# Twisted Basic Dolbeault cohomology on transverse Kähler foliations

Seoung Dal Jung  Department of Mathematics
Jeju National University
Jeju 690-756
Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the twisted basic Dolbeault cohomology and transverse
hard Lefschetz theorem on a transverse Kähler foliation. And we give some
properties for $\Delta_{\kappa}$-harmonic forms and prove the Kodaira-Serre
type duality and Dolbeault isomorphism for the twisted basic Dolbeault
cohomology.

###### Key words and phrases:

Riemannian foliation, transverse Kähler foliation, basic Dolbeault cohomology,
twisted basic Dolbeault cohomology, Kodaira-Serre duality, hard Lefschetz
theorem

###### 2010 Mathematics Subject Classification:

53C12; 53C21; 53C55; 57R30; 58J50

††The author was supported by the National Research Foundation of Korea (NRF)
grant funded by the Korea government (MSIP) (NRF-2018R1A2B2002046).

##  1\. Introduction

Let $(M,\mathcal{F})$ be a smooth manifold with a foliation $\mathcal{F}$. One
of smooth invariants of $\mathcal{F}$ is the basic cohomology. The basic forms
of $(M,\mathcal{F})$ are locally forms on the leaf space; that is, forms
$\phi$ satisfying $X\lrcorner\phi=X\lrcorner d\phi=0$ for any vector $X$
tangent to the leaves, where $X\lrcorner$ denotes the interior product with
$X$. Basic forms are preserved by the exterior derivative and are used to
define basic de Rham cohomology groups $H_{B}^{*}(\mathcal{F})$, which is
defined by

$\displaystyle H_{B}^{r}(\mathcal{F})={\ker d_{B}\over{\rm Im}\,d_{B}},$

where $d_{B}$ is the restriction of $d$ to the basic forms. In general, the
basic de Rham cohomology group does not necessarily satisfy Poincaré duality,
in fact, it satisfies the twisted Poincaré duality [18]:

$\displaystyle H_{B}^{r}(\mathcal{F})\cong H_{T}^{q-r}(\mathcal{F}),$

where $q=codim(\mathcal{F})$ and $H_{T}^{r}(\mathcal{F})={\ker d_{T}\over{\rm
Im}\;d_{T}}$ is the cohomology of $d_{T}=d_{B}-\kappa_{B}\wedge$. Here
$\kappa_{B}$ is the basic part of the mean curvature form $\kappa$ of
$\mathcal{F}$.

It is well-known [2, 7, 19, 22] that on a compact oriented manifold $M$ with a
transversally oriented Riemannian foliation $\mathcal{F}$,
$H_{B}^{r}(\mathcal{F})\cong{\rm ker}\Delta_{B}$ is finite dimensional, where
$\Delta_{B}$ is the basic Laplacian. Because of this Hodge theorem,
researchers have been able to show relationships between curvature bounds and
basic cohomology. In [10], J. Hebda proved that a lower bound on transversal
Ricci curvature for a Riemannian foliation of a compact manifold causes the
space of leaves to be compact and the first basic cohomology group to be
trivial. Another example relating the geometry and the topology came in 1991,
when M. Min-Oo et al. [20] proved that if the transversal curvature operator
of $(M,\mathcal{F})$ is positive definite, then the cohomology
$H_{B}^{r}(\mathcal{F})=0\ (0<r<q)$; that is, any basic harmonic $r$-form is
trivial. There are many other examples of known relationships between
transversal curvature and basic cohomology.

Recently, G. Habib and K. Richardson [8] introduced the twisted basic de Rham
cohomology

$H_{\kappa}^{r}(\mathcal{F})={\ker d_{\kappa}\over{\rm Im}\,d_{\kappa}}$

of the twisted differential operator
$d_{\kappa}=d_{B}-\frac{1}{2}\kappa_{B}\wedge$ and proved the Poincaré duality
of the twisted basic de Rham cohomology $H_{\kappa}^{*}(\mathcal{F})$ on
foliations, that is,

$\displaystyle H_{\kappa}^{r}(\mathcal{F})\cong
H_{\kappa}^{q-r}(\mathcal{F}).$

Moreover, $H_{\kappa}^{r}(\mathcal{F})\cong\ker\Delta_{\kappa}$, where
$\Delta_{\kappa}=d_{\kappa}\delta_{\kappa}+\delta_{\kappa}d_{\kappa}$ is the
twisted basic Laplacian (cf. Section 2.2). From the Weitzenböck formula for
the twisted basic Laplacian, if the transversal Ricci curvature ${\rm
Ric}^{Q}$ is non-negative and either $\mathcal{F}$ is nontaut or ${\rm
Ric}^{Q}$ is positive at some point, then
$H_{\kappa}^{1}(\mathcal{F})=\\{0\\}$. Also, the authors [8] proved that
$\mathcal{F}$ is taut if and only if $H_{\kappa}^{0}(\mathcal{F})\cong
H_{\kappa}^{q}(\mathcal{F})=\\{0\\}$ (Tautness theorem).

On a transverse Kähler foliation of codimension $q=2n$, the basic Dolbeault
cohomology group of $\bar{\partial}_{B}$ is defined by

$H^{r,s}_{B}(\mathcal{F})={\ker\bar{\partial}_{B}\over{\rm
Im}\,\bar{\partial}_{B}},$

where $0\leq r,s\leq n$. The Kodaira-Serre duality for the basic Dolbeault
cohomology does not necessarily hold unless $\mathcal{F}$ is taut, but we
exhibit a version of Kodaira-Serre duality [15] that actually does hold in all
cases. That is,

$\displaystyle H_{B}^{r,s}(\mathcal{F})\cong H_{T}^{n-r,n-s}(\mathcal{F}),$

where $H_{T}^{r,s}(\mathcal{F})={\ker\bar{\partial}_{T}\over{\rm
Im}\;\bar{\partial}_{T}}$ is the Dolbeault cohomology group of
$\bar{\partial}_{T}$ (cf. Section 3.2). Recently, G. Habib and L. Vezzoni [9]
studied the basic Dolbeault cohomology and proved some vanishing theorem for
basic Dolbeault cohomology by using the Weitzenböck formula for the twisted
basic Laplacian.

In this paper, we define the twisted basic Dolbeault cohomolgy
$H_{\kappa}^{r,s}(\mathcal{F})$ of the twisted operator
$\bar{\partial}_{\kappa}$ (cf. Section 3.2) acting on the basic forms of type
$(r,s)$ by

$\displaystyle
H_{\kappa}^{r,s}(\mathcal{F})={\ker\bar{\partial}_{\kappa}\over{\rm
Im}\,\bar{\partial}_{\kappa}}.$

The twisted basic Dolbeault cohomology satisfies the Kodaira-Serre duality
(Theorem 3.11), that is,

$\displaystyle H_{\kappa}^{r,s}(\mathcal{F})\cong
H_{\kappa}^{n-r,n-s}(\mathcal{F}).$

Trivially, if $\mathcal{F}$ is taut, then $H_{B}^{r,s}(\mathcal{F})\cong
H_{T}^{r,s}(\mathcal{F})\cong H_{\kappa}^{r,s}(\mathcal{F})$. Also, we prove
the Hodge decomposition for the twisted basic Dolbeault cohomology (Theorem
3.13) and give some properties for $\Delta_{\kappa}$-harmonic forms.

######  Remark 1.1.

The twisted basic de Rham cohomology is the special cohomology of Lichnerowicz
basic cohomology on foliations [1]. The Lichnerowicz cohomology is the
cohomology of the complex $(\Omega^{*}(M),d_{\theta})$ of differential forms
on a smooth manifold $M$ with the de Rham differential operator
$d_{\theta}=d+\theta\wedge$ deformed by a closed 1-form $\theta$. The
Lichnerowicz cohomology is a proper tool of locally conformal symplectic
geometry [8].

######  Remark 1.2.

The Lichnerowicz basic cohomology depends only on the basic class of a closed
basic 1-form [1, Proposition 3.0.11]. In particular, the twisted basic de Rham
cohomology group depends only on the Álvarez class $[\kappa_{B}]\in
H_{B}^{1}(\mathcal{F})$. That is, for any bundle-like metrics associated with
the same transverse structure, the twisted basic de Rham cohomologies are
isomorphic [8, Theorem 2.11].

##  2\. The basic cohomology

###  2.1. The basic cohomology

Let $(M,\mathcal{F},g_{Q})$ be a $(p+q)$-dimensional Riemannian foliation of
codimension $q$ with a holonomy invariant metric $g_{Q}$ on the normal bundle
$Q=TM/T\mathcal{F}$, meaning that $L_{X}g_{Q}=0$ for all $X\in T\mathcal{F}$
on a Riemannian manifold $(M,g_{M})$ with a bundle-like metric $g_{M}$ adapted
to $g_{Q}$, where $T\mathcal{F}$ is the tangent bundle of $\mathcal{F}$ and
$L_{X}$ denotes the Lie derivative. Let $\nabla$ be the transverse Levi-Civita
connection on the normal bundle $Q$, which is torsion-free and metric with
respect to $g_{Q}$ [23, 24]. Let $R^{Q}$ and ${\rm Ric}^{Q}$ be the
transversal curvature tensor and the transversal Ricci operator of
$\mathcal{F}$ with respect to $\nabla$, respectively. The mean curvature
vector $\tau$ of $\mathcal{F}$ is given by

$\tau=\sum_{i=1}^{p}\pi(\nabla^{M}_{f_{i}}f_{i}),$

where $\\{f_{i}\\}_{i=1,\cdots,p}$ is a local orthonormal basis of
$T\mathcal{F}$ and $\pi:TM\to Q$ is a natural projection. Then the mean
curvature form $\kappa$ of $\mathcal{F}$ is given by

$\kappa(X)=g_{Q}(\tau,\pi(X))$

for any tangent vector $X\in\Gamma TM$. Let $\Omega_{B}^{r}(\mathcal{F})$ be
the space of all basic $r$-forms, i.e., $\phi\in\Omega_{B}^{r}(\mathcal{F})$
if and only if $X\lrcorner\phi=0$ and $L_{X}\phi=0$ for any vector $X\in\Gamma
T\mathcal{F}$. The foliation $\mathcal{F}$ is said to be minimal if
$\kappa=0$. It is well-known [23] that on a compact manifold, $\kappa_{B}$ is
closed, i.e., $d\kappa_{B}=0$, where $\kappa_{B}$ is the basic part of
$\kappa$. And the mean curvature form satisfies the Rummler’s formula:

$d\chi_{\mathcal{F}}=-\kappa\wedge\chi_{\mathcal{F}}+\varphi_{0},\quad\chi_{\mathcal{F}}\wedge*\varphi_{0}=0,$

where $\chi_{\mathcal{F}}$ is the characteristic form of $\mathcal{F}$ and $*$
is the Hodge star operator associated to $g_{M}$.

