Adiabatic Limit and the Bott Connection

Background
Let $M$ be an even dimensional oriented closed spin manifold. We define the $\hat{A}$-genus of $M$, denote by $\hat{A}(M)$, by
\[
\hat{A}(M)=\big\langle \hat{A}(TM),[M] \big\rangle= \int_M\hat{A}(TM,\nabla^{TM})\in\Z.
\]
If there exists a Riemannian metric $g^{TM}$ such that the related section curvature $K>0$, then $\hat{A}(M)=0$.

证明 . Since
\[
D^2=-\triangle+\frac{K}{4}
\]
is a non-negative defined operator and $-\triangle$, $\frac{K}{4}$ too. Then if $D^2s=0$ for any $s\in C^{\infty}(M)$, \[
0=\langle\langle D^2s,s\rangle\rangle=\langle\langle Ds,Ds \rangle\rangle=\langle\langle-\triangle s,s \rangle\rangle+\langle\langle \frac{K}{4}s,s \rangle\rangle \geq0.
\]
Thus $s=0$.

In fact, if we change the condition $K>0$ into $K\geq0$, the result is still right. Because there is an other Riemannian metric $\widetilde{g}^{TM}$ such that the related section curvature $\widetilde{K}>0$ by the theory of Yamabe problem.
We consider the above question under a weak condition. If the integral manifold of any integrable sub-bundle of $TM$ is a spin manifold, could I still obtain $\hat{A}(M)=0$? Recently, Weiping Zang settles this question when the integral manifold of any integrable sub-bundle of $TM$ is almost riemannian. This section comes from his related research.

Adiabatic Limit
On $F^{\bot}$, there are two connection $\nabla^{F^{\bot}}$, $\widetilde{\nabla}^{F^{\bot}}$. Obviously, the connection $\nabla^{F^{\bot}}$ is preserved metric, connection $\widetilde{\nabla}^{F^{\bot}}$ is not preserved metric by definition1.13(i). In fact, by passing $g^{TM}$ to its adiabatic limit, one sees that underlying limit of $\nabla^{F^{\bot}}$ and the Bott connection $\widetilde{\nabla}^{F^{\bot}}$ are ultimately related.

For any $\epsilon>0$, let $g^{TM,\epsilon}$ be the metric on $TM$ defined by
\[
g^{TM,\epsilon}=g^{F}\oplus \frac{1}{\epsilon}g^{F^{\bot}}.
\]
Let $\nabla^{TM,\epsilon}$ be the Levi-Civita connection of $g^{TM,\epsilon}$. Let $\nabla^{F,\epsilon}$ (resp. $\nabla^{F^{\bot},\epsilon}$) be the restriction of $\nabla^{TM,\epsilon}$ to $F$ (resp. $F^{\bot}$). The process of taking the limit $\epsilon\rightarrow0$ is called taking the adiabatic limit.

In fact, as $\epsilon\rightarrow0 $ the distance between leafs of foliation foliated by $F$ in direction of $F^{\bot}$ increases gradually. We will examine the behavior of $\nabla^{f^{\bot},\epsilon}$ as $\epsilon$. Let $\widetilde{\nabla}^{f^{\bot},*}$ be the connection on $F{\bot}$ which is dual to $\widetilde{\nabla}^{F^{\bot}}$. That is , for any sections $U,V$$\in\Gamma(F^{\bot})$,
\[
d\langle U,V \rangle_{g^{TM}}=\left\langle \widetilde{\nabla}^{F^{\bot}}U,V\right\rangle_{g^{TM}} + \left\langle \widetilde{\nabla}^{F^{\bot},*}U,V\right\rangle_{g^{TM}}.
\]

练习 1. Validate $\widetilde{\nabla}^{F^{\bot},*}$ is a connection on $F^{\bot}$.

