stabilizer $H$. Let $H=RQ$ be a Levi decomposition and ${\cal
H},{\cal Q}$ and ${\cal R}$ be the Lie algebras of $H,Q$ and $R$,
respectively. Then there is a submanifold $M\times N\subset G\times_{H}V$ in
the neighbourhood of $(e,0)$ with an action of ${\cal G}$. The map
$\mu:M\times N\rightarrow V$ given by $\mu(g,n)=g\cdot(x+n)$ is a ${\cal
G}$-equivariant diffeomorphism which maps a neighborhood of $(e,0)\in M\times
N$ to a neighborhood of $x\in V$. We then have:

  1. 1.

Let $\mathfrak{g}$ be a typical element of ${\cal G}$. Let
$\mathfrak{g}=\mathfrak{s}+\mathfrak{r}+\mathfrak{q}$, with
$\mathfrak{s}\in{\cal S},\mathfrak{r}\in{\cal R}$ and $\mathfrak{q}\in{\cal
Q}$. Then the action $\gamma^{\prime}$ of $\mathfrak{g}$ on $M\times N$ is
given as:

$\mathfrak{g}\circ(e,n)=(\mathfrak{s}+\lambda_{\cal
S}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot n),\>\mathfrak{r}\cdot
n+\lambda_{N}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot n))$

  2. 2.

If ${\cal G}_{n}\subseteq{\cal G}$ is the stabilizer of $x+n\in V$, then there
is a subspace ${\cal H}_{n}\subseteq{\cal H}$ of the same dimension as ${\cal
G}_{n}$ such that for every $\mathfrak{g}\in{\cal G}_{n}$, there is a unique
$\mathfrak{h}\in{\cal H}_{n}$ with $\mathfrak{h}=\mathfrak{q}+\mathfrak{r}$
such that:

$\mathfrak{r}\cdot n+\lambda_{N}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot
n)=0$

Moreover, $\mathfrak{g}=\mathfrak{h}+\mathfrak{s}$, where $\mathfrak{s}$ is as
given below:

$\mathfrak{s}=-\lambda_{\cal S}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)$

The element $\mathfrak{g}$ is called the ${\cal S}$-completion of
$\mathfrak{h}$ at the point $x+n$.

Proof: Let us express $\mathfrak{g}=\mathfrak{s}+\mathfrak{h}$ and apply Prop.
2.5, part (ii). We then have that
$\gamma^{\prime}(\mathfrak{g})(e,n)=(\mathfrak{s}+\mathfrak{s^{\prime}},n^{\prime})$,
where $\mathfrak{s}^{\prime}$ and $n^{\prime}$ solve the equation:

$\mathfrak{h}\cdot n=\mathfrak{s}^{\prime}(x+n)+n^{\prime}$

By lemma 2.6, these are precisely $\lambda_{\cal
S}(1+\theta(n)^{-1}(\mathfrak{h}\cdot n)$ and
$n^{\prime}=\lambda_{N}(1+\theta(n))^{-1}(\mathfrak{h}\cdot n)$. Now, breaking
$\mathfrak{h}$ as $\mathfrak{h}=\mathfrak{q}+\mathfrak{r}$, and recalling that
$\mathfrak{r}\cdot N\subseteq N$, we see $\lambda_{N}(\mathfrak{r}\cdot
n)=\mathfrak{r}\cdot n$, and $\lambda_{\cal S}(\mathfrak{r}\cdot n)=0$. Thus
$\theta(n)(\mathfrak{r}\cdot n)=0$ and thus
$\lambda_{N}(1+\theta(n))^{-1}(\mathfrak{r}\cdot n+\mathfrak{q}\cdot
n)=\lambda_{N}(\mathfrak{r}\cdot
n)+\lambda_{N}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)=\mathfrak{r}\cdot
n+\lambda_{N}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)$. Similarly,
$\lambda_{\cal S}(1+\theta(n))^{-1}(\mathfrak{r}\cdot n+\mathfrak{q}\cdot
n)=\lambda_{\cal S}(\mathfrak{r}\cdot n)+\lambda_{\cal
S}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)=\lambda_{\cal
S}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)$.

This proves (1).

The hypothesis that $\mathfrak{g}\in{\cal G}_{n}$ is equivalent to the claim
that $\gamma^{\prime}(\mathfrak{g})(e,n)=0$. This requires both the components
of the action of $\mathfrak{g}\cdot(e,n)$ to vanish. This implies that
$\mathfrak{r}\cdot n+\lambda_{N}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot
n)=0$. This is the first assertion.

Now, given that $\mathfrak{h}=\mathfrak{q}+\mathfrak{r}$ satisfies the above,
$\mathfrak{s}$ is uniquely determined by the requirement that the first
component vanishes, i.e., $\mathfrak{s}+\lambda_{\cal
S}\circ(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)$ must vanish. This determines
$\mathfrak{s}$ as required.

Finally, if there were $\mathfrak{g}^{\prime}=\mathfrak{h}+\mathfrak{s}$ and
$\mathfrak{g}^{\prime}=\mathfrak{h}+\mathfrak{s}^{\prime}$ both elements of
${\cal G}_{n}$, then we would have $\mathfrak{s}-\mathfrak{s^{\prime}}\in{\cal
G}_{n}$. Whence,
$(\mathfrak{s}-\mathfrak{s}^{\prime})\cdot(e,n)=(\mathfrak{s}-\mathfrak{s}^{\prime},0)\neq
0$ which violates the requirement that the first component vanishes. $\Box$

###### Remark 2.8

Consider the case when the stabilizer $H$ is reductive, whence $R=H$ and
$Q=0$. We may select an $N$ which is $H$-invariant as above. Since ${\cal
Q}=0$, we see that $\mathfrak{q}\cdot n=0$, and thus, the ${\cal
S}$-completion of any element $\mathfrak{h}\in H$ is $\mathfrak{h}$ itself.
Thus the stabilizer ${\cal G}_{n}$ of $x+n$ must be a subalgebra of ${\cal
H}$. This is the Lie algebraic version of Luna’s slice theorem.

We now prove certain equivariant properties of maps closely related to
$\theta$. We continue with the Levi decomposition of $H=RQ$ and the choice of
$N$ and ${\cal S}$.

The Lie algebra action defined as a map $\rho:{\cal G}\rightarrow End(V)$ may
well be defined as $\hat{\rho}:{\cal G}\otimes V\rightarrow V$. Note that
${\cal G}$ is an $R$-module under the adjoint action, and thus so is ${\cal
G}\otimes V$. Then we have the following lemma:

###### Lemma 2.9

The map $\hat{\rho}$ is $R$-equivariant. Next, let $T_{x}O$ be the tangent
space of the orbit of $x$ under the action of $G$ and ${\cal S}$ be an
$R$-invariant complement to ${\cal H}\subseteq{\cal G}$. Then the map
$\alpha:{\cal S}\rightarrow T_{x}O$ given by
$\alpha(\mathfrak{s})=\rho(\mathfrak{s})\cdot x$ is an isomorphism of
$R$-modules.

Proof. For the first assertion, note that for $r\in R,v\in V$ and
$\mathfrak{g}\in{\cal G}$, we have:

$r\cdot\hat{\rho}(\mathfrak{g}\otimes v)=r\cdot\rho(\mathfrak{g})\cdot
r^{-1}\otimes r\cdot v=r\cdot(\rho(\mathfrak{g})\cdot v)$

Thus $\hat{\rho}$ is indeed $R$-equivariant. Next, the map $\alpha:{\cal
S}\rightarrow T_{x}O$ is merely the restriction to $\hat{\rho}$ to the tensor
product of two $R$-submodules, viz., ${\cal S}\subset{\cal G}$ and
$\mathbb{C}\cdot x\subseteq V$. This proves the second assertion. $\Box$

Recall that $\theta(n):V\rightarrow V$ is defined as the composite:

$\theta:V\stackrel{{\scriptstyle\lambda_{\cal S}}}{{\rightarrow}}{\cal
S}\stackrel{{\scriptstyle\cdot n}}{{\rightarrow}}V$

We define two associated maps: $\Theta:V\otimes N\rightarrow V$ and
$\Phi:{\cal S}\times N\rightarrow{\cal S}$ as:

$\Theta(v\otimes n)=\theta(n)(v)=\lambda_{\cal S}(v)\cdot n$

and:

$\Phi(\mathfrak{s}\otimes n)=\lambda_{\cal S}(\mathfrak{s}\cdot n)$

Then, we have the following:

###### Proposition 2.10

The maps $\Theta$ and $\Phi$ are $R$-equivariant.

Proof: $\Theta$ is merely the composite:

$V\otimes N\stackrel{{\scriptstyle\lambda_{\cal S}\otimes{\tt
id}}}{{\longrightarrow}}{\cal S}\otimes
N\stackrel{{\scriptstyle\hat{\rho}}}{{\longrightarrow}}V$

As for $\Phi$, it is:

${\cal S}\otimes
N\stackrel{{\scriptstyle\hat{\rho}}}{{\longrightarrow}}V\stackrel{{\scriptstyle\lambda_{\cal
S}}}{{\longrightarrow}}{\cal S}$

This proves the proposition. $\Box$

###### Remark 2.11

The map $\Phi$ above is the algebraic analogue of the Second Fundamental
Tensor at the point $x$ for the orbit $O(x)$ regarded as a submanifold of $V$.
Indeed, suppose that $G$ is the orthogonal group and $V$ is equipped with a
$G$-invariant inner product. Let $N$ be chosen orthogonal to $T_{x}O$, the
tangent space of the orbit $O(x)$ at $x$. Let $v_{1},\ldots,v_{K}$ be an
orthonormal basis for $N$. Let $\mathfrak{s}_{1},\ldots,\mathfrak{s}_{L}$ be
chosen such that the elements $\\{v_{K+i}=(\mathfrak{s}_{i}\cdot
x)|i=1,\ldots,L\\}$ is an orthonormal basis for $T_{x}O$. The data for $\Phi$
is the tuple $\\{(\alpha_{ij}^{r})|i,j=1,\ldots,L,r=1,\ldots,K\\}$, where we
have:

$\lambda_{\cal S}(\mathfrak{s}_{i}\cdot
v_{r})=\sum_{j}\alpha_{ij}^{r}(\mathfrak{s}_{j}\cdot x)$

Now, one must recall that $\rho(\mathfrak{s}_{i})$ may be given as a linear
vector field as follows:

$\rho(\mathfrak{s}_{i})=\sum_{j=1}^{K+L}\sum_{k=1}^{K+L}S_{i}(j,k)v_{k}\frac{\partial}{\partial
v_{j}}$

where $S_{i}\in\mathbb{C}^{(K+L)\times(K+L)}$. We abbreviate this as
$\rho(\mathfrak{s}_{i})=\nabla S_{i}\overline{v}$, where $\nabla$ is the row
vector $[(\partial/\partial v_{1}),\ldots,(\partial/\partial v_{K+L})]$ and
$\overline{v}$ is the column vector $[v_{1},\ldots,v_{K+L}]^{T}$.

Note that $\rho$ preserves inner products implies that $S_{i}^{T}=-S_{i}$. The
data $\alpha_{ij}^{r}$ may then be obtained as:

$\alpha_{ij}^{r}=(S_{j}x)^{T}(S_{i}v_{r})=-x^{T}S_{j}S_{i}v_{r}$

We also have the Second Fundamental Form of the orbit $O(x)$ as a map
$\Pi:T_{x}O\times T_{x}O\rightarrow N$ defined as follows. For $v_{i},v_{j}\in
T_{x}O$, let $X_{i},X_{j}$ be vector fields on $O(x)$ such $X_{i}(x)=v_{i}$
and $X_{j}(x)=v_{j}$. Then
$\Pi(v_{i},v_{j})=\lambda_{N}(D_{X_{i}}(X_{j}))=\sum_{r=1}^{K}\beta_{ij}^{r}v_{r}$.
Now, we have $\rho(\mathfrak{s}_{i})$ available to us as the $X_{i}$ above.
Then, it is easily checked that $D_{X_{i}}(X_{j})$ is given $\nabla
S_{j}S_{i}\overline{v}$. Whence the evaluation of $\Pi(v_{i},v_{j})$ at the
point $x$ and in the direction $v_{r}$ is given by
$\beta_{ij}^{r}=v_{r}^{T}S_{j}S_{i}x$. However, this is equal to
$v_{r}^{T}S_{i}S_{j}x+v_{r}^{T}[S_{j},S_{i}]x$. Since there is a
$\mathfrak{g}=[\mathfrak{s}_{i},\mathfrak{s}_{j}]\in{\cal G}$ such that
$\rho(\mathfrak{g})=\nabla[S_{j},S_{i}]\overline{v}$, this evaluated at $x$
must be in the tangent space $T_{x}O$, and thus $v_{r}^{T}[S_{j},S_{i}]x=0$.
Whence

$\beta_{ij}^{r}=v_{r}^{T}S_{i}S_{j}x=x^{T}S_{j}^{T}S_{i}^{T}v_{r}=x^{T}S_{j}S_{i}v_{r}=-\alpha_{ij}^{r}$

This proves the claim. $\Box$

We explore further uses of the Reimannian curvature tensor to study projective
orbit closures in Section 8.2.

###  2.3 Applying the local model

We close this section with a general application of the local model to
projective limits of a particular type, and a simple concrete example.

###### Proposition 2.12

Let $V$ be a $G$-module, where $G\subseteq GL(X)$, for a suitable $X$ and let
$p,q\in V$. Let $I$ be the identity matrix and $tI$ act on $p$ as $tI\cdot
p=t^{d}p$. Let $A(t)\subseteq G$ be an algebraic family of elements in $G$
such that:

$A(t)\cdot q=t^{k}\cdot p+\mbox{ higher degree terms }$

Suppose that the stabilizer of $p$ is $H$. Then the local model at $p$
contains a scalar multiple of a $G$-conjugate of $q$ which is of the form
$p+n$, i.e., appears on the normal slice $N$ at $p$.

Proof. Consider the family $A^{\prime}(t)=t^{-k/d}\cdot A(t)$, i.e., whose
elements are scalar multiples of $A(t)$ and thus are elements of $G$. Applying
this family to $q$, we see that:

$A^{\prime}(t)\cdot q=p+\mbox{ positive powers of $t$}$

Thus, $\lim_{t\rightarrow 0}A^{\prime}(t)q=p$, and for every $\epsilon>0$,
there is a $q_{0}=A^{\prime}(t_{0})\cdot q$, i.e., a scalar multiple of a
$G$-conjugate of $q$ such $\|q_{0}-p\|<\epsilon$. Thus, in the local model
$\mu:M\times N\rightarrow V$, we have a $(g,n)$ such that $g\cdot(p+n)=q_{0}$.
Applying $g^{-1}$ on both sides, we see that $g^{-1}q_{0}=p+n$. This proves
the claim. $\Box$

Let us consider a small example to illustrate the computational ease of the
local model.

###### Example 2.13

Consider $sl(2)$ given as the matrix below acting of $\mathbb{C}\cdot X$,
where $X=\\{x,y\\}$. Let $V=Sym^{2}(X)$ with the ordered basis
$B=\\{x^{2},xy,y^{2}\\}$. The action of the generic group element is also
shown below:

$sl(2)=\left\\{g_{a,b,c}=\left[\begin{array}[]{cr}a&b\\\
c&-a\end{array}\right]\mbox{ with }a,b,c\in\mathbb{C}\right\\}$

On $Sym^{2}(X)$, it may be useful to interpret this as the operation:

$a(x\frac{\partial}{\partial x}-y\frac{\partial}{\partial
y})+by\frac{\partial}{\partial x}+cx\frac{\partial}{\partial y}$

Whence:

$g_{a,b,c}\cdot\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]=\left[\begin{array}[]{ccc}2a&2b&0\\\ c&0&b\\\
0&2c&-2a\end{array}\right]\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]$

Let us consider three forms $p_{0}=0,p_{1}=x^{2}$ and $p_{2}=x^{2}+y^{2}$ and
apply the local model to each.

  1. 1.

For $p_{0}=0$, we see that ${\cal H}_{0}$, the stabilizer of $p_{0}$ is the
complete group $sl(2)$. Thus ${\cal S}=0$, the orbit is $0$-dimensional, and
the maps $\theta$ is $0$. The only choice for $N$ is $V$, and for any $n$, the
stabilizer ${\cal G}_{n}\subseteq{\cal H}_{0}=sl(2)$.

  2. 2.

Consider next $p_{2}=x^{2}+y^{2}$. We see that the stabilizer ${\cal H}_{2}$
is the group $O\subset sl(2)$, given below. A complementary subspace is also
shown.

