# Minimal Integral Models for principal series Weil characters

Luca Candelori, Yatin Patel<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of
Mathematics, Wayne State University, 656 W. Kirby, Detroit, MI 48202

###### Abstract.

We prove a conjecture of Udo Riese about the minimal ring of definition for
principal series Weil characters of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$, for $p$ an odd prime. More
precisely, we show that the $(p+1)/2$-dimensional Weil characters can be
realized over the ring of integers of $\mathbb{Q}(\sqrt{\varepsilon p})$,
where $\varepsilon=(-1)^{(p-1)/2}$, and we provide explicit integral models
over these quadratic rings. We do so by studying the Galois action on the
integral models of Weil characters recently discovered by Yilong Wang.

##  1\. Introduction

Let $\rho:G\rightarrow\operatorname{GL}(V)$ be an irreducible complex
representation of a finite group $G$ of exponent $e$. A famous theorem of
Brauer states that there is a choice of basis for $V$ (i.e. a model for
$\rho$) so that $\rho(g)$ is a matrix with entries in the cyclotomic field
$\mathbb{Q}(\zeta_{e})$, for all $g\in G$. Since the entries of the character
table for $G$ are always algebraic integers, it is natural to formulate a much
stronger conjecture stating that, in fact, the basis can be chosen so that the
matrix entries lie in the ring of integers
$\mathbb{Z}[\zeta_{e}]\subseteq\mathbb{Q}(\zeta_{e})$ [2],[5]. In this case,
we say that $\rho$ has an integral model over the ring
$R=\mathbb{Z}[\zeta_{e}]$. Proving the existence of such integral models is a
notoriously difficult problem in integral representation theory. Even when
existence can be ascertained, the arguments often cannot be directly adapted
to explicitly construct such integral models.

In the case $G=\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$, for $p\geq 3$ a
prime, Udo Riese [7] proved the existence of integral models over
$\mathbb{Z}[\zeta_{e}]$. The question remains however whether the result is
best possible, or whether for this particular family of groups there are
integral models defined over proper subrings $R\subset\mathbb{Z}[\zeta_{e}]$.
For each irreducible character $\chi$, the entries of the character table
provide a minimal ring of definition $R_{min}(\chi)$, but it is not clear a
priori whether such a minimal integral model over $R_{min}(\chi)$ actually
exists. For example, the character of the Steinberg representation $\chi=St$
(the unique irreducible character of dimension $p$) has entries in
$\mathbb{Z}$, thus $R_{min}(St)=\mathbb{Z}$. It can be shown that an integral
model for $St$ exists over $\mathbb{Z}$ [7, Prop. 1], [3], thus providing a
minimal integral model for $St$. Similarly, minimal integral models can easily
be found for the characters belonging to the irreducible principal series of
$G$.

The purpose of this article is to explicitly provide minimal integral models
for the Weil characters arising from the reducible principal series of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$. For each $p\geq 3$, there are
precisely two non-isomorphic such irreducible characters $\xi_{1}$ and
$\xi_{2}$, each of dimension $(p+1)/2$. The entries in the character table
give

$R_{min}(\xi_{i})=\mathbb{Z}\left[\frac{1+\sqrt{\varepsilon
p}}{2}\right],\quad\varepsilon=\left\\{\begin{array}[]{cl}1&\text{ if }p\equiv
1\mod 4\\\ -1&\text{ if }p\equiv 3\mod 4\end{array}\right.$

for both $i=1,2$. We therefore seek explicit integral models defined over this
quadratic ring $R=R_{min}(\xi_{i})$, which in all cases coincides with the
ring of integers of $\mathbb{Q}(\sqrt{\varepsilon p})$. In [7], it is
conjectured that such minimal integral models should always exist, and
existence is proved under the restriction $p\equiv 5\mod 8$ [7, Prop. 4]. The
methods of [7] are based on class-field theory and do not provide explicit
integral models even under the more restrictive assumptions. In this article
(Theorems 4.1 and 4.3) we prove the existence of minimal integral models for
any prime $p$, with no restrictions, and we provide them explicitly. For
example, for the prime $p=7$ and $\xi=\xi_{1}$ we obtain the following model

$\xi_{1}(\mathfrak{s})=\begin{bmatrix}-1&\frac{1}{2}(1-\sqrt{-p})&0&0\\\
-\frac{1}{2}(1+\sqrt{-p})&1&0&0\\\
\frac{1}{2}(1-\sqrt{-p})&\frac{1}{2}(1+\sqrt{-p})&0&-1\\\
1&-1&1&0\end{bmatrix}$

and

$\xi_{1}(\mathfrak{t})=\begin{bmatrix}0&0&0&-1\\\
1&0&0&\frac{1}{2}(1-\sqrt{-p})\\\ 0&1&0&1\\\
0&0&1&\frac{1}{2}(1+\sqrt{-p})\end{bmatrix},$

where $\mathfrak{s}=\left(\begin{smallmatrix}0&-1\\\
1&0\end{smallmatrix}\right)$ and $\mathfrak{t}=\left(\begin{smallmatrix}1&1\\\
0&1\end{smallmatrix}\right)$ are the standard choices of generators for
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$. Our methods are based on recent
work of Yilong Wang [9], who provided explicit integral models for $\xi_{i}$
over $\mathbb{Z}[\zeta_{p}]$. By extensive explicit calculations, the authors
observed that in fact Wang’s models are defined over the minimal rings
$R_{min}(\xi_{i})$. This is what we prove in this article, by studying the
action of the Galois group
$\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ on Wang’s integral
models.

There are also two irreducible cuspidal Weil characters in the character table
of $\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$, each of dimension $(p-1)/2$.
Explicit integral models for these have recently been provided over
$\mathbb{Z}[\zeta_{p}]$ by Shaul Zemel [11], who was also motivated by Wang’s
work. However, minimal integral models are not known for such cuspidal Weil
characters, and the methods of this article do not directly extend to the
cuspidal case. Riese proves existence whenever $p\equiv 3\mod 4$ [7, Prop. 3],
and provides evidence for non-existence in the case $p\equiv 1\mod 4$, but no
explicit models are known in either case. This is certainly a subject for
further investigation. In addition, it would be interesting to extend these
results to the case $q=p^{r}$ a prime power, which was the original scope of
Riese’s conjectures.

We now summarize the contents of this article. In Section 2, we recall the
definition of the Weil representation, which provides explicit models for
representations of $\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ defined over
the ring $\mathbb{Z}[1/p,\zeta_{p}]$. Since conventions about the Weil
representation vary between different sources, we collect all the necessary
formulas in the section. In Section 3, we explain how to obtain explicit
models for the Weil characters of $\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$
from the Weil representation, and we recall Wang’s construction of integral
models for $\xi_{1}$ and $\xi_{2}$ [9]. In Section 4 we prove our main results
Thms. 4.1, 4.2 and 4.3, which show that Wang’s integral models are in fact
minimal. Finally, in Section 5 we provide explicit examples of the minimal
integral models for $p=7$ and $p=13$ and for both $\xi_{1}$ and $\xi_{2}$.

We would like to thank Siu-Hung Ng, Yilong Wang and Shaul Zemel for sharing
many great ideas and conversations about this problem. The first author is
supported by the U.S. Department of Energy, Office of Science, Basic Energy
Sciences, under Award Number DE-SC-SC0022134. The second author is supported
by a 2021-2022 Thomas C. Rumble University Graduate Fellowship through the
Department of Mathematics at Wayne State University.

##  2\. The Weil representation

In this section we recall the construction of the Weil representation of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$, for $p>2$ a prime, first
introduced in [10]. This construction will be used to provide explicit models
for the Weil characters $\xi_{1}$ and $\xi_{2}$ over the ring
$\mathbb{Z}[1/p,\zeta_{p}]$, and it serves as the point of departure for
providing integral models over $\mathbb{Z}[\zeta_{p}]$ [9]. There are many
formulas and conventions about the Weil representation in the literature, and
some references are hard to locate or translate. We thus find it necessary to
state explicitly all the facts, conventions, and formulas that we will be
using in this article, while providing references for the proofs.

Let $p>2$ be a prime, let $C_{p}=(\mathbb{Z}/p\mathbb{Z},+)$ be the cyclic
group of order $p$ and let $K_{p}:=C_{p}\times C_{p}$. Let
$\nu_{p}=\\{\zeta\in\mathbb{C}:\zeta^{p}=1\\}$ be the group of $p$-th roots of
unity in $\mathbb{C}$ and let $\zeta_{p}:=e^{2\pi i/p}$ be the principal
$p$-th root of unity. Let $Q(x):C_{p}\rightarrow\mathbb{Q}/\mathbb{Z}$ be a
quadratic form on $C_{p}$ and let $B(x,y)=Q(x+y)-Q(x)-Q(y)$ be its associated
bilinear form.

######  Definition 2.1.

The Heisenberg group $H_{Q}$ is the set of elements
$(\lambda,x,y)\in\nu_{p}\times K_{p}$, together with multiplication given by

$(\lambda_{1},x_{1},y_{1})(\lambda_{2},x_{2},y_{2})=(\lambda_{1}\lambda_{2}\,e^{2\pi
iB(x_{1},y_{2})},x_{1}+x_{2},y_{1}+y_{2}).$

Note that $\nu_{p}=Z(H_{p})$ and $H_{Q}$ is a (non-trivial) central extension:

$1\to\nu_{p}\to H_{Q}\to K_{p}\to 0.$

The generators of $H_{Q}$ are $(\zeta_{p},0,0)$, $(1,1,0)$, and $(1,0,1)$. The
subgroup of $K_{p}$ that is generated by $(x,0)$ where $x\in\mathbb{F}_{p}$ is
the called the modulation subgroup. Let
$A=\\{(\lambda,x,0),\lambda\in\nu_{p},x\in C_{p}\\}.$ Then $A$ is an index $p$
subgroup of $H_{Q}$, projecting onto the modulation subgroup. Since any
isomorphism $C_{p}\cong\nu_{p}$ corresponds to a choice of a primitive $p$-th
root of unity, we choose the one that sends $1\mapsto\zeta_{p}$. With this
choice, given characters $\phi_{1}$ and $\phi_{2}$ of $C_{p}$, we define a
one-dimensional character $\psi:A\rightarrow\mathbb{C}^{\times}$ by

$\psi(\lambda,x,0)=\phi_{1}(x)\phi_{2}(\lambda).$

We induce the representation of the subgroup $A$ to obtain an irreducible
$p$-dimensional representation $\sigma:=\operatorname{Ind}_{A}^{H_{p}}(\psi)$.
The underlying vector space for this representation is

$V=\left\\{\text{functions: }C_{p}\to\mathbb{C}\right\\}$

which has a canonical basis of delta functions
$\\{\delta_{0},\cdots,\delta_{p-1}\\}$, defined by $\delta_{j}(k)=\delta_{jk}$
(Kronecker’s $\delta$-function). With respect to this basis, we obtain an
explicit model for the irreducible representation

$\sigma:H_{Q}\to\operatorname{GL}(V)\simeq\operatorname{GL}_{p}(\mathbb{C}),$

such that an element $(\zeta,0,0)\in\nu_{p}\subset H_{Q}$ acts by scalar
multiplication via $\phi_{2}(\zeta)$. In addition, the generator $(1,1,0)$
acts via the diagonal matrix with entries

$\sigma(1,1,0)_{jj}=\phi_{1}(1)\phi_{2}(e^{2\pi iB(1,j)}),$

while the generator $(1,0,1)$ acts by a permutation matrix:

$\begin{split}\sigma(1,0,1)&=\begin{bmatrix}0&0&0&\cdots&0&1\\\
1&0&0&\cdots&0&0\\\ 0&1&0&\cdots&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&\cdots&1&0\end{bmatrix}.\end{split}$

The “natural character” of $\nu_{p}$ is defined to be the inclusion of
$\nu_{p}$ into $\mathbb{C}$. In our conventions, this corresponds to
$\phi_{2}$ being the identity on $\nu$:
$\phi_{2}(\zeta_{p}^{j})=\zeta_{p}^{j}$. For each prime $p$, there exists a
unique (up to isomorphism) irreducible representation of $H_{Q}$ such that
$\nu_{p}$ acts by its natural character. This is the Mackey-Stone-von Neumann
Theorem [1]:

######  Theorem 2.2 (Mackey-Stone-von Neumann).

For fixed non-trivial (unitary) central character, up to isomorphism there is
a unique irreducible (unitary) representation of the Heisenberg group with
that central character. Further, any (unitary) representation with that
central character is a multiple of that irreducible.

The Theorem is usually stated for more general Heiseberg groups, including
infinite ones, arising as central extensions of symplectic vector spaces. Note
however that for the finite Heisenberg group $H_{Q}$ this result can be easily
proved by computing the character table.

######  Remark 2.3.

The character table of $H_{Q}$ can be computed by inducing characters of $A$
as above. It follows that every irreducible character of $H_{Q}$ has a minimal
integral model defined over $\mathbb{Z}[\zeta_{p}]$, verifying the stronger
form of Brauer’s theorem stated in the Introduction. Since $H_{Q}$ is solvable
of exponent $p$, this also follows from the more general result of [2].

In order to construct the Weil representation of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$, we need to consider
automorphisms of $H_{Q}$, as follows.

######  Definition 2.4.

The Clifford group $\mathscr{C}$ is the group of automorphisms of the
Heisenberg group that fix its center $\nu_{p}$. An automorphism
$\psi\in\mathscr{C}$ is symmetric if $\psi(\lambda,-x,-y)=\psi(\lambda,x,y)$.
Let $\mathscr{C}^{\operatorname{Sym}}$ be the subgroup of $\mathscr{C}$
consisting of all symmetric automorphisms that fix $\nu_{p}$.

