## Lecture 6. Transverse Weitzenbock formula and heat equation on one-forms

Let $\mathbb M$ be a smooth, connected manifold with dimension $n+m$. We assume that $\mathbb M$ is equipped with a Riemannian foliation $\mathcal{F}$ with bundle like metric $g$ and totally geodesic $m$-dimensional leaves.

We define the canonical variation of $g$ as the one-parameter family of Riemannian metrics:

$g_{\varepsilon}=g_\mathcal{H} \oplus \frac{1}{\varepsilon }g_{\mathcal{V}}, \quad \varepsilon >0.$
We now introduce some tensors and definitions that will play an important role in the sequel.

For $Z \in \Gamma^\infty(T\mathbb M)$, there is a unique skew-symmetric endomorphism $J_Z:\mathcal{H}_x \to \mathcal{H}_x$ such that for all horizontal vector fields $X$ and $Y$,

$g_\mathcal{H} (J_Z (X),Y)= g_\mathcal{V} (Z,T(X,Y)).$
where $T$ is the torsion tensor of $\nabla$. We then extend $J_{Z}$ to be 0 on $\mathcal{V}_x$. If $Z_1,\cdots,Z_m$ is a local vertical frame, the operator $\sum_{l=1}^m J_{Z_l}J_{Z_l}$ does not depend on the choice of the frame and shall concisely be denoted by $\mathbf{J}^2$. For instance, if $\mathbb M$ is a K-contact manifold equipped with the Reeb foliation, then $\mathbf{J}$ is an almost complex structure, $\mathbf{J}^2=-\mathbf{Id}_{\mathcal{H}}$.
The horizontal divergence of the torsion $T$ is the $(1,1)$ tensor which is defined in a local horizontal frame $X_1,\cdots,X_n$ by

$\delta_\mathcal{H} T (X)=- \sum_{j=1}^n(\nabla_{X_j} T) (X_j,X), \quad X \in \Gamma^\infty(\mathbb M).$
The $g$-adjoint of $\delta_\mathcal{H}T$ will be denoted $\delta_\mathcal{H} T^*$.

In the sequel, we shall need to perform computations on one-forms. For that purpose we introduce some definitions and notations on the cotangent bundle.

We say that a one-form to be horizontal (resp. vertical) if it vanishes on the vertical bundle $\mathcal{V}$ (resp. on the horizontal bundle $\mathcal{H}$). We thus have a splitting of the cotangent space

$T^*_x \mathbb M= \mathcal{H}^*(x) \oplus \mathcal{V}^*(x)$

The metric $g_\varepsilon$ induces then a metric on the cotangent bundle which we still denote $g_\varepsilon$. By using similar notations and conventions as before we have for every $\eta$ in $T^*_x \mathbb M$,

$\| \eta \|^2_{\varepsilon} =\| \eta \|_\mathcal{H}^2+\varepsilon \| \eta \|_\mathcal{V}^2.$

By using the duality given by the metric $g$, $(1,1)$ tensors can also be seen as linear maps on the cotangent bundle $T^* \mathbb M$. More precisely, if $A$ is a $(1,1)$ tensor, we will still denote by $A$ the fiberwise linear map on the cotangent bundle which is defined as the $g$-adjoint of the dual map of $A$. The same convention will be made for any $(r,s)$ tensor.

We define then the horizontal Ricci curvature $\mathfrak{Ric}_{\mathcal{H}}$ as the fiberwise symmetric linear map on one-forms such that for every smooth functions $f,g$,

$\langle \mathfrak{Ric}_{\mathcal{H}} (df), dg \rangle=\mathbf{Ricci} (\nabla_\mathcal{H} f ,\nabla_\mathcal{H} g),$
where $\mathbf{Ricci}$ is the Ricci curvature of the connection $\nabla$.
If $V$ is a horizontal vector field and $\varepsilon >0$, we consider the fiberwise linear map from the space of one-forms into itself which is given for $\eta \in \Gamma^\infty(T^* \mathbb M)$ and $Y \in \Gamma^\infty(T \mathbb M)$ by

$\mathfrak{T}^\varepsilon_V \eta (Y) = \begin{cases} \frac{1}{\varepsilon} \eta (J_Y V), \quad Y \in \Gamma^\infty(\mathcal{V}) \\ -\eta (T(V,Y)), Y \in \Gamma^\infty(\mathcal{H}) \end{cases}$
We observe that $\mathfrak{T}^\varepsilon_V$ is skew-symmetric for the metric $g_\varepsilon$ so that $\nabla -\mathfrak{T}^\varepsilon$ is a $g_\varepsilon$-metric connection.