$O=\left\\{\left[\begin{array}[]{cr}0&b\\\
-b&0\end{array}\right]|b\in\mathbb{C}\right\\}\>\>\>{\cal
S}=\left\\{\left[\begin{array}[]{cr}a&c\\\
c&-a\end{array}\right]|a,c\in\mathbb{C}\right\\}$

Since ${\cal S}$ is two dimensional, so is the orbit. Applying ${\cal S}$ to
$p_{2}$ gives $T_{p_{2}}O=\mathbb{C}\cdot(x^{2}-y^{2})+\mathbb{C}\cdot xy$.
Thus a normal $n$ to the orbit is $x^{2}+y^{2}$, i.e., $p_{2}$ itself. Next,
we compute $\theta(n):V\rightarrow V$. For this, we must first compute
$\lambda_{S}$, the ${\cal S}$-parametrization of the projection of
$V\rightarrow T_{p_{2}}O$, and then apply this to $n$. But since $n=p_{2}$,
this makes $\theta(n)$ a projection onto $T_{p_{2}}O$. On the other hand,
since ${\cal H}_{2}$ is reductive, ${\cal Q}=0$, and the stabilizer of
$p_{2}+n$, is $O_{2}$.

  3. 3.

Let us consider next $p_{1}=x^{2}$. The stabilizer ${\cal H}_{1}$ and ${\cal
S}$ are given below:

${\cal H}_{1}=\left\\{\left[\begin{array}[]{cr}0&0\\\
c&0\end{array}\right]|c\in\mathbb{C}\right\\}\>\>\>{\cal
S}=\left\\{\left[\begin{array}[]{cr}a&b\\\
0&-a\end{array}\right]|a,c\in\mathbb{C}\right\\}$

We see that ${\cal S}\cdot p_{1}=2ax^{2}+2bxy$, and thus, we may choose
$n=y^{2}$. Note that ${\cal R}=0$ and ${\cal Q}={\cal H}$. Let us choose
$\mathfrak{q}$ as the element $g_{0,0,1}$, and note that $\mathfrak{q}\cdot
n=2xy$. Let us compute $\theta(n)$ next. Since $\lambda_{\cal S}(y^{2})=0$, we
see:

$\lambda_{\cal S}\cdot\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]=\left[\begin{array}[]{c}g_{1/2,0,0}\\\ g_{0,1/2,0}\\\
0\end{array}\right]\>\>\mbox{and thus
}\theta(n)\cdot\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]=\left[\begin{array}[]{ccc}0&0&-1\\\ 0&0&0\\\
0&0&0\end{array}\right]\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]$

This gives us:

$(1+\theta(n))^{-1}\cdot\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]=\left[\begin{array}[]{ccc}1&0&1\\\ 0&1&0\\\
0&0&1\end{array}\right]\left[\begin{array}[]{c}x^{2}\\\ xy\\\
y^{2}\end{array}\right]$

Whence $\lambda_{N}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)=0$ and thus ${\cal
H}_{n}={\cal H}_{1}$. The ${\cal S}$-completion is given by the condition that
$\mathfrak{s}+\lambda_{\cal S}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)=0$.
Since $\mathfrak{q}\cdot n=2xy$, we get $\lambda_{\cal
S}(1+\theta(n))^{-1}(\mathfrak{q}\cdot n)=g_{0,1,0}$. So
$\mathfrak{s}=-g_{0,-1,0}$, and thus, the ${\cal S}$-completion of the element
$g_{0,0,1}$ is $g_{0,-1,1}$, the generator of $O$. Thus, the stabilizer of
$p_{2}=p_{1}+y^{2}$ matches the one computed earlier.

##  3 Forms and their limits

We will now specialize the above local model to the space of forms, and
examine projective limits when they arise as leading terms under a 1-parameter
family. This question is central to the framing of several questions in
computational complexity theory. In Section 3.1 we start the investigation
with a general 1-parameter family. Let $g$ be the limit of a form $f$ under
the action of such a 1-parameter family $A(t)$. So $g$ is the leading term of
$A(t)\cdot f$. We show a connection (Theorem 3.13) between the stabilizer of
$f$ at a generic value of the parameter $t$, the stabilizer of $g$ and the
stabilizer of second leading term of $A(t)\cdot f$, what we call the tangent
of approach. In Section 4 we consider the case when the 1-parameter family has
more structure, and is in fact a 1-parameter subgroup (1-PS). In this case one
can exploit the fact that the stabilizers come with a natural grading to get a
tighter relation between the stabizers of $f,g$ and the stabilizers of the
tangent of approach and the tangent of exit.

###  3.1 Stabilizers of Limits of Forms under a 1-parameter family

Let $X$ be a set of indeterminates and let $V$ be the $GL(X)$-module
$Sym^{d}(X)$ and $f,g\in V$ be non-zero forms. Let ${\cal G}$ be the algebra
$gl(X)$, ${\cal K}\subseteq{\cal G}$ be the stabilizer of $f$, and its
dimension be $k$. Let ${\cal H}$ be the Lie algebra stabilizing $g$ and its
dimension be $r$.

Suppose that we have an algebraic family $A(t)\subseteq GL(X)$, parametrized
by $t$ such that $A(1)=e$, the identity element. Let

$f(t)=A(t).f=t^{a}g+t^{b}f_{b}+t^{b+1}f_{b+1}+\ldots+t^{D}f_{D},$

By normalizing the family $A(t)$, as in the proof of Prop. 2.12, and using
suitable powers of $t$, we may assume that:

$f(t)=A(t).f=g+t^{b}f_{b}+t^{b+1}f_{b+1}+\ldots+t^{D}f_{D},$

with $f_{b}\neq 0$. We will write $f^{+}(t)$ for the sum
$\sum_{i=b}^{D}t^{i}f_{i}$. Thus $\lim_{t\rightarrow 0}f(t)=g$ and
$\lim_{t\rightarrow 0}t^{-b}f^{+}(t)=f_{b}$. We call $f_{b}$ as the tangent of
approach.

###### Assumption 3.1

Transversality Assumption. We assume that the vector space spanned by
$f_{b},\ldots,f_{D}$ intersects $T_{g}O(g)$ trivially, where $T_{g}O(g)$ is
the tangent space of the $GL(X)$-orbit $O(g)$ at point $g$.

Let ${\cal S}$ be a complement to ${\cal H}$ in ${\cal G}$. Note that the
tangent space $T_{g}O(g)$ of the $GL(X)$-orbit of $g$ is precisely ${\cal
S}\cdot g$ and is of the same dimension as ${\cal S}$. Let $N$ be a complement
to $T_{g}O(g)$ in $TV_{g}\cong V$. By the transversality assumption, we may
assume that $f^{+}(t)\in N$ for all $t$. This completes the data required to
compute the local model.

For a Lie algebra element $\mathfrak{g}=\mathfrak{s}+\mathfrak{h}$, with
$\mathfrak{s}\in{\cal S}$ and $\mathfrak{h}\in{\cal H}$, let us compute the
action of $\mathfrak{g}$ on $f(t)=g+f^{+}(t)$. Identifying the tangent space
$TV_{g}$ with ${\cal S}\oplus N$ as in Section 2.2, we have:

$(\mathfrak{s}+\mathfrak{h})\cdot f(t)=(\mathfrak{s}+\lambda_{\cal
S}((1+\theta(f^{+}(t)))^{-1}(\mathfrak{h}\cdot
f^{+}(t)),\lambda_{N}((1+\theta(f^{+}(t)))^{-1}(\mathfrak{h}\cdot f^{+}(t)).$
(1)

Thus, for a generic $t_{0}\in\mathbb{C}$, every element $\mathfrak{k}\in{\cal
K}(t_{0})$, the Lie algebra of the stabilizer of $f(t_{0})$, may be expressed
as $\mathfrak{k}=\mathfrak{h}+\mathfrak{s}$ with $\mathfrak{h}$ satisfying the
condition below:

$\lambda_{N}((1+\theta(f^{+}(t_{0})))^{-1}((\mathfrak{h}\cdot
f^{+}(t_{0}))=0.$ (2)

and $\mathfrak{s}$, the ${\cal S}$-completion of $\mathfrak{h}$ at the point
$f^{+}(t_{0})$. We then have the following:

###### Lemma 3.2

For a generic $t_{0}\in\mathbb{C}$ the stabilizer ${\cal K}(t_{0})$ of
$f(t_{0})$ is conjugate to the stabilizer ${\cal K}$ of $f$. The dimension of
${\cal K}(t_{0})$ is also $k$.

Proof: If $t_{0}$ is so chosen so that $A(t_{0})$ is invertible, then
$f(t_{0})$ is in the $GL(X)$-orbit of $f$ and the assertion follows. $\Box$

Our objective is to construct a uniform basis
$\\{\mathfrak{k}_{1}(t),\ldots,\mathfrak{k}_{k}(t)\\}$ so that for a generic
$t_{0}$, the elements $\mathfrak{k}_{1}(t_{0}),\ldots,\mathfrak{k}_{k}(t_{0})$
is a $\mathbb{C}$-basis for ${\cal K}(t_{0})$.

Let $u_{1},\ldots,u_{m}$ be a basis of $N$ and let us extend it by the
elements $u_{m+1}=\mathfrak{s}_{1}\cdot g,\ldots,u_{m+p}=\mathfrak{s}_{p}\cdot
g$ to a basis of $TV_{g}=V$, where $\mathfrak{s}_{1},\ldots,\mathfrak{s}_{p}$
is a basis of ${\cal S}$. This basis implements the isomorphism
$TV_{g}=N\oplus TO_{g}$. Let us denote these bases by $\mathfrak{u}_{N}$ and
$\mathfrak{u}_{\cal S}$.

Note that $(1+\theta(f^{+}(t)))^{-1}$ is a map $V\rightarrow V$ and we may
represent it as the $(m+p)\times(m+p)$ matrix ${\cal I}(\theta,t)$ with
entries in $\mathbb{C}(t)$ such that
$(1+\theta(f^{+}(t)))^{-1}[\mathfrak{u}_{N},\mathfrak{u}_{\cal
S}]=[\mathfrak{u}_{N},\mathfrak{u}_{\cal S}]\cdot{\cal I}(\theta,t)$.

###### Lemma 3.3

Let $\Delta(t)$ be the determinant of the matrix $I+\theta(f^{+}(t))$ . Then
$\Delta=1+c.t^{2b}+\mbox{ higher degree terms}$ (where $c\in\mathbb{C}$). The
matrix $\Delta\cdot{\cal I}(\theta,t)$ has entries in $\mathbb{C}[t]$ and is
of the following form:

$\Delta\cdot{\cal I}(\theta,t)=I-t^{b}\theta(f_{b})+\text{ higher degree
terms}.$

Proof: By linearity of $\theta$, we have:

$\theta(f^{+}(t))=t^{b}\theta(f_{b})+t^{b+1}\theta(f_{b+1})+\text{ higher
degree terms}.$

Let $J=I+\theta(f^{+}(t))$ in the above basis, then $J_{ij}\in\mathbb{C}[t]$
with the following properties: (i) for $i\neq j$, we have $t^{b}$ divides
$J_{ij}$, and (ii) $t^{b}$ divides $J_{ii}-1$. The first assertion on
$\Delta(t)$, the determinant of $J$, follows. For the second assertion, again
for $i\neq j$ each minor $\tilde{J}_{ji}(t)$ of J is
$-t^{b}\theta(f_{b})_{ij}+\mbox{ higher terms}$, while for $i=j$, we have
$\tilde{J}_{ii}=1-t^{b}\theta(f_{b})_{ii}$ +higher terms. This proves the
second part. $\Box$

Thus, if we express ${\cal I}(\theta,t)$ as a matrix with entries in
$\mathbb{C}[[t]]$, then:

${\cal I}(\theta,t)=I-t^{b}\theta(f_{b})+\mbox{higher terms}.$

Let ${\cal M}_{N}(t):{\cal H}\rightarrow N$ be the map
$\mathfrak{h}\rightarrow\lambda_{N}\circ(1+\theta(f^{+}(t)))^{-1}(\mathfrak{h}\cdot
f^{+}(t))$ and ${\cal M}_{\cal S}(t):{\cal H}\rightarrow{\cal S}$ be the map
$\mathfrak{h}\rightarrow\lambda_{\cal
S}\circ(1+\theta(f^{+}(t)))^{-1}(\mathfrak{h}\cdot f^{+}(t))$. Note that
${\cal M}_{N}$ and ${\cal M}_{\cal S}$ are linear maps which depend on the
parameter $t$. The kernel $ker({\cal M}_{N}(t))$ determines the ${\cal
H}$-part of every element of ${\cal K}(t)$, while ${\cal M}_{\cal S}$ is the
${\cal S}$-completion at the point $f^{+}(t)$.

Let us now compute matrices for $M_{N}$ and $M_{\cal S}$ for ${\cal M}_{N}$
and ${\cal M}_{\cal S}$, respectively.

Let $\mathfrak{h}_{1},\ldots,\mathfrak{h}_{r}$ be a basis of ${\cal H}$ and
let us denote this sequence by $\overline{\mathfrak{b}}$. Let us form the
matrix $M_{N}$, an $m\times r$ matrix with entries in $\mathbb{C}(t)$ as
follows. Let ${\cal M}_{N}(\mathfrak{h}_{i})=\sum_{k=1}^{m}(M_{N})_{ki}u_{k}$.
Similarly, let the $p\times r$-matrix $M_{\cal S}$ be defined as follows: Let
${\cal M}_{\cal S}(\mathfrak{h}_{i})=\sum_{k=1}^{p}(M_{\cal
S})_{ki}\mathfrak{s}_{k}$. We collect this information in the lemma below:

###### Lemma 3.4

In the above notation, we have:

  1. 1.

The matrices $M_{N}(t)$ and $M_{\cal S}(t)$ satisfy the following equations:

$\mathfrak{u}_{N}M_{N}(t)=\lambda_{N}\cdot{\cal
I}(\theta,t)\cdot(\overline{\mathfrak{b}}\cdot f^{+}(t))=[{\cal
M}_{N}(\overline{\mathfrak{b}})]\>\>\mbox{ and }\>\mathfrak{u}_{\cal S}M_{\cal
S}(t)=\lambda_{\cal S}\cdot{\cal
I}(\theta,t)\cdot(\overline{\mathfrak{b}}\cdot f^{+}(t))$

  2. 2.

Let ${\cal K}(t_{0})$ be the stabilizer of $f(t_{0})$ for some generic
$t_{0}\in\mathbb{C}$ and let $\mathfrak{k}\in{\cal K}(t_{0})$. Express
$\mathfrak{k}$ as $\mathfrak{k}=\mathfrak{h}+\mathfrak{s}$ and let
$\mathfrak{h}=\overline{\mathfrak{b}}\cdot\overline{\alpha}$, i.e., a linear
combination $\overline{\alpha}$ of the basis elements
$\overline{\mathfrak{b}}$. Then

$\mathfrak{u}_{N}M_{N}(t_{0})\overline{\alpha}=0\mbox{ and }\mathfrak{s}\cdot
g=-\mathfrak{u}_{\cal S}M_{\cal S}(t_{0})\overline{\alpha}.$

  3. 3.

Conversely, any element $\mathfrak{k}=\mathfrak{h}+\mathfrak{s}$ with
$\mathfrak{h}=\overline{\mathfrak{b}}\cdot\overline{\alpha}$, satisfying the
above conditions, is an element of ${\cal K}(t_{0})$.

Proof: All the assertions follow from the construction of $M_{N}$ and $M_{\cal
S}$ and the requirements of Eqn. 1. $\Box$

###### Lemma 3.5

  1. (i)

Let $\mathfrak{R}\subseteq\mathbb{C}(t)$ be the localization of
$\mathbb{C}[t]$ at $\Delta$. In other words,
$\mathfrak{R}=\\{p/\Delta^{d}|p\in\mathbb{C}[t]\mbox{ and }d\geq 0\\}$. Then,
the entries of $M_{N}(t)$ and $M_{\cal S}(t)$ are elements of $\mathfrak{R}$.

  2. (ii)

$t^{b}$ divides the matrix $M_{\cal S}(t)$ and $M_{N}(t)$ in $\mathfrak{R}$.
In particular, $M_{\cal S}(0)=0$.

  3. (iii)

For any generic $t_{0}\in\mathbb{C}$, the rank of $M_{N}(t_{0})$ is a constant
and equals the $\mathbb{C}(t)$-rank of $M_{N}(t)$. This equals $dim({\cal
H})-dim({\cal K})$, i.e., $r-k$.

  4. (iv)

The $\mathbb{C}(t)$-rank of $M_{\cal S}(t)$ is zero iff ${\cal H}\cdot
f(t)\subseteq N$.