Now we recall the transversal star operator
$\bar{*}:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{q-r}(\mathcal{F})$ given by

$\displaystyle\bar{*}\phi=(-1)^{p(q-r)}*(\phi\wedge\chi_{\mathcal{F}}).$

Trivially, $\bar{*}^{2}\phi=(-1)^{r(q-r)}\phi$ for any basic form
$\phi\in\Omega_{B}^{r}(\mathcal{F})$. Let $\nu$ be the transversal volume
form, that is, $*\nu=\chi_{\mathcal{F}}$. Then the pointwise inner product
$\langle\cdot,\cdot\rangle$ on $\Omega_{B}^{*}(\mathcal{F})$ is defined by
$\langle\phi,\psi\rangle\nu=\phi\wedge\bar{*}\psi$.

Let $d_{B}=d|_{\Omega_{B}^{*}(\mathcal{F})}$ and
$d_{T}=d_{B}-\kappa_{B}\wedge$. Then the formal adjoint operators $\delta_{B}$
and $\delta_{T}$ of $d_{B}$ and $d_{T}$ on basic forms are given by

$\delta_{B}=(-1)^{q(r+1)+1}\bar{*}d_{T}\bar{*}=\delta_{T}+\kappa_{B}^{\sharp}\lrcorner,\quad\delta_{T}=(-1)^{q(r+1)+1}\bar{*}d_{B}\bar{*},$
(2.1)

respectively [2, 19, 24], where $\kappa_{B}^{\sharp}$ be the dual vector of
$\kappa_{B}$. For a local orthonormal frame $\\{E_{a}\\}_{a=1,\cdots,q}$ of
the normal bundle $Q$, $\delta_{T}$ is given [16] by

$\delta_{T}=-\sum_{a=1}^{q}E_{a}\lrcorner\nabla_{E_{a}}.$ (2.2)

Now we define two Laplacians $\Delta_{B}$ and $\Delta_{T}$ acting on
$\Omega_{B}^{*}(\mathcal{F})$ by

$\Delta_{B}=d_{B}\delta_{B}+\delta_{B}d_{B},\quad\Delta_{T}=d_{T}\delta_{T}+\delta_{T}d_{T},$

respectively. The Laplacian $\Delta_{B}$ is said to be basic Laplacian. Then

$\Delta_{B}\bar{*}=\bar{*}\Delta_{T}.$ (2.3)

Also, we have the generalization of the usual de Rham-Hodge decompositon.

######  Theorem 2.1.

([19],[24]) Let $(M,\mathcal{F},g_{Q})$ be a transversally oriented Riemannian
foliation $\mathcal{F}$ on a compact oriented manifold $M$ with a bundle-like
metric. Then

$\displaystyle\Omega_{B}^{r}(\mathcal{F})$
$\displaystyle=\mathcal{H}_{B}^{r}(\mathcal{F})\oplus{\rm Im}\,d_{B}\oplus{\rm
Im}\,\delta_{B}$ $\displaystyle=\mathcal{H}_{T}^{r}(\mathcal{F})\oplus{\rm
Im}\,d_{T}\oplus{\rm Im}\,\delta_{T}$

with finite dimensional $\mathcal{H}_{B}^{r}(\mathcal{F})=\ker\Delta_{B}$ and
$\mathcal{H}_{T}^{r}(\mathcal{F})=\ker\Delta_{T}$.

On a compact manifold, $d_{T}^{2}=0$ because of $d\kappa_{B}=0$. So the basic
de Rham cohomology groups $H_{B}^{r}(\mathcal{F})$ and
$H_{T}^{r}(\mathcal{F})$ are defined by

$\displaystyle H_{B}^{r}(\mathcal{F})={\ker d_{B}\over{\rm Im}\,d_{B}},\quad
H_{T}^{r}(\mathcal{F})={\ker d_{T}\over{\rm Im}\,d_{T}},$

respectively. Then it is well-known [24] that

$\displaystyle
H_{B}^{r}(\mathcal{F})\cong\mathcal{H}_{B}^{r}(\mathcal{F}),\quad
H_{T}^{r}(\mathcal{F})\cong\mathcal{H}_{T}^{r}(\mathcal{F}).$

From (2.3), we have the twisted duality [18]

$\displaystyle H_{B}^{r}(\mathcal{F})\cong H_{T}^{q-r}(\mathcal{F}).$ (2.4)

If the foliation $\mathcal{F}$ is taut (means that there exists a Riemannian
metric on $M$ for which all leaves are minimal), then the Poincaré duality
holds [17]. That is,

$\displaystyle H_{B}^{r}(\mathcal{F})\cong H_{B}^{q-r}(\mathcal{F}).$

Now, we introduce the operator $\nabla_{\rm tr}^{*}\nabla_{\rm tr}$, which is
defined by

$\displaystyle\nabla_{\rm tr}^{*}\nabla_{\rm
tr}\phi=-\sum_{a}\nabla_{E_{a}}\nabla_{E_{a}}\phi+\nabla_{\kappa_{B}^{\sharp}}\phi.$

It is well-known ([12, Proposition 3.1]) that the operator $\nabla_{\rm
tr}^{*}\nabla_{\rm tr}$ is non-negative and formally self-adjoint. Then we
have the following generalized Weitzenböck formula [12]: for any basic form
$\phi\in\Omega_{B}^{r}(\mathcal{F})$

$\Delta_{B}\phi=\nabla_{\rm tr}^{*}\nabla_{\rm
tr}\phi+A_{\kappa_{B}^{\sharp}}(\phi)+F(\phi),$ (2.5)

where $A_{Y}(\phi)=L_{Y}\phi-\nabla_{Y}\phi$ and

$\displaystyle F(\phi)=\sum_{a,b}\theta^{a}\wedge E_{b}\lrcorner
R^{Q}(E_{b},E_{a})\phi.$

Here, $\theta^{a}$ is the dual basic 1-form of $E_{a}$. In particular, if
$\phi$ is a basic 1-form, then $F(\phi)^{\sharp}={\rm
Ric}^{Q}(\phi^{\sharp})$.

######  Remark 2.2.

Observe that, from the preceding results, the dimensions of
$H_{B}^{r}(\mathcal{F})$ and $H_{T}^{r}(\mathcal{F})$ are smooth invariants of
$\mathcal{F}$ and do not depend on the choices of bundle-like metric $g_{M}$
or on the transversal metric $g_{Q}$, even though the spaces of harmonic forms
do depend on these choices.

######  Theorem 2.3.

[15] Let $(M,\mathcal{F},g_{M})$ be a compact Riemannian manifold with a
foliation $\mathcal{F}$ of codimension $q$ and a bundle-like metric $g_{M}$.
If the endomorphism $F$ is positive-definite, then there are no nonzero basic
harmonic forms, that is, $H_{B}^{r}(\mathcal{F})=\\{0\\}.$ In particular, if
${\rm Ric}^{Q}$ is positive-definite, then $H_{B}^{1}(\mathcal{F})=\\{0\\}$.

###  2.2. The twisted basic cohomology

Now, we recall the twisted differential operators $d_{\kappa}$ and
$\delta_{\kappa}$ [8], which are given by

$\displaystyle
d_{\kappa}=d_{B}-\frac{1}{2}\kappa_{B}\wedge,\quad\delta_{\kappa}=\delta_{B}-\frac{1}{2}\kappa_{B}^{\sharp}\lrcorner.$
(2.6)

The operator $\delta_{\kappa}$ is a formal adjoint operator of $d_{\kappa}$
with respect to the global inner product. Let
$\Delta_{\kappa}=d_{\kappa}\delta_{\kappa}+\delta_{\kappa}d_{\kappa}$ be the
twisted basic Laplacian. Then we have the following relation: for any basic
form $\phi$,

$\displaystyle\Delta_{\kappa}\phi=\Delta_{B}\phi-\frac{1}{2}\Big{(}L_{\kappa_{B}^{\sharp}}+(L_{\kappa_{B}^{\sharp}})^{*}\Big{)}\phi+\frac{1}{4}|\kappa_{B}|^{2}\phi.$
(2.7)

Also, we have the following relations.

######  Lemma 2.4.

[8] On $\Omega_{B}^{r}(\mathcal{F})$, the following equations hold:

  1. (1)

$d_{\kappa}^{2}=\delta_{\kappa}^{2}=0,\quad[d_{\kappa},\Delta_{\kappa}]=[\Delta_{\kappa},\delta_{\kappa}]=[\Delta_{\kappa},\bar{*}]=0$.

  2. (2)

$d_{\kappa}\bar{*}=(-1)^{r}\bar{*}\delta_{\kappa},\quad\bar{*}d_{\kappa}=(-1)^{r+1}\delta_{\kappa}\bar{*}$.

Since $d_{\kappa}^{2}=0$, we can define the twisted basic de Rham cohomology
group $H_{\kappa}^{r}(\mathcal{F})$ by

$\displaystyle H_{\kappa}^{r}(\mathcal{F})={\ker d_{\kappa}\over{\rm
Im}\,d_{\kappa}}.$

Then we have the Hodge decomposition.

######  Theorem 2.5.

[8] Let $(M,\mathcal{F},g_{Q})$ be as in Theorem2.1. Then

$\displaystyle\Omega_{B}^{r}(\mathcal{F})={\mathcal{H}}_{\kappa}^{r}(\mathcal{F})\oplus{\rm
Im}\,d_{\kappa}\oplus{\rm Im}\,\delta_{\kappa}$

with finite dimensional
${\mathcal{H}}_{\kappa}^{r}(\mathcal{F})=\ker\Delta_{\kappa}$. Moreover,
${\mathcal{H}}_{\kappa}^{r}(\mathcal{F})\cong H_{\kappa}^{r}(\mathcal{F})$.