Let $\psi\in\mathscr{C}^{\operatorname{Sym}}(H_{Q})$ and let
$\sigma^{\psi}:=\sigma\circ\psi$. Since $\psi$ fixes the center, the
representation $\sigma^{\psi}$ is irreducible, and $\nu_{p}$ acts by its
natural character, so that $\sigma\cong\sigma^{\psi}$ by the Mackey-Stone-von
Neumann Theorem. Therefore there exists a linear transformation
$M(\psi)\in\operatorname{GL}(V)$ such that
$M(\psi)\sigma=\sigma^{\psi}M(\psi)$, that is, making the following diagram

${H_{Q}}$${\operatorname{GL}(V)}$${H_{Q}}$${\operatorname{GL}(V)}$$\scriptstyle{\sigma}$$\scriptstyle{\psi}$$\scriptstyle{M(\psi)}$$\scriptstyle{\sigma^{\psi}}$

commutative. The linear map $M(\psi)$ is only defined up to scalars, but it
can be shown [10] that such scalars can be chosen for each $\psi$ so that the
function $\psi\mapsto M(\psi)$ gives a group homomorphism
$\mathscr{C}^{\operatorname{Sym}}\rightarrow\operatorname{GL}(V)$. Since $p>2$
is an odd prime, there is an isomorphism
$\mathscr{C}^{\operatorname{Sym}}(H_{Q})\simeq\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$
(see [4]) and we obtain the following:

######  Definition 2.5.

The Weil representation

$\rho_{Q}:\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})\longrightarrow\operatorname{GL}(V),$

is the representation of $\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ obtained
by composing the isomorphism
$\mathscr{C}^{\operatorname{Sym}}(H_{Q})\simeq\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$
with the homomorphism $\psi\mapsto M(\psi)$.

We now write an explicit model for the Weil representation using the canonical
basis of delta functions for $V$. The generator
$\mathfrak{s}\in\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ maps
isomorphically to the symmetric automorphism
$\psi_{\mathfrak{s}}\in\mathscr{C}^{\operatorname{Sym}}$ given by

$\psi_{\mathfrak{s}}(\lambda,x,y)=(\lambda e^{2\pi iB(-x,y)},-y,x).$

The corresponding linear transformation $M(\psi_{s})$ is a discrete Fourier
transform (DFT) given by

(1)
$\rho_{Q}({\mathfrak{s}})=M(\psi_{\mathfrak{s}})=\dfrac{1}{\sqrt{\varepsilon
p}}\begin{bmatrix}1&1&1&\cdots&1\\\ 1&e^{2\pi iB(1,1)}&e^{2\pi
iB(1,2)}&\cdots&e^{2\pi iB(1,p-1)}\\\ 1&e^{2\pi iB(2,1)}&e^{2\pi
iB(2,2)}&\cdots&e^{2\pi iB(2,p-1)}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
1&e^{2\pi iB(p-1,1)}&e^{2\pi iB(p-1,2)}&\cdots&e^{2\pi
iB(p-1,p-1)}\end{bmatrix}.$

The other generator
${\mathfrak{t}}\in\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ maps to the
symmetric automorphism
$\psi_{\mathfrak{t}}\in\mathscr{C}^{\operatorname{Sym}}$ given by

$\psi_{\mathfrak{t}}(\lambda,x,y)=(\lambda e^{2\pi iQ(y)},x+y,y)$

and

(2)
$\rho_{Q}({\mathfrak{t}})=M(\psi_{\mathfrak{t}})=\begin{bmatrix}1&0&0&\cdots&0&0\\\
0&e^{2\pi iQ(1)}&0&\cdots&0&0\\\ 0&0&e^{2\pi iQ(2)}&\cdots&0&0\\\
\vdots&\vdots&\vdots&\dots&\vdots\\\ 0&0&0&\cdots&0&e^{2\pi
iQ(p-1)}\end{bmatrix}.$

It is easily checked that the identities
$\rho_{Q}({\mathfrak{s}})^{2}=(\rho_{Q}({\mathfrak{s}})\rho_{Q}({\mathfrak{t}}))^{3}$
and $\rho_{Q}({\mathfrak{s}})^{4}=\mathds{1}$ are satisfied, so that
$\rho_{Q}$ is indeed a representation of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ dimension $p$.

The isomorphism class of each Weil representation only depends on the
equivalence class of the quadratic form $Q$. Note that there are two
nonequivalent quadratic forms on $C_{p}$:

(3) $Q_{1}(x)=x^{2}/p\quad\text{ and }\quad Q_{2}(x)=cx^{2}/p$

where $c$ is a quadratic non-residue modulo $p$. Therefore we obtain two non-
isomorphic Weil representations

(4)
$\rho_{1},\rho_{2}:\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})\longrightarrow\operatorname{GL}(V),$

along with their explicit (non-integral) models defined over the ring
$\mathbb{Z}[1/p,\zeta_{p}]$.

##  3\. Integral models for Weil characters

The Weil representations $\rho_{1},\rho_{2}$ defined by (4) decompose into
irreducible representations $\rho_{i}=\xi_{i}\oplus\pi_{i}$, $i=1,2$. The
characters $\xi_{1},\xi_{2}$ are the two Weil characters of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$ of dimension $(p+1)/2$; they
belong to the principal series of
$\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$. The other two Weil characters
$\pi_{1},\pi_{2}$ are of dimension $(p-1)/2$ and they belong to the cusipidal
series.

We now construct an explicit integral model over $\mathbb{Z}[\zeta_{p}]$ for
the principal series Weil characters $\xi_{1},\xi_{2}$, following [9]. Let
$V^{+}\subseteq V=\left\\{\text{functions: }C_{p}\to\mathbb{C}\right\\}$ be
the subspace of even functions, satisfying $f(-x)=f(x)$ for all $x\in C_{p}$.
This subspace is an invariant subspace for both $\rho_{1}$ and $\rho_{2}$, and
the principal series Weil characters $\xi_{1},\xi_{2}$ are the restrictions of
$\rho_{1},\rho_{2}$ to $V^{+}$. To write down an explicit model for these
characters, consider the basis for $V^{+}$ given by the even delta functions:

$b^{+}_{0}:=\delta_{0},\ b^{+}_{1}:=\delta_{1}+\delta_{p-1},\ \cdots,\ \
b^{+}_{(p-1)/2}:=\delta_{(p-1)/2}+\delta_{(p+1)/2}.$

Then the explicit model for the Weil representation given by (1) and (2) can
be used to give an explicit model for $\xi_{1}$ and $\xi_{2}$ over the ring
$\mathbb{Z}[1/p,\zeta_{p}]$. Let $S$ and $T$ denote the matrices of the Weil
representations $\rho(\mathfrak{s})$ and $\rho(\mathfrak{t})$ respectively,
restricted to the basis of even functions. Then

(5) $\begin{split}S_{jk}&=\left\\{\begin{array}[]{lr}\dfrac{\sqrt{\varepsilon
p}}{\varepsilon p}&0\leq j\leq\frac{p-1}{2}\text{ and }k=0\\\
\dfrac{2\sqrt{\varepsilon p}}{\varepsilon p}&j=0\text{ and }1\leq
k\leq\frac{p-1}{2}\\\ \dfrac{1}{\sqrt{\varepsilon p}}\left(e^{-2\pi
iB(j,k)}+e^{2\pi
iB(j,k)}\right)&\text{otherwise}\end{array}\right.\end{split}$

and

(6) $T_{jk}=\left\\{\begin{array}[]{rl}e^{2\pi iQ(j)}&\text{if }j=k\\\
0&\text{otherwise.}\end{array}\right.$

Note that these models are not integral, since they can only be defined over
the ring $\mathbb{Z}[1/p,\zeta_{p}]$.

######  Remark 3.1.

The $(p-1)/2$-dimensional cuspidal series representations $\pi_{1}$ and
$\pi_{2}$ can be constructed similarly by restricting $\rho$ to the subspace
$V^{-}\subseteq V$ of odd functions, those satisfying $f(-x)=-f(x)$. This
subspace has basis

$b^{-}_{0}:=\delta_{1}-\delta_{(p-1)},\ \ \cdots,\ \
b^{-}_{(p-3)/2}:=\delta_{(p-1)/2}-\delta_{(p+1)/2}.$

Let $S$ and $T$ denote the matrices of the Weil representations
$\rho(\mathfrak{s})$ and $\rho(\mathfrak{t})$ respectively, restricted to the
basis of odd delta functions. Then $S$ and $T$ are given by

$S_{jk}=\dfrac{1}{\sqrt{\varepsilon p}}\left(e^{-2\pi iB(j+1,k+1)}-e^{2\pi
iB(j+1,k+1)}\right),$

and

$T_{jk}=\left\\{\begin{array}[]{rcl}e^{2\pi iQ(j+1)}&\text{if}&j=k\\\
0&&\text{otherwise.}\end{array}\right.$

These formulas provide explicit models for $\pi_{1}$ and $\pi_{2}$ over
$\mathbb{Z}[1/p,\zeta_{p}]$.

Wang [9] provides an integral model for $\xi_{2}$ over the ring
$\mathbb{Z}[\zeta_{p}]$, by using a basis consisting of circulant vectors. His
construction can also be employed to provide an integral model for $\xi_{1}$,
also defined over the ring $\mathbb{Z}[\zeta_{p}]$. Note that the existence of
such integral models for $\xi_{1},\xi_{2}$ was proved by Riese in [7], but as
far as the authors know Wang was the first to provide explicit integral
models. To recall Wang’s construction, let $Q=Q_{1},Q_{2}$ be one of the
quadratic forms (3), corresponding to each Weil character $\xi_{1},\xi_{2}$.
Let $r=(p-1)/2$ and let $\theta_{j}=e^{2\pi iQ(j)}$ for $0\leq j\leq r$, be
the eigenvalues of the $T$-matrix (2). Note that $\theta_{j}\neq\theta_{k}$
for all $j\neq k$. Consequently, the Vandermonde matrix

(7) $V_{Q}=\begin{bmatrix}1&1&1&1&\cdots&1\\\
1&\theta_{1}&\theta_{1}^{2}&\theta_{1}^{3}&\cdots&\theta_{1}^{r}\\\
1&\theta_{2}&\theta_{2}^{2}&\theta_{2}^{3}&\cdots&\theta_{2}^{r}\\\
1&\theta_{3}&\theta_{3}^{2}&\theta_{3}^{3}&\cdots&\theta_{3}^{r}\\\
\vdots&\vdots&\vdots&\vdots&\cdots&\vdots\\\
1&\theta_{r}&\theta_{r}^{2}&\theta_{r}^{3}&\cdots&\theta_{r}^{r}\end{bmatrix}$

is invertible. Wang proves the following:

######  Theorem 3.2 ([9], Thm. 1).

Suppose $p\geq 5$ is an odd prime. Let $S,T$ be the matrices (5) and (6). Then
the matrices $V_{Q}^{-1}SV_{Q}$ and $V_{Q}^{-1}TV_{Q}$ have entries in
$\mathbb{Z}[\zeta_{p}]$.

By Wang’s theorem, setting

$\xi_{i}(\mathfrak{s})=V_{Q_{i}}^{-1}SV_{Q_{i}},\quad\xi_{i}(\mathfrak{t})=V_{Q_{i}}^{-1}TV_{Q_{i}}$

yields explicit integral models for $\xi_{i}$, $i=1,2$ over the ring
$\mathbb{Z}[\zeta_{p}]$.

######  Remark 3.3.

In his proof, Wang shows that the representation $\xi_{2}$ is defined over
$\mathbb{Z}[\zeta_{p}]$ for $p\equiv 1\mod 4$ and $\mathbb{Z}[\zeta_{p},i]$
for $p\equiv 3\mod 4$. However, upon further inspection of Proposition 2.4,
the proof readily generalizes to show integrality over $\mathbb{Z}[\zeta_{p}]$
for all odd primes and for both $\xi_{1}$ and $\xi_{2}$.

######  Remark 3.4.

Zemel [11] has recently constructed integral models over
$\mathbb{Z}[\zeta_{p}]$ for the cuspidal series Weil characters $\pi_{1}$ and
$\pi_{2}$.

##  4\. Minimal integral models and proof of the Main Theorem

In this section we prove that the integral models given by Theorem 3.2 are in
fact defined over the ring of integers of $\mathbb{Q}(\sqrt{\varepsilon p})$,
thus providing minimal integral models for $\xi_{1}$ and $\xi_{2}$. We do so
by analyzing the action of the Galois group
$\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ on the integral models.
The key observation in this analysis is a surprising compatibility between the
Galois actions on the Vandermonde matrix $V_{Q}$ and on the entries of the
matrices of the Weil representations $\rho_{Q}$, for $Q=Q_{1},Q_{2}$.
Investigating this compatibility for more general quadratic forms $Q$ might be
of independent interest.

Recall that for an odd prime $p$, the quadratic Gauss sums can be evaluated as
follows:

$\sum_{x\in\mathbb{Z}/p\mathbb{Z}}\zeta_{p}^{x^{2}}=\left\\{\begin{array}[]{ll}\sqrt{p}&\text{for
}\ p\equiv 1\mod 4\\\ \\\ \sqrt{-p}&\text{for }\ p\equiv 3\mod
4,\end{array}\right.$

and therefore the quadratic field $\mathbb{Q}(\sqrt{\varepsilon p})$ is always
contained in $\mathbb{Q}(\zeta_{p})$. By the fundamental theorem of Galois
theory, this subfield must correspond to the subgroup
$H\subseteq\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ of index 2,

(8) ${\mathbb{Q}(\zeta_{p})}$${\\{e\\}}$${\mathbb{Q}(\sqrt{\varepsilon
p})}$${\Longleftrightarrow}$${H=\langle\gamma^{2}\rangle}$${\mathbb{Q}}$${\langle\gamma\rangle}$$\scriptstyle{r}$$\scriptstyle{r}$$\scriptstyle{2}$$\scriptstyle{2}$

where in the diagram we chose a generator
$\langle\gamma\rangle=\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})\simeq\mathbb{Z}/(2r)\mathbb{Z}$
and $r=(p-1)/2$. The generator $\gamma$ can be written down as an automorphism
$\gamma:\zeta_{p}\to\zeta_{p}^{j}$ where $\gcd(j,2r)=1$. Then
$\gamma^{2}:\zeta_{p}\to\zeta_{p}^{j^{2}}$ acts on the exponents of
$\zeta_{p}$ by sending squares to squares and non-squares to non-squares. We
can write the sums of square exponents and sums of nonsquares exponents in
terms of the quadratic Gauss sum:

$\sum_{k\in(\mathbb{F}_{p})^{\times},\ k\ a\
square}\zeta_{p}^{k}=\dfrac{-1+\sqrt{\varepsilon
p}}{2}\quad\text{and}\quad\sum_{j\in(\mathbb{F}_{p})^{\times},\ j\ not\ a\
square}\zeta_{p}^{j}=\dfrac{-1-\sqrt{\varepsilon p}}{2}$

and it is clear that they are always contained in the ring of integers
$\mathbb{Z}[\frac{1}{2}(1+\sqrt{\varepsilon
p})]\subset\mathbb{Q}(\sqrt{\varepsilon p})$.