Proof: The matrix $M_{N}(t)$ is the product of operations expressed in
appropriate bases, viz., (a) $\lambda_{N}:V\rightarrow N$, (b),
$(1+\theta(f^{+}(t))^{-1}:V\rightarrow V$ and (c)
$(\overline{\mathfrak{b}}\cdot f^{+}(t)):{\cal H}\rightarrow V$. Let us look
at each one in turn. The first is a projection and is a matrix over
$\mathbb{C}$ of rank $m+p$. For (b), this is precisely the matrix ${\cal
I}(\theta,t)$ which we have shown has entries in $\mathfrak{R}$. Moreover, we
know that ${\cal I}(\theta,0)$ is the identity matrix. Finally, since ${\cal
H}$ acts linearly on $V$ and $t^{b}$ divides $f^{+}(t)$, the matrix
$(\overline{\mathfrak{b}}\cdot f^{+}(t))$ has entries in
$t^{b}\cdot\mathbb{C}[t]$. Thus the product, viz., $M_{N}(t)$ has entries in
$\mathfrak{R}$ and $t^{b}$ divides it (in $\mathfrak{R}$). The same reasoning
holds for $M_{\cal S}$ as well. This proves (i) and (ii).

For (iii), by lemma 3.4, the dimension of ${\cal K}(t_{0})$ for a generic
$t_{0}$ equals $k$ and this must be the nullity of $M_{N}(t_{0})$. But this is
the same as nullity of the matrix $M_{N}(t)$ as a matrix over $\mathbb{C}(t)$.
Thus $k=r-rank_{\mathbb{C}(t)}(M_{N}(t))=r-rank_{\mathbb{C}}(M_{N}(t_{0}))$.
This proves (iii).

Finally, for (iv), the condition that $M_{\cal S}$ is identically zero is
equivalent to ${\cal M}_{\cal S}$ being zero, which is equivalent to ${\cal
H}\cdot f(t)\subset N$. $\Box$

Lemma 3.5 now allows us to construct ${\cal K}(t)$ as the column annihilator
space of $M_{N}(t)$. Since $rank(M_{N}(t))=r-k$, we may choose column vectors
$\overline{\alpha}_{1}(t),\ldots,\overline{\alpha}_{k}(t)\in\mathbb{C}(t)^{r}$
such that $M_{N}\cdot\overline{\alpha}_{i}=0$. Since the entries of $M_{N}$
are $\mathfrak{R}$, we may assume that these column annihilators may be chosen
over $\mathbb{C}[t]$ and in a standard form as given by the following lemma:

###### Lemma 3.6

Let $A=[\overline{\alpha}_{1},\ldots,\overline{\alpha}_{k}]$ be the $r\times
k$-matrix with entries in $\mathbb{C}(t)$ and the columns
$\overline{\alpha}_{1},\ldots,\overline{\alpha}_{k}$ as above. Then we may
assume that there is another basis (over $\mathbb{C}(t)$)
$\overline{\alpha}^{\prime}_{1},\ldots,\overline{\alpha}^{\prime}_{k}$, for
the column space of $A$ and a matrix
$A^{\prime}=[\overline{\alpha}^{\prime}_{1},\ldots,\overline{\alpha}^{\prime}_{k}]$
such that all entries of $A^{\prime}$ are in $\mathbb{C}[t]$ and
$A^{\prime}(0)$ is of rank $k$. Moreover, if $c$ is any column in the column
space of $A$, such that $c(0)$ is defined, then $c(0)$ is in the column space
of $A^{\prime}(0)$.

Proof: We first note that elementary column operations on $A$ do not change
the corresponding ${\cal K}(t)$. Next, by multiplying by a suitable polynomial
$p(t)$ and pulling out an appropriate power of $t$ from each column, we may
assume that $A$ has entries in $\mathbb{C}[t]$ and that $A(0)$ exists and each
of its column is non-zero. Finally, if a column, say, $\overline{\alpha}_{k}$
of $A$ is such that
$\overline{\alpha}_{k}(0)=\sum_{i=1}^{k-1}\mu_{i}\overline{\alpha}_{i}(0)$,
with $\mu_{i}\in\mathbb{C}$, then we may replace $\overline{\alpha}_{k}$ by
$t^{-a}(\overline{\alpha}_{k}-\sum_{i=1}^{k-1}\mu_{i}\overline{\alpha}_{i})$
for some suitable $a>0$, to get a new column of a lower degree. This process
may be repeated till the assertions in the lemma are satisfied. Let us come to
the second part. Let $R=\\{r_{1},\ldots,r_{k}\\}$ be the row indices such that
$det(A^{\prime}(0)[R])\neq 0$ and let $\Delta(t)=det(A^{\prime}(t)[R])$. Since
$c$ is in the column space of $A^{\prime}$, there are polynomials
$p_{1},\ldots,p_{k}\in\mathbb{C}[t]$ such that
$\Delta(t)c=\sum_{i=1}^{k}p_{i}\overline{\alpha}^{\prime}_{i}$. The result
follows. $\Box$

Henceforth, we assume that the chosen
$\overline{\alpha_{1}},\ldots,\overline{\alpha_{k}}$ have the properties
assured to us by lemma 3.6. This has an immediate corollary.

###### Proposition 3.7

  1. 1.

There is a $\mathbb{C}(t)$-basis $\\{\mathfrak{k}_{i}(t)\\}_{i=1}^{k}$ of
${\cal K}(t)$, the stabilizer Lie algebra of $f(t)$ and a large number $D$
such that

$\mathfrak{k}_{i}(t)=\sum_{j=0}^{D}(\mathfrak{s}_{ij}+\mathfrak{h}_{ij})t^{j}$

for suitable elements $\mathfrak{s}_{ij}\in{\cal S},\mathfrak{h}_{ij}\in{\cal
H}$.

  2. 2.

The elements $\mathfrak{s}_{ij}$ are zero for all $i$, and for all
$j=0,\ldots,b-1$. As a result, we also have:

$\mathfrak{k}_{i}(t)=\mathfrak{h}_{i}(t)+t^{b}\mathfrak{s}_{i}(t)$

where $\mathfrak{h}_{i}(t)\in\mathbb{C}[t]\otimes{\cal H}$ and
$\mathfrak{s}_{i}(t)\in\mathbb{C}[t]\otimes{\cal S}$.

  3. 3.

The space ${\cal
K}_{0}=\mathbb{C}\cdot\\{\mathfrak{k}_{1}(0),\ldots,\mathfrak{k}_{k}(0)\\}$,
the subspace formed by $\mathbb{C}$-linear combinations of the leading terms
$\mathfrak{k}_{i}(0)=\mathfrak{k}_{i0}=\mathfrak{h}_{i0}$ is a Lie subalgebra
of ${\cal H}$ and of dimension $k$. Moreover, if $\mathfrak{k}(t)\in{\cal
K}(t)$ is any element such that $\mathfrak{k}(0)$ is defined, then
$\mathfrak{k}(0)\in{\cal K}_{0}$.

  4. 4.

For any element $\mathfrak{h}\in{\cal K}_{0}$, we have
$\lambda_{N}(\mathfrak{h}\cdot f_{b})=0$ and thus, there is an
$\mathfrak{s}\in{\cal S}$ such that $\mathfrak{s}\cdot g+\lambda_{\cal
S}(\mathfrak{h}\cdot f_{b})=0$

Proof: For (1), let us assume that
$\overline{\alpha}_{1}(t),\ldots,\overline{\alpha}_{k}(t)$ are as in Lemma
3.6. We may then compute the element
$\mathfrak{h}_{i}(t)=\mathfrak{b}\cdot\overline{\alpha}_{i}(t)$. Next, we
define $\mathfrak{s}_{i}(t)$ such that $\mathfrak{s}_{i}(t)\cdot
g=-\mathfrak{u}_{\cal S}\cdot M_{\cal S}\overline{\alpha}_{i}(t)$. Then
$\mathfrak{k}_{i}(t)=\mathfrak{s}_{i}(t)+\mathfrak{h}_{i}(t)$ stabilizes
$f(t)$. Note that these are elements of ${\cal G}\otimes\mathbb{C}[t]$, i.e.,
linear combinations of Lie algebra elements with coefficients in
$\mathbb{C}[t]$. Assertion (1) is an expansion of these as a polynomial in $t$
and $D$ is a bound on the degree of these polynomials. Assertion (2) follows
from the fact that the element
$\mathfrak{s}_{i}(t)=\sum_{j=0}^{D}\mathfrak{s}_{ij}t^{j}$ arises from the
matrix equation $\mathfrak{s}_{i}(t)\cdot g=-\mathfrak{u}_{\cal S}M_{\cal
S}\cdot\overline{\alpha_{i}}(t)$, and that $t^{b}$ divides $M_{\cal S}$ as in
lemma 3.5 (ii).

Let us now come to (3). Since ${\cal K}(t)$ is a Lie algebra, for any two
elements $\mathfrak{k}_{i}(t),\mathfrak{k}_{j}(t)$, we have:

$\begin{array}[]{rcl}[\mathfrak{k}_{i}(t),\mathfrak{k}_{j}(t)]&=&[\mathfrak{k}_{i0},\mathfrak{k}_{j0}]+([\mathfrak{k}_{i0},\mathfrak{k}_{j1}][\mathfrak{k}_{i1},\mathfrak{k}_{j0}])t^{1}+\ldots\\\
\end{array}$

This implies that $\lim_{t\rightarrow
0}[\mathfrak{k}_{i}(t),\mathfrak{k}_{j}(t)]$ exists and equals
$[\mathfrak{k}_{i0},\mathfrak{k}_{j0}]$. Next, note that lemma 3.6 implies
that ${\cal K}(t)$ is a $k$-dimensional subspace of
$gl(X)\otimes\mathbb{C}(t)$ with the basis
$\\{\mathfrak{k}_{1}(t),\ldots,\mathfrak{k}_{k}(t)\\}$ such that
$\\{\mathfrak{k}_{1}(0),\ldots,\mathfrak{k}_{k}(0)\\}$ continue to be linearly
independent over $\mathbb{C}$ in $gl(X)$. Thus we may express
$[\mathfrak{k}_{i}(t),\mathfrak{k}_{j}(t)]$ in this basis as

$\begin{array}[]{rcl}[\mathfrak{k}_{i}(t),\mathfrak{k}_{j}(t)]&=&\sum_{u=1}^{k}\beta^{ij}_{u}(t)\mathfrak{k}_{u}(t),\\\
\end{array}$

where $\beta^{ij}_{u}(t)\in\mathbb{C}(t)$. The condition on $\Delta(t)$ in the
proof of lemma 3.6 give us that $\beta_{ij}(0)$ exist. Taking limits on both
sides gives us:

$\begin{array}[]{rcl}[\mathfrak{k}_{i}(0),\mathfrak{k}_{j}(0)]&=&\sum_{u=1}^{k}\beta^{ij}_{u}(0)\mathfrak{k}_{u}(0).\\\
\end{array}$

This proves that ${\cal K}_{0}$ is a Lie algebra. Note that since
$\mathfrak{s}_{i0}=0$ for all $i$, the elements $\mathfrak{k}_{i0}$ are
elements of ${\cal H}$ and thus ${\cal K}_{0}\subseteq{\cal H}$. The assertion
that $\\{h_{i0}\\}$ are linearly independent, and the second assertion, follow
from Lemma 3.6.

Finally, let us come to the (4). Recall that $\mathfrak{b}$ is the ordered
basis $\\{\mathfrak{h}_{i},\ldots,\mathfrak{h}_{r}\\}$. We see that the
condition that $\mathfrak{h}_{i}(t)=\mathfrak{b}\cdot\overline{\alpha}_{i}(t)$
where $M_{N}\cdot\overline{\alpha}_{i}(t)=0$ is tantamount to asserting that

$\mathfrak{u}_{N}M_{N}\overline{\alpha}_{i}(t)=\lambda_{N}\cdot(1+\theta(f^{+}))^{-1}\cdot(\mathfrak{h}_{i}(t)\cdot
f^{+}(t))=0,$

and, setting $\mathfrak{s}_{i}\cdot g=-\mathfrak{u}_{\cal S}M_{\cal
S}\overline{\alpha}_{i}$ that,

$\mathfrak{s}_{i}(t)\cdot g+\lambda_{\cal
S}\cdot(1+\theta(f^{+}))^{-1}\cdot(\mathfrak{h}_{i}(t)\cdot f^{+}(t))=0.$

Both together imply that $\mathfrak{k}_{i}(t)$ stabilizes $f(t)$. Now
comparing coefficient of the lowest degree, i.e., $t^{b}$, on both sides of
the two equations, gives us:

$\lambda_{N}\cdot\mathfrak{h}_{i0}\cdot f_{b}=0\>\mbox{ and
}\>\mathfrak{s}_{i,b}\cdot g+\lambda_{\cal S}\cdot\mathfrak{h}_{i0}\cdot
f_{b}=0.$

This finishes the proof of the proposition. $\Box$

###### Remark 3.8

Surprisingly, the fact that the next lowest degree term in
$(1+\theta(f^{+}))^{-1}$ is $-t^{b}\theta(f_{b})$ is neither used nor
required.

We now have the following proposition.

###### Proposition 3.9

Let ${\cal K}_{0}$ be as above. Then:

  1. 1.

Let $N\subseteq V$ be as above. Define the $\star$ action of ${\cal H}$ on $N$
as below:

$\mathfrak{h}\star n=\lambda_{N}(\mathfrak{h}\cdot n)$

Then $\star:{\cal H}\times N\rightarrow N$ is a Lie algebra action on $N$ and
matches the action of the quotient module $\overline{N}=V/TO_{g}$. The
isomorphism from $N$ to $\overline{N}$ is given by
$n\rightarrow\overline{n}=n+TO_{g}$.

  2. 2.

The subalgebra ${\cal K}_{0}\subseteq{\cal H}$ stabilizes the element
$\overline{f_{b}}\in\overline{N}$. In other words, ${\cal K}_{0}\subseteq{\cal
H}_{b}\subseteq{\cal H}$, the stabilizer of $\overline{f_{b}}\in\overline{N}$.

Proof: Part (1) is straightforward.

For part (2), note that Prop. 3.7 (4) tells us that
$\lambda_{N}(\mathfrak{h}\cdot f_{b})=0$, for any $\mathfrak{h}\in{\cal
K}_{0}$. Thus, $\mathfrak{h}$ stabilizes $\overline{f_{b}}$. $\Box$

###### Definition 3.10

Let ${\cal K}(t)$ be the stabilizer of $f(t)$ and let
$\mathfrak{k}_{i}(t)=\mathfrak{h}_{i}(t)+\mathfrak{s}_{i}(t)$, for
$i=1,\ldots,k$, be a basis for ${\cal K}(t)$ as above. We define the space
${\cal H}(t)$ as the $\mathbb{C}[t]$-space spanned by
$\\{\mathfrak{h}_{i}(t)|i=1,\ldots,k\\}$. For a $t_{0}\in\mathbb{C}$, let
${\cal H}(t_{0})$ be the $\mathbb{C}$-space spanned by the vectors
$\\{\mathfrak{h}_{i}(t_{0})|i=1,\ldots,k\\}$. Note that, by definition, ${\cal
H}(0)={\cal K}_{0}$.

###### Lemma 3.11

Let $t_{0}\in\mathbb{C}$ be generic and
$\mathfrak{k}=\mathfrak{s}+\mathfrak{h}$ be an element of ${\cal K}(t_{0})$,
then $\mathfrak{h}\in{\cal H}(t_{0})$.

Proof: The assertion is straightforward and follows from the fact that for
generic $t_{0}$, the rank of the matrix $A(t_{0})$ of lemma 3.6 is $k$. $\Box$

We now come to the main theorem of this section which summarizes the above
results. We will recall the definition of ${\cal S}$-completion.

###### Definition 3.12

For any element $\mathfrak{h}\in{\cal H}$, and $n\in N$, we define its ${\cal
S}$ completion as the element $\mathfrak{h}+\mathfrak{s}$, with
$\mathfrak{s}\in{\cal S}$ such that $\lambda_{\cal S}(\mathfrak{h}\cdot
n)=\mathfrak{s}\cdot g$. For a set ${\cal J}\subseteq{\cal H}$, we define the
${\cal S}$-completion of ${\cal J}$ as the collection of all ${\cal
S}$-completions of every element $\mathfrak{j}\in{\cal J}$.

Note that since the map ${\cal S}\rightarrow{\cal S}\cdot g$ is a bijection,
the ${\cal S}$-completion of an element $\mathfrak{h}\in{\cal H}$ always
exists and is unique.

###### Theorem 3.13

Let $A(t)\cdot f=f(t)=g+t^{b}f_{b}+\mbox{ higher terms},$ as above, with
${\cal K}(t)$ of generic dimension $k$ as its stabilizer, and $f_{b}$, the
tangent of approach. Then

  1. (i)

There is a subalgebra ${\cal K}_{0}$ of ${\cal H}_{b}\subseteq{\cal H}$ the
stabilizer of $\overline{f_{b}}$ for the $\star$-action of ${\cal H}$ whose
dimension is the same as the dimension of ${\cal K}(t_{0})$ for a generic
$t_{0}\in\mathbb{C}$.

  2. (ii)

There is a basis $\\{\mathfrak{k}_{i}(t)|i=1,\ldots,k\\}$ of ${\cal K}(t)$
such that $\\{\mathfrak{k}_{i}(0)|i=1,\ldots,k\\}$ is a basis for ${\cal
K}_{0}$. Moreover, ${\cal K}_{0}={\cal H}(0)$.