And we have the Poincaré duality for $d_{\kappa}$-cohomology. That is,

######  Theorem 2.6.

[8] (Poincaré duality for $d_{\kappa}$-cohomology) On a compact Riemannian
manifold with a Riemannian foliation $\mathcal{F}$, we have

$\displaystyle H_{\kappa}^{r}(\mathcal{F})\cong
H_{\kappa}^{q-r}(\mathcal{F}).$

######  Remark 2.7.

Theorem 2.6 resolves the problem of the failure of Poincaré duality to hold
for standard basic de Rham cohomology $H_{B}^{r}(\mathcal{F})$ (cf. (2.4)).

######  Theorem 2.8.

[8] (Tautness Theorem) Let $(M,\mathcal{F},g_{Q})$ be as in Theorem 2.1. Then
$\mathcal{F}$ is taut if and only if $H_{\kappa}^{0}(\mathcal{F})\cong
H_{\kappa}^{q}(\mathcal{F})\neq\\{0\\}$.

From (2.5) and (2.7), we have the Weitzenböck formula [8] for the twisted
basic Laplacian $\Delta_{\kappa}$. Namely, for any basic form $\phi$,

$\displaystyle\Delta_{\kappa}\phi=\nabla_{\rm tr}^{*}\nabla_{\rm
tr}\phi+F(\phi)+\frac{1}{4}|\kappa_{B}|^{2}\phi.$

Then we have the following theorem.

######  Theorem 2.9.

[8] Let $(M,\mathcal{F},g_{Q})$ be a Riemannian foliation on a compact,
connected manifold $M$ with a bundle-like metric such that the mean curvature
form $\kappa$ is basic-harmonic. Then:

(1) if the operator $F+{\frac{1}{4}}|\kappa|^{2}$ is strictly positive, then
$H_{\kappa}^{r}(\mathcal{F})=\\{0\\}$.

(2) if the transversal Ricci curvature ${\rm Ric}^{Q}$ is non-negative and
either $M$ is nontaut or ${\rm Ric}^{Q}$ is positive at least one point, then
$H_{\kappa}^{1}(\mathcal{F})=\\{0\\}$.

(3) suppose that the transversal sectional curvatures are nonnegative and
positive at least one point. If $\mathcal{F}$ is nontaut, then
$H_{\kappa}^{r}(\mathcal{F})=\\{0\\}$ for $1<r<q$.

##  3\. The twisted basic Dolbealt cohomology

###  3.1. The basic Dolbeault cohomology

In this section, we generally use the same notation and cite elementary
results from [13, Section 3] and [21]. Let $(M,\mathcal{F},g_{Q},J)$ be a
transverse Kähler foliation of codimension $q=2n$ on a Riemannian manifold $M$
with a holonomy invariant transverse Hermitian metric $g_{Q}$ and an almost
complex structure $J$ on $Q$ such that $\nabla J=0$, with $\nabla$ being the
transversal Levi-Civita connection on $Q$, extended in the usual way to
tensors [21]. In some of what follows, we will merely need the fact that the
foliation is transverse Hermitian (all of the above, merely requiring $J$ is
integrable and not that $\nabla J=0$), and other times we will need the full
power of the Kähler condition $\nabla J=0$. The basic Kähler form $\omega$ is
given by

$\omega(X,Y)=g_{Q}(\pi(X),J\pi(Y))$

for any vecgor fields $X,Y\in TM$. Locally, the basic Kähler form $\omega$ may
be expressed by

$\displaystyle\omega=-\frac{1}{2}\sum_{a=1}^{2n}\theta^{a}\wedge J\theta^{a},$

where $\\{\theta^{a}\\}_{a=1,\cdots,2n}$ is a local orthonormal frame of
$Q^{*}$. Here, we extend $J$ to elements of $Q^{*}$ by setting
$(J\phi)(X)=-\phi(JX)$ for any $X\in Q_{x}$ and $\phi\in Q_{x}^{*}$. When it
is convenient, we will also refer to the bundle map $J^{\prime}:TM\to TM$
defined by $J^{\prime}(v)=J(\pi(v))$ and abuse notation by denoting
$J=J^{\prime}$. Similarly, we sometimes will act on all of $T^{*}M$ using the
symbol $J$. We note that all of the above is true for transverse Hermitian
foliations, but the form $\omega$ is not closed unless the foliation is
Kähler.

Let $Q^{C}=Q\otimes\mathbb{C}$ be the complexified normal bundle and let

$\displaystyle Q^{1,0}=\\{Z\in Q^{C}|JZ=iZ\\},\quad Q^{0,1}=\\{Z\in
Q^{C}|JZ=-iZ\\}.$

The element of $Q^{1,0}$ (resp. $Q^{0,1}$) is called a complex normal vector
field of type $(1,0)$ (resp. (0,1)). Then $Q^{C}=Q^{1,0}\oplus Q^{0,1}$. Now,
let $Q^{*}_{C}$ be the real dual bundle of $Q^{C}$, defined at each $x\in M$
to be the $\mathbb{C}$-linear maps from $Q_{x}^{C}$ to $\mathbb{C}$. Then
$Q_{C}^{*}=Q_{1,0}\oplus Q_{0,1}$, where

$\displaystyle Q_{1,0}=\\{\theta+iJ\theta|\ \theta\in
Q^{*}\\}\quad\text{and}\quad Q_{0,1}=\\{\theta-iJ\theta|\ \theta\in Q^{*}\\}.$

Let $\Lambda^{r,s}_{C}Q^{*}$ be the subspace of $\Lambda Q_{C}^{*}$ spanned by
$\xi\wedge\eta$, where $\xi\in\Lambda^{r}Q_{1,0}$ and
$\eta\in\Lambda^{s}Q_{0,1}$. The sections of $\Lambda^{r,s}_{C}Q^{*}$ are said
to be forms of type $(r,s)$. Let $\Omega_{B}^{r,s}(\mathcal{F})$ be the set of
the basic forms of type $(r,s)$. Let $\\{E_{a},JE_{a}\\}_{a=1,\cdots,n}$ be a
local orthonormal frame of $Q$ and
$\\{\theta^{a},J\theta^{a}\\}_{a=1,\cdots,n}$ be their dual basic forms on
$Q^{*}$. Let $V_{a}={1\over\sqrt{2}}(E_{a}-iJE_{a})$ and
$\omega^{a}={1\over\sqrt{2}}(\theta^{a}+iJ\theta^{a})$. Then

$\displaystyle\omega^{a}(V_{b})=\overline{\omega}^{a}(\overline{V}_{b})=\delta_{ab},\
\omega^{a}(\overline{V}_{b})=\overline{\omega}^{a}(V_{b})=0.$

A frame fields $\\{V_{a}\\}_{a=1,\cdots,n}$ is a local orthonormal frame of
$Q^{1,0}$, which is called a normal frame field of type $(1,0)$, and
$\\{\omega^{a}\\}_{a=1,\cdots,n}$ is a dual frame of $Q_{1,0}$.

Now, we extend the connection $\nabla$ on $Q$ in the natural way so that
$\nabla_{X}Y$ is defined for any $X\in\Gamma(TM\otimes\mathbb{C})$ and any
$Y\in\Gamma(Q^{C})$. We further extend it to differential forms, requiring
that $\nabla$ is a Hermitian connection, i.e., for any $V\in Q^{C}$ and any
$\phi,\psi\in\Omega_{B}^{r,s}(\mathcal{F})$,

$\displaystyle
V\langle\phi,\psi\rangle=\langle\nabla_{V}\phi,\psi\rangle+\langle\phi,\nabla_{\overline{V}}\psi\rangle.$

It is an easy exercise to show that for any complex vector field $X$,
$\nabla_{X}$ preserves the $(r,s)$ type of the form or vector field.

Now, the transversal star operator
$\bar{*}:\Omega_{B}^{r,s}(\mathcal{F})\to\Omega_{B}^{n-s,n-r}(\mathcal{F})$ on
$\Omega_{B}^{*,*}(\mathcal{F})$ is given by

$\displaystyle\phi\wedge\bar{*}\bar{\psi}=\langle\phi,\psi\rangle\nu$

for any $\phi,\psi\in\Omega_{B}^{r,s}(\mathcal{F})$, where
$\nu={\omega^{n}\over n!}$ is the transversal volume form. Then for any
$\phi\in\Omega_{B}^{r,s}(\mathcal{F})$,

$\displaystyle\overline{\bar{*}\phi}=\bar{*}\bar{\phi},\quad\bar{*}^{2}\phi=(-1)^{r+s}\phi.$

From (2.1), the adjoint operators $\delta_{B}$ and $\delta_{T}$ of $d_{B}$ and
$d_{T}$ are given by

$\displaystyle\delta_{B}=-\bar{*}d_{T}\bar{*},\quad\delta_{T}=-\bar{*}d_{B}\bar{*},$
(3.1)

respectively. Note that $d_{B}=\partial_{B}+\bar{\partial}_{B}$ and
$d_{T}=\partial_{T}+\bar{\partial}_{T}$, where

$\displaystyle\partial_{T}\phi=\partial_{B}\phi-\kappa_{B}^{1,0}\wedge\phi,\quad\bar{\partial}_{T}\phi=\bar{\partial}_{B}\phi-\kappa_{B}^{0,1}\wedge\phi,$
(3.2)

where
$\kappa_{B}^{1,0}=\frac{1}{2}(\kappa_{B}+iJ\kappa_{B})\in\Omega_{B}^{1,0}(\mathcal{F})$
and $\kappa_{B}^{0,1}=\overline{\kappa_{B}^{1,0}}$ [15].