Let $S$ and $T$ be the matrices given in (5) and (6), let $V_{Q}$ be the
Vandermonde matrix (7), and let $T^{\prime}=V_{Q}^{-1}TV_{Q}$. We first prove
the integrality of $T^{\prime}$ over the quadratic ring:

######  Theorem 4.1.

The matrix entries of $T^{\prime}$ lie in
$\mathbb{Z}[\frac{1}{2}(1+{\sqrt{\varepsilon p}})]$, the ring of integers of
$\mathbb{Q}(\sqrt{\varepsilon p})$.

###### Proof.

Note that $T^{\prime}$ is the following circulant matrix

$\begin{split}T^{\prime}&=\begin{bmatrix}0&0&0&\cdots&0&-a_{r+1}\\\
1&0&0&\cdots&0&-a_{r}\\\ 0&1&0&\cdots&0&-a_{r-1}\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&0&\cdots&1&-a_{1}\\\
\end{bmatrix},\end{split}$

whose characteristic polynomial $m(x)\in\mathbb{Q}(\zeta_{p})[x]$ is given by

$\begin{split}m(x)=x^{r+1}+a_{1}x^{r}+a_{2}x^{r-1}+\cdots+a_{r}x+a_{r+1}.\end{split}$

Thus to prove the Theorem it suffices to show that the coefficients of $m(x)$
are contained in $\mathbb{Z}[\frac{1}{2}(1+{\sqrt{\varepsilon p}})]$. Since
$T$ and $T^{\prime}$ are conjugate matrices, we know that $m(x)$ splits as
$(x-1)(x-\theta_{1})(x-\theta_{2})\cdots(x-\theta_{r})$ in
$\mathbb{Q}(\zeta_{p})$, where the $\theta_{i}$’s are the eigenvalues/diagonal
entries of $T$. In the case $Q=Q_{1}$, each $\theta_{j}$ is of the form
$\zeta_{p}^{s}$, where $s$ is a square mod $p$, while in the case $Q=Q_{2}$,
each $\theta_{i}$ is of the form $\zeta_{p}^{n}$, where $n$ is not a square
mod $p$. In each case, the set of roots of the polynomial $m(x)$ is permuted
by the index 2 subgroup
$H\subseteq\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ defined in
(8). So $H$ fixes the coefficients of $m(x)$, since those are expressible in
terms of elementary symmetric polynomials in the $\theta_{j}$’s. Therefore the
coefficients $a_{i}$ of $m(x)$ lie in the quadratic extension
$\mathbb{Q}(\sqrt{\varepsilon p})$. In addition, since all the roots of unity
$\zeta_{p}^{j}$ belong to the ring of integers $\mathbb{Z}[\zeta_{p}]$ and
since the symmetric polynomials have integral coefficients, it must follow
that the coefficients $a_{j}$ actually belong to the ring of integers of
$\mathbb{Q}(\sqrt{\varepsilon p})$. ∎

For any element
$\alpha\in\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ and any matrix
$M=(m_{jk})$ with entries $m_{jk}\in\mathbb{Q}(\zeta_{p})$, we define
$\alpha(M)=(\alpha(m_{jk}))$ to be the matrix obtained from $M$ by applying
the field automorphism $\alpha$ to each entry. More generally, if
$\rho:G\rightarrow\operatorname{GL}_{n}(\mathbb{Q}(\zeta_{p}))$ is a model for
a group representation, we denote by $\rho^{\alpha}$ the model for the group
representation $\rho^{\alpha}(g)=\alpha(\rho(g))$, obtained by applying the
field automorphism $\alpha$ to each matrix entry $\rho(g)$, $g\in G$.

In particular, let $\tau=\gamma^{2}$ be the generator of the subgroup $H$ of
the Galois group of $\mathbb{Q}(\zeta_{p})$ defined in (8). Clearly we have
$|\tau|=|H|=r$. Consider the action of $\tau$ on the Vandermonde matrix
$V_{Q}$. Since $\tau(\zeta_{p}^{x})=\zeta_{p}^{y}$ where $x,y$ are either both
squares or both non-squares mod $p$, and since $\tau$ fixes the number 1, it
follows that $\tau(V_{Q})$ is a matrix obtained by permuting the rows of
$V_{Q}$. Let ${P}$ denote the permutation matrix corresponding to this
permutation of the rows of $V_{Q}$:

(9) $\tau(V_{Q})={P}\cdot V_{Q}.$

Since $|\tau|=r$, the order of $P$ is also $r$. $P$ fixes the first row and
therefore $P^{-1}$ also fixes the first row. We can consider $P$ as a $r-1$
cycle. We also note that $P^{-1}=P^{t}$, since the inverse of a permutation
matrix is its transpose.

Interestingly, it turns out that the permutation matrix $P$ also gives the
Galois action of $\tau$ on the Weil representation models for
$\xi_{1},\xi_{2}$ given by the matrices $S,T$ defined in (5) and (6). This
non-trivial compatibility between the Galois action on the Vandermonde matrix
$V_{Q}$ and the Weil representation models is the key observation of this
article:

######  Theorem 4.2.

Let
$\xi_{1},\xi_{2}:\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})\rightarrow\operatorname{GL}_{r+1}(\mathbb{Q}(\zeta_{p}))$
be the explicit models for the principal series Weil characters determined by
$\xi_{i}(\mathfrak{s})=S,\xi_{i}(\mathfrak{t})=T$, where $S,T$ are given in
(5) and (6). Then for each $i=1,2$ and all primes $p>2$, the permutation
matrix $P$ given in (9) satisfies

$\xi_{i}^{\tau}(g)=P\,\xi_{i}(g)\,P^{-1}$

for all $g\in\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$.

###### Proof.

As is well-known, the character table entries for $\xi_{i}$ are defined over
the quadratic field $\mathbb{Q}(\sqrt{\varepsilon p})$. Since this the fixed
field of the subgroup
$H=\langle\tau\rangle\subseteq\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$,
it follows that the character table entries of $\xi_{i}$ and $\xi_{i}^{\tau}$
are the same, so $\xi_{i}^{\tau}\cong\xi_{i}$. This implies that there exists
a matrix $M\in\operatorname{GL}(V)$ (uniquely defined up to multiplication by
a scalar) such that

(10) $\xi_{i}^{\tau}(g)=M^{-1}\cdot\xi_{i}(g)\cdot M$

for all $g\in\operatorname{\rm{SL}}_{2}(\mathbb{F}_{p})$. We want to show that
$M=\lambda P^{-1}$ where $\lambda\in\mathbb{C}$ and $P$ is the permutation
matrix (9), determined by the relation $\tau(V_{Q})=PV_{Q}$. First, recall
that the matrix $T^{\prime}=V_{Q}^{-1}TV_{Q}$ has entries in
$\mathbb{Q}(\sqrt{\varepsilon p})$, by Theorem 4.1, and therefore
$\tau(T^{\prime})=T^{\prime}$. It follows that

$\tau(T)=\tau(V_{Q})\cdot T^{\prime}\cdot\tau(V_{Q})^{-1}=PV_{Q}\cdot
T^{\prime}\cdot V_{Q}^{-1}P^{-1}=PTP^{-1}.$

On the other hand, we know from (10) that

$\tau(T)=M^{-1}TM$

so that the matrix $T$ must commute with the matrix $PM$. Since $T$ is a
diagonal matrix whose entries $\theta_{i}$ are all distinct, it follows that
$PM=D$ is also a diagonal matrix, and $M=P^{-1}D$. We want to show that
$D=\lambda\cdot\mathds{1}$ in fact scalar. To show this, we also need to
employ the $S$-matrix. Let

$D=\begin{bmatrix}\lambda_{1}&0&0&\cdots&0\\\ 0&\lambda_{2}&0&\cdots&0\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\ 0&0&0&\cdots&\lambda_{r}\end{bmatrix}.$

Again recall from (10) that $\tau(S)=M^{-1}\cdot S\cdot M$, so that

$\tau(S)=M\cdot S\cdot M^{-1}=P^{-1}D\cdot S\cdot D^{-1}P$

Recall that $S$ has the form

$S=\dfrac{\sqrt{\varepsilon p}}{\varepsilon p}\begin{bmatrix}1&2&2&\cdots&2\\\
1&s_{11}&s_{12}&\cdots&s_{1r}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
1&s_{r1}&s_{r2}&\cdots&s_{rr}\\\ \end{bmatrix},$

for some entries $s_{jk}$. In particular, since the first row and first column
of $S$ consists of elements of $\mathbb{Q}(\sqrt{\varepsilon p})$, the action
of $\tau$ leaves them fixed. In addition, by definition the permutation matrix
$P$ also fixes the first row and the first column, and so does $P^{-1}=P^{t}$.
Therefore we can write

$P\cdot\tau(S)\cdot P^{-1}=\dfrac{\sqrt{\varepsilon p}}{\varepsilon
p}\begin{bmatrix}1&2&2&\cdots&2\\\ 1&x_{11}&x_{12}&\cdots&x_{1r}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\ 1&x_{r1}&x_{r2}&\cdots&x_{rr}\\\
\end{bmatrix},$

for some entries $x_{jk}$’s. So

$\displaystyle D^{-1}\cdot P\cdot\tau(S)\cdot P^{-1}\cdot D$
$\displaystyle=\dfrac{\sqrt{\varepsilon p}}{\varepsilon
p}\begin{bmatrix}1&2\lambda_{1}\lambda_{2}^{-1}&2\lambda_{1}\lambda_{3}^{-1}&\cdots&2\lambda_{1}\lambda_{r}^{-1}\\\
\lambda_{2}\lambda_{1}^{-1}&*&*&\cdots&*\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
\lambda_{r}\lambda_{1}^{-1}&*&*&\cdots&*\\\ \end{bmatrix}$
$\displaystyle=\dfrac{\sqrt{\varepsilon p}}{\varepsilon
p}\begin{bmatrix}1&2&2&\cdots&2\\\ 1&s_{11}&s_{12}&\cdots&s_{1r}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\ 1&s_{r1}&s_{r2}&\cdots&s_{rr}\\\
\end{bmatrix}=S.$

Equating the first column and row of these matrices we conclude that
$\lambda_{j}=\lambda_{k}=\lambda$ are all equal, and thus $M=\lambda P^{-1}$.
∎

Let now $S^{\prime}:=V_{Q}^{-1}\cdot S\cdot V_{Q}$. Theorem 4.2 enables us to
prove the following theorem.

######  Theorem 4.3.

$S^{\prime}$ has matrix entries in
$\mathbb{Z}[\frac{1}{2}(1+{\sqrt{\varepsilon p}})]$, the ring of integers of
$\mathbb{Q}(\sqrt{\varepsilon p})$.

###### Proof.

Again let $\tau$ be the generator of the subgroup
$H\subseteq\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})$ defined in
(8). We want to show that $\tau(S^{\prime})=S^{\prime}$. By Theorem 4.3, we
have:

$\begin{split}\tau(S^{\prime})&=\tau(V_{Q}^{-1}\cdot S\cdot V_{Q})\\\
&=\tau(V_{Q}^{-1})\cdot\tau(S)\cdot\tau(V_{Q})\\\ &=V_{Q}^{-1}\cdot
P^{-1}\cdot P\cdot S\cdot P^{-1}\cdot P\cdot V_{Q}\\\ &=V_{Q}^{-1}\cdot S\cdot
V_{Q}\\\ &=S^{\prime}.\end{split}$

Since $\tau$ fixes $S^{\prime}$, the entries of $S^{\prime}$ lie in the
quadratic extension $\mathbb{Q}(\sqrt{\varepsilon p})$. But we also know from
Thm. 3.2 that the entries of $S^{\prime}$ belong to the ring of algebraic
integers $\mathbb{Z}[\zeta_{p}]$, therefore they must lie in the ring of
integers of $\mathbb{Q}(\sqrt{\varepsilon p})$. ∎

By Theorems 4.1 and 4.3, setting

$\xi_{i}(\mathfrak{s})=S^{\prime},\xi_{i}(\mathfrak{t})=T^{\prime}$

gives integral models for the principal series Weil characters over the ring
of integers of $\mathbb{Q}(\sqrt{\varepsilon p})$, therefore giving minimal
integral models for these characters.

######  Remark 4.4.

Recently Zemel [11] has generalized the integral models of Wang to integral
models for the cuspidal Weil characters $\pi_{1},\pi_{2}$ defined over the
ring $\mathbb{Z}[\zeta_{p}]$. It does not appear however that these models are
defined over smaller rings. In particular, minimal integral models for
cuspidal Weil characters remain unknown.

##  5\. Examples

We now show the computations giving the explicit minimal integral models for
$\xi_{1}$ and $\xi_{2}$ for the primes $p=7$ and $p=13$. We used SageMath[8]
and MATLAB[6] to make and verify the computations.