If $\eta$ is a one-form, we define the horizontal gradient of $\eta$ in a local frame as the $(0,2)$ tensor

$\nabla_\mathcal{H} \eta =\sum_{i=1}^n \nabla_{X_i} \eta \otimes \theta_i.$

Similarly, we will use the notation

$\mathfrak{T}^\varepsilon_\mathcal{H} \eta =\sum_{i=1}^n \mathfrak{T}^\varepsilon_{X_i} \eta \otimes \theta_i.$

Finally, we will still denote by $\Delta_\mathcal{H}$ the covariant extension on one-forms of the horizontal Laplacian. In a local horizontal frame, we have thus

$\Delta_{\mathcal{H}}=-\nabla_{\mathcal{H}}^* \nabla_{\mathcal{H}}=\sum_{i=1}^n \nabla_{X_i}\nabla_{X_i} -\nabla_{\nabla_{X_i} X_i}.$
For $\varepsilon >0$, we consider the following operator which is defined on one-forms by

$\square_\varepsilon=-(\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)^* (\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)-\frac{1}{ \varepsilon}\mathbf{J}^2+\frac{1}{\varepsilon} \delta_\mathcal{H} T- \mathfrak{Ric}_{\mathcal{H}},$
where the adjoint is understood with respect to the metric $g_{\varepsilon}$. It is easily seen that, in a local horizontal frame,
$-(\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)^* (\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon) =\sum_{i=1}^n (\nabla_{X_i} -\mathfrak{T}^\varepsilon_{X_i})^2 - ( \nabla_{\nabla_{X_i} X_i}- \mathfrak{T}^\varepsilon_{\nabla_{X_i} X_i}),$

We can also consider the operator which is defined on one-forms by

$\square_\infty:=\sum_{i=1}^n (\nabla_{X_i} -\mathfrak{T}^\infty_{X_i})^2 - ( \nabla_{\nabla_{X_i} X_i}- \mathfrak{T}^\infty_{\nabla_{X_i} X_i}) - \mathfrak{Ric}_{\mathcal{H}}$

It is clear that for every smooth one-form $\alpha$ on $\mathbb M$ and every $x \in \mathbb M$ the following holds

$\lim_{\varepsilon \to \infty} \square_\varepsilon \alpha (x)=\square_\infty \alpha (x).$

The following theorem that was proved in this paper is the main result of the lecture:

Theorem: Let $0 < \varepsilon \le +\infty$. For every $f \in C^\infty(\mathbb M)$, we have
$d \Delta_{\mathcal{H}} f=\square_\varepsilon df.$

Proof:
We only sketch the proof and refer to the original paper for the details. If $Z_1,\cdots,Z_m$ is a local vertical frame of the leaves, we denote

$\mathfrak J(\eta)=-\sum_{l=1}^mJ_{Z_l}(\iota_{Z_l}d\eta_{\mathcal V}),$
where $\eta_{\mathcal V}$ is the the projection of $\eta$ onto the vertical cotangent bundle. It does not depend on the choice of the frame and therefore defines a globally defined tensor.
Also, let us consider the map $\mathcal{T} \colon \Gamma^\infty(\wedge^2 T^*\mathbb M)\to \Gamma^\infty( T^*\mathbb M)$ which is given in a local coframe $\theta_i \in \Gamma^\infty(\mathcal{H}^*)$, $\nu_k \in \Gamma^\infty(\mathcal{V}^*)$

$\mathcal{T}(\theta_i\wedge\theta_j)=-\gamma_{ij}^l \nu_l,\quad \mathcal{T}(\theta_i\wedge\nu_k)=\mathcal{T}(\nu_k\wedge\nu_l)=0.$
A direct computation shows then that