Let $\partial_{T}^{*}$, $\bar{\partial}_{T}^{*}$, $\partial_{B}^{*}$ and
$\bar{\partial}_{B}^{*}$ be the formal adjoint operators of $\partial_{T},\
\bar{\partial}_{T},\ \partial_{B}$ and $\bar{\partial}_{B}$, respectively, on
the space of basic forms. Then

$\displaystyle\delta_{B}=\partial_{B}^{*}+\bar{\partial}_{B}^{*},\quad\delta_{T}=\partial_{T}^{*}+\bar{\partial}_{T}^{*},$
(3.3)

and from (3.1),

$\displaystyle\partial_{T}^{*}\phi=-\bar{*}\bar{\partial}_{B}\bar{*}\phi,\quad\bar{\partial}_{T}^{*}\phi=-\bar{*}\partial_{B}\bar{*}\phi,$
(3.4)
$\displaystyle\partial_{B}^{*}\phi=-\bar{*}\bar{\partial}_{T}\bar{*}\phi,\quad\bar{\partial}_{B}^{*}\phi=-\bar{*}\partial_{T}\bar{*}\phi.$
(3.5)

Since $\bar{*}(\kappa_{B}^{0,1}\wedge)\bar{*}=H^{1,0}\lrcorner$ [15], from
(3.2) and (3.5), we have [15]

$\displaystyle\partial^{*}_{B}\phi=\partial_{T}^{*}\phi+H^{1,0}\lrcorner\,\phi,\quad\bar{\partial}_{B}^{*}\phi=\bar{\partial}_{T}^{*}\phi+H^{0,1}\lrcorner\,\phi,$
(3.6)

where $H^{1,0}=\frac{1}{2}(\kappa_{B}^{\sharp}-iJ\kappa_{B}^{\sharp})$ and
$H^{0,1}=\overline{H^{1,0}}$. Then from (2.2) and (3.3)

$\displaystyle\partial_{T}^{*}\phi=-\sum_{a=1}^{n}V_{a}\lrcorner\nabla_{\overline{V}_{a}}\phi,\quad\bar{\partial}_{T}^{*}\phi=-\sum_{a=1}^{n}\overline{V}_{a}\lrcorner\nabla_{V_{a}}\phi.$
(3.7)

Since $\bar{\partial}_{B}^{2}=0$, we can define the basic Dolbeault cohomology
group $H_{B}^{r,s}(\mathcal{F})$ by

$H_{B}^{r,s}(\mathcal{F})={{\ker\bar{\partial}_{B}}\over{\rm
Im}\>\bar{\partial}_{B}}.$

Now, let
$\square_{B}=\partial_{B}\partial_{B}^{*}+\partial_{B}^{*}\partial_{B}$ and
$\overline{\square}_{B}=\bar{\partial}_{B}\bar{\partial}_{B}^{*}+\bar{\partial}_{B}^{*}\bar{\partial}_{B}$.
Then we have the basic Dolbeault decomposition.

######  Theorem 3.1.

[6, 15] Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a
compact Riemannian manifold $M$ with a bundle-like metric. Then

$\Omega_{B}^{r,s}(\mathcal{F})=\mathcal{H}_{B}^{r,s}(\mathcal{F})\otimes{\rm
Im}\,\bar{\partial}_{B}\otimes{\rm Im}\,\bar{\partial}_{B}^{*},$

where $\mathcal{H}_{B}^{r,s}(\mathcal{F})=\ker\overline{\square}_{B}$ is
finite dimensional. Moreover, $\mathcal{H}_{B}^{r,s}(\mathcal{F})\cong
H_{B}^{r,s}(\mathcal{F})$.

Generally, the basic Laplacians do not satisfies the properties which hold on
a ordinary Kähler manifold such as $\Delta=2\square=2\overline{\square}$. But
if $\mathcal{F}$ is taut, then
$\Delta_{B}=2\square_{B}=2\overline{\square}_{B}$ [15].

###  3.2. The twisted basic Dolbeault cohomology

Let
$\partial_{\kappa}:\Omega_{B}^{r,s}(\mathcal{F})\to\Omega_{B}^{r+1,s}(\mathcal{F})$
and
$\bar{\partial}_{\kappa}:\Omega_{B}^{r,s}(\mathcal{F})\to\Omega_{B}^{r,s+1}(\mathcal{F})$
be defined by

$\displaystyle\partial_{\kappa}\phi=\partial_{B}\phi-\frac{1}{2}\kappa_{B}^{1,0}\wedge\phi,\quad\bar{\partial}_{\kappa}\phi=\bar{\partial}_{B}\phi-\frac{1}{2}\kappa_{B}^{0,1}\wedge\phi,$
(3.8)

respectively. From (2.6), it is trivial that
$d_{\kappa}=\partial_{\kappa}+\bar{\partial}_{\kappa}.$ Let
$\partial_{\kappa}^{*}$ and $\bar{\partial}_{\kappa}^{*}$ be the formal
adjoint operators of $\partial_{\kappa}$ and $\bar{\partial}_{\kappa}$,
respectively. Then we have the following.

######  Proposition 3.2.

On a transverse Kähler foliation, we have

$\displaystyle\partial_{\kappa}^{*}=\partial_{B}^{*}-\frac{1}{2}H^{1,0}\lrcorner,\quad\bar{\partial}_{\kappa}^{*}=\bar{\partial}_{B}^{*}-\frac{1}{2}H^{0,1}\lrcorner,\quad\delta_{\kappa}=\partial_{\kappa}^{*}+\bar{\partial}_{\kappa}^{*}.$

###### Proof.

From (2.6) and (3.8), the proofs are easy. ∎

Let $L:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{r+2}(\mathcal{F})$ and
$\Lambda:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{r-2}(\mathcal{F})$ be given
by

$\displaystyle
L(\phi)=\omega\wedge\phi,\quad\Lambda(\phi)=\omega\lrcorner\phi,$

respectively, where
$(\xi_{1}\wedge\xi_{2})\lrcorner=\xi_{2}^{\sharp}\lrcorner\xi_{1}^{\sharp}\lrcorner$
for any basic 1-forms $\xi_{i}(i=1,2)$. Trivially, $\langle
L\phi,\psi\rangle=\langle\phi,\Lambda\psi\rangle$ and
$\Lambda=-\bar{*}L\bar{*}$ [5]. Also, it is well-known [13] that

$\displaystyle[L,X\lrcorner]=JX^{b}\wedge,\quad[\Lambda,X^{b}\wedge]=-JX\lrcorner,\quad[L,X^{b}\wedge]=[\Lambda,X\lrcorner]=0$
(3.9)

for any vector field $X\in Q$. From (3.9), we have the following.

######  Proposition 3.3.

[13] On a transverse Kähler foliation, we have

$\displaystyle[L,d_{B}]=[\Lambda,\delta_{B}]=[L,\partial_{B}]=[L,\bar{\partial}_{B}]=[\Lambda,\partial_{B}^{*}]=[\Lambda,\bar{\partial}_{B}^{*}]=0,$
$\displaystyle[L,\partial_{B}^{*}]=-i\bar{\partial}_{T},\
[L,\bar{\partial}_{B}^{*}]=i\partial_{T},\
[\Lambda,\partial_{B}]=-i\bar{\partial}_{T}^{*},\
[\Lambda,\bar{\partial}_{B}]=i\partial_{T}^{*}.$

From (3.9) and Proposition 3.3, we have the following.

######  Proposition 3.4.

On a transverse Kähler foliation, we have

$\displaystyle[L,d_{\kappa}]=[\Lambda,\delta_{\kappa}]=[L,\partial_{\kappa}]=[L,\bar{\partial}_{\kappa}]=[\Lambda,\partial_{\kappa}^{*}]=[\Lambda,\bar{\partial}_{\kappa}^{*}]=0,$
(3.10) $\displaystyle[L,\partial_{\kappa}^{*}]=-i\bar{\partial}_{\kappa},\
[L,\bar{\partial}_{\kappa}^{*}]=i\partial_{\kappa},\
[\Lambda,\partial_{\kappa}]=-i\bar{\partial}_{\kappa}^{*},\
[\Lambda,\bar{\partial}_{\kappa}]=i\partial_{\kappa}^{*}.$ (3.11)

Let $\square_{\kappa}$ and $\overline{\square}_{\kappa}$ be Laplace operators,
which are defined by

$\displaystyle\square_{\kappa}=\partial_{\kappa}\partial_{\kappa}^{*}+\partial_{\kappa}^{*}\partial_{\kappa}\quad{\rm
and}\quad\overline{\square}_{\kappa}=\bar{\partial}_{\kappa}\bar{\partial}_{\kappa}^{*}+\bar{\partial}_{\kappa}^{*}\bar{\partial}_{\kappa},$

respectively. Trivially, $\square_{\kappa}$ and $\overline{\square}_{\kappa}$
preserve the types of the forms.

######  Theorem 3.5.

On a transverse Kähler foliation, we have

$\displaystyle\square_{\kappa}=\overline{\square}_{\kappa},\quad\Delta_{\kappa}=2\square_{\kappa}=2\overline{\square}_{\kappa}.$

###### Proof.

Since $d_{\kappa}^{2}=0$, it is trivial that
$\partial_{\kappa}^{2}=\bar{\partial}_{\kappa}^{2}=\partial_{\kappa}\bar{\partial}_{\kappa}+\bar{\partial}_{\kappa}\partial_{\kappa}=0$.
From Proposition 3.4 (3.11), we have

$\displaystyle
i(\partial_{\kappa}\partial_{\kappa}^{*}+\partial_{\kappa}^{*}\partial_{\kappa})$
$\displaystyle=\partial_{\kappa}\Lambda\bar{\partial}_{\kappa}+\Lambda\bar{\partial}_{\kappa}\partial_{\kappa}-\partial_{\kappa}\bar{\partial}_{\kappa}\Lambda-\bar{\partial}_{\kappa}\Lambda\partial_{\kappa}$
$\displaystyle=[\partial_{\kappa},\Lambda]\bar{\partial}_{\kappa}+\bar{\partial}_{\kappa}[\partial_{\kappa},\Lambda]$
$\displaystyle=i(\bar{\partial}_{\kappa}^{*}\bar{\partial}_{\kappa}+\bar{\partial}_{\kappa}\bar{\partial}_{\kappa}^{*})$

and

$\displaystyle
i(\bar{\partial}_{\kappa}\partial_{\kappa}^{*}+\partial_{\kappa}^{*}\bar{\partial}_{\kappa})=i(\partial_{\kappa}\bar{\partial}_{\kappa}^{*}+\bar{\partial}_{\kappa}^{*}\partial_{\kappa})=0.$

Hence $\square_{\kappa}=\overline{\square}_{\kappa}$ and by a direct
calculation,

$\displaystyle\Delta_{\kappa}=\square_{\kappa}+\overline{\square}_{\kappa}=2\square_{\kappa}=2\overline{\square}_{\kappa}.$

∎

######  Remark 3.6.

Recall that if the transverse Kähler foliation is minimal, then a basic form
of type $(r,0)$ is basic-harmonic if and only if it is basic holomorphic [15].
But, if the foliation is not minimal, then the relation does not hold.