######  Example 5.1.

Let $p=7$. For the equivalent representation of $\xi_{1}$, let $c=1$. So,
$Q_{1}(x)=x^{2}/7$, $B_{1}(x,y)=2xy/7$ and the Gauss sum is given by

$\sum_{x\in\mathbb{Z}/7\mathbb{Z}}\zeta_{7}^{x^{2}}=2\zeta_{7}^{4}+2\zeta_{7}^{2}+2\zeta_{7}+1=\sqrt{-7}.$

Then we have

$S=\dfrac{-\sqrt{7}}{7}\begin{bmatrix}1&2&2&2\\\
1&\zeta_{7}^{5}+\zeta_{7}^{2}&\zeta_{7}^{4}+\zeta_{7}^{3}&\zeta_{7}^{6}+\zeta_{7}\\\
1&\zeta_{7}^{4}+\zeta_{7}^{3}&\zeta_{7}^{6}+\zeta_{7}&\zeta_{7}^{5}+\zeta_{7}^{2}\\\
1&\zeta_{7}^{6}+\zeta_{7}&\zeta_{7}^{5}+\zeta_{7}^{2}&\zeta_{7}^{4}+\zeta_{7}^{3}\end{bmatrix},$
$T=\begin{bmatrix}1&0&0&0\\\ 0&\zeta_{7}&0&0\\\ 0&0&\zeta_{7}^{4}&0\\\
0&0&0&\zeta_{7}^{2}\end{bmatrix},$ $V_{Q}=\begin{bmatrix}1&1&1&1\\\
1&\zeta_{7}&\zeta_{7}^{2}&\zeta_{7}^{3}\\\
1&\zeta_{7}^{4}&\zeta_{7}&\zeta_{7}^{5}\\\
1&\zeta_{7}^{2}&\zeta_{7}^{4}&\zeta_{7}\end{bmatrix},$
$S^{\prime}=V_{Q}^{-1}SV_{Q}=\begin{bmatrix}-1&\frac{1}{2}(1-\sqrt{-7})&0&0\\\
\frac{1}{2}(-1-\sqrt{-7})&1&0&0\\\
\frac{1}{2}(1-\sqrt{-7})&\frac{1}{2}(1+\sqrt{-7})&0&-1\\\
1&-1&1&0\end{bmatrix},$

and

$T^{\prime}=V_{Q}^{-1}TV_{Q}=\begin{bmatrix}0&0&0&-1\\\
1&0&0&\frac{1}{2}(1-\sqrt{-7})\\\ 0&1&0&1\\\
0&0&1&\frac{1}{2}(1+\sqrt{-7})\end{bmatrix}.$

For the equivalent representation of $\xi_{2}$, we choose $c=3$. So,
$Q_{2}(x)=3x^{2}/7$, $B_{2}(x,y)=6xy/7$ and the Gauss sum is given by

$\sum_{x\in\mathbb{Z}/7\mathbb{Z}}\zeta_{7}^{3x^{2}}=-2\zeta_{7}^{4}-2\zeta_{7}^{2}-2\zeta_{7}-1=-\sqrt{-7}.$

Then we have

$S=\dfrac{\sqrt{-7}}{7}\begin{bmatrix}1&2&2&2\\\
1&\zeta_{7}^{6}+\zeta_{7}&\zeta_{7}^{5}+\zeta_{7}^{2}&\zeta_{7}^{4}+\zeta_{7}^{3}\\\
1&\zeta_{7}^{5}+\zeta_{7}^{2}&\zeta_{7}^{4}+\zeta_{7}^{3}&\zeta_{7}^{6}+\zeta_{7}\\\
1&\zeta_{7}^{4}+\zeta_{7}^{3}&\zeta_{7}^{6}+\zeta_{7}^{1}&\zeta_{7}^{5}+\zeta_{7}^{2}\end{bmatrix},$
$T=\begin{bmatrix}1&0&0&0\\\ 0&\zeta_{7}^{3}&0&0\\\ 0&0&\zeta_{7}^{5}&0\\\
0&0&0&\zeta_{7}^{6}\end{bmatrix},$ $V_{Q}=\begin{bmatrix}1&1&1&1\\\
1&\zeta_{7}^{3}&\zeta_{7}^{6}&\zeta_{7}^{2}\\\
1&\zeta_{7}^{5}&\zeta_{7}^{3}&\zeta_{7}\\\
1&\zeta_{7}^{6}&\zeta_{7}^{5}&\zeta_{7}^{4}\end{bmatrix},$
$S^{\prime}=V_{Q}^{-1}SV_{Q}=\begin{bmatrix}-1&\frac{1}{2}(1+\sqrt{-7})&0&0\\\
\frac{1}{2}(-1+\sqrt{-7})&1&0&0\\\
\frac{1}{2}(1+\sqrt{-7})&\frac{1}{2}(1-\sqrt{-7})&0&-1\\\
1&-1&1&0\end{bmatrix},$

and

$T^{\prime}=V_{Q}^{-1}TV_{Q}=\begin{bmatrix}0&0&0&-1\\\
1&0&0&\frac{1}{2}(1+\sqrt{7})\\\ 0&1&0&1\\\
0&0&1&\frac{1}{2}(1-\sqrt{7})\end{bmatrix}.$

######  Example 5.2.

Let $p=13$. For the equivalent representation of $\xi_{1}$, let $c=1$. So,
$Q_{1}(x)=x^{2}/13$, $B_{1}(x,y)=2xy/13$ and the Gauss sum is given by

$\sum_{x\in\mathbb{Z}/13\mathbb{Z}}\zeta_{13}^{x^{2}}=-2\zeta_{13}^{11}-2\zeta_{13}^{8}-2\zeta_{13}^{7}-2\zeta_{13}^{6}-2\zeta_{13}^{5}-2\zeta_{13}^{2}-1=\sqrt{13}.$

Then we have

$S=\dfrac{\sqrt{13}}{13}\begin{bmatrix}1&2&2&2&2&2&2\\\
1&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{12}+\zeta_{13}\\\
1&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{11}+\zeta_{13}^{2}\\\
1&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{10}+\zeta_{13}^{3}\\\
1&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{9}+\zeta_{13}^{4}\\\
1&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{8}+\zeta_{13}^{5}\\\
1&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{7}+\zeta_{13}^{6}\end{bmatrix},$
$T=\begin{bmatrix}1&0&0&0&0&0&0\\\ 0&\zeta_{13}&0&0&0&0&0\\\
0&0&\zeta_{13}^{4}&0&0&0&0\\\ 0&0&0&\zeta_{13}^{9}&0&0&0\\\
0&0&0&0&\zeta_{13}^{3}&0&0\\\ 0&0&0&0&0&\zeta_{13}^{12}&0\\\
0&0&0&0&0&0&\zeta_{13}^{10}\end{bmatrix},$
$V_{Q}=\begin{bmatrix}1&1&1&1&1&1&1\\\
1&\zeta_{13}&\zeta_{13}^{2}&\zeta_{13}^{3}&\zeta_{13}^{4}&\zeta_{13}^{5}&\zeta_{13}^{6}\\\
1&\zeta_{13}^{4}&\zeta_{13}^{8}&\zeta_{13}^{12}&\zeta_{13}^{3}&\zeta_{13}^{7}&\zeta_{13}^{11}\\\
1&\zeta_{13}^{9}&\zeta_{13}^{5}&\zeta_{13}&\zeta_{13}^{10}&\zeta_{13}^{6}&\zeta_{13}^{2}\\\
1&\zeta_{13}^{3}&\zeta_{13}^{6}&\zeta_{13}^{9}&\zeta_{13}^{12}&\zeta_{13}^{2}&\zeta_{13}^{5}\\\
1&\zeta_{13}^{12}&\zeta_{13}^{11}&\zeta_{13}^{10}&\zeta_{13}^{9}&\zeta_{13}^{8}&\zeta_{13}^{7}\\\
1&\zeta_{13}^{10}&\zeta_{13}^{7}&\zeta_{13}^{4}&\zeta_{13}&\zeta_{13}^{11}&\zeta_{13}^{8}\end{bmatrix},$
$S^{\prime}=V_{Q}^{-1}SV_{Q}=\begin{bmatrix}\frac{1}{2}(3+\sqrt{13})&\frac{1}{2}(1+\sqrt{13})&0&0&0&0&0\\\
-\frac{1}{2}(5+\sqrt{13})&-\frac{1}{2}(3+\sqrt{13})&0&0&0&0&0\\\
3+\sqrt{13}&\frac{1}{2}(5+\sqrt{13})&0&0&0&0&-1\\\
-4-\sqrt{13}&-\frac{1}{2}(5+\sqrt{13})&0&0&1&0&0\\\
3+\sqrt{13}&\frac{1}{2}(3+\sqrt{13})&0&1&0&0&0\\\
-\frac{1}{2}(5+\sqrt{13})&-\frac{1}{2}(1+\sqrt{13})&0&0&0&-1&0\\\
\frac{1}{2}(3+\sqrt{13})&1&-1&0&0&0&0\end{bmatrix},$

and

$T^{\prime}=V_{Q}^{-1}TV_{Q}=\begin{bmatrix}0&0&0&0&0&0&1\\\
1&0&0&0&0&0&-\frac{1}{2}(1+\sqrt{13})\\\
0&1&0&0&0&0&\frac{1}{2}(3+\sqrt{13})\\\
0&0&1&0&0&0&-\frac{1}{2}(5+\sqrt{13})\\\
0&0&0&1&0&0&\frac{1}{2}(5+\sqrt{13})\\\
0&0&0&0&1&0&-\frac{1}{2}(3+\sqrt{13})\\\
0&0&0&0&0&1&\frac{1}{2}(1+\sqrt{13})\end{bmatrix}.$

For the equivalent representation of $\xi_{2}$, let $c=2$. So,
$Q_{2}(x)=2x^{2}/13$, $B_{2}(x,y)=4xy/13$ and the Gauss sum is given by

$\sum_{x\in\mathbb{Z}/13\mathbb{Z}}\zeta_{13}^{x^{2}}=2\zeta_{13}^{11}+2\zeta_{13}^{8}+2\zeta_{13}^{7}+2\zeta_{13}^{6}+2\zeta_{13}^{5}+2\zeta_{13}^{2}+1=-\sqrt{13}.$

Then we have

$S=\dfrac{-\sqrt{13}}{13}\begin{bmatrix}1&2&2&2&2&2&2\\\
1&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{11}+\zeta_{13}^{2}\\\
1&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{9}+\zeta_{13}^{4}\\\
1&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{7}+\zeta_{13}^{6}\\\
1&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{8}+\zeta_{13}^{5}\\\
1&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{12}+\zeta_{13}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{10}+\zeta_{13}^{3}\\\
1&\zeta_{13}^{11}+\zeta_{13}^{2}&\zeta_{13}^{9}+\zeta_{13}^{4}&\zeta_{13}^{7}+\zeta_{13}^{6}&\zeta_{13}^{8}+\zeta_{13}^{5}&\zeta_{13}^{10}+\zeta_{13}^{3}&\zeta_{13}^{12}+\zeta_{13}\end{bmatrix},$
$T=\begin{bmatrix}1&0&0&0&0&0&0\\\ 0&\zeta_{13}^{2}&0&0&0&0&0\\\
0&0&\zeta_{13}^{8}&0&0&0&0\\\ 0&0&0&\zeta_{13}^{5}&0&0&0\\\
0&0&0&0&\zeta_{13}^{6}&0&0\\\ 0&0&0&0&0&\zeta_{13}^{11}&0\\\
0&0&0&0&0&0&\zeta_{13}^{7}\end{bmatrix},$
$V_{Q}=\begin{bmatrix}1&1&1&1&1&1&1\\\
1&\zeta_{13}^{2}&\zeta_{13}^{4}&\zeta_{13}^{6}&\zeta_{13}^{8}&\zeta_{13}^{10}&\zeta_{13}^{12}\\\
1&\zeta_{13}^{8}&\zeta_{13}^{3}&\zeta_{13}^{11}&\zeta_{13}^{6}&\zeta_{13}&\zeta_{13}^{9}\\\
1&\zeta_{13}^{5}&\zeta_{13}^{10}&\zeta_{13}^{2}&\zeta_{13}^{7}&\zeta_{13}^{12}&\zeta_{13}^{4}\\\
1&\zeta_{13}^{6}&\zeta_{13}^{12}&\zeta_{13}^{5}&\zeta_{13}^{11}&\zeta_{13}^{4}&\zeta_{13}^{10}\\\
1&\zeta_{13}^{11}&\zeta_{13}^{9}&\zeta_{13}^{7}&\zeta_{13}^{5}&\zeta_{13}^{3}&\zeta_{13}\\\
1&\zeta_{13}^{7}&\zeta_{13}&\zeta_{13}^{8}&\zeta_{13}^{2}&\zeta_{13}^{9}&\zeta_{13}^{3}\end{bmatrix},$
$S^{\prime}=V_{Q}^{-1}SV_{Q}=\begin{bmatrix}\frac{1}{2}(3-\sqrt{13})&\frac{1}{2}(1-\sqrt{13})&0&0&0&0&0\\\
\frac{1}{2}(-5+\sqrt{13})&\frac{1}{2}(-3+\sqrt{13})&0&0&0&0&0\\\
3-\sqrt{13}&\frac{1}{2}(5-\sqrt{13})&0&0&0&0&-1\\\
-4+\sqrt{13})&\frac{1}{2}(-5+\sqrt{13})&0&0&1&0&0\\\
3-\sqrt{13}&\frac{1}{2}(3-\sqrt{13})&0&1&0&0&0\\\
\frac{1}{2}(-5+\sqrt{13})&\frac{1}{2}(-1+\sqrt{13})&0&0&0&-1&0\\\
\frac{1}{2}(3-\sqrt{13})&1&-1&0&0&0&0\end{bmatrix},$

and

$T^{\prime}=V_{Q}^{-1}TV_{Q}=\begin{bmatrix}0&0&0&0&0&0&1\\\
1&0&0&0&0&0&\frac{1}{2}(-1+\sqrt{13})\\\
0&1&0&0&0&0&\frac{1}{2}(3-\sqrt{13})\\\
0&0&1&0&0&0&\frac{1}{2}(-5+\sqrt{13})\\\
0&0&0&1&0&0&\frac{1}{2}(5-\sqrt{13})\\\
0&0&0&0&1&0&\frac{1}{2}(-3+\sqrt{13})\\\
0&0&0&0&0&1&\frac{1}{2}(1-\sqrt{13})\end{bmatrix}.$

## References

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# Unified $(r,s)$-relative entropy††thanks: This project is supported by
Natural Science Foundation of China (10771191 and 10471124) and Natural
Science Foundation of Zhejiang Province of China (Y6090105).

Wang Jiamei, Wu Junde Department of Mathematics, Zhejiang University, Hangzhou
310027, P. R. China. E-mail<EMAIL_ADDRESS>
###### Abstract

In this paper, we introduce and study unified $(r,s)$-relative entropy and
quantum unified $(r,s)$-relative entropy, in particular, our main results of
quantum unified $(r,s)$-relative entropy are established on the infinite
dimensional separable complex Hilbert spaces.

Key Words. Hilbert space, unified $(r,s)$-relative entropy, state.