$-(\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)^* (\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon) =$
$\Delta_{\mathcal H} +2\mathfrak J-\frac{2}{\varepsilon}\mathcal{T}\circ d+\delta_\mathcal{H} T^*-\frac{1}{\varepsilon}\delta_\mathcal{H} T+\frac{1}{\varepsilon}\mathbf{J}^2.$
Thus, we just need to prove that if $\square_\infty$ is the operator defined on one-forms by

$\square_\infty=\Delta_{\mathcal H}+2\mathfrak J-\mathfrak{Ric}_{\mathcal{H}}+\delta_\mathcal{H} T^* ,$
then for any $f\in C^\infty(\mathbb M)$,

$d\Delta_{\mathcal H} f=\square_\infty df.$
A computation in local frame shows that

$d\Delta_{\mathcal H} f- d\Delta_{\mathcal H} f = 2\mathfrak J(df) -\mathfrak{Ric}_{\mathcal{H}}(df) +\delta_\mathcal{H} T^* (df),$
which completes the proof $\square$

We also have the following Bochner’s type identity.

Theorem: For any $\eta \in \Gamma^\infty(T^* \mathbb M)$,
$\frac{1}{2} \Delta_\mathcal{H} \| \eta \|_{\varepsilon}^2 -\langle \square_\varepsilon \eta , \eta \rangle_{\varepsilon}= \| \nabla_{\mathcal{H}} \eta -\mathfrak{T}^\varepsilon_{\mathcal{H}} \eta \|_{\varepsilon}^2 + \left\langle \mathfrak{Ric}_{\mathcal{H}} (\eta), \eta \right\rangle_\mathcal{H} -\left \langle \delta_\mathcal{H} T (\eta) , \eta \right\rangle_\mathcal{V} +\frac{1}{\varepsilon} \langle \mathbf{J}^2 (\eta) , \eta \rangle_\mathcal{H}.$

We now turn to probabilistic applications.

We denote by $(X_t)_{t\geq 0}$ the horizontal Brownian motion on $\mathbb M$. The lifetime of the process is denoted by $\mathbf{e}$. We assume that the metric $g$ is complete and $\mathcal{H}$ is bracket generating. As a consequence, one can define $(P_t)_{t \ge 0}$ the heat semigroup associated to $(X_t)_{t\geq 0}$ as being the semigroup generated by the self-adjoint extension of $\frac{1}{2} \Delta_\mathcal{H}$.

We define a process $\tau^\varepsilon_t:T_{X_t}^*\mathbb{M}\rightarrow T^*_{X_0}\mathbb{M}$ by the formula
$\tau^{\varepsilon}_t=\mathcal{M}_{t}^{\varepsilon}\Theta_{t}^{\varepsilon}, \quad t < \mathbf{e}$
where the process $\Theta_t^{\varepsilon}: T_{X_t}^{*}\mathbb{M}\rightarrow T_{X_0}^{*}\mathbb{M}$ is the stochastic parallel transport with respect to the connection $\nabla -\mathfrak{T}^\varepsilon$ along the paths of $(X_t)_{t\geq 0}$. The multiplicative functional $(\mathcal{M}_t^{\varepsilon})_{t\geq 0}$ is defined as the solution of the following ordinary differential equation

$\frac{d\mathcal{M}_t^{\varepsilon}}{dt}=-\frac{1}{2}\mathcal{M}_t^{\varepsilon}\Theta_t^{\varepsilon}\left(\frac{1}{\varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+\mathfrak{Ric}_{\mathcal{H}} \right)(\Theta_t^{\varepsilon})^{-1}, ~~\mathcal{M}_0^{\varepsilon}=\mathbf{Id}.$

Observe that the process $\tau^\varepsilon_t:T_{X_t}^*\mathbb{M}\rightarrow T^*_{X_0}\mathbb{M}$ is a solution of the following covariant Stratonovitch stochastic differential equation:

$d[\tau^\varepsilon_t \alpha(X_t)]=\tau^\varepsilon_t\left( \nabla_{\circ dX_t}-\mathfrak{T}_{\circ dX_t}^{\varepsilon}-\frac{1}{2} \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+ \mathfrak{Ric}_{\mathcal{H}}\right)dt\right) \alpha(X_t),~~\tau_0=\mathbf{Id},$
where $\alpha$ is any smooth one-form.