######  Definition 3.7.

On a transverse Kähler foliation, a basic form $\phi$ is said to be
$\bar{\partial}_{\kappa}$-holomorphic if $\bar{\partial}_{\kappa}\phi=0$.

######  Theorem 3.8.

On a transverse Kähler foliation, a $\bar{\partial}_{\kappa}$-holomorphic form
of type $(r,0)$ is $\Delta_{\kappa}$-harmonic. In addition, if $M$ is compact,
then the converse holds.

###### Proof.

Let $\phi$ be a $\bar{\partial}_{\kappa}$-holomorphic form of type $(r,0)$.
Since $\bar{\partial}^{*}_{\kappa}\phi=0$ automatically,
$\bar{\partial}_{\kappa}\phi$ implies
$\Delta_{\kappa}\phi=2\overline{\square}_{\kappa}\phi=0$. Conversely, if $M$
is compact, then $\Delta_{\kappa}\phi=0$ implies that
$\int_{M}|\bar{\partial}_{\kappa}\phi|^{2}=0$, i.e., $\phi$ is
$\bar{\partial}_{\kappa}$-holomorphic. ∎

Now, we consider $\bar{\partial}_{\kappa}$-complex

$\displaystyle\cdots\overset{\bar{\partial}_{\kappa}}{\longrightarrow}\Omega_{B}^{r,s-1}(\mathcal{F})\overset{\bar{\partial}_{\kappa}}{\longrightarrow}\Omega_{B}^{r,s}(\mathcal{F})\overset{\bar{\partial}_{\kappa}}{\longrightarrow}\Omega_{B}^{r,s+1}(\mathcal{F})\overset{\bar{\partial}_{\kappa}}{\longrightarrow}\cdots.$

Since $\bar{\partial}_{\kappa}^{2}=0$, the twisted basic Dolbeault cohomology
group is defined by

$\displaystyle
H_{\kappa}^{r,s}(\mathcal{F})={{\ker\bar{\partial}_{\kappa}}\over{\rm
Im}\>\bar{\partial}_{\kappa}}.$

Then we have the generalization of the Dolbeault decomposition.

######  Theorem 3.9.

Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a compact
manifold $M$ with a bundle-like metric. Then

$\displaystyle\Omega_{B}^{r,s}(\mathcal{F})=\mathcal{H}_{\kappa}^{r,s}(\mathcal{F})\oplus{\rm
Im}\,\bar{\partial}_{\kappa}\oplus{\rm Im}\,\bar{\partial}_{\kappa}^{*},$

where
$\mathcal{H}_{\kappa}^{r,s}(\mathcal{F})=\ker\overline{\square}_{\kappa}$ is
finite dimensional.

###### Proof.

The proof is similar to the one in Theorem 2.2. See [19] precisely. ∎

As a Corollary of Theorem 3.9, we have the Dolbeault isomorphism.

######  Corollary 3.10.

(Dolbeault isomorphism) Let $(M,\mathcal{F},g_{Q},J)$ be as in Theorem 3.9.
Then

$\mathcal{H}_{\kappa}^{r,s}(\mathcal{F})\cong H_{\kappa}^{r,s}(\mathcal{F}).$

###### Proof.

The proof is similar to the proof of the Hodge isomorphism. ∎

Then we have the Kodaira-Serre duality.

######  Theorem 3.11.

(Kodaira-Serre duality) Let $(M,\mathcal{F},g_{Q},J)$ be as in Theorem 3.9.
Then

$\displaystyle H_{\kappa}^{r,s}(\mathcal{F})\cong
H_{\kappa}^{n-r,n-s}(\mathcal{F}).$

###### Proof.

We define the operator
$\sharp:\Omega_{B}^{r,s}(\mathcal{F})\to\Omega_{B}^{n-r,n-s}(\mathcal{F})$ by

$\displaystyle\sharp\phi:=\bar{*}\bar{\phi},$

which is an isomorphism. Since
$\bar{*}(\kappa_{B}^{1,0}\wedge)\bar{*}=H^{0,1}\lrcorner$ [15], we have that
for $\phi\in\Omega_{B}^{r,s}(\mathcal{F})$,

$\bar{*}(\kappa_{B}^{0,1}\wedge\phi)=(-1)^{r+s}H^{1,0}\lrcorner\bar{*}\phi$
(3.12)

and from (3.4), we have

$\bar{*}\bar{\partial}_{B}\phi=(-1)^{r+s+1}\partial_{T}^{*}\bar{*}\phi.$
(3.13)

From (3.12) and (3.13), we get that on $\Omega_{B}^{r,s}(\mathcal{F})$,

$\bar{*}\bar{\partial}_{\kappa}=(-1)^{r+s+1}\partial_{\kappa}^{*}\bar{*}.$
(3.14)

Also, from (3.5), we get that on $\Omega_{B}^{r,s}(\mathcal{F})$,

$\bar{*}\bar{\partial}_{\kappa}^{*}=(-1)^{r+s}\partial_{\kappa}\bar{*}.$
(3.15)

From (3.14) and (3.15), we have that on $\Omega_{B}^{r,s}(\mathcal{F})$,

$\displaystyle\bar{*}\bar{\partial}_{\kappa}\bar{\partial}_{\kappa}^{*}=-\partial_{\kappa}^{*}\partial_{\kappa}\bar{*},\quad\bar{*}\bar{\partial}_{\kappa}^{*}\bar{\partial}_{\kappa}=-\partial_{\kappa}\partial_{\kappa}^{*}\bar{*},$

which implies

$\displaystyle\bar{*}\overline{\square}_{\kappa}=\square_{\kappa}\bar{*}.$

Hence for any basic form $\phi\in\Omega_{B}^{r,s}(\mathcal{F})$, we get

$\displaystyle\sharp\overline{\square}_{\kappa}\phi=\bar{*}\square_{\kappa}\bar{\phi}=\overline{\square}_{\kappa}\bar{*}\bar{\phi}=\overline{\square}_{\kappa}\sharp\phi.$

That is, $\sharp$ preserves
$\mathcal{H}_{\kappa}^{r,s}(\mathcal{F})=\ker\overline{\square}_{\kappa}$.
From the Dolbeault isomorphism (Corollary 3.10), the proof follows. ∎

######  Remark 3.12.

In general, the Kodaira-Serre duality does not hold for
$\bar{\partial}_{B}$-cohomology. In fact,

$\displaystyle H_{B}^{r,s}(\mathcal{F})\cong H_{T}^{n-r,n-s}(\mathcal{F}),$

where $H_{T}^{r,s}(\mathcal{F})={{\ker\bar{\partial}_{T}}\over{{\rm
Im}\,\bar{\partial}_{T}}}$ is the $\bar{\partial}_{T}$-cohomology.
$H_{T}^{r,s}(\mathcal{F})$ is a type of Lichnerowicz basic cohomology. The
interested reader may consult [3, Section 3] and [25, Section 3, called
“adapted cohomology” here] for information about ordinary Lichnerowicz
cohomology and [1] for the basic case. Theorem 3.11 resolves the problem of
the failure of Kodaira-Serre duality to hold for
$\bar{\partial}_{B}$-cohomology.

From Theorem 3.5, we have the following Hodge decomposition for the twisted
basic Dolbeault cohomology.

######  Proposition 3.13.

Let $(M,\mathcal{F},g_{Q},J)$ be as in Theorem 3.9. Then

$H_{\kappa}^{l}(\mathcal{F})=\oplus_{r+s=l}H_{\kappa}^{r,s}(\mathcal{F})$
(3.16)

for $0\leq l\leq 2n$ and

${\rm dim}_{\mathbb{C}}H_{\kappa}^{r,s}(\mathcal{F})={\rm
dim}_{\mathbb{C}}H_{\kappa}^{s,r}(\mathcal{F}).$ (3.17)

###### Proof.

Since $\Delta_{\kappa}=2\overline{\square}_{\kappa}$ and
$\overline{\square}_{\kappa}$ preserves the space
$\Omega_{B}^{r,s}(\mathcal{F})$, the proof of (3.16) follows by Hodge
isomorphism (Theorem 3.11). The proof of (3.17) follows from that the map
$\mathcal{H}_{\kappa}^{r,s}(\mathcal{F})\to\mathcal{H}_{\kappa}^{s,r}(\mathcal{F})$
is a conjugate linear isomorphism. ∎

######  Remark 3.14.

The Hodge decomposition for $\bar{\partial}_{B}$-cohomology does not hold
unless the mean curvature of $\mathcal{F}$ is automorphic [16, Corollary 6.7].

Now let $\Omega_{B,P}^{r}(\mathcal{F})$ be the set of all primitive basic
$r$-forms $\phi$, that is, $\Lambda\phi=0$. Then by
$\mathfrak{sl}_{2}(\mathbb{C})$ representation theory, we have the following
proposition.

######  Proposition 3.15.

[16] Let $(M,\mathcal{F},g_{Q},J)$ be as in Theorem 3.9. Then we have the
following.

(1) $\Omega_{B,P}^{r}(\mathcal{F})=0$ if $r>n$.

(2) If $\phi\in\Omega_{B,P}^{r}(\mathcal{F})$, then $L^{s}\phi\neq 0$ for
$0\leq s\leq n-r$ and $L^{s}\phi=0$ for $s>n-r$.

(3) The map
$L^{s}:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{r+2s}(\mathcal{F})$ is
injective for $0\leq s\leq n-r$.

(4) The map
$L^{s}:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{r+2s}(\mathcal{F})$ is
surjective for $s\geq n-r$.

(5) $\Omega_{B}^{r}(\mathcal{F})=\oplus_{s\geq
0}L^{s}\Omega_{B,P}^{r-2s}(\mathcal{F})$.