1\. Introduction

In 1991, Rathie and Taneja introduced the unified $(r,s)$-entropy which
generalized many classical entropies ([1]), that is, let
$A=(a_{1},a_{2},\cdots,a_{n})$ be a discrete probability distribution
satisfies that $0<a_{i}\leq 1$ and $\sum_{i=1}^{n}a_{i}=1$. If we denote
$p(r)=\sum_{i=1}^{n}a_{i}^{r}$, then for any $r>0$ and any real number $s$,
the unified $(r,s)-$entropy is defined by

$\displaystyle E_{r}^{s}(A)=\left\\{\begin{array}[]{ll}H_{r}^{s}(A),&{\rm if\
}r\neq 1,s\neq 0,\\\ H_{r}(A),&{\rm if\ }r\neq 1,s=0,\\\ H^{r}(A),&{\rm if\
}r\neq 1,s=1,\\\ _{\frac{1}{r}}H(A),&{\rm if\ }r\neq 1,s=1/r,\\\ H(A),&{\rm
if\ }r=1,\end{array}\right.$

where

$H_{r}^{s}(A)=[(1-r)s]^{-1}[p(r)^{s}-1],$ $H_{r}(A)=(1-r)^{-1}\ln p(r),$
$H^{r}(A)=(1-r)^{-1}(p(r)-1),$ ${}_{r}H(A)=(r-1)^{-1}[p(\frac{1}{r})^{r}-1],$
$H(A)=-\sum_{i=1}^{n}a_{i}\ln{a_{i}}$

are the $(r,s)$-entropy, R$\acute{e}$nyi entropy of order $r$, the Tsallis
entropy, the entropy of type $r$ and the well-known Shannon entropy,
respectively.

In 2006, Hu and Ye introduced the quantum version of the unified
$(r,s)$-entropy ([2]), that is, let $H$ be a complex Hilbert space and $\rho$
a state (see [3]) on $H$. If we denote $P(r)=tr(\rho^{r})$, then for any $r>0$
and any real number $s$, the quantum unified $(r,s)-$entropy is defined by

$\displaystyle
E_{r}^{s}(\rho)=\left\\{\begin{array}[]{ll}S_{r}^{s}(\rho),&{\rm if\ }r\neq
1,s\neq 0,\\\ S_{r}(\rho),&{\rm if\ }r\neq 1,s=0,\\\ S^{r}(\rho),&{\rm if\
}r\neq 1,s=1,\\\ _{\frac{1}{r}}S(\rho),&{\rm if\ }r\neq 1,s=1/r,\\\
S(\rho),&{\rm if\ }r=1,\end{array}\right.$

where

$S_{r}^{s}(\rho)=[(1-r)s]^{-1}\left[P(r)^{s}-1\right],$
$S_{r}(\rho)=(1-r)^{-1}\ln P(r),$ $S^{r}(\rho)=(1-r)^{-1}\left(P(r)-1\right),$
${}_{r}S(\rho)=(r-1)^{-1}\left[P(\frac{1}{r})^{r}-1\right],$
$S(\rho)=-tr(\rho\ln\rho)$

are the quantum $(r,s)$-entropy, the quantum R$\acute{e}$nyi entropy of order
$r$, the quantum Tsallis entropy, the quantum entropy of type $r$ and the
well-known Von Neumann entropy, respectively.

On the other hand, although the R$\acute{e}$nyi relative entropy of order $r$
([4]), the Tsallis relative entropy of degree $r$ (([5]), the relative entropy
([3]), even the quantum R$\acute{e}$nyi relative entropy ([4]) and quantum
Tsallis relative entropy of degree $r$ ([5-6]) were studied, respectively,
nevertheless, until now, we do not find the works of unified $(r,s)$-relative
entropy and quantum unified $(r,s)$-relative entropy. In this paper, we fill
this gap.

2\. The unified $(r,s)$-relative entropy

Let $A=(a_{1},a_{2},\cdots,a_{n})$, $B=(b_{1},b_{2},\cdots,b_{n})$ be two
discrete probability distributions satisfying $0<a_{i},b_{i}<1$ and
$\sum\limits_{i=1}^{n}a_{i}=\sum\limits_{i=1}^{n}b_{i}=1$. Then for any $r>0$
and any real number $s$, the unified $(r,s)-$relative entropy is defined by

$\displaystyle
E_{r}^{s}(A\|B)=\left\\{\begin{array}[]{ll}H_{r}^{s}(A\|B),&{\rm if\ }r\neq
1,s\neq 0,\\\ H_{r}(A\|B),&{\rm if\ }r\neq 1,s=0,\\\ H^{r}(A\|B),&{\rm if\
}r\neq 1,s=1,\\\ _{\frac{1}{r}}H(A||B),&{\rm if\ }r\neq 1,s=\frac{1}{r},\\\
H(A\|B),&{\rm if\ }r=1,\end{array}\right.$

where

$\displaystyle
H_{r}^{s}(A\|B)=-[(1-r)s]^{-1}\left[\left(\sum_{i=1}^{n}a_{i}\frac{a_{i}^{r-1}}{b_{i}^{r-1}}\right)^{s}-1\right],r>0,r\neq
1,s\neq 0,$ $\displaystyle
H_{r}(A\|B)=-(1-r)^{-1}\ln\left(\sum_{i=1}^{n}a_{i}\frac{a_{i}^{r-1}}{b_{i}^{r-1}}\right),r>0,r\neq
1,$ $\displaystyle
H^{r}(A\|B)=-(1-r)^{-1}\left(\sum_{i=1}^{n}a_{i}\frac{a_{i}^{r-1}}{b_{i}^{r-1}}-1\right),r>0,r\neq
1,$
${}_{r}H(A\|B)=-(r-1)^{-1}\left[\left(\sum_{i=1}^{n}a_{i}\frac{a_{i}^{\frac{1}{r}-1}}{b_{i}^{\frac{1}{r}-1}}\right)^{r}-1\right],r>0,r\neq
1,$ $\displaystyle H(A\|B)=\sum_{i=1}^{n}a_{i}\ln\frac{a_{i}}{b_{i}}$

are the $(r,s)$-relative entropy, the R$\acute{e}$nyi relative entropy of
order $r$, the Tsallis relative entropy of degree $r$, the relative entropy of
type $r$ and the relative entropy, respectively ([3-5]).

Now, we discuss some elementary properties of the unified $(r,s)$-relative
entropy. First, we point out an important unified $(r,s)$-directed divergence
${\cal F}_{r}^{s}(A\|B)$ which was studied in [7], note that when $r\neq 1$,
$E^{\frac{s-1}{r-1}}_{r}(A\|B)={\cal F}_{r}^{s}(A\|B)$, so by using Theorem 1
in [7], we can prove the nonnegativity, nonadditivity and convexity of
$E_{r}^{s}(A\|B)$ directly:

(i) Let
$\Delta_{n}=\\{A=(a_{1},a_{2},\cdots,a_{n}):a_{i}>0,\sum\limits_{i=1}^{n}a_{i}=1\\}$.
If $A,B\in\Delta_{n}$, then $E_{r}^{s}(A\|B)\geq 0$, and the equality holds
iff $A=B$.

(ii) Let
$\Delta_{m}=\\{B=(b_{1},b_{2},\cdots,b_{m}):b_{i}>0,\sum\limits_{i=1}^{m}b_{i}=1\\}$.
If $A_{1},A_{2}\in\Delta_{n}$, $B_{1},B_{2}\in\Delta_{m}$, and denote
$A*B=(a_{1}b_{1},\cdots,a_{1}b_{m},a_{2}b_{1},\cdots,a_{2}b_{m},\cdots,a_{n}b_{m})$,
then

$E_{r}^{s}(A_{1}*B_{1}\|A_{2}*B_{2})=E_{r}^{s}(A_{1}\|A_{2})+E_{r}^{s}(B_{1}\|B_{2})+(r-1)sE_{r}^{s}(A_{1}\|A_{2})E_{r}^{s}(B_{1}\|B_{2}).$

(iii) If $r=1$ or $r>1,s\geq 1$ or $0<r<1,s\leq 1$, then $E_{r}^{s}(A\|B)$ is
a convex function of $(A,B)$.

Next, we prove the following:

Theorem 2.1. If $r=1$ or $0<r<1,s\geq 0$, then

$E_{r}^{s}(A\|B)\leq H(A\|B)\leq E_{2-r}^{s}(A\|B).$

Proof. That $r=1$ is clear. Let $0<r<1,s=0$. By the convexity of the function
$f(x)=\frac{1}{1-r}\ln x$, we get that

$\displaystyle H_{r}(A\|B)$ $\displaystyle=$
$\displaystyle-(1-r)^{-1}\ln\left(\sum_{i}a_{i}\frac{{a_{i}}^{r-1}}{{b_{i}}^{r-1}}\right)$
$\displaystyle\leq$
$\displaystyle\sum_{i}a_{i}\left[-(1-r)^{-1}\ln\frac{{a_{i}}^{r-1}}{{b_{i}}^{r-1}}\right]$
$\displaystyle=$ $\displaystyle\sum_{i}a_{i}\ln\frac{a_{i}}{b_{i}}=H(A\|B).$

By a similar way, we get that $H(A\|B)\leq H_{2-r}(A\|B)$. Thus, we have

$\displaystyle H_{r}(A\|B)\leq H(A\|B)\leq H_{2-r}(A\|B).$ (4)

If $0<r<1$ and $s>0,$ let
$x_{0}=\sum\limits_{i=1}^{n}a_{i}\frac{b_{i}^{1-r}}{a_{i}^{1-r}}.$ Then
$\sum\limits_{i=1}^{n}a_{i}\frac{b_{i}^{1-r}}{a_{i}^{1-r}}\leq[\sum\limits_{i=1}^{n}(a_{i}\frac{b_{i}}{a_{i}})]^{1-r}=1$.
Note that when $0<x\leq 1$ and $s>0$, we have $\ln x\leq\frac{x^{s}-1}{s}$, so
for any $0<r<1,s>0$, we have $-\frac{x_{0}^{s}-1}{(1-r)s}\leq-\frac{1}{1-r}\ln
x_{0}$, thus,

$-\frac{{\left(\sum\limits_{i=1}^{n}a_{i}\frac{b_{i}^{1-r}}{a_{i}^{1-r}}\right)}^{s}-1}{(1-r)s}\leq-\frac{1}{1-r}\ln\left(\sum\limits_{i=1}^{n}a_{i}\frac{b_{i}^{1-r}}{a_{i}^{1-r}}\right),$

that is,

$\displaystyle H_{r}^{s}(A\|B)\leq H_{r}(A\|B).$ (5)

It follows from (1) and (2) that $H_{r}^{s}(A\|B)\leq H(A\|B)$. By a similar
way, we can prove $H(A\|B)\leq H_{2-r}^{s}(A\|B).$ Thus, we proved the
theorem.

3\. The quantum unified $(r,s)$-relative entropy

Let $H$ be a separable complex Hilbert space and $\rho,\sigma$ be two states
on $H$. Then for any $0\leq r\leq 1$ and any real number $s$, the quantum
unified $(r,s)-$relative entropy is defined by

$\displaystyle
E_{r}^{s}(\rho\|\sigma)=\left\\{\begin{array}[]{ll}H_{r}^{s}(\rho\|\sigma),&{\rm
if\ }0\leq r<1,s\neq 0,\\\ H_{r}(\rho\|\sigma),&{\rm if\ }0\leq r<1,s=0,\\\
H^{r}(\rho\|\sigma),&{\rm if\ }0\leq r<1,s=1,\\\
_{\frac{1}{r}}H(\rho\|\sigma),&{\rm if\ }0<r<1,s=\frac{1}{r},\\\
H(\rho\|\sigma),&{\rm if\ }r=1,\end{array}\right.$

where

$\displaystyle
H_{r}^{s}(\rho\|\sigma)=-[(1-r)s]^{-1}[\left(tr(\rho^{r}\sigma^{1-r})\right)^{s}-1],$
$\displaystyle
H_{r}(\rho\|\sigma)=-(1-r)^{-1}\ln\left(tr(\rho^{r}\sigma^{1-r})\right),$
$\displaystyle H^{r}(\rho\|\sigma)=-(1-r)^{-1}[tr(\rho^{r}\sigma^{1-r})-1],$
${}_{r}H(\rho\|\sigma)=-(r-1)^{-1}[\left(tr({\rho^{\frac{1}{r}}\sigma^{1-\frac{1}{r}}})\right)^{r}-1],$
$\displaystyle H(\rho\|\sigma)=tr(\rho\ln\rho)-tr(\rho\ln\sigma)$

are the quantum $(r,s)$-relative entropy, the quantum R$\acute{e}$nyi relative
entropy of order $r$, the quantum Tsallis relative entropy, the quantum
relative entropy of type $r$ and the quantum relative entropy ([3-6]),
respectively.

We point out that if the state $\sigma$ is invertible, then the definition of
quantum unified $(r,s)-$relative entropy can be extended to $r>1$. Moreover,
we have the following important equalities:

$\displaystyle H^{1}_{r}(\rho\|\sigma)$ $\displaystyle=$ $\displaystyle
H^{r}(\rho\|\sigma),$ (7) $\displaystyle H^{r}_{\frac{1}{r}}(\rho\|\sigma)$
$\displaystyle=$ ${}_{r}H(\rho\|\sigma),$ (8)

(3) and (4) showed that the quantum Tsallis relative entropy and quantum
relative entropy of type $r$ are the particular cases of the quantum
$(r,s)$-relative entropy.

In order to study the properties of quantum unified $(r,s)$-relative entropy,
we need the following lemma.

Lemma 3.1. Let $H$ be a separable complex Hilbert spaces, $A$ and $B$ two
positive trace class operators on $H$. Then for any $\lambda,\mu>0$, we have
$R(\lambda A+\mu B)=R(A+B)$, where $R(A)$ is the range of $A$.

Proof. In fact, if $0\leq A\leq B$, that is, $0\leq
A^{\frac{1}{2}}A^{\frac{1}{2}}\leq B^{\frac{1}{2}}B^{\frac{1}{2}}$, then it
follows from Theorem 1 in [8] that $R({A}^{\frac{1}{2}})\subseteq
R({B}^{\frac{1}{2}})$. Note that $R({A}^{\frac{1}{2}})=R({A})$,
$R({B}^{\frac{1}{2}})=R(B)$, so $R(A)\subseteq R(B)$. Thus, we have
$R(A+B)\subseteq R(A+(1+\alpha)B)$ for any $\alpha>0$. On the other hand, take
$n$ such that $0\leq\frac{A+(1+\alpha)B}{n}\leq A+B$, then
$R(A+(1+\alpha)B)=R(\frac{A+(1+\alpha)B}{n})\subseteq R(A+B).$ Thus,
$R(A+B)=R(A+(1+\alpha)B)$, i.e., $R(A+B)=R(A+\beta B)$ for any $\beta>1.$
Replace $B$ with $\frac{1}{\beta}B$, we have $R(A+\frac{1}{\beta}B)=R(A+B)$
for any $\beta>1$. Hence, $R(A+\mu B)=R(A+B)$ for any $\mu>0$. Furthermore,
$R(\lambda A+\mu B)=R(A+\mu B)=R(A+B)$ for any $\lambda>0$ and $\mu>0$.