From Gronwall’s lemma and the fact that $\Theta_t^{\varepsilon}$ is an isometry, we easily deduce that

Lemma: Let $\varepsilon >0$. Assume that there exists a constant $C_\varepsilon \ge 0$ such that for every $\alpha \in \Gamma^\infty (T^*\mathbb{M})$,

$\left \langle \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+ \mathfrak{Ric}_{\mathcal{H}}\right) \alpha , \alpha \right\rangle_\varepsilon \ge -C_\varepsilon \| \alpha \|^2_\varepsilon$
Then, there exists a constant $\tilde{C}_\varepsilon \ge 0$, such that for every $t \ge 0$,

$\| \tau^\varepsilon_t \alpha(X_t) \|_\varepsilon \le e^{ \tilde{C}_\varepsilon t }\| \alpha(X_t) \|_\varepsilon$

For $x \in \mathbb M$, as usual we will denote

$\mathbb{P}_x=\mathbb{P} \left( \cdot \mid X_0=x \right).$

Theorem: Assume that there exists a constant $C_\varepsilon \ge 0$ such that for every $\alpha \in \Gamma^\infty (T^*\mathbb{M})$,

$\left \langle \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+ \mathfrak{Ric}_{\mathcal{H}}\right) \alpha , \alpha \right\rangle_\varepsilon \ge -C_\varepsilon \| \alpha \|^2_\varepsilon$
Let $\eta$ be a one-form on $\mathbb{M}$ which is smooth and compactly supported. The unique solution in $L^2$ of the Cauchy problem:

$\begin{cases} \phi (0,x)=\eta (x) \\ \frac{\partial \phi}{\partial t} =\frac{1}{2} \square_\varepsilon \phi \end{cases}$
is given by

$\phi(t,x)=\mathbb{E}_x \left( \tau^\varepsilon_t \eta (X_t) 1_{t <\mathbf{e}}\right) .$

Proof:

This is Feynman-Kac formula.
Sketch of the proof.

It is proved in this course, that the operator

$-(\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)^* (\nabla_\mathcal{H} -\mathfrak{T}_\mathcal{H}^\varepsilon)$

is essentially self-adjoint on the space of smooth and compactly supported one-forms. Thus, from the assumption, $\frac{1}{2} \square_\varepsilon$ is the generator of a bounded semigroup $Q_t^\varepsilon$ in $L^2$ that uniquely solves the above Cauchy problem.

From the Bochner’s identity, one has

$\frac{1}{2} \Delta_\mathcal{H} \| \eta \|_{\varepsilon}^2 -\langle \square_\varepsilon \eta , \eta \rangle_{\varepsilon} \ge -C_\varepsilon \| \eta \|^2_\varepsilon$.

From Shigekawa (L^p contraction for vector valued semigroups), this implies the a priori pointwise bound

$\| Q_t^\varepsilon \eta \|^2_\varepsilon \le e^{2C_\varepsilon t} P_t (\| \eta\|^2_\varepsilon)$.

We now claim that the process

$N_s=\tau^\varepsilon_s(Q_{T-s}^\varepsilon \eta) (X_s)1_{T <\mathbf{e}}, \quad 0 \le s \le T$,

is a local martingale. Indeed, from Ito’s formula and the definition of $\tau^\varepsilon$, we have

$dN_s=\tau^\varepsilon_s \left( \nabla_{\circ dX_s}-\mathfrak{T}_{\circ dX_s}^{\varepsilon}-\frac{1}{2} \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+\mathfrak{Ric}_{\mathcal{H}}\right)ds\right) (Q_{T-s}^\varepsilon \eta) (X_s)$
$+\tau^\varepsilon_s \frac{d}{ds} (Q_{T-s}^\varepsilon \eta) (X_s) ds.$
We now  conclude from the fact that the bounded variation part of

$\tau^\varepsilon_s \left( \nabla_{\circ dX_s}-\mathfrak{T}_{\circ dX_s}^{\varepsilon}-\frac{1}{2} \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+ \mathfrak{Ric}_{\mathcal{H}}\right)ds\right)(Q_{T-s}^\varepsilon \eta) (X_s)$

is given by $\frac{1}{2} \tau^\varepsilon_s \square_\varepsilon (Q_{T-s}^\varepsilon \eta) (X_s) (X_s) ds$.