######  Theorem 3.16.

(Hard Lefschetz theorem) Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler
foliation on a compact manifold $M$ with a bundle-like metric. Then the Hard
Lefschetz theorem holds for twisted basic cohomology. That is, the map

$L^{s}:H_{\kappa}^{r}(\mathcal{F})\to H_{\kappa}^{r+2s}(\mathcal{F})$ (3.18)

is injective for $0\leq s\leq n-r$ and surjective for $s\geq n-r$, $s\geq 0$.
Moreover,

$\displaystyle H_{\kappa}^{r}(\mathcal{F})$ $\displaystyle=\oplus_{s\geq
0}L^{s}H_{\kappa,P}^{r-2s}(\mathcal{F}),$ (3.19) $\displaystyle
H_{\kappa}^{r,s}(\mathcal{F})$ $\displaystyle=\oplus_{t\geq
0}L^{t}H_{\kappa,P}^{r-t,s-t}(\mathcal{F}),$ (3.20)

where
$H_{\kappa,P}^{r}(\mathcal{F})\cong\Omega_{B,P}^{r}(\mathcal{F})\cap\ker\Delta_{\kappa}$
and
$H_{\kappa,P}^{r,s}(\mathcal{F})\cong\Omega_{B,P}^{r,s}(\mathcal{F})\cap\ker\Delta_{\kappa}$.

###### Proof.

Since
$\bar{\partial}_{\kappa}\partial_{\kappa}+\partial_{\kappa}\bar{\partial}_{\kappa}=0$,
$[L,\overline{\square}_{\kappa}]=[L,\square_{\kappa}]=0$ by Proposition 3.4,
and so $[L,\Delta_{\kappa}]=0$. Hence by Proposition 3.15 and Hodge
isomorphism (Theorem 2.5), the proofs of (3.18) and (3.19) follow. The proof
of (3.20) follows from the Dolbeault isomorphism (Corollary 3.10). ∎

######  Remark 3.17.

Generally, Hard Lefschetz theorem for basic cohomology does not hold unless
$[\partial_{B}\kappa_{B}^{0,1}]$ is trivial. (cf. [16, Theorem 5.11]).

######  Example 3.18.

We consider the Carrière example from [4]. Also, see [8, Section 7.1] and [16,
Example 9.1]. Let $A$ be a matrix in $\mathrm{SL}_{2}(\mathbb{Z})$ of trace
strictly greater than $2$. We denote respectively by $w_{1}$ and $w_{2}$ unit
eigenvectors associated with the eigenvalues $\lambda$ and $\frac{1}{\lambda}$
of $A$ with $\lambda>1$ irrational. Let the hyperbolic torus
$\mathbb{T}_{A}^{3}$ be the quotient of $\mathbb{T}^{2}\times\mathbb{R}$ by
the equivalence relation which identifies $(m,t)$ to $(A(m),t+1)$, where
$\mathbb{T}^{2}=\mathbb{R}^{2}/\mathbb{Z}^{2}$ is the torus. The flow
generated by the vector field $W_{2}$ is a Riemannian foliation with bundle-
like metric (letting $\left(x,s,t\right)$ denote local coordinates in the
$w_{2}$ direction, $w_{1}$ direction, and $\mathbb{R}$ direction,
respectively)

$g=\lambda^{-2t}dx^{2}+\lambda^{2t}ds^{2}+dt^{2}.$

Since $A$ preserves the integral lattice $\mathbb{Z}^{2}$, it induces a
diffeomorphism $A_{0}$ of the torus $\mathbb{T}^{2}$. So the flow generated by
$W_{2}$ is invariant under the diffeomorphism $A_{0}$ of $\mathbb{T}^{2}$.
Note that the mean curvature of the flow is
$\kappa=\kappa_{B}=\log\left(\lambda\right)dt$, since
$\chi_{\mathcal{F}}=\lambda^{-t}dx$ is the characteristic form and
$d\chi_{\mathcal{F}}=-\log\left(\lambda\right)\lambda^{-t}dt\wedge
dx=-\kappa\wedge\chi_{\mathcal{F}}$. It is well known that all twisted de Rham
cohomology groups vanish, that is, $H_{\kappa}^{r}(\mathcal{F})=\\{0\\}$ for
all $r=0,1,2$ in [8].

Now we will show that all twisted basic Dolbeault cohomology grous satisfy
Kodaira-Serre duality. First, we note that an orthonormal frame field for this
manifold is
$\\{X=\lambda^{t}\partial_{x},S=\lambda^{-t}\partial_{s},T=\partial_{t}\\}$
corresponding to the orthonormal coframe
$\\{X^{\ast}=\chi_{\mathcal{F}}=\lambda^{-t}dx,S^{\ast}=\lambda^{t}ds,T^{\ast}=dt\\}$.
Then, letting $J$ be defined by $J(S)=T,J(T)=-S$, the Nijenhuis tensor

$N_{J}(S,T)=[S,T]+J\left([JS,T]+[S,JT]\right)-[JS,JT]$

clearly vanishes, so that $J$ is integrable. The corresponding transverse
Kähler form is seen to be $\omega=T^{\ast}\wedge S^{\ast}=\lambda^{t}dt\wedge
ds=d(\frac{1}{\log\lambda}S^{\ast})$, an exact form in basic cohomology. From
the above,

$\kappa_{B}=-i\left(\log\lambda\right)Z^{\ast}+i\left(\log\lambda\right)\bar{Z}^{\ast},$

where
$Z^{\ast}=\frac{1}{2}(S^{\ast}+iT^{\ast})\in\Omega_{B}^{1,0}(\mathcal{F})$.
Then

$\displaystyle\kappa_{B}^{1,0}$ $\displaystyle=$
$\displaystyle-i\log\left(\lambda\right)Z^{\ast}=-\frac{i}{2}\left(\log\lambda\right)\left(\lambda^{t}ds+idt\right)$
$\displaystyle\bar{\partial}_{B}\kappa_{B}^{1,0}$ $\displaystyle=$
$\displaystyle
d\kappa_{B}^{1,0}=\left(\log\lambda\right)^{2}\bar{Z}^{\ast}\wedge Z^{\ast}$
$\displaystyle\bar{\partial}_{\kappa}\kappa_{B}^{1,0}$ $\displaystyle=$
$\displaystyle\bar{\partial}_{B}\kappa_{B}^{1,0}-\frac{1}{2}\kappa_{B}^{0,1}\wedge\kappa_{B}^{1,0}=\frac{9}{8}\bar{\partial}_{B}\kappa_{B}^{1,0}.$

It is impossible to change the metric so that
$\bar{\partial}_{B}\kappa_{B}^{1,0}=0$. Hence $\mathcal{F}$ is nontaut [16,
Corollary 5.12].

The basic Dolbeault cohomolog groups are given by $H_{B}^{0,0}=\mathbb{R},\
H_{B}^{1,0}=\\{0\\},\ H_{B}^{0,1}=\mathbb{R}$ and $H_{B}^{1,1}=\\{0\\}$. Then
observe that the ordinary basic cohomology Betti numbers for this foliation
are $h_{B}^{0}=h_{B}^{1}=1$, $h_{B}^{2}=0$, we see that the basic Dolbeault
Betti numbers satisfy

$h_{B}^{0,0}=h_{B}^{0,1}=1,\quad h_{B}^{1,0}=h_{B}^{1,1}=0.$

So even though it is true that

$h_{B}^{j}=\sum_{r+s=j}h_{B}^{r,s},$

and the foliation is transversely Kähler, we also have (with $n=1$)

$h_{B}^{r,s}\neq h_{B}^{s,r},\quad h_{B}^{r,s}\neq h_{B}^{n-r,n-s}.$

Thus, for a nontaut, transverse Kähler foliation, it is not necessarily true
that the odd basic Betti numbers are even, and the basic Dolbeault numbers do
not have the same kinds of symmetries as Dolbeault cohomology on Kähler
manifolds.

Now we compute the twisted basic Dolbeault cohomology groups
$H_{\kappa}^{\ast,\ast}\left(\mathcal{F}\right)$. Let $f\in
H_{\kappa}^{0,0}(\mathcal{F})$, that is, $f$ is a periodic function alone $t$
and $\bar{\partial}_{\kappa}f=0$. Equivalently,

$\displaystyle\bar{\partial}_{B}f=\frac{1}{2}f\kappa_{B}^{0,1}.$ (3.21)

On the other hand, since
$\bar{\partial}_{B}f\in\Omega_{B}^{0,1}(\mathcal{F})$, by a direct
calculation, $\bar{\partial}_{B}f=if^{\prime}(t)\overline{Z}^{*}$. Hence from
(3.34),
$f^{\prime}(t)\overline{Z}^{*}=\frac{1}{2}f(\log\lambda)\overline{Z}^{*}$.
That is, $f^{\prime}=\frac{1}{2}(\log\lambda)f$. Then $f=c\lambda^{-t}$ for
some constant. Since $f$ is periodic, $f(t)=0$. Hence

$\displaystyle H_{\kappa}^{0,0}(\mathcal{F})=\\{0\\}.$

Let $\varphi\in H_{\kappa}^{1,0}$. That is,
$\varphi=f(t)Z^{*}\in\Omega_{B}^{1,0}(\mathcal{F})$ and
$\bar{\partial}_{\kappa}\varphi=0$ for a periodic function $f$. Hence

$\displaystyle\bar{\partial}_{B}\varphi=\frac{1}{2}\kappa_{B}^{0,1}\wedge\varphi.$
(3.22)

By a direct calculation, $\bar{\partial}_{B}\varphi={i\over
2}(\log\lambda)f\overline{Z}^{*}\wedge Z^{*}$. So from (3.4)

$\displaystyle f^{\prime}=-\frac{3}{2}(\log\lambda)f$

and so $f(t)=c\lambda^{-{3\over 2}t}$ for some $c\in\mathbb{R}$. Since $f$ is
periodic, it is zero. Thus, $\varphi=0$. That is,

$\displaystyle H_{\kappa}^{1,0}(\mathcal{F})=\\{0\\}.$

By complex conjugatation, we get

$\displaystyle H_{\kappa}^{0,1}(\mathcal{F})=\\{0\\}.$

Let $\varphi\in H_{\kappa}^{1,1}(\mathcal{F})$. Then
$\varphi\in\Omega_{B}^{1,1}(\mathcal{F})$ is of the form
$\varphi=f(t)Z^{*}\wedge\overline{Z}^{*}$, where $f$ is a periodic function.
Trivially, $\bar{\partial}_{\kappa}\varphi=0$. Now, let
$\bar{\partial}_{\kappa}^{*}\varphi=0$. Since

$\displaystyle\delta_{\kappa}\varphi$
$\displaystyle=\delta_{T}\varphi+\frac{1}{2}\kappa^{\sharp}\lrcorner\varphi$
$\displaystyle=-{i\over
2}(f^{\prime}-\frac{3}{4}(\log\lambda)f)(Z^{*}+\overline{Z}^{*}),$

we have $\bar{\partial}_{\kappa}^{*}\varphi=-{i\over
2}(f^{\prime}-\frac{3}{4}(\log\lambda)f)Z^{*}$. Thus the solution of
$f^{\prime}-\frac{3}{4}(\log\lambda)f=0$ reduced to zero for periodic
functions $f$. Hence

$\displaystyle H_{\kappa}^{1,1}(\mathcal{F})=\\{0\\}.$

This shows that the Kodaira-Serre duality and Hodge isomorphism are satisfied
for the twisted basic Dolbeault cohomology.