Theorem 3.1. Let $H$, $H_{1}$ and $H_{2}$ be separable complex Hilbert spaces.

(I) If $\rho$ and $\sigma$ are two states on $H$, then
$E_{r}^{s}(\rho\|\sigma)\geq 0$. Furthermore, when $0<r\leq 1$,
$E_{r}^{s}(\rho\|\sigma)=0$ iff $\rho=\sigma$; when $r=0$,
$E_{r}^{s}(\rho\|\sigma)=0$ iff $Ker(\rho)\subseteq Ker(\sigma)$.

(II) If $\rho_{j}$ and $\sigma_{j}$ are states on $H$, $\lambda_{i}>0$,
$j=1,2,\cdots,n$, and $\sum_{j=1}^{n}\lambda_{j}=1$, then when $r=1$ or $0\leq
r<1$ and $s\leq 1$, we have

$E_{r}^{s}(\sum_{j}\lambda_{j}\rho_{j}\|\sum_{j}\lambda_{j}\sigma_{j})\leq\sum_{j}\lambda_{j}E_{r}^{s}(\rho_{j}\|\sigma_{j}).$

(III) If $\rho$ and $\sigma$ are two states, $U$ is a unitary operator on $H$,
then

$E_{r}^{s}(U\rho U^{*}\|U\sigma U^{*})=E_{r}^{s}(\rho\|\sigma).$

(IV) If $\rho_{1}$ and $\sigma_{1}$ are two states on $H_{1}$, $\rho_{2}$ and
$\sigma_{2}$ are two states on $H_{2}$, then

$E_{r}^{s}(\rho_{1}\otimes\rho_{2}\|\sigma_{1}\otimes\sigma_{2})=E_{r}^{s}(\rho_{1}\|\sigma_{1})+E_{r}^{s}(\rho_{2}\|\sigma_{2})+(r-1)sE_{r}^{s}(\rho_{1}\|\sigma_{1})E_{r}^{s}(\rho_{2}\|\sigma_{2}).$

Proof. For $r=1$, the conclusion had been proved (see [3], [9-12]). Note that
(3) and (4), we only need to prove the cases of
$E^{s}_{r}(\rho\|\sigma)=H^{s}_{r}(\rho\|\sigma)$ and
$E^{s}_{r}(\rho\|\sigma)=H_{r}(\rho\|\sigma)$.

(I) Note that when $0\leq r<1$ and $s\neq 0$, $h(x)=\frac{1-x^{s}}{(1-r)s}$
and $g(x)=\frac{\ln x}{r-1}$ are monotone decreasing, so it is sufficient to
prove that $0\leq tr(\rho^{r}\sigma^{1-r})\leq 1$.

Let $\rho=0P_{0}+\sum\limits_{i}\lambda_{i}P_{i}$ and
$\sigma=0Q_{0}+\sum\limits_{j}\mu_{j}Q_{j}$ be the spectral decompositions of
states $\rho$ and $\sigma$, where $i,j\in{\bf N}=\\{1,2,\cdots\\}$, $P_{i}$
and $Q_{j}$ are the one dimension projection operators, $P_{0}$ and $Q_{0}$
are the projections on the kernel spaces of $\rho$ and $\sigma$, respectively,
and $\lambda_{i}>0,\mu_{j}>0$. Then $P_{i}P_{0}=0$, $Q_{j}Q_{0}=0$,
$P_{i}P_{j}=Q_{i}Q_{j}=0$ if $i\neq j$,
$P_{0}+\sum\limits_{i}P_{i}=Q_{0}+\sum\limits_{j}Q_{j}=I$ and
$\sum\limits_{i}\lambda_{i}=\sum\limits_{j}\mu_{j}=1$. So, we have
$tr(P_{0}Q_{j})+\sum\limits_{i}tr(P_{i}Q_{j})=tr(Q_{j})=1$ and
$tr(Q_{0}P_{i})+\sum\limits_{j}tr(P_{i}Q_{j})=tr(P_{i})=1$. Thus, when $0\leq
r<1$,

$\displaystyle tr(\rho^{r}\sigma^{1-r})$ $\displaystyle=$
$\displaystyle\sum_{i}\sum_{j}\lambda_{i}^{r}\mu_{j}^{1-r}tr(P_{i}Q_{j})$
$\displaystyle=$
$\displaystyle\sum_{i}\sum_{j}\lambda_{i}^{r}\mu_{j}^{1-r}tr(Q_{j}P_{i}Q_{j})$
$\displaystyle\geq$ $\displaystyle 0.$

When $r=0,$

$\displaystyle tr(\rho^{0}\sigma^{1-0})$ $\displaystyle=$
$\displaystyle\sum_{ij}\mu_{j}tr(P_{i}Q_{j})$ $\displaystyle=$
$\displaystyle\sum_{j}\mu_{j}tr(\sum_{i}P_{i}Q_{j})$ $\displaystyle\leq$
$\displaystyle\sum_{j}\mu_{j}=1,$

and with equality iff for any $j,\sum\limits_{i}tr(P_{i}Q_{j})=1$ iff for any
$j,tr(P_{0}Q_{j})=0$ iff $P_{0}\leq Q_{0}$ iff $Ker(\rho)\subseteq
Ker(\sigma)$.

When $0<r<1$, note that $\sum_{i}tr(P_{i}Q_{j})+tr(P_{0}Q_{j})=1,$ by the
concavity of $f(x)=x^{r}$, we have

$\displaystyle tr(\rho^{r}\sigma^{1-r})$ $\displaystyle=$
$\displaystyle\sum_{i}\sum_{j}\lambda_{i}^{r}\mu_{j}^{1-r}tr(P_{i}Q_{j})$ (9)
$\displaystyle=$
$\displaystyle\sum_{j}\mu_{j}[\sum_{i}(\frac{\lambda_{i}}{\mu_{j}})^{r}tr(P_{i}Q_{j})+(\frac{0}{\mu_{j}})^{r}tr(P_{0}Q_{j})]$
(10) $\displaystyle\leq$
$\displaystyle\sum_{j}\mu_{j}[\sum_{i}\frac{\lambda_{i}}{\mu_{j}}tr(P_{i}Q_{j})+\frac{0}{\mu_{j}}tr(P_{0}Q_{j})]^{r}$
(11) $\displaystyle\leq$
$\displaystyle(\sum_{j}\mu_{j}{\sum_{i}\frac{\lambda_{i}}{\mu_{j}}tr(P_{i}Q_{j})})^{r}$
(12) $\displaystyle=$
$\displaystyle(\sum_{i}\lambda_{i}\sum_{j}tr(P_{i}Q_{j}))^{r}$ (13)
$\displaystyle\leq$ $\displaystyle 1.$ (14)

Thus, we proved that when $0\leq r<1$, $0\leq tr(\rho^{r}\sigma^{1-r})\leq 1$,
and when $r=0$, $tr(\rho^{0}\sigma^{1-0})=1$ iff $Ker(\rho)\subseteq
Ker(\sigma)$. Note that when $0<r<1$, $E_{r}^{s}(\rho\|\sigma)=0$ iff
$tr(\rho^{r}\sigma^{1-r})=1$, so, we only need to prove that if $0<r<1$ and
$tr(\rho^{r}\sigma^{1-r})=1$, then $\rho=\sigma$.

First, if $tr(\rho^{r}\sigma^{1-r})=1$, it follows from (9) and (10) that for
each $i\in\bf N$, $\sum\limits_{j}tr(P_{i}Q_{j})=1$, so
$tr(P_{i}Q_{0})=0=tr(P_{i}Q_{0}P_{i})$, it is easily to know that
$P_{i}Q_{0}P_{i}=0$, so for each $i\in\bf N$, $P_{i}Q_{0}=0$, thus we have
$Q_{0}\leq P_{0}$. Moreover, if $tr(\rho^{r}\sigma^{1-r})=1$, then (7) takes
equality, we get that (i) or (ii) as follows:

(i) For each given $j$, there exists a $i_{j}\in\bf N$ such that
$tr(P_{i_{j}}Q_{j})=1$ and $tr(P_{i}Q_{j})=0$ for all $i\neq i_{j}$.

(ii) For each $j$, we have
$\frac{\lambda_{1}}{\mu_{j}}=\frac{\lambda_{2}}{\mu_{j}}=\cdots$ and
$\sum\limits_{i}tr(P_{i}Q_{j})=1$.

If (i) is satisfied, then for each $j$, we have $P_{0}Q_{j}=0$, so $P_{0}\leq
Q_{0}$, combining this and $Q_{0}\leq P_{0}$ proved before, we get
$Q_{0}=P_{0}$. Moreover, note that $P_{i_{j}}$ and $Q_{j}$ are both one
dimensional projections and $tr(P_{i_{j}}Q_{j})=1$, so, it is easy to know
that $P_{i_{j}}=Q_{j}$. It also follows from $tr(\rho^{r}\sigma^{1-r})=1$ that
$\frac{\lambda_{i_{j}}}{\mu_{j}}=1$, thus, we can prove that $\rho=\sigma$.

If (ii) is satisfied, then for each $j$, we have
$\frac{\lambda_{1}}{\mu_{j}}=\frac{\lambda_{2}}{\mu_{j}}=\cdots$ and
$\sum_{i}tr(P_{i}Q_{j})=1$, so we have $\lambda_{1}=\lambda_{2}=\cdots$ and
for each $j$, $tr(P_{0}Q_{j})=0$, so we can prove that $P_{0}\leq Q_{0}$,
thus, $P_{0}=Q_{0}$. Moreover, it follows from $\frac{\lambda_{i}}{\mu_{j}}$
is a constant, $\sum_{i}tr(P_{i}Q_{j})=1$ and (5)-(10) that
$\mu_{1}=\mu_{2}=\cdots=\lambda_{1}=\lambda_{2}=\cdots$, thus, we have
$\rho=\sigma$, (I) is proved.

(II) Let $\rho$ and $\sigma$ be two states on $H$ and
$f(\rho,\sigma)=tr({\rho}^{r}{\sigma}^{1-r})$. If $0<r<1$, then it follows
from [13, Corollary 1.1] that $f(\rho,\sigma)=tr({\rho}^{r}{\sigma}^{1-r})$ is
a joint concave functional with respect to the states $\rho$ and $\sigma$,
that is, for any states $\rho_{1}$, $\rho_{2}$, $\sigma_{1}$ and $\sigma_{2}$,
when $0<\lambda<1$, we have

$\displaystyle
f(\lambda\rho_{1}+(1-\lambda)\rho_{2},\lambda\sigma_{1}+(1-\lambda)\sigma_{2})\geq\lambda
f(\rho_{1},\sigma_{1})+(1-\lambda)f(\rho_{2},\sigma_{2}).$ (15)

If $r=0$, let $P_{1}$, $P_{2}$ and $P$ be the projection operators on
$R(\rho_{1})$, $R(\rho_{2})$ and $R(\lambda\rho_{1}+(1-\lambda)\rho_{2})$,
respectively, then $\rho_{1}^{0}=P_{1}$, $\rho_{2}^{0}=P_{2}$,
$(\lambda\rho_{1}+(1-\lambda)\rho_{2})^{0}=P$. It follows from Lemma 3.1 that
$P\geq P_{1}$ and $P\geq P_{2}.$ Therefore, we have

$\displaystyle
tr((\lambda\rho_{1}+(1-\lambda)\rho_{2})^{0}(\lambda\sigma_{1}+(1-\lambda)\sigma_{2})^{1})$
$\displaystyle=$ $\displaystyle
tr(P(\lambda\sigma_{1}+(1-\lambda)\sigma_{2}))$ $\displaystyle=$
$\displaystyle\lambda tr(P\sigma_{1})+(1-\lambda)tr(P\sigma_{2})$
$\displaystyle\geq$ $\displaystyle\lambda
tr(P_{1}\sigma_{1})+(1-\lambda)tr(P_{2}\sigma_{2})$ $\displaystyle=$
$\displaystyle\lambda tr(P_{1}\sigma_{1})+(1-\lambda)tr(P_{2}\sigma_{2})$
$\displaystyle=$ $\displaystyle\lambda
tr((\rho_{1})^{0}\sigma_{1})+(1-\lambda)tr((\rho_{2})^{0}\sigma_{2}).$

This shows that the inequality (11) also holds when $r=0$.

If $0\leq r<1,s=0$, by the monotone decreasing property and convexity of the
function $g(x)=\frac{\ln x}{r-1}$, we have

$\displaystyle
H_{r}(\lambda\rho_{1}+(1-\lambda)\rho_{2}\|\lambda\sigma_{1}+(1-\lambda)\sigma_{2})$
$\displaystyle=$
$\displaystyle\frac{1}{r-1}\ln(f(\lambda\rho_{1}+(1-\lambda)\rho_{2},\lambda\sigma_{1}+(1-\lambda)\sigma_{2}))$
$\displaystyle\leq$ $\displaystyle\frac{1}{r-1}\ln(\lambda
f(\rho_{1},\sigma_{1})+(1-\lambda)f(\rho_{2},\sigma_{2}))$ $\displaystyle\leq$
$\displaystyle\lambda\frac{1}{r-1}\ln(f(\rho_{1},\sigma_{1})+(1-\lambda)\frac{1}{r-1}\ln(f(\rho_{2},\sigma_{2}))$
$\displaystyle=$ $\displaystyle\lambda
H_{r}(\rho_{1}\|\sigma_{1})+(1-\lambda)H_{r}(\rho_{2}\|\sigma_{2}).$

If $0\leq r<1,s\neq 0$ and $s\leq 1$, then $h(x)=\frac{1-x^{s}}{(1-r)s}$ is
also a monotone decreasing convex function, so

$\displaystyle
H_{r}^{s}(\lambda\rho_{1}+(1-\lambda)\rho_{2}\|\lambda\sigma_{1}+(1-\lambda)\sigma_{2})$
$\displaystyle=$
$\displaystyle[(1-r)s]^{-1}[1-f^{s}(\lambda\rho_{1}+(1-\lambda)\rho_{2},\lambda\sigma_{1}+(1-\lambda)\sigma_{2})]$
$\displaystyle\leq$ $\displaystyle[(1-r)s]^{-1}[1-(\lambda
f^{s}(\rho_{1},\sigma_{1})+(1-\lambda)f^{s}(\rho_{2},\sigma_{2}))]$
$\displaystyle=$ $\displaystyle\lambda
H_{r}^{s}(\rho_{1}\|\sigma_{1})+(1-\lambda)H_{r}^{s}(\rho_{2}\|\sigma_{2}).$

Thus, (II) is proved. (III) and (IV) can be proved easily, we omit them.