  3. (iii)

Let ${\cal H}(t)$ be the $\mathbb{C}[t]$-span of the vectors
$\mathfrak{h}_{1}(t),\ldots,\mathfrak{h}_{k}(t)$. Then, for any generic
$t_{0}\in\mathbb{C}$, the subspace ${\cal H}(t_{0})\subseteq{\cal H}$ has
dimension $k$ and ${\cal K}(t_{0})$ is the ${\cal S}$-completion of ${\cal
H}(t_{0})$ for the point $f^{+}(t_{0})$.

Proof: Part (i) and (ii) are shown in Prop. 3.7. Part (iii) follows from lemma
3.11. This proves the theorem. $\Box$

##  4 Projective limits under the action of 1-ps

We specialize the results of the previous section to the case when the
1-parameter family is a 1-parameter subgroup $\lambda(t)$ (1-PS). Each
$\lambda$ effectively partitions the variable set $X$ into several parts
$X=X_{1}\cup\ldots\cup X_{k}$ with $\lambda$ operating on $X_{i}$ simply as
$\lambda(t)(X_{i})=t^{d_{i}}X_{i}$. In Section 4.1 we tackle the two block
case, when the variable set $X$ is partitioned into two parts $Y$ and $Z$,
with the 1-ps acting trivially on $Y$ and scaling by $t$ the variables in $Z$.
Proposition 4.10 shows that in this case either there are elements which
stabilizes $f,g$ and the tangent of approach and exit, i.e., a triple
stabilizer, or the limiting Lie algebra ${\cal K}_{0}$ is nilpotent. Next, we
study the general 1-ps case in Section 4.2. We prove Proposition 4.15, which
is similar in spirit to that proved in the two block 1-ps.

###  4.1 Two block $\lambda$

Let us partition $X$ as $X=Y\cup Z$. Let $\lambda:\mathbb{C}^{*}\rightarrow
GL(X)$ be a 1-PS such that $\lambda(t)y=y$ for all $y\in Y$ but
$\lambda(t)z=tz$. Let us denote by $V_{i}$ the space $Sym^{d-i}(Y)\otimes
Sym^{i}(Z)$ and note that $Sym^{d}(X)=\oplus_{i=0}^{d}V_{i}$. Moreover, for
any $v_{i}\in V_{i}$, we have $\lambda(t)v_{i}=t^{i}v_{i}$. Let $f$ and $g$ be
forms so that:

$\lambda(t)f=f(t)=t^{a}g+t^{b}f_{b}+\ldots+t^{d}f_{d}$

with $f_{b}\neq 0$. In other words, $g\in V_{a}$ is the leading term for the
action of $\lambda$ on $f$ and $f_{b}$ is the tangent of approach. The family
$A(t)=t^{-a/d}\lambda(t)$ is a suitable family and satisfies the conditions
specified in the previous section. Whence, if the transversality assumption
holds, then all the results, including Theorem 3.13 are available. The special
structure of $A(t)$ as above allows some more observations. We continue to use
the same notation ${\cal K}(t)$, ${\cal H}(t)$ from the previous section.

First note that $\lambda(t)$ acts on ${\cal G}$ by conjugation whence, we have
the weight space decomposition ${\cal G}_{-1}\oplus{\cal G}_{0}\oplus{\cal
G}_{1}$, where ${\cal G}_{-1}=Hom(Z,Y)$, ${\cal G}_{0}=Hom(Y,Y)\oplus
Hom(Z,Z)$ and ${\cal G}_{1}=Hom(Y,Z)$. Note that
$\lambda(t)\mathfrak{g}_{i}\lambda(t)^{-1}=t^{i}\mathfrak{g}_{i}$ for any
$\mathfrak{g}_{i}\in{\cal G}_{i}$, and that ${\cal G}_{i}\cdot V_{j}=V_{i+j}$.

Since $g\in V_{a}$ and is homogeneous in the degree in $Y$, we have
$\lambda(t)g=t^{a}g$ is the leading term of $\lambda(t)f$. Since the Lie
algebra stabilizers of $t^{a}g$ and $g$ are the same, we have:

###### Lemma 4.1

In the above notation, the stabilizer ${\cal H}$ of $g$ is graded, i.e.,
${\cal H}=\oplus_{i=-1,0,1}{\cal H}_{i}$ such that $\lambda(t){\cal
H}_{i}\lambda(t)^{-1}=t^{i}{\cal H}_{i}$. Moreover, we may choose ${\cal
S}_{i}$ as the complement of ${\cal H}_{i}\subseteq{\cal G}_{i}$, then ${\cal
S}=\oplus_{i=-1,0,1}{\cal S}_{i}$ is a complement of ${\cal H}$ in ${\cal G}$,
and $\lambda(t){\cal S}_{i}\lambda(t)^{-1}\subseteq t^{i}{\cal S}_{i}$.
Moreover, let $\pi_{\cal S}$ and $\pi_{\cal H}$ be the projections from ${\cal
G}={\cal S}\oplus{\cal H}$, to the first and second components, then
$\lambda(t)$ intertwines with these projections. In other words, for any
$\mathfrak{g}\in{\cal G}$, we have $\pi_{\cal
S}((\lambda(t)\mathfrak{g}\lambda(t)^{-1})=\lambda(t)\pi_{\cal
S}(\mathfrak{g})\lambda(t)^{-1}$ and $\pi_{\cal
H}((\lambda(t)\mathfrak{g}\lambda(t)^{-1})=\lambda(t)\pi_{\cal
H}(\mathfrak{g})\lambda(t)^{-1}$

Thus, $T_{g}O(g)={\cal S}\cdot g\subseteq V_{a-1}\oplus V_{a}\oplus V_{a+1}$.
The transversality condition translates into the requirement that
$f_{b}\not\in T_{g}O(g)$.

In view of Lemma 4.1 and the resulting discussion, the stabilization
conditions of Theorem 3.13 can be further studied. We present this analysis in
the Appendix B.

###### Lemma 4.2

Let $u,v\in\mathbb{C}$ be generic. Then $\lambda(u){\cal
H}(v)\lambda(u)^{-1}\subseteq{\cal H}(uv)$.

Proof: Let ${\cal K}$ be the stabilizer of $f=f(1)$. Now ${\cal
K}(u)=\lambda(u){\cal K}\lambda(u)^{-1}$ and that
$\lambda(uv)=\lambda(u)\lambda(v)$, tells us that ${\cal
K}(uv)=\lambda(u){\cal K}(v)\lambda(u)^{-1}$. Moreover, note that $\pi_{\cal
H}({\cal K}(t_{0}))={\cal H}(t_{0})$, for any generic $t_{0}\in\mathbb{C}$,
and thus may be applied to ${\cal K}(uv)$ above. This proves the lemma. $\Box$

We also have the important:

###### Proposition 4.3

The algebra ${\cal K}_{0}$ has a weight space decomposition ${\cal
K}_{0}={\cal K}_{0,-1}\oplus{\cal K}_{0,0}\oplus{\cal K}_{0,1}$, where ${\cal
K}_{0,i}\subseteq{\cal H}_{i}$.

Proof: Let $\\{\mathfrak{h}_{i}|i=1,\ldots,k\\}$ be a basis of ${\cal H}(1)$
and let $\mathfrak{h}_{i}=\oplus_{j}\mathfrak{h}_{ij}$ be its the weight space
decomposition. We may then write:

$\mathfrak{b}_{i}(t)=\sum_{j}t^{j}\mathfrak{h}_{ij}$

The previous lemma implies that $\\{\mathfrak{b}_{i}(t)|i=1,\ldots,k\\}$ is
another $\mathbb{C}(t)$-basis for ${\cal H}(t)$. The leading terms of this
basis would decompose by weights and generate ${\cal H}(0)={\cal K}_{0}$.
$\Box$

###### Lemma 4.4

  1. 1.

Let $\ell=\log_{t}(\lambda(t))\in gl(X)$, i.e.,
$\ell=diag([0,\ldots,0,1,\ldots,1])$ in a suitable basis.Then for any
$\mathfrak{g}\in{\cal G}_{i}$, $[\ell,\mathfrak{g}]=i\mathfrak{g}$. For any
$v\in V_{i}$, $\ell\cdot v=i\cdot v$.

  2. 2.

The algebra ${\cal H}_{b}$ is graded. Let
$\ell^{\prime}=\log_{t}(t^{-a/d}\lambda(t))=\ell-a/d\cdot I$. Then
$\ell^{\prime}\in{\cal H}$, $\ell^{\prime}$ normalizes both ${\cal H}_{b}$ and
${\cal K}_{0}$, but $\ell^{\prime}\not\in{\cal H}_{b}$.

Proof: The first part of (1) is straightforward. For the action of $\ell$ on
$V$, note that $\ell$ is also the differential operator $\sum_{z\in
Z}z\frac{\partial}{\partial z}$. Thus $\ell v=i\cdot v$ for $v\in V_{i}$.
Coming to (2), note that $\overline{f_{b}}$ is homogeneous and ${\cal H}$ is
graded, hence ${\cal H}_{b}$ is graded. $[\ell^{\prime},{\cal
H}_{b,i}]\subseteq{\cal H}_{b,i}$. By the same token $[\ell^{\prime},{\cal
K}_{0,i}]\subseteq{\cal K}_{0,i}$. By construction,
$\ell^{\prime}\cdot\overline{f_{b}}=(b-a)\overline{f_{b}}\neq 0$. $\Box$

The graded case also allows us to analyse the tangent of exit.

###### Definition 4.5

Let $\lambda(t)$ be a 1-PS acting on the form $f$ and let:

$f(t)=\lambda(t)\cdot f=t^{a}g+t^{b}f_{b}+\ldots+t^{D}f_{D}$

Then the tangent of exit is the form $\lim_{t\rightarrow
1}\frac{f(t)-f(1)}{t-1}$.

Since

$\ell f=af_{a}+bf_{b}+\ldots+Df_{D}$

the tangent of exit is $\ell f-f$. Note that both $f$ and $\ell f$ are
elements of $TO_{f}$, the tangent space of the orbit $O(f)$ at the point $f$.
Thus, the tangent of exit is given (upto addition of a suitable multiple of
the form $f$) by the action $\ell f$, where $\ell\not\in{\cal K}$.

An important question is the stabilizer of $\ell f$ within $TO_{f}$ under the
action of ${\cal K}$, the stabilizer of $f$. This is answered by the following
lemma:

###### Lemma 4.6

$\Pi_{i}:{\cal G}\rightarrow{\cal G}_{i}$ be the weight projections as per
$\lambda$. Then the stabilizer ${\cal K}_{\ell f}\subseteq{\cal K}$ within the
stabilizer of $f$ consists of precisely those elements $\mathfrak{k}\in{\cal
K}$ such that $[\mathfrak{k},\ell]\in{\cal K}$. In particular, it contains all
elements of ${\cal K}$ which are pure in weight, i.e., elements $\mathfrak{k}$
such that $\Pi_{i}(\mathfrak{k})=i\cdot\mathfrak{k}$ for some $i$. Thus, these
elements $\mathfrak{k}\in{\cal K}$ of pure weight are triple stabilizers,
i.e., members of ${\cal K}\cap{\cal H}\cap{\cal K}_{\ell f}$.

Proof: Let $\mathfrak{k}\in{\cal K}$ stabilize $\ell f$. Then, since
$\mathfrak{k}f=0$, we have $\mathfrak{k}\ell
f=[\mathfrak{k},\ell]f+\ell\mathfrak{k}f=[\mathfrak{k},\ell]f$. This, in turn,
is equivalent to $[\mathfrak{k},\ell]\in{\cal K}$. This proves the first
assertion. For the second, note that if $\mathfrak{k}\in{\cal K}$ is of pure
weight, then $[\mathfrak{k},\ell]=i\cdot\mathfrak{k}$ is certainly an element
of ${\cal K}$. The last assertion comes from the fact that pure weight
elements are also elements of ${\cal K}_{0}\subseteq{\cal H}$. Note $\Box$

###### Lemma 4.7

Pure elements $\mathfrak{k}\in{\cal K}$ stabilize all components in the
decomposition:

$f=f_{a}+f_{b}+\ldots+f_{D}$

Proof: Let $[\mathfrak{k},\ell]=r\mathfrak{k}$. We claim that
$\mathfrak{k}\ell^{k}f=0$ for all $k\geq 0$, which we prove by induction on
$k$. The assertion for $k=0$ is obvious. Then, for general $k$, we have:

$\begin{array}[]{rcl}0=\mathfrak{k}\ell^{k}f&=&\ell\mathfrak{k}\ell^{k-1}f+r\mathfrak{k}\ell^{k-1}f\\\
&=&0+0\end{array}$

This proves the assertion. Next, we apply $\mathfrak{k}\ell^{k}$ to the above
equation to get:

$\begin{array}[]{rcl}0=\mathfrak{k}(\ell^{k}f)&=&\mathfrak{k}(\ell^{k}f_{a})+\mathfrak{k}(\ell^{k}f_{b})+\ldots+\mathfrak{k}(\ell^{k}f_{D})\\\
&=&\mathfrak{k}(a^{k}f_{a}+b^{k}f_{b}+\ldots+D^{k}f_{D})\\\
&=&a^{k}(\mathfrak{k}f_{a})+b^{k}(\mathfrak{k}f_{b})+\ldots+D^{k}(\mathfrak{k}f_{D})\end{array}$

Since all of these are zero, we must have $\mathfrak{k}f_{c}=0$ for all $c$.
$\Box$

Thus, pure elements stabilize limits, tangents of approach and exit and all
components in between. That there is an element $\mathfrak{k}\in{\cal K}$
which is pure with respect $\ell$ is a significant alignment between ${\cal
K}$ and $\ell$.

###### Remark 4.8

The above condition also connects with the weight structure of ${\cal K}_{0}$.
Let $\mathfrak{k}\in{\cal K}$ be such that its weight space decomposition is
$\mathfrak{k}=\mathfrak{k}^{-1}+\mathfrak{k}^{0}+\mathfrak{k}^{1}$. We define

${\cal K}^{\geq 0}=\\{\mathfrak{k}\in{\cal K}|\mathfrak{k}^{-1}=0\\}\mbox{ and
}{\cal K}^{\geq 1}=\\{\mathfrak{k}\in{\cal
K}|\mathfrak{k}^{0}=0,\mathfrak{k}^{-1}=0\\}$

Note that ${\cal K}^{\geq 1}\subseteq{\cal K}^{\geq 0}\subseteq{\cal
K}^{\geq(-1)}={\cal K}$ is a nested sequence of subalgebras such that ${\cal
K}^{\geq i}/{\cal K}^{\geq(i+1)}\cong({\cal K}_{0})_{i}$. Since $({\cal
K}_{0})_{i}\subseteq{\cal H}_{i}$, we have $dim(({\cal K}_{0})_{i})\leq
dim({\cal H}_{i})$.

In Section 5.2 we work out this decomposition for various 1-PS acting on the
form $det_{3}$, the $3\times 3$-determinant.

###### Proposition 4.9

Let $f,g,\lambda$ and $\ell$ be as above. Let $P(\lambda)$ be defined as
below:

$P(\lambda)=\\{g\in G|\lim_{t\rightarrow 0}\lambda(t)g\lambda(t)^{-1}\mbox{
exists}\\}$

and let $U(\lambda)$ be its unipotent radical. Suppose that
$\mathfrak{k}\in{\cal P}(\lambda)\cap{\cal K}$ is a semi-simple element, then
there is a unipotent element $u\in U(\lambda)$ such that:

  1. 1.

If $f^{u}=u\cdot f$, then:

$\lambda(t)f^{u}=t^{a}g+\mbox{ higher terms }$ (3)

Thus, $g$ is the limit of $f^{u}$ under $\lambda$.

  2. 2.

The element $\mathfrak{k}^{u}=u\mathfrak{k}u^{-1}$ stabilizes $f^{u}$, $g$ and
$\ell f^{u}$. Thus, $\mathfrak{k}^{u}$ is a triple stabilizer of the limit
$g$, $f^{u}$ and the tangent of exit $\ell f^{u}$.