Form the previous estimates, we conclude that $N$ is a martingale $\square$

Corollary:  Let $\varepsilon >0$. Assume that there exists a constant $C_\varepsilon \ge 0$ such that for every $\alpha \in \Gamma^\infty (T^*\mathbb{M})$,

$\left \langle \left( \frac{1}{ \varepsilon}\mathbf{J}^2-\frac{1}{\varepsilon} \delta_\mathcal{H} T+ \mathfrak{Ric}_{\mathcal{H}}\right) \alpha , \alpha \right\rangle_\varepsilon \ge -C_\varepsilon \| \alpha \|^2_\varepsilon$

Then, for $f \in C^\infty_0(\mathbb M)$, and $t \ge 0$

$dP_tf (x)=\mathbb{E}_x ( \tau^\varepsilon_t df (X_t) 1_{t <\mathbf{e}})$
As a consequence, $\mathbb{P}_x(\mathbf{e}=+\infty)=1$.

Proof: Let $\phi(t,x)=dP_t f(x)$.
We have

$\frac{\partial \phi}{\partial t} =d\Delta_{\mathcal{H}} P_t f= \frac{1}{2} \square_\varepsilon \phi.$
Thus, from the previous theorem

$dP_tf (x)=\mathbb{E}_x ( \tau^\varepsilon_t df (X_t) 1_{t <\mathbf{e}})$
This representation implies the bound

$\| dP_t f (x) \|_\varepsilon \le e^{\frac{C_\varepsilon}{2} t} (P_t \| df \|_\varepsilon) (x).$

It is well-known that this type of gradient bound implies the stochastic completeness of $P_t$. More precisely, we can adapt an argument of Bakry. Let $f,g \in C^\infty_0(\mathbb M)$, we have

$\int_{\mathbb M} (P_t f -f) g d\mu = \int_0^t \int_{\mathbb M}\left( \frac{\partial}{\partial s} P_s f \right) g d\mu ds$
$=\frac{1}{2} \int_0^t \int_{\mathbb M}\left(\Delta_{\mathcal{H}} P_s f \right) g d\mu ds$
$- \int_0^t \int_{\mathbb M} \langle \nabla P_s f , \nabla g\rangle_\mathcal{H} d\mu ds.$

By means of Cauchy-Schwarz inequality we
find
$\left| \int_{\mathbb M} (P_t f -f) g d\mu \right| \le 2 \left(\int_0^t e^{\frac{C_\varepsilon}{2} s} ds\right) \| df \|_{\varepsilon,\infty} \int_{\mathbb M}\| \nabla_\mathcal{H} g \|^{\frac{1}{2}}d\mu.$

We now apply the previous inequality with $f = h_n$, where $h_n$ is an increasing sequence in $C_0^\infty(\mathbb M)$, $0 \le h_n \le 1$, such that $h_n\nearrow 1$ on $\mathbb{M}$, and $||\Gamma (h_n)||_{\infty} \to 0$, as $n\to \infty$.

By monotone convergence theorem we have $P_t h_k(x)\nearrow P_t 1(x)$ for every $x\in \mathbb M$. We conclude that the
left-hand side converges to $\int_{\mathbb M} (P_t 1 -1) g d\mu$. Since the right-hand side converges to zero, we reach the conclusion

$\int_{\mathbb M} (P_t 1 -1) g d\mu=0,\ \ \ g\in C^\infty_0(\mathbb M).$
Since it is true for every $g\in C^\infty_0(\mathbb M)$, it follows that $P_t 1 =1$.

$\square$

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