Adiabatic Limit and the Bott Connection

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Background
Let $M$ be an even dimensional oriented closed spin manifold. We define the 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

$\hat{A}(M)=0$ .

Proof . 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 $U,V$ $\in\Gamma(F^{\bot})$ ,

$d\langle U,V \rangle_{g^{TM}}=\left\langle \widetilde{\nabla}^{F^{\bot}}U,V\right\rangle_{g^{TM}} + \left\langle \widetilde{\nabla}^{F^{\bot},*}U,V\right\rangle_{g^{TM}}.$

Exercise 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

$X\in\Gamma(F)$ ,

$\omega^{F^{\bot}}(X)=0$ .

Theorem 1. For any smooth section

$X\in\Gamma(F)$ . one has

$\lim_{\epsilon\rightarrow0}\nabla^{F^{\bot},\epsilon}_{X}=\hat{\nabla}^{F^{\bot}}_{X}.$

Proof . 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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