Proof: Let $L(\lambda)$ be the elements of $GL(X)$ which commute with
$\lambda$. Then $P(\lambda)=L(\lambda)U(\lambda)=U(\lambda)L(\lambda)$ is a
Levi factorization, with $L(\lambda)$ as a reductive complement. Applying this
to the corresponding Lie algebras, for any semisimple $\mathfrak{k}$ as above,
there is a $u\in U(\lambda)$ so that
$\mathfrak{k}^{u}=u\mathfrak{k}u^{-1}\in{\cal L}(\lambda)$ and thus is an
element of pure weight. Moreover $\mathfrak{k}^{u}$ stabilizes $f^{u}=u\cdot
f$. Since $u\in U(\lambda)$ is unipotent, for any $h_{c}\in V_{c}$, the
element $u\cdot h$ has the following weight decomposition with respect to
$\lambda(t)$:

$uh_{c}=h_{c}+h^{\prime}_{c+1}+\mbox{higher terms}$

Now since $f$ has the weight decomposition:

$f=g_{a}+f_{b}+\mbox{ higher terms }$ (4)

on applying $u$ to Eq. 4 we have:

$\begin{array}[]{rcl}uf&=&u\cdot g+u\cdot f_{b}+\mbox{higher order terms}\\\
&=&g+f^{\prime}_{a+1}+\mbox{ higher terms}\\\ \end{array}$

Thus, $g$ continues to be the leading term of $f^{u}$. Whence, we have:

$\lambda(t)(uf)=t^{a}g+t^{a+1}f^{\prime}_{a+1}+\mbox{higher terms}$

Thus for the data $f^{u},\lambda,\ell$ and $g$, we have that $g$ is the limit
of $f^{u}$ under the action of $\lambda$. Now note that
$\mathfrak{k}^{u}\in{\cal K}_{f^{u}}$ and is of pure weight. Thus, by lemma
4.6, $\mathfrak{k}^{u}$ stabilizes the tangent of exit $\ell f^{u}$ and $g$.
$\Box$

###### Proposition 4.10

Let $f,g,\lambda$ and $\ell$ be as above. Then at least one of the following
hold:

  1. (A)

${\cal K}_{0}$ is a nilpotent algebra, or

  2. (B)

there is a unipotent element $u\in U(\lambda)$ and an element
$\mathfrak{k}\in{\cal K}$ such that $g$ is a limit of $f^{u}$ under $\lambda$
and $\mathfrak{k}^{u}$ is a triple stabilizer for the data $(f^{u},\ell
f^{u},g)$.

Proof: Note that ${\cal G}_{0}\oplus{\cal G}_{1}={\cal P}(\lambda)$ and ${\cal
G}_{1}={\cal U}(\lambda)$, the Lie algebras of $P(\lambda)$ and $U(\lambda)$,
and that the above decomposition is a Levi decomposition of ${\cal
P}(\lambda)$ with ${\cal L}(\lambda)={\cal G}_{0}$ as a semisimple complement.

Case 1: Suppose that $dim(\Pi_{-1}({\cal K}))=dim({\cal K})$, i.e., for every
$\mathfrak{k}\in{\cal K}$, if $\mathfrak{k}\neq 0$ then so is
$\Pi_{-1}(\mathfrak{k})$. If this happens, then ${\cal K}_{0}\subseteq{\cal
G}_{-1}$ and ${\cal K}_{0}$ is nilpotent. Thus (A) holds.

Case 2: On the other hand, if $dim(\Pi_{-1}({\cal K}))<dim({\cal K})$, then
either:

  1. 2a

There is an element $\mathfrak{k}\in{\cal U}(\lambda)\cap{\cal K}$. In this
case, $\mathfrak{k}$ is a pure element and by Lemma 4.6, $\mathfrak{k}$
stabilizes $f,g$ and $\ell f$. Thus (B) holds with $u=1$.

  2. 2b

All elements $\mathfrak{k}\in{\cal P}(\lambda)$ may be written as
$\mathfrak{k}=\mathfrak{k}_{0}+\mathfrak{k}_{1}$ with $\mathfrak{k}_{0}\neq
0$. Let us use Jordan’s decomposition to the element $\mathfrak{k}$ to express
$\mathfrak{k}=\mathfrak{k}_{ss}+\mathfrak{k}_{n}$, where $\mathfrak{k}_{ss}$
is semisimple and $\mathfrak{k}_{n}$ is nilpotent. Both are elements of ${\cal
P}(\lambda)\cap{\cal K}$. If there is a $\mathfrak{k}_{ss}\neq 0$ as above,
then by Prop. 4.9, (B) holds.

  3. 2c

We come to the final case where for all $\mathfrak{k}\in{\cal
P}(\lambda)\cap{\cal K}$, we have
$\mathfrak{k}=\mathfrak{k}_{n}=\mathfrak{k}_{0}+\mathfrak{k}_{1}$ with
$\mathfrak{k}_{0}\neq 0$. Since $\mathfrak{k}_{n}^{k}=0$ for some $k$, we have
$\mathfrak{k}_{0}^{k}=0$ as well and thus $\mathfrak{k}_{0}$ is nilpotent.
Now, note that $\mathfrak{k}_{0}$ is the leading term of $\mathfrak{k}$ and
thus $\mathfrak{k}_{0}\in{\cal K}_{0}$. Whence we have ${\cal K}_{0}=({\cal
K}_{0})_{0}+({\cal K}_{0})_{-1}$ where the general element
$\mathfrak{k}^{\prime}\in{\cal K}_{0}$ may be written as
$\mathfrak{k}^{\prime}_{0}+\mathfrak{k}^{\prime}_{-1}$, where both are
individually nilpotent. The grading ensures that $\mathfrak{k}^{\prime}$ is
nilpotent too. Thus ${\cal K}_{0}$ is nilpotent, and (A) holds.

$\Box$

###### Remark 4.11

The condition that ${\cal K}_{0}\not\subseteq{\cal G}_{1}$ is tantamount to
saying that $dim(\Pi_{-1}({\cal K}))<dim({\cal K})$, i.e., ${\cal K}$ is not
generically placed with respect to $\ell$. If indeed that is so, then all
leading terms of ${\cal K}$ have weight $-1$ and ${\cal K}_{0}$ is nilpotent.

###  4.2 General 1-PS

We extend the results of the previous section to a general 1-PS $\lambda$,
beyond the two-component case considered earlier. Thus, let
$X=X_{1}\cup\ldots\cup X_{k}$, with $|X_{i}|=n_{i}$, and let $\lambda(t)\in
GL(X)=G$ be defined so that $\lambda(t)(x_{i})=t^{d_{i}}x_{i}$. We call
$\overline{d}=(d_{1}^{n_{1}},\ldots,d_{k}^{n_{k}})$ as the type of $\lambda$.
Thus, in the chosen basis of $(x_{i})=X$, $\lambda(t)$ is the diagonal matrix
$diag(t^{\overline{d}})$. As before, $V$ is a $G$-module via the map
$\rho:GL(X)\rightarrow GL(V)$. We also have the Lie algebra map (also)
$\rho:gl(X)\rightarrow gl(V)$. We now have $V=\oplus_{\chi}V^{\chi}$, the
weight space decomposition of $V$. Note that $\chi\in\mathbb{Z}$. Let
$\ell=\log_{t}(\lambda(t))$ as before. Then for $v^{\chi}\in V^{\chi}$, we
have $\rho(\ell)\cdot v^{\chi}=\chi\cdot v^{\chi}$. We also have ${\cal
G}=\oplus_{\chi}{\cal G}^{\chi}$.

###### Definition 4.12

For any ${\cal G}$-module $M$, we say that $m\in M$ is pure if $\ell\cdot
m=\chi\cdot m$ for some $\chi\in\mathbb{Z}$. We say that $M$ is graded if
$M=\oplus_{\chi}M^{\chi}$, the sum of its pure elements.

###### Proposition 4.13

For the above situation, we have the following:

  1. 1.

The stabilizer ${\cal H}\subseteq{\cal G}$ of $x$ is graded.

  2. 2.

Let ${\cal K}$ be the stabilizer of $y$, and ${\cal K}_{0}$ be the limit of
${\cal K}$ under the action of $\lambda$, then ${\cal K}_{0}$ is graded.

  3. 3.

The tangent space $T_{y}O(y)$ of $y$ as an element of its $G$-orbit $O(y)$ is
a ${\cal K}$-module. Let ${\cal K}_{\ell y}\subseteq{\cal K}$ be the
stabilizer of the tangent vector $\ell y$. Then $\mathfrak{k}\in{\cal K}$
stabilizes $\ell y$ if and only if $[\ell,\mathfrak{k}]\in{\cal K}$.
Consequently, if $\mathfrak{k}\in{\cal K}$ is of pure weight, then
$\mathfrak{k}\in{\cal K}_{\ell y}$ as well.

Proof: The arguments of Section 4.1 are easily modified for the general case.
$\Box$

Prop. 4.9 holds in the general case as well and Prop. 4.14 below is a
restatement in the notation of this section. However, Prop. 4.10, on the
structure of ${\cal K}_{0}$ does not; its modification is Prop. 4.15.

###### Proposition 4.14

Let $y,x,\lambda$ and $\ell$ be as above. Let $P(\lambda)$ be defined as
below:

$P(\lambda)=\\{g\in G|\lim_{t\rightarrow 0}\lambda(t)g\lambda(t)^{-1}\mbox{
exists}\\}$

and let $U(\lambda)$ be its unipotent radical. Suppose that
$\mathfrak{k}\in{\cal P}(\lambda)\cap{\cal K}$ is a semi-simple element, then
there is a unipotent element $u\in U(\lambda)$ such that:

  1. 1.

If $y^{u}=u\cdot y$, then:

$\lambda(t)y^{u}=t^{a}x+\mbox{ higher terms }$ (5)

Thus, there is a $y^{u}$ in the orbit of $y$ such that $x$ is the limit of
$y^{u}$ under $\lambda$.

  2. 2.

The element $\mathfrak{k}^{u}=u\mathfrak{k}u^{-1}$ stabilizes $y^{u}$, $x$ and
$\ell y^{u}$. Thus, $\mathfrak{k}^{u}$ is a triple stabilizer of the limit
$x$, $y^{u}$ and the tangent of exit $\ell y^{u}$.

The proof of this is a straightforward adaptation of the proof of Prop. 4.9.

###### Proposition 4.15

Let $y,x,\lambda$ and $\ell$ be as above. Then at least one of the following
holds:

  1. (A)

Let ${\cal K}^{\prime}={\cal K}_{0}\oplus\mathbb{C}\ell$, then ${\cal
K}^{\prime}$ is a Lie algebra of rank 1, i.e., the dimension of any maximal
torus in ${\cal K}^{\prime}$ is $1$, $[\ell,{\cal K}_{0}]\subseteq{\cal
K}_{0}$, and every graded component $({\cal K}_{0})_{i}$ is composed of
nilpotent elements or

  2. (B)

there is a unipotent element $u\in U(\lambda)$ and an element
$\mathfrak{k}\in{\cal K}$ such that $x$ is a limit of $y^{u}$ under $\lambda$
and $\mathfrak{k}^{u}$ is a triple stabilizer for the data $(y^{u},\ell
y^{u},x)$.

Proof: Note that ${\cal P}(\lambda)=\oplus_{\chi\geq 0}\>{\cal G}_{\chi}$ and
${\cal U}(\lambda)=\oplus_{\chi>0}\>{\cal G}_{\chi}$, the Lie algebras of
$P(\lambda)$ and $U(\lambda)$, and that the above decomposition is a Levi
decomposition of ${\cal P}(\lambda)$ with ${\cal G}_{0}$ as a semisimple
complement. Next note that ${\cal K}^{\prime}={\cal
K}_{0}\oplus\mathbb{C}\ell$ is indeed a Lie algebra since $[\ell,{\cal
K}_{0}]\subseteq{\cal K}_{0}$. Now we do a case analysis.

Case 1: Let $\Pi_{-}:{\cal G}\rightarrow\oplus_{\chi<0}\>{\cal G}_{\chi}$ be
the projection. Suppose that $dim(\Pi_{-}({\cal K}))=dim({\cal K})$. If this
happens, then ${\cal K}_{0}\subseteq\oplus_{\chi<0}{\cal G}_{\chi}$ and ${\cal
K}_{0}$ is nilpotent. Even more, ${\cal K}^{\prime}={\cal
K}_{0}\oplus\mathbb{C}\ell$ is a Levi decomposition of ${\cal K}^{\prime}$.
Thus (A) holds.

Case 2: On the other hand, if $dim(\Pi_{-1}({\cal K}))<dim({\cal K})$, then
${\cal K}\cap{\cal P}(\lambda)\neq 0$.

  1. 2a

If there is an $\ell^{\prime}\in{\cal P}(\lambda)\cap{\cal K}$ such that
$[\ell^{\prime},\ell]=0$, then it must be that $\ell^{\prime}\in{\cal G}_{0}$,
i.e., it must be of pure weight and therefore a triple stabilizer, and
therefore (B) holds.

  2. 2b

If there is a semi-simple element $\mathfrak{k}\in{\cal P}(\lambda)\cap{\cal
K}$ then, by Prop. 4.15, there is a $u$ and a triple stabilizer as required.
Thus (B) holds.

  3. 2d

We come to the final case where all $\mathfrak{k}\in{\cal P}(\lambda)\cap{\cal
K}$ are nilpotent. Let $\mathfrak{k}_{\chi_{0}}$ be the leading term of
$\mathfrak{k}=\sum_{\chi\leq 0}\mathfrak{k}_{\chi}$, i.e., the smallest degree
$d$ such that $\mathfrak{k}_{d}\neq 0$. If $\chi_{0}>0$ then
$\mathfrak{k}_{\chi_{0}}\in{\cal K}_{0}$ is nilpotent. Otherwise, we may write
$\mathfrak{k}=\mathfrak{k}_{0}+\sum_{\chi>0}\mathfrak{k}_{\chi}$. Since
$\mathfrak{k}^{k}=0$ for some $k$, we have $\mathfrak{k}_{0}^{k}=0$ as well
and thus $\mathfrak{k}_{0}$ is nilpotent. Thus, for every $\chi$, ${\cal
K}_{0}\cap{\cal G}_{\chi}$ consists of nilpotent elements. From Case 2a above,
we already know that the rank of ${\cal K}^{\prime}={\cal
K}\oplus\mathbb{C}\ell$ is $1$

This proves the proposition. $\Box$

##  5 Closures of affine orbits

We apply the results of the last two sections to the closure of forms whose
$SL(X)$-orbit is affine. As mentioned in the introduction (see also Appendix
A), the motivation for studying comes from geometric complexity theory.
Examples of such forms are $det(X)$, the determinant of the matrix $X$, and
$perm(X)$, its permanent. In the next Section, we cover some general results.
In Section 5.2 we apply these and earlier results to the $3\times 3$
determinant form.

###  5.1 General Results

Let $f(X)\in Sym^{d}(X)$ be such that the stabilizer $K\subseteq GL(X)$ is
reductive (of dimension $k$). This implies that the $SL(X)$-orbit of $f$ is an
affine variety. Let $O(f)$ be the $GL(X)$-orbit of $f$ and $\overline{O(f)}$
be its projective closure. It is known that the closed set
$\overline{O(f)}-O(f)$ is of co-dimension one. Moreover, there are forms
$g_{1},\ldots,g_{m}$ such that it is contained in the union of the projective
closures $\\{\overline{O(g_{i})}\\}_{i=1}^{m}$.

Separately, it is known that for any form $f$, forms $g$ which appear in the
closure of the orbit $O(f)$ appear as leading terms under a 1-parameter
substitutions $A(t)\cdot X$, where $A(t)\in GL(X)$ has entries which are
polynomials in $t$. Thus, for the above $g_{i}$, there are linear
transformations $A_{i}(t)$, with polynomial entries, such that
$f(A_{i}(t)\cdot X)=t^{d}g_{i}+$ higher degree terms. It is also clear from
dimension counting that for all $i$, the stabilizer $H_{i}$ of $g_{i}$ is of
dimension $k+1$.

Our focus will be on one such those $g_{i}=g$. We begin with a lemma:

###### Lemma 5.1

Given a transformation $A(t)\in gl(X)$ such that $g$ is the leading term of
$f(t)=f(A(t)\cdot X)$, we may assume that $A(t)\in GL(X)$, i.e., $A(t)$ is
invertible for most $t\in\mathbb{C}$.

Proof: Let us consider $A^{\prime}(t)=A(t)+t^{D}I$ for a large power $D>0$.
Then $g$ continues to be the leading term of
$f^{\prime}(t)=f(A^{\prime}(t)\cdot X)$. Moreover, $A^{\prime}(t)$ is
invertible for most $t\in\mathbb{C}$. $\Box$

By the above lemma, we may assume that $A(t)$ is generically invertible and
that:

$f(t)=f(A(t)\cdot X)=t^{a}g+t^{b}f_{b}+\ldots+t^{d}f_{d}$

Let ${\cal H}$ be the Lie algebra stabilizer of $g$ as before and ${\cal
K}(t)=A(t){\cal K}A(t)^{-1}$ that of $f(t)$, with $dim({\cal K}(t))=k$. Let
${\cal K}_{0}$ be the limit of ${\cal K}(t)$ and note that $dim({\cal
K}_{0})=k$ and that ${\cal K}_{0}\subseteq{\cal H}_{b}$, the stabilizer for
$\star$-action of ${\cal H}$ on $\overline{N}$. We have the following lemma:

###### Lemma 5.2

The algebra ${\cal K}_{0}$ is precisely ${\cal H}_{b}$, the stabilizer of
$\overline{f_{b}}\in\overline{N}$ under the $\star$-action of ${\cal H}$.