###  3.3. The $d_{\kappa}d_{\kappa}^{c}$ Lemma

Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation of codimension
$2n$ on a compact Riemannian manifold $M$ with bundle-like metric. First, we
recall the operator
$C:\Omega_{B}^{*}(\mathcal{F})\to\Omega_{B}^{*}(\mathcal{F})$ defined by [16]

$C=\sum_{0\leq r,s\leq n}(\sqrt{-1})^{r-s}P_{r,s},$

where $P_{r,s}:\Omega_{B}^{*}(\mathcal{F})\to\Omega_{B}^{r,s}(\mathcal{F})$ is
the projection. Then

$C^{*}=C^{-1}=\sum_{0\leq r,s\leq n}(\sqrt{-1})^{s-r}P_{r,s}.$

Now, we define
$d_{\kappa}^{c}:\Omega_{B}^{r}(\mathcal{F})\to\Omega_{B}^{r+1}(\mathcal{F})$
by

$d_{\kappa}^{c}=C^{*}d_{\kappa}C=C^{-1}d_{\kappa}C.$

Then $d_{\kappa}^{c}=\sqrt{-1}(\bar{\partial}_{k}-\partial_{k})$ and

$d_{\kappa}d_{\kappa}^{c}|_{\Omega_{B}^{*}(\mathcal{F})}=-d_{\kappa}^{c}d_{\kappa}|_{\Omega_{B}^{*}(\mathcal{F})}.$

Let $\delta_{\kappa}^{c}$ be the adjoint operator of $d_{\kappa}^{c}$, which
is given by

$\delta_{\kappa}^{c}=C^{*}\delta_{\kappa}C=C^{-1}\delta_{\kappa}C.$

Let
$\Delta_{\kappa}^{c}=d_{\kappa}^{c}\delta_{\kappa}^{c}+\delta_{\kappa}^{c}d_{\kappa}^{c}$.
Since $\Delta_{\kappa}$ preserves the type of differential form, we have

$\Delta_{\kappa}^{c}=C^{-1}\Delta_{\kappa}C=\Delta_{\kappa}.$

######  Lemma 3.19.

($d_{\kappa}d_{\kappa}^{c}$ Lemma) Let $(M,\mathcal{F},g_{Q},J)$ be a
transverse Kähler foliation on a compact manifold $M$ with a bundle-like
metric. Then on $\Omega_{B}^{*}(\mathcal{F})$,

$\ker d_{\kappa}\cap{\rm Im}\;d_{\kappa}^{c}={\rm
Im}\;d_{\kappa}d_{\kappa}^{c}.$

###### Proof.

Let $\alpha\in\ker d_{\kappa}\cap{\rm
Im}\;d_{\kappa}^{c}\cap\Omega_{B}^{r}(\mathcal{F})$. That is, for some basic
$(r-1)-$form $\beta$, $\alpha=d_{\kappa}^{c}\beta$ and
$\beta=\gamma+d_{\kappa}\gamma_{1}+\delta_{\kappa}\gamma_{2}$ with
$\gamma\in\mathcal{H}_{\kappa}^{r-1}(\mathcal{F})$ by the Hodge decomposition
(Theorem 2.5). Since $M$ is compact, by Theorem 3.5, $\Delta_{\kappa}\gamma=0$
implies $\square_{\kappa}\gamma=\overline{\square}_{\kappa}\gamma=0$.
Therefore, $\partial_{\kappa}\gamma=\bar{\partial}_{\kappa}\gamma=0$ and so
$d_{\kappa}^{c}\gamma=0$. Hence

$\displaystyle\alpha$ $\displaystyle=d_{\kappa}^{c}\beta$
$\displaystyle=d_{\kappa}^{c}\gamma+d_{\kappa}^{c}d_{\kappa}\gamma_{1}+d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}$
$\displaystyle=d_{\kappa}^{c}d_{\kappa}\gamma_{1}+d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}$
$\displaystyle=d_{\kappa}d_{\kappa}^{c}(-\gamma_{1})+d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}.$
(3.23)

Moreover, since $d_{\kappa}\alpha=0$, by the equation above,

$\displaystyle
0=d_{\kappa}d_{\kappa}^{c}\beta=d_{\kappa}d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}=-d_{\kappa}\delta_{\kappa}d_{\kappa}^{c}\gamma_{2}.$

The last equality follows from
$d_{\kappa}^{c}\delta_{\kappa}+\delta_{\kappa}d_{\kappa}^{c}=0$ (Theorem 3.5).
By integrating,

$\displaystyle 0=\int_{M}\langle
d_{\kappa}\delta_{\kappa}d_{\kappa}^{c}\gamma_{2},d_{\kappa}^{c}\gamma_{2}\rangle=\int_{M}\|\delta_{\kappa}d_{\kappa}^{c}\gamma_{2}\|^{2}=\int_{M}\|d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}\|^{2}.$

That is, $d_{\kappa}^{c}\delta_{\kappa}\gamma_{2}=0$. So from (3.23),

$\displaystyle\alpha=d_{\kappa}d_{\kappa}^{c}(-\gamma_{1}),$

which implies that $\alpha\in{\rm Im}\;d_{\kappa}d_{\kappa}^{c}$. ∎

######  Remark 3.20.

If $\mathcal{F}$ is taut, then the $d_{\kappa}d_{\kappa}^{c}$ Lemma implies
that $dd_{c}$ Lemma [14, Lemma 7.3].

###  3.4. $\Delta_{\kappa}$-harmonic forms

Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a compact
Riemannian manifold $M$ with a bundle-like metric. We define two operators

$\displaystyle\nabla_{T}^{*}\nabla_{T}\phi$
$\displaystyle=-\sum_{a}\nabla_{V_{a}}\nabla_{\bar{V}_{a}}\phi+\nabla_{H^{0,1}}\phi,$
$\displaystyle\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}\phi$
$\displaystyle=-\sum_{a}\nabla_{\bar{V}_{a}}\nabla_{V_{a}}\phi+\nabla_{H^{1,0}}\phi.$

Then by a direct calculation, we have

$\displaystyle\nabla_{T}^{*}\nabla_{T}\phi=\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}\phi+\nabla_{H^{0,1}-H^{1,0}}\phi-\sum_{a}R^{Q}(V_{a},\bar{V}_{a})\phi$
(3.24)

for any basic form $\phi$. Then the operators $\nabla_{T}^{*}\nabla_{T}$ and
$\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}$ are formally self-adjoint and positive-
definite [15].

######  Proposition 3.21.

[15] Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a
Riemannian manifold $M$ with a bundle-like metric. Then for all
$\phi\in\Omega_{B}^{r,s}(\mathcal{F})$,

$\displaystyle\overline{\square}_{B}\phi$
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi+\sum_{a,b}\bar{\omega}^{a}\wedge\bar{V}_{b}\lrcorner
R^{Q}(V_{b},\bar{V}_{a})\phi+\sum_{a}\bar{\omega}^{a}\wedge(\nabla_{\bar{V}_{a}}H^{0,1})\lrcorner\,\phi,$
(3.25)

From (3.24), we have the following.

######  Proposition 3.22.

[15] Let $(M,\mathcal{F},g_{Q},J)$ be as in Proposition 3.21.

$(1)$ If $\phi$ is a basic form of type $(r,0)$, then

$\displaystyle\overline{\square}_{B}\phi$
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi.$ (3.26)
$\displaystyle=\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}\phi+\nabla_{H^{0,1}-H^{1,0}}\phi-\sum_{a}R^{Q}(V_{a},\bar{V}_{a})\phi.$
(3.27)

$(2)$ If $\phi$ is a basic form of type $(r,n)$, then

$\displaystyle\overline{\square}_{B}\phi$
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi+\sum_{a}R^{Q}(V_{a},\bar{V}_{a})\phi+\operatorname{div}_{\nabla}(H^{0,1})\phi$
(3.28)
$\displaystyle=\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}\phi+\nabla_{H^{0,1}-H^{1,0}}\phi+\operatorname{div}_{\nabla}(H^{0,1})\phi.$
(3.29)

On the other hand, by a direct calculation, we have the following.

######  Proposition 3.23.

On a transverse Kähler foliation, we have

$\displaystyle\overline{\square}_{\kappa}$
$\displaystyle=\overline{\square}_{B}-\frac{1}{2}\Big{(}\epsilon(\kappa_{B}^{0,1})\bar{\partial}_{B}^{*}+\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\Big{)}-\frac{1}{2}\Big{(}\bar{\partial}_{B}H^{0,1}\lrcorner+H^{0,1}\lrcorner\bar{\partial}_{B}\Big{)}+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}.$

From Proposition 3.21 and Proposition 3.23, we have the following.

######  Proposition 3.24.

On a transverse Kähler foliation, we have

$\displaystyle\overline{\square}_{\kappa}\phi$
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi+\sum_{a,b}\overline{\omega}^{a}\wedge\overline{V}_{b}\lrcorner
R^{Q}(V_{b},\overline{V}_{a})\phi+\sum_{a}\overline{\omega}^{a}\wedge(\nabla_{\overline{V}_{a}}H^{0,1})\lrcorner\phi$
$\displaystyle-\frac{1}{2}\Big{(}\epsilon(\kappa_{B}^{0,1})\bar{\partial}_{B}^{*}+\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\Big{)}\phi-\frac{1}{2}\Big{(}\bar{\partial}_{B}H^{0,1}\lrcorner+H^{0,1}\lrcorner\bar{\partial}_{B}\Big{)}\phi+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi.$

######  Remark 3.25.

Proposition 3.24 was also shown in [9, Theorem 3.1], but the expression is
little bit different. The authors in [9] proved the vanishing theorem of
transversally holomorphic basic $(r,0)$-form by using the Weitzenböck formula
for the twisted basic Laplacian $\overline{\square}_{\kappa}$ [9, Theorem
3.4]. In this research, we deal with the vanishing theorem of
$\bar{\partial}_{\kappa}$-holomorphic $(r,0)$-form (Corollary 3.31 below).