In order to study the other properties of quantum unified $(r,s)$-relative
entropy, we need the following:

Let $H_{1}$ and $H_{2}$ be two separable complex Hilbert spaces and
$H_{1}\otimes H_{2}$ their tensor product. The set of all trace class
operators on $H_{1}\otimes H_{2}$ is denoted by $T(H_{1}\otimes H_{2})$, the
set of all trace class positive operators on $H_{1}\otimes H_{2}$ is denoted
by $T_{+}(H_{1}\otimes H_{2})$. If $A\in T_{+}(H_{1}\otimes H_{2})$, by the
following form, we can define a trace class positive operator $A_{1}$ on
$H_{1}$:

$(x,A_{1}y)=\sum_{i}(x\otimes e_{i},A(y\otimes e_{i})),$

where $x,y\in H_{1}$, $\\{e_{i}\\}$ is an orthonormal basis of $H_{2}$. We
call $A_{1}$ to be the partial trace of $A$ on $H_{1}$ and denoted by
$A_{1}=tr_{2}A$. Similarly, we can define the partial trace $A_{2}$ of $A$ on
$H_{2}$. Note that when $A$ is a state, $A_{1}$ and $A_{2}$ are also states.

It follows from Theorem 3.1(III), Theorem 3.1(IV) and the methods in the proof
of [3, Theorem 11.17], we have

Lemma 3.2. Let $H_{1},H_{2}$ be two finite dimensional complex Hilbert space.
If $r=1$ or $0\leq r<1$ and $s\leq 1$, then for any states $\rho$ and $\sigma$
on $H_{1}\otimes H_{2}$,

$E_{r}^{s}(\rho_{1}||\sigma_{1})\leq E_{r}^{s}(\rho||\sigma),$

where $\rho_{1}$ and $\sigma_{1}$ are the partial traces of $\rho$ and
$\sigma$ on $H_{1}$, respectively.

Lemma 3.3. Let $H$ be a finite dimensional complex Hilbert space, $\Phi$ a
trace-preserving completely positive map of $T(H)$ into itself. If $r=1$ or
$0\leq r<1$ and $s\leq 1$, then for any states $\rho$ and $\sigma$ on $H$,

$E_{r}^{s}(\Phi(\rho)||\Phi(\sigma))\leq E_{r}^{s}(\rho||\sigma).$

Proof. Taking a finite dimensional complex Hilbert space $H_{0}$ such that the
dimension of $H_{0}$ is bigger than 1. Then it follows from ([9,11-12]) that
there are a unitary operator $U$ on $H\otimes H_{0}$ and a projection operator
$P$ on $H_{0}$ such that for any state $\rho$ on $H$, we have

$\Phi(\rho)=tr_{2}(U(\rho\otimes P)U^{*}),$

thus, it follows from Lemma 3.2 that

$E_{r}^{s}(\Phi(\rho)||\Phi(\sigma))\leq E_{r}^{s}(U(\rho\otimes
P)U^{*}||U(\sigma\otimes P)U^{*})=E_{r}^{s}(\rho\otimes P||\sigma\otimes
P)=E_{r}^{s}(\rho||\sigma).$

Lemma 3.4 ([11]). Let $H$ be a separable complex Hilbert space, $\Phi$ a
trace-preserving completely positive map of $T(H)$ into itself, and
$\\{P_{n}\\}$ a family of finite-dimensional projections such that $P_{m}\leq
P_{n}$ for $m\leq n$ and $P_{n}\rightarrow I$ strongly when
$n\rightarrow\infty$. Then there is a family $\\{\Phi_{n}\\}$ of completely
positive maps such that $\\{\Phi_{n}\\}$ is trace-preserving on $P_{n}(H)$ and
$\Phi_{n}(A)\rightarrow\Phi(A)$ uniformly for each $A\in T_{+}(H)$.

Theorem 3.2. Let $H$ be a separable complex Hilbert space and $\Phi$ a trace-
preserving completely positive map of $T(H)$ into itself. If $r=1$ or $0\leq
r<1$ and $s\leq 1$, then for any state $\rho$ and $\sigma$, we have

$\displaystyle E_{r}^{s}(\Phi(\rho)\|\Phi(\sigma))\leq
E_{r}^{s}(\rho\|\sigma).$

Proof. Let $P_{n}$ and $\Phi_{n}$ satisfy the conditions of Lemma 3.4. Then
$\Phi_{n}(\rho)\rightarrow\Phi(\rho)$ uniformly for each state $\rho$. Since
function $x^{r}y^{1-r}$ is continuous, we have
$(\Phi_{n}(\rho))^{r}(\Phi_{n}(\sigma))^{1-r}\rightarrow(\Phi(\rho))^{r}(\Phi(\sigma))^{1-r}$
uniformly, hence $tr((\Phi_{n}(\rho))^{r}(\Phi_{n}(\sigma))^{1-r})\rightarrow
tr((\Phi(\rho))^{r}(\Phi(\sigma))^{1-r}).$ This shows that

$\displaystyle
E_{r}^{s}(\Phi(\rho)\|\Phi(\sigma))=\lim_{n\rightarrow\infty}E_{r}^{s}(\Phi_{n}(\rho)\|\Phi_{n}(\sigma)).$

Let $\rho_{n}=\frac{P_{n}\rho P_{n}}{tr(\rho
P_{n})},\sigma_{n}=\frac{P_{n}\sigma P_{n}}{tr(\sigma P_{n})}$. By the proof
of Lemma 4 in [10], $P_{n}\rho P_{n}\rightarrow\rho$ and $P_{n}\sigma
P_{n}\rightarrow\sigma$ uniformly. Hence $tr(P_{n}\rho P_{n})\rightarrow
tr(\rho)=1$ and $tr(P_{n}\sigma P_{n})\rightarrow tr(\sigma)=1.$ Therefore

$\displaystyle\lim_{n\rightarrow\infty}(\Phi_{n}(\rho))^{r}(\Phi_{n}(\sigma))^{1-r}$
$\displaystyle=$ $\displaystyle\lim_{n\rightarrow\infty}(\Phi_{n}(P_{n}\rho
P_{n}))^{r}(\Phi_{n}(P_{n}\sigma P_{n}))^{1-r}$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{(\Phi_{n}(P_{n}\rho
P_{n}))^{r}(\Phi_{n}(P_{n}\sigma P_{n}))^{1-r}}{(tr(P_{n}\rho
P_{n}))^{r}(tr(P_{n}\sigma P_{n}))^{1-r}}$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}(\Phi_{n}(\rho_{n}))^{r}(\Phi_{n}(\sigma_{n}))^{1-r}.$

Hence we get that
$\lim\limits_{n\rightarrow\infty}E_{r}^{s}(\Phi_{n}(\rho)\|\Phi_{n}(\sigma))=\lim\limits_{n\rightarrow\infty}E_{r}^{s}(\Phi_{n}(\rho_{n})\|\Phi_{n}(\sigma_{n})).$
By Lemma 3.2, $E_{r}^{s}(\Phi_{n}(\rho_{n})\|\Phi_{n}(\sigma_{n}))\leq
E_{r}^{s}(\rho_{n}\|\sigma_{n})$. Again
$\rho_{n}\rightarrow\rho,\sigma_{n}\rightarrow\sigma$ uniformly, we get that
$\lim\limits_{n\rightarrow\infty}E_{r}^{s}(\rho_{n}\|\sigma_{n})=E_{r}^{s}(\rho\|\sigma).$
Therefore $E_{r}^{s}(\Phi(\rho)\|\Phi(\sigma))\leq E_{r}^{s}(\rho\|\sigma).$
That completes the proof.

Theorem 3.3 (Monotonicity). Let $H_{1},H_{2}$ be separable complex Hilbert
space, $H=H_{1}\otimes H_{2}$. If $r=1$ or $0\leq r<1$ and $s\leq 1$, then for
any state $\rho$ and $\sigma$ on $H$,

$E_{r}^{s}(\rho_{1}\|\sigma_{1})\leq E_{r}^{s}(\rho\|\sigma),$

where $\rho_{1}$ and $\sigma_{1}$ are the partial traces of $\rho$ and
$\sigma$ on $H_{1}$, respectively.

Proof. Since $H_{2}$ is a separable complex Hilbert space, so there is a
sequence of $\\{P_{n}\\}$ of finite-dimensional projection operators on
$H_{2}$ such that $P_{m}\leq P_{n}$ for $m\leq n$ and $P_{n}\rightarrow I$
strongly when $n\rightarrow\infty$. Let
$H_{2}^{n}=P_{n}(H_{2}),H^{n}=H_{1}\otimes H_{2}^{n}$. It follows from the
proof of Lemma 4 in [10] again that

$\rho_{n}=\frac{(I\otimes P_{n})\rho(I\otimes P_{n})}{tr(\rho(I\otimes
P_{n}))}\rightarrow\rho,$ $\sigma_{n}=\frac{(I\otimes P_{n})\sigma(I\otimes
P_{n})}{tr(\sigma(I\otimes P_{n}))}\rightarrow\sigma,$
$\rho_{1n}=tr_{2}\rho_{n}\rightarrow\rho_{1},$
$\sigma_{1n}=tr_{2}\sigma_{n}\rightarrow\sigma_{1}$

uniformly. Hence $E_{r}^{s}(\rho_{1n}\|\sigma_{1n})\rightarrow
E_{r}^{s}(\rho_{1}\|\sigma_{1}),$ and
$E_{r}^{s}(\rho_{n}\|\sigma_{n})\rightarrow E_{r}^{s}(\rho\|\sigma).$

Define $\Phi:B(H^{n})\rightarrow B(H_{1})\otimes\\{\lambda I_{2}^{n}\\}$ by
$\Phi(\rho)=(tr_{2}\rho)\otimes C_{2n},$ where $I_{2}^{n}$ is the identity
operator on $H_{2}^{n}$) and $C_{2n}=(dimH_{2}^{n})^{-1}I^{n}_{2}.$ Then

$E_{r}^{s}(\Phi(\rho_{n})\|\Phi(\sigma_{n}))=E_{r}^{s}(\rho_{1n}\otimes
C_{2n}\|\sigma_{1n}\otimes C_{2n})=E_{r}^{s}(\rho_{1n}\|\sigma_{1n}).$

It is obvious that $\Phi$ is a trace-preserving completely positive map from
$B(H^{n})$ into itself. By Theorem 3.2,
$E_{r}^{s}(\Phi(\rho_{n})\|\Phi(\sigma_{n}))\leq
E_{r}^{s}(\rho_{n}\|\sigma_{n})$, so

$E_{r}^{s}(\rho_{1n}\|\sigma_{1n})\leq E_{r}^{s}(\rho_{n}\|\sigma_{n}).$

Note that $E_{r}^{s}(\rho_{1n}\|\sigma_{1n})\rightarrow
E_{r}^{s}(\rho_{1}\|\sigma_{1}),$ $E_{r}^{s}(\rho_{n}\|\sigma_{n})\rightarrow
E_{r}^{s}(\rho\|\sigma),$ thus we have

$E_{r}^{s}(\rho_{1}\|\sigma_{1})\leq E_{r}^{s}(\rho\|\sigma)$

and the theorem is proved.

Theorem 3.4. Let $H$ be a separable complex Hilbert space, $\rho$ and $\sigma$
two states on $H$ and $\sigma$ invertible. Then for $r=1$ or $0\leq r<1,s\geq
0,$ we have

$\displaystyle E_{r}^{s}(\rho\|\sigma)\leq H(\rho\|\sigma)\leq
E_{2-r}^{s}(\rho\|\sigma).$ (16)

Proof. That $r=1$ is clear. If $0\leq r<1,s=0$, we need to prove that

$\displaystyle H_{r}(\rho\|\sigma)\leq H(\rho\|\sigma)\leq
H_{2-r}(\rho\|\sigma).$ (17)

Let $\rho=0P_{0}+\sum\limits_{i}\lambda_{i}P_{i}$ and
$\sigma=\sum\limits_{j}\mu_{j}Q_{j}$ be the spectral decompositions of $\rho$
and $\sigma$, where $P_{i}$ and $Q_{j}$ be the one dimension projection
operators, $P_{0}$ be the projection operator on the kernel space of $\rho$,
and $P_{i}P_{0}=0$, $\lambda_{i}>0,\mu_{j}>0$ when $i,j\in\bf N$, and
$P_{i}P_{j}=Q_{i}Q_{j}=0$ if $i\neq j$,
$\sum\limits_{i}\lambda_{i}=\sum\limits_{j}\mu_{j}=1$,
$P_{0}+\sum\limits_{i}P_{i}=\sum\limits_{j}Q_{j}=I$. Then

$\displaystyle H_{2-r}(\rho\|\sigma)$ $\displaystyle=$
$\displaystyle-\frac{\ln tr(\rho^{2-r}\sigma^{r-1})}{r-1}$ $\displaystyle=$
$\displaystyle-\frac{1}{r-1}\ln\sum_{ij}\lambda_{i}^{2-r}\mu_{j}^{r-1}tr(P_{i}Q_{j})$
$\displaystyle=$
$\displaystyle-\frac{1}{r-1}\ln\sum_{ij}\lambda_{i}tr(P_{i}Q_{j})(\frac{\lambda_{i}}{\mu_{j}})^{1-r}$

Let $g(x)=-\frac{1}{r-1}\ln x,\ \alpha_{ij}=\lambda_{i}tr(P_{i}Q_{j})$ and
$x_{ij}=(\frac{\lambda_{i}}{\mu_{j}})^{1-r}.$ Then
$\sum\limits_{ij}\alpha_{ij}=\sum\limits_{ij}\lambda_{i}tr(P_{i}Q_{j})=\sum\limits_{j}tr(\rho
Q_{j})=tr(\rho)=1.$ By the concavity of the function $g(x)=-\frac{1}{r-1}\ln
x,$ we have

$\displaystyle H_{2-r}(\rho\|\sigma)$ $\displaystyle=$ $\displaystyle
g(\sum_{ij}\alpha_{ij}x_{ij})$ $\displaystyle\geq$
$\displaystyle\sum_{ij}\alpha_{ij}g(x_{ij})$ $\displaystyle=$
$\displaystyle\sum_{ij}\lambda_{i}tr(P_{i}Q_{j})\left(-\frac{1}{r-1}\ln(\frac{\lambda_{i}}{\mu_{j}})^{1-r}\right)$
$\displaystyle=$
$\displaystyle\sum_{ij}\lambda_{i}tr(P_{i}Q_{j})(\ln\lambda_{i}-\ln\mu_{j})$
$\displaystyle=$ $\displaystyle H(\rho\|\sigma).$

The left-hand side inequality of (13) is proven by a similar way.