Proof: We have the containment ${\cal K}_{0}\subseteq{\cal H}_{b}\subset{\cal
H}$. Next, the orbit of $g$ is of co-dimension 1 within the orbit of $f(t)$.
This, gives us that $dim({\cal H})=dim({\cal K}(t))-1=k+1$. Since ${\cal
H}_{b}\neq{\cal H}$, we must have ${\cal H}_{b}={\cal K}_{0}$. $\Box$

###### Lemma 5.3

Suppose further that $A(t)$ is a one-parameter subgroup $\lambda(t)$. So we
have

$\lambda(t)\cdot f=f(t)=t^{a}\cdot g+t^{b}\cdot f_{b}+\mbox{higher terms}$

Let $\ell=\log(t^{-a/d}\lambda(t))\in gl(X)$ be the Lie algebra element
generating the 1-PS $t^{-a}\lambda(t)$. Suppose that $\overline{f_{b}}\neq 0$.
Then (i) $\ell\in{\cal H}$ but does not stabilize $\overline{f_{b}}$, (ii)
${\cal K}_{0}={\cal H}_{b}$, and (iii) ${\cal H}={\cal
K}_{0}\oplus\mathbb{C}\cdot\ell$.

Proof: Since $t^{-a}\lambda(t)\cdot g=g$ but
$t^{-a}\lambda(t)f_{b}=t^{b-a}f_{b}$, (i) is clear. Thus $\ell\in{\cal
H}-{\cal K}_{0}$. Since the dimension of ${\cal K}_{0}$ is one less than the
dimension of ${\cal H}$, we must have ${\cal H}={\cal K}_{0}\oplus\ell$. This
proves the lemma. $\Box$

When $g$ is obtained as the limit of $\lambda(t)$ where $\ell$ has only two
distinct entries (i.e., the 2-block case), we have the following result:

###### Proposition 5.4

Let $g$ be a projective limit of codimension $1$ of $f$ under $\lambda$ as
above. Then at least one of the following conditions hold:

  1. 1.

${\cal K}_{0}$ is nilpotent and ${\cal H}={\cal K}_{0}\oplus\mathbb{C}\ell$ is
a Levi factorization of ${\cal H}$.

  2. 2.

There is a unipotent element $u\in U(\lambda)$ (see 4.10) and an element
$\mathfrak{k}\in{\cal K}$ such that $u\mathfrak{k}u^{-1}$ is a stabilizer of
$g,f^{u}$ and $\ell f^{u}$, and $g$ is the limit of $f^{u}$ under $\lambda$.

Proof: This is a consequence of Prop 4.10 and the fact that ${\cal H}={\cal
K}_{0}\oplus\mathbb{C}\ell$. $\Box$

###  5.2 The analysis of $det_{3}$, the $3\times 3$ \- determinant.

We consider here the set $X=\\{x_{1},\ldots,x_{9}\\}$ and the
$165$-dimensional space $Sym^{3}(X)$ of homogeneous forms of degree $3$, acted
upon by $GL(X)$. The $3\times 3$-determinant, $det_{3}(X)$ is a special
element and its definition is given below:

$det_{3}(X)=det\left(\left[\begin{array}[]{ccc}x_{1}&x_{2}&x_{3}\\\
x_{4}&x_{5}&x_{6}\\\ x_{7}&x_{8}&x_{9}\\\ \end{array}\right]\right)$

The obvious question are the codimension one forms in the orbit closure of the
orbit of $det_{3}(X)$.

These have been identified by Hüttenhain[Hüt17] and are given by the two forms
$Q_{1}$ and $Q_{2}$ below:

$\begin{array}[]{rcl}Q_{1}(X)&=&det\left(\left[\begin{array}[]{ccc}x_{1}&x_{2}&x_{3}\\\
x_{4}&x_{5}&x_{6}\\\ x_{7}&x_{8}&-x_{5}-x_{1}\\\ \end{array}\right]\right)\\\
Q_{2}(X)&=&x_{4}x_{1}^{2}+x_{5}x_{2}^{2}+x_{6}x_{3}^{2}+x_{7}x_{1}x_{2}+x_{8}x_{2}x_{3}+x_{9}x_{1}x_{3}\end{array}$

We will study each in turn. Before we begin, let us look at the stabilizer
${\cal K}$ of $det_{3}(X)$. This will serve as our form $f$, in the notation
of this section. Within $Sym^{3}(X)$ the dimension of the $GL(X)$-orbit
$O(det_{3}(X))$ of $det_{3}(X)$ is given by $dim(GL(X))-dim({\cal K}))$. It is
known that this is composed of the transformations $X\rightarrow AXB$ with
$det(AB)=1$ and the $X\rightarrow X^{T}$. The condition $det(AB)=1$ and that
$(cA)X(c^{-1}B)$ lead to the same linear transformation on $X$, give us two
independent constraints on $GL_{3}\times GL_{3}\rightarrow{\cal K}$. This
gives us that $dim({\cal K}(1))=18-2=16$ and $dim(O(det_{3}(X))=65$.

Coming to $Q_{1}$, we see that:

$det_{3}(X)=Q_{1}(X)+(x_{1}+x_{5}+x_{9})(x_{1}x_{5}-x_{2}x_{4})$

We denote by $Q_{1}^{\prime}$ the form
$(x_{1}+x_{5}+x_{9})(x_{1}x_{5}-x_{2}x_{4})$. This motivates us to define
$Y=\\{x_{1},\ldots,x_{8}\\}$ and $Z=\\{x_{1}+x_{5}+x_{9}\\}$. We define
$\lambda_{1}(t)\in GL(X)$ as $\lambda_{1}(t)x_{i}=x_{i}$ for $i=1,\ldots,8$
and $\lambda(t)(z)=tz$, where $z=(x_{1}+x_{5}+x_{9})$. We thus see that:

$\lambda_{1}(t)\cdot det_{3}(X)=Q_{1}+t\cdot Q_{1}^{\prime}$

Thus in the notation of this section, $f=det_{3}(X)$, $g=Q_{1}$ and
$f_{b}=Q_{1}^{\prime}$, with $a=0$ and $b=1$.

###### Lemma 5.5

The stabilizer of ${\cal H}_{1}$ of $Q_{1}$ within $gl(X)$ has dimension 17.
Moreover, the stabilizer ${\cal H}_{1}$ has a direct sum decomposition into
degree 0 and degree 1 parts, as in Section 4. The tangent of exit is the form
$Q^{\prime}_{1}$ and the dimension of its stabilizer within ${\cal K}$, the
stabilizer of $det^{3}(X)$, is $4$.

Proof: Let us identify $Y$ as the space of all trace zero matrices $y$
labelled with their 9 entries minus the entry $y(3,3)$. Then, $Q_{1}$ is
precisely the determinant of $y\in Y$. For any element $A\in GL_{3}$, the
action of $A$ on $y$ given by $y\rightarrow AyA^{-1}$ preserves the space $Y$
of trace zero matrices as well as $Q_{1}$. Whence, we get a map
$GL_{3}\rightarrow{\cal H}_{1}$, the stabilizer of $Q_{1}$. Since this map has
a 1-dimensional kernel (viz., the matrices $cI$, with $c\neq 0$), we have an
$8$-dimensional image ${\cal R}_{Y}\subseteq{\cal H}_{1}$. Next, let
$Hom(Z,Y\oplus Z)$ be the collection of all linear substitutions for $z$ in
terms of the variables of $Y$and $z$. Since $Q_{1}$ does not involve $z$,
these substitutions leave $Q_{1}$ invariant. This gives us a map
$Hom(Z,Y)\oplus Hom(Z,Z)\rightarrow{\cal H}_{1}$. We call this image as the
spaces ${\cal Q}_{1}\oplus{\cal R}_{Z}$. It can be shown that ${\cal
Q}_{1},{\cal R}_{Y}$ and ${\cal R}_{Z}$ intersect trivially and thus ${\cal
R}_{Y}\oplus{\cal R}_{Z}\oplus{\cal Q}_{1}\subseteq{\cal H}_{1}$ to obtain a
$17$-dimensional subalgebra. The result of [Hüt17] implies that the dimension
can not be more, since we know that the dimension of the orbit of $Q_{1}$ is
only one less than that of $det_{3}(X)$, and hence its stabilizer dimension
must be $17$.

We then see that ${\cal H}_{1}={\cal R}_{Y}\oplus{\cal R}_{Z}\oplus{\cal
Q}_{1}={\cal R}\oplus{\cal Q}_{1}$ and splits by degree, i.e., $({\cal
H}_{1})_{0}={\cal R}$ and $({\cal H}_{1})_{-1}={\cal Q}_{1}$.

The stabilizer dimension of $Q^{\prime}_{1}$ is through a direct computation.
$\Box$

###### Lemma 5.6

We have $H_{b}={\cal R}_{Y}\oplus{\cal Q}_{1}$ and ${\cal K}_{0}={\cal
H}_{b}$. Let $\ell_{1}=\log(\lambda(t))$, then $\ell_{1}\in{\cal H}_{1}$ so
that ${\cal H}_{1}={\cal H}_{b}\oplus\ell_{1}$ and $[\ell_{1},{\cal
H}_{b}]\subseteq{\cal H}_{b}$. Moreover, $\ell_{1}\cdot
det^{3}(X)=Q^{\prime}_{1}$, the tangent of exit, and ${\cal H}_{b}$ is an
ideal of ${\cal H}_{1}$. Finally, as in Case (i), Section B with $b-a=1$, (i)
for every element $\mathfrak{r}\in{\cal R}_{Y}$, we have an element
$\mathfrak{s}$ in ${\cal S}_{1}$ such that $\mathfrak{r}\cdot
f_{b}=\mathfrak{s}\cdot Q_{1}$, and (ii) for any $\mathfrak{q}\in{\cal
Q}_{1}$, there is an $\mathfrak{s}\in{\cal S}_{0}$ such that
$\mathfrak{q}\cdot f_{b}=\mathfrak{s}\cdot Q_{1}$.

Proof: The equality of ${\cal K}_{0}$ and ${\cal H}_{b}$ comes because $Q_{1}$
is one of the co-dimension 1 forms in the closure of $det_{3}(X)$. The second
assertion follows from Lemma 5.3. The last set of assertions follow from the
fact that ${\cal H}_{b}\cdot f_{b}\subseteq{\cal S}\cdot Q_{1}$. $\Box$

###### Remark 5.7

The statement that in this case ${\cal H}_{b}$ is an ideal in ${\cal H}$
follows from a characterization of codimension 1 Lie subalgebras of a Lie
algebra due to Hoffman [Hof65]. In Section 6 when we discuss connections to
Lie algebra cohomology, Hoffman’s theorem is stated as Theorem 6.20.

###### Example 5.8

We do an example of an $\mathfrak{r}$ and a corresponding $\mathfrak{s}$ such
that $\mathfrak{r}\cdot f_{b}=\mathfrak{s}\cdot Q_{1}$. However, constructing
elements $\mathfrak{r}\in{\cal R}_{Y}$ needs some care. Consider the matrix
$E_{ij}$ such that $E_{ij}(i,j)=1$ and $E_{ij}(k,l)=0$ for all other tuples
$(k,l)$. Let $Y$ be the matrix such that $det(Y)=Q_{1}$, as given below:

$Y=\left[\begin{array}[]{ccc}x_{1}&x_{2}&x_{3}\\\ x_{4}&x_{5}&x_{6}\\\
x_{7}&x_{8}&-x_{1}-x_{5}\end{array}\right]$

We construct $[E_{23},Y]=E_{23}Y-YE_{23}$ as below:

$[E_{23},Y]=R_{23}=\left[\begin{array}[]{ccc}0&x_{3}&0\\\
-x_{7}&x_{6}-x_{8}&x_{1}+x_{5}\\\ 0&-x_{1}-x_{5}&0\end{array}\right]$

The corresponding element $\mathfrak{r}$ is to be constructed as the linear
operator below:

$\mathfrak{r}=x_{3}\frac{\partial}{\partial
x_{2}}-x_{7}\frac{\partial}{\partial
x_{4}}+(x_{6}-x_{8})\frac{\partial}{\partial
x_{5}}+(x_{1}+x_{5})\frac{\partial}{\partial
x_{6}}-(x_{1}+x_{5})\frac{\partial}{\partial x_{8}}$

We may check that $\mathfrak{r}\cdot Q_{1}=0$. On the other hand, we have:

$\mathfrak{r}\cdot
f_{b}=z((x_{2}x_{7}-x_{1}x_{8})-(x_{3}x_{4}-x_{1}x_{6}))=(-z\frac{\partial}{\partial
x_{6}}+z\frac{\partial}{\partial x_{8}})\cdot Q_{1}=\mathfrak{s}\cdot Q_{1}$

The second operator $\mathfrak{s}$ is clearly an element of ${\cal S}_{1}$.

Let us now come to $Q_{2}$.

That $Q_{2}$ is in the orbit closure of $det_{3}$ is shown by the following
lemma.

###### Lemma 5.9

[Hüt17] Let $Y,Z$ be the generic matrices below and let $X=Y\oplus Z$.

$Y=\left[\begin{array}[]{ccc}0&x_{1}&-x_{2}\\\ -x_{1}&0&x_{3}\\\
x_{2}&-x_{3}&0\\\
\end{array}\right]\>\>\>\>Z=\left[\begin{array}[]{ccc}2x_{6}&x_{8}&x_{9}\\\
x_{8}&2x_{5}&x_{7}\\\ x_{9}&x_{7}&2x_{4}\\\ \end{array}\right]$

Let $\lambda_{2}(t)$ be such that $\lambda_{2}(t)\cdot Y=Y$ and
$\lambda_{2}(t)\cdot Z=tZ$. Let us define $det^{3}(X)$ as the determinant of
the matrix $Y+Z$. Then:

$det^{3}(\lambda_{2}(t)\cdot X))=det(Y+tZ)=tQ_{2}+t^{3}Q_{3}$

where:

$\begin{array}[]{rcl}Q_{2}(X)&=&x_{4}x_{1}^{2}+x_{5}x_{2}^{2}+x_{6}x_{3}^{2}+x_{7}x_{1}x_{2}+x_{8}x_{2}x_{3}+x_{9}x_{1}x_{3}\\\
Q_{3}(X)&=&8x_{4}x_{5}x_{6}-2x_{6}x_{7}^{2}-2x_{4}x_{8}^{2}-2x_{5}x_{9}^{2}+2x_{7}x_{8}x_{9}\end{array}$

The proof is a verification. Note that this parametrization of the generic
matrix $X$, instead of the one in the case of $Q_{1}$, is merely a reordering
of the variables and hence the expression $det^{3}(X)$ of the determinant is
in the orbit of $det_{3}(X)$, and hence the orbit closures of $det(X)$ and
$det^{3}(X)$ match. The expansion of $det^{3}(\lambda_{2}(t)\cdot X)$ puts us
into the familiar situation where $Q_{2}$ is the limit, $f_{b}=Q_{3}$ with and
$a=1$ and $b=3$.

###### Lemma 5.10

The stabilizer of ${\cal H}_{2}$ of $Q_{2}$ within $gl(X)$ has dimension 17.
The tangent of exit is the form $Q_{3}$ and the dimension of its stabilizer
with ${\cal K}$, the stabilizer of $det^{3}(X)$, is $8$.

We may write $Q_{2}$ as the inner product:

$Q_{2}(X)=R\cdot
C=\left[\begin{array}[]{cccccc}x_{1}^{2}&x_{2}^{2}&x_{3}^{2}&x_{1}x_{2}&x_{2}x_{3}&x_{1}x_{3}\\\
\end{array}\right]\left[\begin{array}[]{c}x_{4}\\\ x_{5}\\\ x_{6}\\\ x_{7}\\\
x_{8}\\\ x_{9}\\\ \end{array}\right]$

Let $Y=\\{x_{1},x_{2},x_{3}\\}$ and $Z=\\{x_{4},\ldots,x_{9}\\}$ and let
$\rho:GL(Y)\rightarrow GL(Sym^{2}(Y))$ be the representation of $GL(Y)$. We
see that for any $A\in GL(Y)$, we have:

$Q_{2}(X)=R\cdot\rho(A)^{T}(\rho(A)^{T})^{-1}\cdot C$

Thus, we have an embedding $\alpha:GL(Y)\rightarrow{\cal H}_{2}\subseteq
GL(Y)\times GL(Z)$ of $GL(Y)$ into ${\cal H}_{2}$, given by $A\rightarrow
A\times(\rho(A)^{T})^{-1}$. The image ${\cal R}=\alpha(GL(Y))$ is the
reductive part of $H_{2}$ and has dimension equal to $dim(GL(Y))=9$. We may
further factor ${\cal R}$ as follows. Let $D\subset gl(Y)$ be multiples of the
identity matrix. We may then write $gl(X)=D\oplus sl(X)$ giving us a
decomposition of ${\cal R}={\cal D}\oplus{\cal R}^{\prime}$, where
$\alpha(D)={\cal D}$ is a $1$-dimensional subspace of ${\cal R}$, and ${\cal
R}^{\prime}=\alpha(sl(X))$.