######  Proposition 3.26.

[16] Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a
closed Riemannian manifold $M$. Then there exists a bundle-like metric
compatible with the Kähler structure such that
$\partial_{B}^{*}\kappa_{B}^{1,0}=0$, or
$\bar{\partial}_{B}^{*}\kappa_{B}^{0,1}=0$.

######  Lemma 3.27.

For any $\phi\in\Omega_{B}^{r,0}(\mathcal{F})$, we get

$\displaystyle\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi=-\nabla_{H^{1,0}}\phi.$

###### Proof.

From (3.12) and (3.13), for any $\phi\in\Omega_{B}^{r,0}(\mathcal{F})$, since
$\overline{V}_{a}\lrcorner\phi=0$, we have

$\displaystyle\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi$
$\displaystyle=-\sum_{a}\overline{V}_{a}\lrcorner\nabla_{V_{a}}\kappa_{B}^{0,1}\wedge\phi-\sum_{a}\kappa_{B}^{0,1}(\overline{V}_{a})\nabla_{V_{a}}\phi+H^{0,1}\lrcorner\epsilon(\kappa_{B}^{0,1})\phi$
$\displaystyle=(\bar{\partial}_{T}^{*}\kappa_{B}^{0,1})\wedge\phi-\nabla_{H^{1,0}}\phi+|\kappa_{B}^{0,1}|^{2}\phi.$

By Proposition 3.26, if we choose the bundle-like metric such that
$\bar{\partial}_{B}^{*}\kappa_{B}^{0,1}=0$, the the proof follows. ∎

From Lemma 3.26 and Proposition 3.27, we get

######  Theorem 3.28.

On a transverse Kähler foliation, the following hold:

(1) If $\phi\in\Omega_{B}^{r,0}(\mathcal{F})$, then

$\displaystyle\overline{\square}_{\kappa}\phi$
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi-\frac{1}{2}\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi-\frac{1}{2}H^{0,1}\lrcorner\bar{\partial}_{B}\phi+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi$
(3.30)
$\displaystyle=\overline{\nabla}_{T}^{*}\overline{\nabla}_{T}\phi+\sum_{a}R^{Q}(\overline{V}_{a},V_{a})\phi+\nabla_{H^{0,1}}\phi+\frac{1}{2}\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi-\frac{1}{2}H^{0,1}\lrcorner\bar{\partial}_{B}\phi+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi.$
(3.31)

(2) If $\phi\in\Omega_{B}^{r,n}(\mathcal{F})$, then

$\displaystyle\overline{\square}_{\kappa}\phi$
$\displaystyle=\overline{\nabla}_{T}^{*}\overline{\nabla}_{T}\phi-\nabla_{H^{1,0}}\phi+\frac{1}{2}\nabla_{H^{0,1}}\phi+\frac{1}{2}{\rm
div}_{\nabla}(H^{0,1})\phi-\frac{1}{2}\kappa_{B}^{0,1}\wedge\bar{\partial}_{B}^{*}\phi+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi$
(3.32)
$\displaystyle=\nabla_{T}^{*}\nabla_{T}\phi+\sum_{a}R^{Q}(V_{a},\overline{V}_{a})\phi-\frac{1}{2}\nabla_{H^{0,1}}\phi-\frac{1}{2}\kappa_{B}^{0,1}\wedge\bar{\partial}_{B}^{*}\phi+\frac{1}{2}{\rm
div}_{\nabla}(H^{0,1})\phi+\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi.$ (3.33)

###### Proof.

Let $\phi$ be a basic form of type $(r,0)$. Since $Z\lrcorner\phi=0$ for any
$Z\in Q^{0,1}$, from Proposition 3.24, (3.30) is proved. Now, we choose the
bundle-like metric such that the mean curvature form is basic harmonic, that
is, $\bar{\partial}_{B}^{*}\kappa_{B}^{0,1}=0$. From Lemma 3.26 and (3.24),
the proof of (3.31) follows. From Proposition 3.21(2) and Proposition 3.22,
the proofs of (3.32) and (3.33) follow. ∎

######  Proposition 3.29.

Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a compact
manifold $M$ with a bundle-like metric. If
$\phi\in\Omega_{B}^{r,0}(\mathcal{F})$ is a $\Delta_{\kappa}$-harmonic form,
then

$\displaystyle\nabla_{V}\phi=0$

for any $V\in\Gamma Q^{0,1}$.

###### Proof.

Since $\Delta_{\kappa}\phi=0$, by Theorem 3.5,
$\overline{\square}_{\kappa}\phi=0$ and so $\bar{\partial}_{\kappa}\phi=0$.
That is, $\bar{\partial}_{B}\phi=\frac{1}{2}\kappa_{B}^{0,1}\wedge\phi$. Since
$\phi$ is type of $(r,0)$, we have

$\displaystyle
H^{0,1}\lrcorner\bar{\partial}_{B}\phi=\frac{1}{2}H^{0,1}\lrcorner(\kappa_{B}^{0,1}\wedge\phi)=\frac{1}{2}|\kappa_{B}^{0,1}|^{2}\phi.$

From (3.30), we get

$\displaystyle\nabla_{T}^{*}\nabla_{T}\phi-\frac{1}{2}\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi+\frac{1}{4}|\kappa_{B}^{0,1}|^{2}\phi=0.$
(3.34)

By integrating $(\ref{3-34})$, we get

$\displaystyle 0$
$\displaystyle=\int_{M}\|\nabla_{T}\phi\|^{2}-\frac{1}{2}\int_{M}\langle\kappa_{B}^{0,1}\wedge\phi,\bar{\partial}_{B}\phi\rangle+\frac{1}{4}\int_{M}|\kappa_{B}^{0,1}|^{2}\|\phi\|^{2}$
$\displaystyle=\int_{M}\|\nabla_{T}\phi\|^{2}-\frac{1}{4}\int_{M}\|\kappa_{B}^{0,1}\wedge\phi\|^{2}+\frac{1}{4}\int_{M}|\kappa_{B}^{0,1}|^{2}\|\phi\|^{2}.$
(3.35)

Since $H^{0,1}\lrcorner\phi=0$ for any $\phi\in\Omega_{B}^{r,0}(\mathcal{F})$,
we get

$\displaystyle\langle\kappa_{B}^{0,1}\wedge\phi,\kappa_{B}^{0,1}\wedge\phi\rangle=\langle\phi,H^{0,1}\lrcorner(\kappa_{B}^{0,1}\wedge\phi)\rangle=|\kappa_{B}^{0,1}|^{2}\|\phi\|^{2}.$

From (3.4), we get

$\displaystyle\int_{M}\|\nabla_{T}\phi\|^{2}=0,$

which implies $\nabla_{T}\phi=0$, that is, for any $V\in\Gamma Q^{0,1}$,
$\nabla_{V}\phi=0$. ∎

######  Theorem 3.30.

Let $(M,\mathcal{F},g_{Q},J)$ be a transverse Kähler foliation on a compact
manifold $M$ with a bundle-like metric.

(1) If the transverse Ricci curvature is nonnegative and positive at some
point, then $\mathcal{H}_{\kappa}^{r,0}(\mathcal{F})=\\{0\\}$ for all $r>0$.

(2) If $\mathcal{F}$ is transversely Ricci-flat and non-taut, then
$\mathcal{H}_{\kappa}^{r,0}(\mathcal{F})=\\{0\\}$ for all $r>0$.

###### Proof.

Let $\phi$ be a $\Delta_{\kappa}$-harmonic form of type $(r,0)$. If we choose
the bundle-like metric such that $\bar{\partial}_{B}^{*}\kappa_{B}^{0,1}=0$,
then from (3.24), Lemma 3.27 and Proposition 3.29,

$\displaystyle\bar{\nabla}_{T}^{*}\bar{\nabla}_{T}\phi+\bar{\partial}_{B}^{*}\epsilon(\kappa_{B}^{0,1})\phi-\sum_{a}R^{Q}(V_{a},\bar{V}_{a})\phi=0.$
(3.36)

Note that
$(\sum_{a}R^{Q}(\bar{V}_{a},V_{a}))\omega^{b}=Ric^{Q}(E_{b},E_{b})\omega^{b}$
[15, Remark 4.7]. By integrating (3.36), we get

$\displaystyle\int_{M}\|\bar{\nabla}_{T}\phi\|^{2}+\int_{M}|\kappa_{B}^{0,1}|^{2}\|\phi\|^{2}+\sum_{i=1}^{r}\int_{M}R^{Q}(E_{a_{i}},E_{a_{i}})\|\phi\|^{2}=0.$

If $Ric^{Q}$ is nonnegative and positive at some point, then $\phi=0$. So the
proof of (1) is proved. If $Ric^{Q}=0$, then

$\displaystyle|\kappa_{B}^{0,1}|\|\phi\|=0.$

So if $\mathcal{F}$ is nontaut, then $\phi=0$, that is, the proof of (2) is
finished. ∎

######  Corollary 3.31.

Let $(M,\mathcal{F},g_{Q},J)$ be as in Theorem 3.30 with a transversely Ricci-
flat foliation. If $\mathcal{F}$ is nontaut, then

$\displaystyle H^{r,0}_{\kappa}(\mathcal{F})\cong
H_{\kappa}^{0,r}(\mathcal{F})\cong H_{\kappa}^{n,s}(\mathcal{F})\cong
H_{\kappa}^{s,n}(\mathcal{F})=\\{0\\}$

for $r,s>0$.

###### Proof.

The proofs follow from complex conjugation, Kodaira-Serra duality and
Dolbeault isomorphism. ∎

The vanishing theorems for the basic harmonic space were proved in [9] and
[15], respectively.

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