Set
\[
\omega^{F^{\bot}}= \widetilde{\nabla}^{F^{\bot},*}-\widetilde{\nabla}^{F^{\bot}};\quad
\hat{\nabla}^{F^{\bot}}=\widetilde{\nabla}^{F^{\bot}}+\frac{\omega^{F^{\bot}}}{2}.
\]
One verifies easily that the connection $\hat{\nabla}^{F^{\bot}}$ preserves $g^{F^{\bot}}$ by the definition of dual connection, and $\widetilde{\nabla}^{F^{\bot}}$ preserves $g^{F^{\bot}}$ when for any $X\in\Gamma(F)$, $\omega^{F^{\bot}}(X)=0$.
定理 1. For any smooth section $X\in\Gamma(F)$. one has
\[
\lim_{\epsilon\rightarrow0}\nabla^{F^{\bot},\epsilon}_{X}=\hat{\nabla}^{F^{\bot}}_{X}.
\]

证明 . We only need to prove that for any $U,V\in\Gamma(F^{\bot})$,
\[
\left\langle \nabla^{F^{\bot},\epsilon}_{X}U,V\right\rangle_{g^{TM}}\rightarrow \left\langle\hat{\nabla}^{F^{\bot}}_{X}U,V\right\rangle_{g^{TM}},\quad as\quad \epsilon\rightarrow0.
\]
By the definition of Bott connection and $\hat{\nabla}^{F^{\bot}}=\frac{1}{2}\left(\widetilde{\nabla}^{F^{\bot},*}+\widetilde{\nabla}^{F^{\bot}}\right)$, we have
\begin{align*}
&\left\langle \nabla^{F^{\bot},\epsilon}_{X}U,V \right\rangle_{g^{TM,\epsilon}}\\
=&\frac{1}{2}\Big\{ X\langle U,V \rangle_{g^{TM,\epsilon}} +U\langle X,V\rangle_{g^{TM,\epsilon}} -V\langle X,U\rangle_{g^{TM,\epsilon}} \\
&+\langle [X,U],V\rangle_{g^{TM,\epsilon}}-\langle[U,V],X\rangle_{g^{TM,\epsilon}}+\langle[V,X],U\rangle_{g^{TM,\epsilon}}\Big\}\\
=&\frac{1}{2\epsilon}\Big\{ X\langle U,V\rangle_{g^{TM}} + \langle [X,U],V\rangle_{g^{TM}} – \langle[X,V],U\rangle_{g^{TM}}\Big\}-\frac{1}{2}\langle[U,V],X\rangle_{g^{TM}}\\
=&\frac{1}{2\epsilon}\Big\{ X\langle U,V\rangle_{g^{TM}} + \langle p^{\bot}[X,U],V\rangle_{g^{TM}} – \langle p^{\bot}[X,V],U\rangle_{g^{TM}} \Big\}-\frac{1}{2}\langle[U,V],X\rangle_{g^{TM}}\\
=&\frac{1}{2\epsilon}\Big\{ X\langle U,V\rangle_{g^{TM}} + \langle \widetilde{\nabla}^{F^{\bot}}_XU,V\rangle_{g^{TM}} – \langle \widetilde{\nabla}^{F^{\bot}}_XV,U\rangle_{g^{TM}} \Big\}-\frac{1}{2}\langle[U,V],X\rangle_{g^{TM}}\\
=&\frac{1}{2\epsilon}\Big\{ \langle \widetilde{\nabla}^{F^{\bot}}_XU,V\rangle_{g^{TM}} + \langle \widetilde{\nabla}^{F^{\bot},*}_XV,U\rangle_{g^{TM}} \Big\}-\frac{1}{2}\langle[U,V],X\rangle_{g^{TM}}\\
=&\frac{1}{\epsilon} \left\langle \hat{\nabla}^{F^{\bot}}_XU,V\right\rangle_{g^{TM}} -\frac{1}{2}\langle[U,V],X\rangle_{g^{TM}},
\end{align*}
and
\[
\left\langle \nabla^{F^{\bot},\epsilon}_{X}U,V \right\rangle_{g^{TM,\epsilon}}=\frac{1}{\epsilon}\left\langle \nabla^{F^{\bot},\epsilon}_{X}U,V \right\rangle_{g^{TM}}.
\]
Hence,
\[
\left\langle \nabla^{F^{\bot},\epsilon}_{X}U,V \right\rangle_{g^{TM}}=\left\langle \hat{\nabla}^{F^{\bot}}_XU,V\right\rangle_{g^{TM}} -\frac{1}{2}\epsilon\langle[U,V],X\rangle_{g^{TM}}.
\]
This ends the proof.

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