Let us now come to the nilpotent part which is a subspace of $Hom(Z,Y)$. Let
$A\in Hom(Z,Y)$ be a $6\times 3$-matrix and let us write the action of $A$ on
$Q_{2}$ as:

$A\cdot Q_{2}(X)=R\cdot
C=\left[\begin{array}[]{cccccc}x_{1}^{2}&x_{2}^{2}&x_{3}^{2}&x_{1}x_{2}&x_{2}x_{3}&x_{1}x_{3}\\\
\end{array}\right]\left(\left[\begin{array}[]{c}x_{4}\\\ x_{5}\\\ x_{6}\\\
x_{7}\\\ x_{8}\\\ x_{9}\\\
\end{array}\right]+\left[\begin{array}[]{ccc}a_{11}&a_{12}&a_{13}\\\ &&\\\
&\vdots&\\\ &&\\\ &\vdots&\\\ a_{61}&a_{62}&a_{63}\\\
\end{array}\right]\left[\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right]\right)$

The requirement that $A$ stabilizes $Q_{2}$ is tantamount to the condition
that:

$p(A,Y)=R\cdot A\cdot\left[\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right]=0$

Since $p(A,Y)\in Sym^{3}(Y)$ and $dim(Sym^{3}(Y))=10$, the condition
$p(A,Y)=0$ translates to $10$ equations. Thus, we define ${\cal A}$ as
follows:

${\cal A}=\\{A\in Hom(Z,Y)|p(A,Y)=0\\}$

Then we have ${\cal A}\rightarrow{\cal H}_{2}$ and that $dim({\cal
A})=dim(Hom(Z,Y))-10=8$. Thus ${\cal H}_{2}={\cal R}\oplus{\cal A}$ and
$dim({\cal H}_{2})=8+9=17$. The stabilizer dimension of $Q_{3}$ is through a
direct computation. $\Box$

###### Lemma 5.11

We have $H_{b}={\cal R}^{\prime}\oplus{\cal A}$ and ${\cal K}_{0}={\cal
H}_{b}$. Let $\ell_{2}=\log(t^{-1/3}\lambda_{2}(t))$, then $\ell_{2}\in{\cal
H}_{2}$ so that ${\cal H}_{2}={\cal H}_{b}\oplus\ell_{2}$ and $[\ell_{2},{\cal
H}_{b}]\subseteq{\cal H}_{b}$. Moreover, $\ell_{2}\cdot det^{3}(X)=Q_{3}$, the
tangent of exit. ${\cal H}_{b}$ is an ideal of ${\cal H}_{2}$ and condition
(3) of Theorem 6.20 holds. Finally, as in Case (ii) Section B, with $b-a=2$,
we have for every element $\mathfrak{a}\in{\cal A}_{Y}$, there is an element
$\mathfrak{s}$ in ${\cal S}_{1}$ such that $\mathfrak{a}\cdot
f_{b}=\mathfrak{s}\cdot Q_{1}$.

###### Remark 5.12

The algebras ${\cal H}_{2}$ and ${\cal K}_{0}$ provide an extremely important
insight into the possible structure of the Levi decomposition of ${\cal
K}_{0}$. The coupling of the $End(Y)$ and $End(Z)$ components through
different representations of the same algebra $gl(3)$, and the representation
on $End(Z,Y)$ as one which interleaves between the two, provides an important
template for the Lie algebra structure of ${\cal K}_{0}$.

Finally, we cover the case of a limit of $det^{3}(X)$ which is not one of the
components of the orbit closure. Let $Y=\\{x_{1},\ldots,x_{4}\\}$,
$Z=\\{x_{5},\ldots,x_{9}\\}$ so that $X=Y\cup Z$. Let
$\lambda_{3}(t):\mathbb{C}^{*}\rightarrow GL(X)$ be such that
$\lambda_{3}(t)\cdot x=x$ for all $x\in Y$, while $\lambda_{3}(t)\cdot x=tx$
for all $x\in Z$. We may write:

$\lambda_{4}(t)\cdot
det^{3}(X)=t[x_{1}(x_{5}x_{9}-x_{6}x_{8})+x_{7}(x_{2}x_{6}-x_{3}x_{5})]+t^{2}[-x_{4}(x_{2}x_{9}-x_{3}x_{8})]$

We may write this as $tQ_{4}+t^{2}Q^{\prime}_{4}$, where
$Q_{4}=x_{1}(x_{5}x_{9}-x_{6}x_{8})+x_{7}(x_{2}x_{6}-x_{3}x_{5})$ and
$Q^{\prime}_{4}=-x_{4}(x_{2}x_{9}-x_{3}x_{8})$. We see that $dim({\cal H})=21$
and exceeds $dim({\cal K})+1$. Moreover, ${\cal K}$ has pure elements of
degree $-1$ and $1$ with respect to $\ell_{4}$ and ${\cal K}_{\ell_{4}f}$ is
of dimension $8$.

###### Remark 5.13

The stabilizer ${\cal H}_{4}$ of $Q_{4}$ is of dimension $21$. The tangent of
exit is $Q^{\prime}_{4}$ and its stabilizer within ${\cal K}$ is of dimension
$8$. Let $\ell_{4}=\log(t^{-1/3}\lambda_{4}(t))$, then $\ell_{4}\in{\cal
H}_{4}$ and $\ell_{4}\cdot det^{3}(X)=Q^{\prime}_{4}$, the tangent of exit.

We end this section with the following table of filtered dimensions for
various 1-PS operating on the form $det_{3}$, the limit $g$ and the dimensions
of the graded components of the stabilizers of the limit $g$ and the tangent
of exit $\ell f$. ${\cal K}_{\ell f}$ is the subalgebra of triple stabilizers.
These calculations illustrate Lemma 4.6 and Proposition 4.3.

$\begin{array}[]{|c|c|c|c|c|c|c|}\hline\cr 1-PS&forms&\vrule\lx@intercol\hfil
dim(({\cal K}_{0})_{i})\hfil\lx@intercol\vrule\lx@intercol&\ell f&dim({\cal
K}_{\ell f})\\\ &&\vrule\lx@intercol\hfil dim(({\cal
H})_{i})\hfil\lx@intercol\vrule\lx@intercol&&\\\ \hline\cr&(f,g)&1&0&-1&&\\\
\hline\cr\hline\cr\ell_{1}&det_{3}&0&8&8&&-\\\
&Q_{1}&0&8&8&Q^{\prime}_{1}&0+4+0\\\ \hline\cr\ell_{2}&det_{3}&0&8&8&&-\\\
&Q_{2}&0&8&8&Q_{3}&0+8+0\\\ \hline\cr\ell_{4}&det_{3}&1&10&5&&-\\\
&Q_{4}&1&13&7&Q^{\prime}_{4}&1+6+1\\\ \hline\cr\end{array}$

##  6 Local Stabilizers and Lie algebra Cohomology obstructions

We begin this section with two examples. Both concern the action of a 1-PS on
a form $f$ with leading term $g$ and direction of approach $f_{b}$. We
continue to use the notation ${\cal K}(t)$, ${\cal K}_{0}$ from the previous
sections. We give an explicit description of the limiting algebra in both
these examples. This motivates the question of recovering ${\cal K}(t)$ from
the ${\cal K}_{0}$.

###### Example 6.1

Let $X=\\{z,y\\}$ and let $f=(y^{2}+z^{2})^{2}$. Let $g=y^{4}$ and note that
with $\lambda(t)\cdot y=y$ and $\lambda(t)\cdot z=t\cdot z$, gives us:

$f(t)=\lambda(t)\cdot f=y^{4}+2t^{2}\cdot y^{2}z^{2}+t^{4}z^{4}$

Thus, $\lim_{t\rightarrow 0}\lambda(t)f(t)=g$, and that $a=0$ and $b=2$ in the
notation of this section.

Let us order $X$ as $x_{1}=z$ and $x_{2}=y$, and let
$e_{11},e_{12},e_{21},e_{22}$ be basis vectors of $gl(X)=gl_{2}$. In other
words, we have:

$e_{11}=\left[\begin{array}[]{cc}1&0\\\ 0&0\\\
\end{array}\right]\>\>e_{12}=\left[\begin{array}[]{cc}0&1\\\ 0&0\\\
\end{array}\right]\>\>e_{21}=\left[\begin{array}[]{cc}0&0\\\ 1&0\\\
\end{array}\right]\>\>e_{22}=\left[\begin{array}[]{cc}0&0\\\ 0&1\\\
\end{array}\right]\>\>$

The stabilizer ${\cal H}$ of $g$ consists of matrices $\mathbb{C}\cdot
e_{11}+\mathbb{C}\cdot e_{12}$, as show below:

${\cal H}=\left[\begin{array}[]{cc}*&*\\\ 0&0\\\ \end{array}\right]$

We may choose ${\cal S}=\mathbb{C}\cdot e_{21}+\mathbb{C}\cdot e_{22}$ and
note that $e_{21}\cdot g=4y^{3}z$ and $e_{22}\cdot g=4g$. Thus $TO_{g}$
consists of the vector $\mathbb{C}\cdot y^{4}+\mathbb{C}\cdot y^{3}z$. We may
decompose $V=TV_{g}$ as:

$V=(\mathbb{C}\cdot y^{4}+\mathbb{C}\cdot y^{3}z)\oplus(\mathbb{C}\cdot
y^{2}z^{2}+\mathbb{C}\cdot yz^{3}+\mathbb{C}\cdot z^{4})=TO_{g}\oplus N.$

We may solve the equation from Proposition 2.5(ii) as follows. As per the
recipe of that equation, we to find $\mathfrak{h}$, $\mathfrak{s}$ which when
applied to the left and right of the expressions below within brackets, gives
$n=0$. We compute the terms of

$\mathfrak{h}(2t^{2}y^{2}z^{2}+t^{4}z^{4})=\mathfrak{s}\cdot(y^{4}+2t^{2}y^{2}z^{2}+t^{4}z^{4})+n,$

in the following table:

$\begin{array}[]{|c|c|c|}\hline\cr\mathfrak{h}&\lambda_{\cal S}&\lambda_{N}\\\
\hline\cr\hline\cr e_{11}&0&4t^{2}y^{2}z^{2}+4t^{4}z^{4}\\\ \hline\cr
e_{12}&t^{2}e_{21}&0\\\ \hline\cr\end{array}$

The stabilizer condition that $\lambda_{N}=0$ gives us the stabilizer of
$f(t)$ as the element $\mathfrak{k}(t)=e_{12}-t^{2}e_{21}$ as shown below:

$\mathfrak{k}(t)=\left[\begin{array}[]{cc}0&1\\\ -t^{2}&0\\\
\end{array}\right]$

One may check, that ${\cal K}(t)=\mathbb{C}\cdot\mathfrak{k}(t)$, and that
${\cal K}_{0}=\mathbb{C}\cdot e_{12}$. We see that $f_{b}=4y^{2}z^{2}$ and
verify that $e_{12}\cdot f_{b}=4y^{3}z\in TO_{g}$, and thus ${\cal
K}_{0}\subseteq{\cal H}_{b}$, the stabilizer of $\overline{f_{b}}$ under the
$\star$-action of ${\cal H}$. $\Box$

###### Example 6.2

Let $X=\\{z,y_{1},y_{2}\\}$ in that order and let
$f=(y_{1}^{2}+y_{2}^{2}+z^{2})^{2}$ and $g=(y_{1}^{2}+y_{2}^{2})^{2}$. Let
$\lambda$ be such that $\lambda(t)\cdot y_{i}=y_{i}$ while $\lambda(t)\cdot
z=tz$. Thus:

$f(t)=(y_{1}^{2}+y_{2}^{2})^{2}+2t^{2}z^{2}(y_{1}^{2}+y_{2}^{2})+t^{4}z^{4}$

Let us have the following elements:

${\cal K}(t)=\left\\{\left[\begin{array}[]{ccc}0&a&b\\\ -at^{2}&0&c\\\
-bt^{2}&-c&0\\\ \end{array}\right]|a,b,c\in\mathbb{C}\right\\}\>\>{\cal
H}=\left\\{=\left[\begin{array}[]{ccc}d&a&b\\\ 0&0&c\\\ 0&-c&0\\\
\end{array}\right]|a,b,c,d\in\mathbb{C}\right\\}$

Whence

${\cal K}_{0}=\\{\left[\begin{array}[]{ccc}0&a&b\\\ 0&0&c\\\ 0&-c&0\\\
\end{array}\right]|\>a,b,c\in\mathbb{C}\\}$

We note that ${\cal K}_{0}\subseteq{\cal H}$ and that ${\cal K}_{0}$
stabilizes $\overline{f_{b}}=\overline{(y_{1}^{2}+y_{2}^{2})z^{2}}$ for the
$\star$-action of ${\cal H}$. In other words, ${\cal K}_{0}\cdot
f_{b}\subseteq TO_{g}$.

It is instructive to form the multiplication table for ${\cal K}(t)$. Let us
put:

$\mathfrak{k}_{1}(t)=\left[\begin{array}[]{ccc}0&1&0\\\ -t^{2}&0&0\\\ 0&0&0\\\
\end{array}\right]\mathfrak{k}_{2}(t)=\left[\begin{array}[]{ccc}0&0&1\\\
0&0&0\\\ -t^{2}&0&0\\\
\end{array}\right]\mathfrak{k}_{3}(t)=\left[\begin{array}[]{ccc}0&0&0\\\
0&0&1\\\ 0&-1&0\\\ \end{array}\right]$

We then have the following structure constants of the Lie bracket:

$\left[\begin{array}[]{c}\mathfrak{k}_{1}(t),\mathfrak{k}_{2}(t)\\\
\mathfrak{k}_{1}(t),\mathfrak{k}_{3}(t)\\\
\mathfrak{k}_{2}(t),\mathfrak{k}_{3}(t)\end{array}\right]=\left[\begin{array}[]{rrr}0&0&-t^{2}\\\
0&1&0\\\ -1&0&0\\\
\end{array}\right]\left[\begin{array}[]{c}\mathfrak{k}_{1}(t)\\\
\mathfrak{k}_{2}(t)\\\ \mathfrak{k}_{3}(t)\end{array}\right]$

Here $\mathfrak{k}_{1}(t),\mathfrak{k}_{2}(t)$ on the left denotes
$[\mathfrak{k}_{1}(t),\mathfrak{k}_{2}(t)]$ and likewise for the other rows on
the left. We note that $\\{\mathfrak{k}_{i}(0)|i=1,2,3\\}$ is a basis for
${\cal K}_{0}$ and the structure constants are precisely those of ${\cal
K}(t)$ specialized to $t=0$.

In the above example the structure constants $(\alpha_{ij}^{k}(t))$ of ${\cal
K}(t)$ were elements of $\mathbb{C}[t]$ and where putting $t=0$ gave us ${\cal
K}_{0}$. Moreover we see that the $\mathbb{C}$-algebra ${\cal K}(t_{0})$,
i.e., ${\cal K}(t)$ instantiated at a $t_{0}\in\mathbb{C}$, may be very
different from ${\cal K}_{0}$, and perhaps “more semisimple” than both ${\cal
H}$ or ${\cal K}_{0}$. We look at the process of constructing ${\cal K}(t)$
from ${\cal K}_{0}$. This naturally leads us to a Lie algebra cohomology
formulation. We continue with the notation and transversality assumption of
Section 3.1.

For any system $(\alpha_{ij}^{k}(t))$ of structure constants of ${\cal K}(t)$,
the verification that these do satisfy the Jacobi conditions is in the
polynomial ring $\mathbb{C}[t]$. Let us consider the ring
$\mathbb{C}[\epsilon_{D}]\cong\mathbb{C}[t]/(t^{D})$ and Lie algebras over
$\mathbb{C}[\epsilon_{D}]$. If $(\alpha_{ij}^{k}(t))$ satisfy the Jacobi
identity, then so do $(\alpha_{ij}^{k}(\epsilon_{D}))$. Let us call the Lie
algebra so generated as ${\cal K}_{D-1}$. Note that ${\cal K}_{D-1}$ is a Lie
algebra, both over $\mathbb{C}$ and over $\mathbb{C}[\epsilon_{D}]$ and that
${\cal K}_{0}$ matches with our earlier definition. Note that if
$dim_{\mathbb{C}[t]}({\cal K}(t))=k$, then each ${\cal K}_{D-1}$ is a free
$\mathbb{C}[\epsilon_{D}]$-module of the same dimension. Moreover, for general
$a>0$, we also have surjective Lie algebra homomorphisms (over $\mathbb{C}$)
${\cal K}_{a}\rightarrow{\cal K}_{a-1}$. Finally, for any ${\cal K}(t)$ over
$\mathbb{C}[t]$, there is a large $D$, that the structure constants
$(\alpha_{ij}^{k}(t))$ may be verified in $\mathbb{C}[\epsilon_{D}]$. However,
the corresponding ${\cal K}_{D-1}$ will not match ${\cal K}(t)$ in structure.
With this remark, let us study the structure of ${\cal K}_{1}$.