If $0\leq r<1,s>0$, we need to prove that

$\displaystyle H_{r}^{s}(\rho\|\sigma)\leq H(\rho\|\sigma)\leq
H_{2-r}^{s}(\rho\|\sigma).$ (18)

Let $tr(\rho^{r}\sigma^{1-r})=x_{0}.$ Since

$\ln x\leq\frac{x^{s}-1}{s},$

for any $x>0,s>0,$ so, for any $0\leq r<1,s>0$, we have

$-\frac{x_{0}^{s}-1}{[(r-1)s]}\geq-(r-1)^{-1}\ln x_{0}.$

That is,

$H_{2-r}^{s}(\rho\|\sigma)\geq H_{2-r}(\rho\|\sigma).$

Combining this with (13), we have $H_{2-r}^{s}(\rho\|\sigma)\geq
H(\rho\|\sigma)$. Similarly, the left-hand side inequality of (14) can be
proven.

Note that when $0\leq r<1,s=1$, the inequalities (12) degenerate into
convexity inequalities for estimating free energy and relative entropy given
by Ruskai and Stillinger in [14].

Theorem 3.5. Let $\rho$ and $\sigma$ be two states on the separable complex
Hilbert space ${\mathcal{H}}$. We have

(1) If $\rho$ is an invertible state, then $E_{0}^{s}(\rho\|\sigma)=0$.

(2) If $s\geq 0$, then $E_{r}^{s}(\rho\|\sigma)$ is monotone increasing with
respect to $r\in[0,1]$; if $s<0$, then $E_{r}^{s}(\rho\|\sigma)$ is monotone
decreasing with respect to $r\in[0,1]$.

(3) For each $0\leq r\leq 1$, $E_{r}^{s}(\rho\|\sigma)$ is monotone decreasing
with respect to $s$.

(4) For each $0\leq r\leq 1$, $E_{r}^{s}(\rho\|\sigma)$ is a convex function
of $s$.

Proof. (1) If $\rho$ is invertible, we have $\rho^{0}=I$, so
$E_{0}^{s}(\rho\|\sigma)=0$.

(2) It follows from Theorem 3.4 that $E_{r}^{s}(\rho\|\sigma)\leq
H(\rho\|\sigma)=E_{1}^{s}(\rho\|\sigma)$, so it is sufficient to prove the
conclusion for $0\leq r<1$ and any $s$.

Let $\rho=0P_{0}+\sum\limits_{i}\lambda_{i}P_{i}$ and
$\sigma=0Q_{0}+\sum\limits_{j}\mu_{j}Q_{j}$ be the spectral decompositions of
$\rho$ and $\sigma$, where $P_{i}$ and $Q_{j}$ are the one dimension
projection operators, $P_{0}$ and $Q_{0}$ are the projection operators on the
zero spaces of $\rho$ and $\sigma$ respectively, and for all $i,j\in\bf N$,
$\lambda_{i}>0,\mu_{j}>0$. Then
$\sum\limits_{i}\lambda_{i}=\sum\limits_{j}\mu_{j}=1$,
$P_{0}+\sum\limits_{i}P_{i}=Q_{0}+\sum\limits_{j}Q_{j}=I$.

Let $f(x)=x\ln x$, $\alpha_{ij}=\lambda_{i}tr(P_{i}Q_{j})$,
$x_{ij}=(\frac{\mu_{j}}{\lambda_{i}})^{1-r}$. Then
$\sum\limits_{i}[\sum\limits_{j}\alpha_{ij}+\lambda_{i}tr(P_{i}Q_{0})]=\sum\limits_{i}\lambda_{i}tr(P_{i}(\sum\limits_{i}Q_{j}+Q_{0}))=tr(\rho)=1.$
Because $f(x)$ is a convex function, we have
$\sum\limits_{ij}\alpha_{ij}f(x_{ij})\geq
f(\sum\limits_{ij}\alpha_{ij}x_{ij}).$ Therefore,

$\sum\limits_{ij}\lambda_{i}tr(P_{i}Q_{j})(\frac{\mu_{j}}{\lambda_{i}})^{1-r}\ln(\frac{\mu_{j}}{\lambda_{i}})^{1-r}\geq\sum\limits_{ij}\lambda_{i}tr(P_{i}Q_{j})(\frac{\mu_{j}}{\lambda_{i}})^{1-r}\ln\sum\limits_{ij}\lambda_{i}tr(P_{i}Q_{j})(\frac{\mu_{j}}{\lambda_{i}})^{1-r},$

that is,

$\displaystyle-(1-r)tr(\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r})\geq
tr(\rho^{r}\sigma^{1-r})\ln tr(\rho^{r}\sigma^{1-r}).$ (19)

(i) If $s=0$, then
$E_{r}^{0}(\rho\|\sigma)=H_{r}(\rho\|\sigma)=-(1-r)^{-1}\ln(tr(\rho^{r}\sigma^{1-r}))$.
Note that $tr(\rho^{r}\sigma^{1-r})=0$ iff for any $i,j\in{\bf
N},P_{i}Q_{j}=0.$ Hence, if for some $0\leq r_{0}<1$ such that
$tr(\rho^{r_{0}}\sigma^{1-{r_{0}}})=0$, then it is easily to see that for any
$0\leq r<1$, $tr(\rho^{r}\sigma^{1-{r}})=0$, so for any $0\leq r<1$,
$E_{r}^{0}(\rho\|\sigma)=H_{r}(\rho\|\sigma)=+\infty$, thus, the conclusion is
also true in this case. If for each $0\leq r<1$, $tr(\rho^{r}\sigma^{1-r})>0$,
then

$\displaystyle\frac{dH_{r}(\rho\|\sigma)}{dr}$ $\displaystyle=$
$\displaystyle\frac{H_{r}(\rho\|\sigma)}{1-r}-\frac{tr(\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r})}{(1-r)tr(\rho^{r}\sigma^{1-r})}$
$\displaystyle=$ $\displaystyle\frac{-\ln
tr(\rho^{r}\sigma^{1-r})}{(1-r)^{2}}-\frac{tr(\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r})}{(1-r)tr(\rho^{r}\sigma^{1-r})}.$

By (15), we know that
$-tr(\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r})\geq\frac{1}{1-r}{tr(\rho^{r}\sigma^{1-r})\ln
tr(\rho^{r}\sigma^{1-r})},$ so

$\frac{dH_{r}(\rho\|\sigma)}{dr}\geq\frac{-\ln
tr(\rho^{r}\sigma^{1-r})}{(1-r)^{2}}+\frac{\ln
tr(\rho^{r}\sigma^{1-r})}{(1-r)^{2}}=0.$

This conclusion is proved when $s=0$.

(ii) If $s\neq 0$, and for some $0\leq r_{0}<1$,
$tr(\rho^{r_{0}}\sigma^{1-{r_{0}}})=0$, then for any $0\leq r<1$,
$tr(\rho^{r}\sigma^{1-{r}})=0$, so for any $0\leq r<1$,
$E_{r}^{s}(\rho\|\sigma)=H_{r}^{s}(\rho\|\sigma)=\frac{1}{(1-r)s}$, thus, the
conclusion is true in this case. If for each $0\leq r<1$,
$tr(\rho^{r}\sigma^{1-r})>0$, then
$E_{r}^{s}(\rho\|\sigma)=H_{r}^{s}(\rho\|\sigma)=-[(1-r)s]^{-1}[(tr(\rho^{r}\sigma^{1-r}))^{s}-1]$,
so by (15) again that

$\displaystyle\frac{dH_{r}^{s}(\rho\|\sigma)}{dr}$ $\displaystyle=$
$\displaystyle\frac{H_{r}^{s}(\rho\|\sigma)}{1-r}-\frac{(tr(\rho^{r}\sigma^{1-r}))^{s-1}tr\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r}}{1-r}$
$\displaystyle=$
$\displaystyle\frac{1}{(1-r)^{2}}\left[\frac{1-(tr(\rho^{r}\sigma^{1-r}))^{s}}{s}-(1-r)(tr(\rho^{r}\sigma^{1-r}))^{s-1}tr(\rho^{r}(\ln\rho-\ln\sigma)\sigma^{1-r})\right]$
$\displaystyle\geq$
$\displaystyle\frac{1}{(1-r)^{2}}\left[\frac{1-(tr(\rho^{r}\sigma^{1-r}))^{s}}{s}+\frac{s(tr(\rho^{r}\sigma^{1-r})\ln(tr\rho^{r}\sigma^{1-r}))(tr(\rho^{r}\sigma^{1-r}))^{s-1}}{s}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{(1-r)^{2}}\left[\frac{1-(tr(\rho^{r}\sigma^{1-r}))^{s}+s(tr(\rho^{r}\sigma^{1-r}))^{s}\ln
tr(\rho^{r}\sigma^{1-r})}{s}\right].$

Let $tr(\rho^{r}\sigma^{1-r})=x,\ f(x)=\frac{sx^{s}\ln x-x^{s}+1}{s}.$ Then
$0<x\leq 1,\ f^{\prime}(x)=sx^{s-1}\ln x.$ Note that If $s>0$, then
$f^{\prime}(x)\leq 0,\ f(x)\geq f(1)=0.$ Thus,
$\frac{dH^{s}_{r}(\rho\|\sigma)}{dr}\geq 0.$ Similarly,
$\frac{dH^{s}_{r}(\rho\|\sigma)}{dr}\leq 0$ if $s<0$.

The conclusion is proved finally.

(3) If $r=1$, then $E_{r}^{s}(\rho\|\sigma)=H(\rho\|\sigma)$ is a constant, so
the conclusion is true in this case.

If for some $0\leq r_{0}<1$, $tr(\rho^{r_{0}}\sigma^{1-{r_{0}}})=0$, then for
any $0\leq r<1$, $tr(\rho^{r}\sigma^{1-{r}})=0$, thus,
$E_{r}^{0}(\rho\|\sigma)=+\infty>\frac{1}{(1-r)s}=E_{r}^{s}(\rho\|\sigma)$ for
any $s$.

If for each $0\leq r<1$, $tr(\rho^{r}\sigma^{1-r})>0$, by the inequality

$\ln x\leq\frac{x^{s}-1}{s}$

for any $x>0,s>0$, we can prove that for any $s>0$ and $0\leq r<1$,
$E_{r}^{s}(\rho\|\sigma)\leq E_{r}^{0}(\rho\|\sigma)$. Similarly, we have
$E_{r}^{s}(\rho\|\sigma)\geq E_{r}^{0}(\rho\|\sigma)$ for any $s<0$ and $0\leq
r<1$.

Thus, in order to prove the conclusion, it is sufficient to show that

$\frac{dE^{s}_{r}(\rho\|\sigma)}{ds}\leq 0$

if $tr(\rho^{r}\sigma^{1-{r}})>0$ for any $s\neq 0,0\leq r<1.$ Note that

$\displaystyle\frac{dE^{s}_{r}(\rho\|\sigma)}{ds}=\frac{[-(tr(\rho^{r}\sigma^{1-r}))^{s}\ln
tr(\rho^{r}\sigma^{1-r})]s-[1-(tr(\rho^{r}\sigma^{1-r}))^{s}]}{(1-r)s^{2}}.$
(20)

Let $tr(\rho^{r}\sigma^{1-r})=x$, $f(x)=\frac{x^{s}-1-sx^{s}\ln
x}{(1-r)s^{2}}.$ Then $0<x\leq 1$,
$\frac{dE^{s}_{r}(\rho\|\sigma)}{ds}=\frac{{x}^{s}-1-s{x}^{s}\ln{x}}{(1-r)s^{2}}$,
$f^{{}^{\prime}}(x)=\frac{-x^{s-1}\ln x}{1-r}\geq 0,$ so $f(x)\leq f(1)=0.$
Thus we get $\frac{dE^{s}_{r}(\rho\|\sigma)}{ds}\leq 0.$

(4) When $r=1$, the conclusion is clear.

When $s=0$, $E_{r}^{s}(\rho\|\sigma)=H_{r}(\rho\|\sigma)$ is a constant
function of $s$. Hence, the conclusion is clear.

When $s\neq 0$, if for some $0\leq r_{0}<1$,
$tr(\rho^{r_{0}}\sigma^{1-{r_{0}}})=0$, then for any $0\leq r<1$,
$tr(\rho^{r}\sigma^{1-{r}})=0$, thus,
$E_{r}^{s}(\rho\|\sigma)=\frac{1}{(1-r)s}$ is a convex function of $s$, hence,
the conclusion is true in this case.

When $s\neq 0$, if $tr(\rho^{r}\sigma^{1-r})>0$ for each $0\leq r<1$. Let
$tr(\rho^{r}\sigma^{1-r})=x$. Then $0<x\leq 1$. Moreover,

$\frac{d^{2}E^{s}_{r}(\rho\|\sigma)}{ds^{2}}=\frac{-s^{2}{x}^{s}\ln^{2}{x}+2s{x}^{s}\ln{x}+2(1-{x}^{s})}{(1-r)s^{3}}.$

Let $g(x)=-s^{2}x^{s}\ln^{2}x+2sx^{s}\ln x+2(1-x^{s}).$ Then
$g^{{}^{\prime}}(x)=-s^{3}x^{s-1}\ln^{2}x.$ Thus, $g^{{}^{\prime}}(x)\leq 0$
if $s>0$, $g^{{}^{\prime}}(x)\geq 0$ if $s<0$. Correspondingly, $g(s)\geq
g(1)=0$ if $s>0$, $g(s)\leq g(1)=0$ if $s<0$. Hence
$\frac{d^{2}E^{s}_{r}(\rho\|\sigma)}{ds^{2}}\geq 0$ for $s\neq 0$. The
conclusion is proved finally.

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