Let $\mathbb{C}[\epsilon]$ denote the ring formed by adding the variable
$\epsilon$ with $\epsilon^{2}=0$. For any $\mathbb{C}$-vector space or algebra
$X$, let $X[\epsilon]$ denote the extension of coefficients, i.e.,
$X[\epsilon]=\mathbb{C}[\epsilon]\otimes_{\mathbb{C}}X$. Note that
$X[\epsilon]=X\oplus\epsilon X$ as vector spaces. We also define the
“evaluation at zero” map $e_{0}:X[\epsilon]\rightarrow X$ as follows. For an
element $x=x_{0}+\epsilon x_{1}\in X[\epsilon]$, we set $e_{0}(x)=x_{0}$. This
will also be denoted as $x(0)$. Let $V[\epsilon]$ and ${\cal G}[\epsilon]$ be
obtained from $V$ and ${\cal G}$ as above. Note that ${\cal G}[\epsilon]$ is a
Lie algebra and $V[\epsilon]$, a ${\cal G}[\epsilon]$ module. By the same
token $V[\epsilon]$ is also an ${\cal H}[\epsilon]$ module under the
$\star$-action.

###### Lemma 6.3

${\cal G}[\epsilon]$ is a Lie algebgra over $\mathbb{C}$ and the map
$e_{0}:{\cal G}[\epsilon]\rightarrow{\cal G}$ defined as
$e_{0}(\mathfrak{g}+\epsilon\mathfrak{g}^{\prime})=\mathfrak{g}$ is a Lie
$\mathbb{C}$-algebra homomorphism.

Let us consider the element $p=g+\epsilon f_{b}\in V[\epsilon]$ and construct
the stabilizer of the point $p$ in ${\cal G}[\epsilon]$. For a typical element
$\mathfrak{g}_{0}+\epsilon\mathfrak{g}_{1}$, we see that the stabilizer
condition implies:

$\mathfrak{g}_{0}\cdot g=0\>\>\mbox{ and }\mathfrak{g}_{0}\cdot
f_{b}+\mathfrak{g}_{1}\cdot g=0$ (6)

The first condition states that $\mathfrak{g}_{0}\in{\cal H}$. The second
condition further implies that $\mathfrak{g}_{0}\in{\cal H}_{b}\subseteq{\cal
H}$, the stabilizer of $\overline{f_{b}}$ under the $\star$-action of ${\cal
H}$ on $V$. This condition may also be interpreted as follows. Let us identify
$TO_{g}$ with the quotient ${\cal G}/{\cal H}$, given simply by
$d_{g}:\mathfrak{g}\rightarrow\mathfrak{g}\cdot g\in TO_{g}$.

###### Lemma 6.4

The map $d_{g}:{\cal G}/{\cal H}\rightarrow TO_{g}\subseteq V$ is an
isomorphism of ${\cal H}$-modules.

Proof: Note that ${\cal H}$ acts on $V$ as maps in $End(V)$ under the Lie
bracket. Let $\mathfrak{s}\in{\cal G}$ and let $\\{\mathfrak{s}\\}$ denote the
coset $\mathfrak{s}+{\cal H}$. The operation of $\mathfrak{h}\in{\cal H}$ on
${\cal G}/{\cal H}$ is given by
$\mathfrak{h}\cdot\\{\mathfrak{s}\\}=\\{[\mathfrak{h},\mathfrak{s}]\\}$.
Whence, we then have:

$\begin{array}[]{rcl}d_{g}(\\{[\mathfrak{h},\mathfrak{s}]\\})&=&[\mathfrak{h},\mathfrak{s}]\cdot
g\\\ &=&(\mathfrak{h}\cdot\mathfrak{s}-\mathfrak{s}\cdot\mathfrak{h})\cdot
g\\\ &=&\mathfrak{h}\cdot\mathfrak{s}\cdot g\mbox{ (since $\mathfrak{h}\cdot
g=0$)}\\\ &=&\mathfrak{h}\cdot d_{g}(\mathfrak{s})\end{array}$

This completes the proof. $\Box$

###### Lemma 6.5

The stabilizer ${\cal H}_{p}\subseteq{\cal G}]\epsilon]$ is given by the set
$\mathfrak{h}_{0}+\epsilon\mathfrak{g}$ such that (i)
$\mathfrak{h}_{0}\in{\cal H}_{b}$, and (ii) $\mathfrak{g}\in
d_{g}^{-1}(-\mathfrak{h}_{0}f_{b})$.

Proof: Let us look at the two parts of Eq. 6. The first implies that
$\mathfrak{h}_{0}\in{\cal H}$. The second implies that, in fact,
$\mathfrak{h}_{0}$ stabilize $\overline{f_{b}}$, and thus
$\mathfrak{h}_{0}\in{\cal H}_{b}$. This proves (i). For (ii), note that the
second condition of Eq. 6, gives us that $\mathfrak{h}_{0}\cdot
f_{b}+d_{g}(\mathfrak{g})=0$. This implies (ii). $\Box$

Let us now define the map $d_{b}:{\cal H}_{b}\rightarrow{\cal G}/{\cal H}$ as
follows. For any element $\mathfrak{s}\in{\cal G}$, let $\\{\mathfrak{s}\\}$
denote the coset $\mathfrak{s}+{\cal H}$ in ${\cal G}/{\cal H}$. For an
$\mathfrak{h}\in{\cal H}_{b}$, let $d_{b}(\mathfrak{h})$ be that element
$\\{\mathfrak{s}\\}\in{\cal G}/{\cal H}$ such that $\mathfrak{s}\cdot
g=\mathfrak{h}\cdot f_{b}$, i.e., $d_{g}(\\{\mathfrak{s}\\})=\mathfrak{h}\cdot
f_{b}$.

###### Definition 6.6

A derivation (see, for example, [Wag10]) $d$ from ${\cal G}$ to a ${\cal
G}$-module W is a ${\mathbb{C}}$-linear map from ${\cal G}$ to $W$ such that
for all $\mathfrak{h}_{1},\mathfrak{h}_{2}\in{\cal G}$ we have

$d([\mathfrak{h}_{1},\mathfrak{h}_{2}])=\mathfrak{h}_{1}\circ
d(\mathfrak{h}_{2})-\mathfrak{h}_{2}\circ d(\mathfrak{h}_{1}).$

###### Lemma 6.7

(1) ${\cal G}/{\cal H}$ is an ${\cal H}_{b}$-module. The map $d_{b}:{\cal
H}_{b}\rightarrow{\cal G}/{\cal H}$ is a derivation. (2) Moreover, the
stabilizer ${\cal H}_{p}\subseteq{\cal G}]\epsilon]$ is also given by the set
$\mathfrak{h}+\epsilon\mathfrak{g}$ such that (i) $\mathfrak{h}_{0}\in{\cal
H}_{b}$, and (ii) $\mathfrak{g}\in d_{b}(-\mathfrak{h}_{0})$.

Proof: Since $[{\cal H}_{b},{\cal H}]\subseteq{\cal H}$, the quotient ${\cal
G}/{\cal H}$ is an ${\cal H}_{b}$-module under the adjoint action. Let
$\mathfrak{h}_{1},\mathfrak{h}_{2}\in{\cal H}_{b}$. Suppose
$\mathfrak{h}_{i}f_{b}=\mathfrak{s}_{i}g$ for some $\mathfrak{s}_{i}\in{\cal
G}$, whence $d_{b}(\mathfrak{h}_{i})=\\{\mathfrak{s}_{i}\\}$, the coset of
$\mathfrak{s}_{i}$. We must show that
$d_{b}([\mathfrak{h}_{1},\mathfrak{h}_{2}])=\mathfrak{h}_{1}\circ
d_{b}(\mathfrak{h}_{2})-\mathfrak{h}_{2}\circ d_{b}(\mathfrak{h}_{1})$ as
elements of ${\cal G}/{\cal H}$. We check this through the isomorphism
$d_{g}$. Applying $d_{g}$ to the RHS, we get:

$\begin{array}[]{rcl}(\mathfrak{h}_{1}\circ
d_{b}(\mathfrak{h}_{2})-\mathfrak{h}_{2}\circ d_{b}(\mathfrak{h}_{1}))\cdot
g&=&(\\{[\mathfrak{h}_{1},\mathfrak{s}_{2}]\\}-\\{[\mathfrak{h}_{2},\mathfrak{s}_{1}]\\})\cdot
g\\\
&=&\mathfrak{h}_{1}\mathfrak{s}_{2}g-\mathfrak{h}_{2}\mathfrak{s}_{1}g\mbox{
(since $\mathfrak{h}_{i}\cdot g=0$)}\end{array}$

We check this with the LHS

$d_{g}([\mathfrak{h}_{1},\mathfrak{h}_{2}])=[\mathfrak{h}_{1},\mathfrak{h}_{2}]\cdot
f_{b}=\mathfrak{h}_{1}\mathfrak{h}_{2}f_{b}-\mathfrak{h}_{2}\mathfrak{h}_{1}f_{b}=(\mathfrak{h}_{1}\mathfrak{s}_{2}-\mathfrak{h}_{2}\mathfrak{s}_{1})\cdot
g$

This proves (1). Concerning (2), the only change from lemma 6.5 is in (ii). It
is easily seen that $\mathfrak{g}\in d_{g}^{-1}(-\mathfrak{h}_{0}f_{b})$ is
equivalent to the condition that $\mathfrak{g}\in d_{b}(-\mathfrak{h}_{0})$.
$\Box$

###### Remark 6.8

The above claim is equivalent to the statement that ${\cal H}_{p}$ is a Lie
subalgebra of ${\cal G}[\epsilon]$.

We collect some properties of ${\cal H}_{p}$.

###### Lemma 6.9

The image $e_{0}({\cal H}_{p})$ is ${\cal H}_{b}$. Moreover, the nilpotent
subalgebra $\epsilon{\cal H}\subseteq{\cal G}[\epsilon]$ is a subset of ${\cal
H}_{p}$.

###### Definition 6.10

Let ${\cal K}\subseteq{\cal H}_{b}$ be a subalgebra and let
$dim_{\mathbb{C}}({\cal K})=K$. Let subalgebra $\overline{{\cal
K}}\subseteq{\cal H}_{p}$ be such that (i) $\overline{\cal K}$ is generated as
a $\mathbb{C}[\epsilon]$-module by $K$ elements
$\mathfrak{k}_{1},\ldots,\mathfrak{k}_{K}$, and (ii) $e_{0}(\overline{\cal
K})={\cal K}$. We call $\overline{\cal K}$ an $\epsilon$-deformation of ${\cal
K}$. Note that $dim_{\mathbb{C}}(\overline{\cal K})=2K$.

The condition that $\overline{\cal K}$ is generated by the $K$ elements as a
$\mathbb{C}[\epsilon]$-module implies that we can write the Lie bracket using
the structure coefficients $(\alpha_{ij}^{k})$, as:

$[\mathfrak{k}_{i},\mathfrak{k}_{j}]=\sum_{k}\alpha_{ij}^{k}\mathfrak{k}_{k}\mbox{
for some $(\alpha_{ij}^{k})\in\mathbb{C}[\epsilon]$}$

###### Lemma 6.11

If $\overline{\cal K}$ is an $\epsilon$-deformation of ${\cal K}$ then
$(\mathfrak{b}_{i})=(e_{0}(\mathfrak{k}_{i}))$ is a $\mathbb{C}$-basis for
${\cal K}$ and its structure coefficients are precisely
$(\alpha_{ij}^{k}(0))$, the evaluations of $(\alpha_{ij}^{k})$ at $0$. Thus,
we have:

$[\mathfrak{b}_{i},\mathfrak{b}_{j}]=\sum_{k}\alpha_{ij}^{k}(0)\mathfrak{k}_{k}$

###### Proposition 6.12

The existence of $\overline{\cal K}$ is equivalent to the existence of a
derivation $\overline{d_{b}}:{\cal K}\rightarrow{\cal G}/{\cal K}$ which
extends the derivation $d_{b}:{\cal K}\rightarrow{\cal G}/{\cal H}$ as shown
in the diagram below:

${{{\cal G}/{\cal K}}}$${{{\cal G}/{\cal H}}}$${{\cal K}}$${{\cal
H}_{b}}$$\scriptstyle{\overline{d_{b}}}$$\scriptstyle{d_{b}}$

Proof: Suppose that $\overline{d_{b}}$ is such an extension. Let us define
$\overline{\cal K}$ as follows:

$\overline{\cal K}=\\{\mathfrak{h}+\epsilon\mathfrak{s}|\mathfrak{h}\in{\cal
K},\mathfrak{s}\in{\cal G}\mbox{ such that
}\mathfrak{s}\in\overline{d_{b}}(-\mathfrak{h})\\}\oplus\epsilon{\cal K}$

Clearly, since $\overline{d_{b}}$ extends $d_{b}$, we have
$\overline{d_{b}}(-\mathfrak{h})\subseteq d_{b}(-\mathfrak{h})$ and hence
every element of $\overline{\cal K}$ stabilizes $p$. Thus $\overline{\cal
K}\subseteq{\cal H}_{p}$. Other properties of Defn. 6.10 are easily verified.
Thus, all that remains to show is that $\overline{\cal K}$ is closed under the
Lie bracket, i.e., it is a subalgebra of ${\cal H}_{p}$.

Let us now verify the Lie bracket condition. For any elements
$\mathfrak{k}_{1}=\mathfrak{h}_{1}+\epsilon\mathfrak{s}_{1}$ and
$\mathfrak{k}_{2}=\mathfrak{h}_{2}+\epsilon\mathfrak{s}_{2}$ we must check
that $[\mathfrak{k}_{1},\mathfrak{k}_{2}]\in\overline{\cal K}$. But

$[\mathfrak{k}_{1},\mathfrak{k}_{2}]=[\mathfrak{h}_{1},\mathfrak{h}_{2}]+\epsilon([\mathfrak{h}_{1},\mathfrak{s}_{2}]+[\mathfrak{s}_{1},\mathfrak{h}_{2}])$

Since $[\mathfrak{h}_{1},\mathfrak{h}_{2}]\in{\cal K}$, there is an element
$[\mathfrak{h}_{1},\mathfrak{h}_{2}]+\epsilon\mathfrak{s}\in\overline{\cal K}$
with the condition that
$\mathfrak{s}\in\overline{d_{b}}(-[\mathfrak{h}_{1},\mathfrak{h}_{2}])$. We
must check that

$\mathfrak{s}-([\mathfrak{h}_{1}\mathfrak{s}_{2}]+[\mathfrak{s}_{1},\mathfrak{h}_{2}])\in{\cal
K}.$

Now, as elements of ${\cal G}/{\cal K}$, we have:

$\begin{array}[]{rcl}[\mathfrak{h}_{1},\mathfrak{s}_{2}]+[\mathfrak{s}_{1},\mathfrak{h}_{2}]&=&[\mathfrak{h}_{1},\mathfrak{s}_{2}]-[\mathfrak{h}_{2},\mathfrak{s}_{1}]\\\
&=&\mathfrak{h}_{1}\cdot\overline{d_{b}}(-\mathfrak{h}_{2})-\mathfrak{h}_{2}\cdot\overline{d_{b}}(-\mathfrak{h}_{1})\\\
&=&\overline{d_{b}}(-[\mathfrak{h}_{1},\mathfrak{h}_{2}])\mbox{ since
$\overline{d_{b}}$ is a derivation}\\\ &=&\mathfrak{s}+{\cal K}\\\
\end{array}$

Thus
$\mathfrak{s}-[\mathfrak{h}_{1},\mathfrak{s}_{2}]-[\mathfrak{s}_{1},\mathfrak{h}_{2}]\in{\cal
K}$. Whence:

$[\mathfrak{h}_{1}+\epsilon\mathfrak{s}_{1},\mathfrak{h}_{2}+\epsilon\mathfrak{s}_{2}]=[\mathfrak{h}_{1},\mathfrak{h}_{2}]+\epsilon\mathfrak{s}+\epsilon\mathfrak{h}\mbox{
with }\mathfrak{h}\in{\cal K}$

This proves that $\overline{\cal K}$ in indeed a Lie subalgebra. Note that
$\epsilon{\cal K}\subseteq\epsilon\overline{\cal K}\subseteq\overline{\cal K}$
and thus if $\mathfrak{h}+\epsilon\mathfrak{s}\in\overline{\cal K}$ then so is
$\mathfrak{h}+\epsilon\mathfrak{s}^{\prime}$, where
$\mathfrak{s}^{\prime}\in\mathfrak{s}+{\cal K}$.

Conversely, if $\overline{\cal
K}=\\{\mathfrak{h}+\epsilon\mathfrak{s}|\mathfrak{h}\in{\cal K}\\}$ is such an
extension, then put $\overline{d_{b}}(\mathfrak{h})=-\mathfrak{s}+{\cal K}$.