Bott and Duistermaat-Herckman Formulas

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In Chapter One we have defined characteristic classes and numbers. A natural question is hoe to compute these characteristic numbers. Let $\omega$ be a characteristic form on an even dimensional smooth closed oriented manifold $M$ . If
$\omega=\omega_{[1]}+\omega_{[2]}+\cdots+\omega_{[\dim M]},\quad \omega_{[i]}\in\Omega^{i}(M), i=1,\cdots,\dim M,$ then the characteristic number associated $\omega$ is defined by $\int_{M}\omega=\int_M\omega_{[\dim M]}$ . The Bott’s result shows that
$\int_M\omega_{[\dim M]}=\sum_{p\in A}\mu(P),$

where $A$ is the fixed point set of $M$ under the lie group $S^1$ action, $\mu$ is a function on $A$ . Let $M$ be an even dimensional smooth closed oriented manifold. We assume that $M$ admits an $S^1$ -(right)action. On $M$ , $S^1$ -(right)action is defined by a smooth map
$\phi:~M\times S^1\rightarrow M, \quad \phi(P,a)=Pa, \text{for any }a\in S^1$ such that for any $a,b\in S^1$ , $P\in M$

$P1=\phi(P,1)=P$ ;
$(Pa)b=P(ab)$ .

One easily verifies the map

$a:P\in M\rightarrow a(P)=Pa$ is a diffeomorphism for any

$a\in S^1$ . We claim a riemannian metric

$g^{TM}$ is

$S^1$ -invariant, if for any

$a\in S^1$ always have

$a^*g^{TM}=g^{TM}$ . Since

$S^1$ is a compact Lie group, there always exists a

$S^1$ -invariant metric on

$M$ . In fact, if

$g^{TM}$ is general riemannian metric, the following metric is

$S^1$ -invariant

$\tilde{g}^{TM}=\int_{a\in S^1}a^*g^{TM}.$

Exercise 1. Please verify

$\tilde{g}^{TM}$ is

$S^1$ -invariant.

Then we can assume that there exists a

$S^1$ -invariant riemannian metric

$g^{TM}$ without loss of generally.

For any $P\in M$ , the map $P:~S^1\rightarrow M$ , $P(a)=Pa$ , gives a curve on $M$ . By $S^1$ is a Lie group, using the exponent map this curve can be parameterize. Let $t\in Lie(S^1)$ be a generator of the Lie algebra of $S^1$ . then the above curve is
$\gamma_{_P}(\epsilon)=\phi(P,\exp(\epsilon t))=P\exp(\epsilon t).$ Take $K(P)=\gamma’_{_P}(0)=(\gamma_{_P})_*\frac{\partial}{\partial \epsilon}|_{\epsilon=0}$ . Then we obtain a vector field $K$ on $M$ , and for any $f\in C^{\infty}(M)$ ,
$K_{_P}f=\left.\frac{\rd}{\rd \epsilon}f\left(P\exp(\epsilon t)\right)\right|_{\epsilon=0}, \quad\text{ for any }P\in M.$

Exercise 2. Prove the integral curve crossing

$P$ of

$K$ is

$\gamma_{_P}(\epsilon)$ .

By the

$S^1$ -action preserves

$g^{TM}$ ,

$K$ is a Killing vector field on

$M$ (see[Chen, Li. Rie.Geo., Beijing university]). Let

$\nabla^E$ be Levi-Civita connection associated

$g^{TM}$ . Then for any

$X,Y\in \Gamma(TM)$ ,

$\begin{equation}\label{eq:2-1} \left\langle \nabla^{TM}_XK,Y \right\rangle + \left\langle \nabla^{TM}_YK,X\right\rangle=0, \end{equation}$ i.e.,

$\left\langle \left(\nabla^{TM}K\right)(X),Y \right\rangle + \left\langle \left(\nabla^{TM}K\right)(Y),X\right\rangle=0.$ It means

$\nabla^{TM}K$ is a antisymmetric operator on

$\Gamma(TM)$ . Now we prove this fact. Let

$\mathcal{L}_K$ denote the Lie derivative of

$K$ on

$\Gamma(TM)$ . Since the

$S^1$ -action preserves

$g^{TM}$ and the integral curve of

$K$ is

$\gamma_{_P}$ ,

$\mathcal{L}_K$ also preserves

$g^{TM}$ , i.e.,

$\mathcal{L}_Kg^{TM}=0$ . For any

$X,Y\in \Gamma(TM)$ , one has

$\begin{align*} \left\langle \nabla^{TM}_XK,Y\right\rangle+\left\langle \nabla^{TM}_YK,X\right\rangle =&\left\langle [X,K]+ \nabla^{TM}_KX,Y\right\rangle +\left\langle X , [Y,K]+\nabla^{TM}_KY \right\rangle \\ =&\left\langle -\mathcal{L}_KX,Y \right\rangle +\left\langle X,-\mathcal{L}_KY \right\rangle +K\left\langle X,Y \right\rangle \\ =&\left\langle -\mathcal{L}_KX,Y \right\rangle +\left\langle X,-\mathcal{L}_KY \right\rangle +\mathcal{L}_K\left\langle X,Y\right\rangle \\ =&\left(\mathcal{L}_Kg^{TM}\right)(X,Y)=0. \end{align*}$
In fact,

$K$ is a Killing vector field if and only if (1) holds.

We still denote the Lei derivative of $K$ on $\Omega^*(M)$ by $\mathcal{L}_K$ . The following on $\Omega^*{M}$ is well-known,
$\mathcal{L}_K=\rd i_{_K}+i_{_K}\rd,$ where $i_{_K}:~\Omega^*{M}\rightarrow \Omega^{*-1}{M}$ ,
$i_{_K}(\omega_1\wedge\cdots\wedge\omega_k)=\sum_{j=1}^k(-1)^{j-1}\omega_1\wedge\cdots\widehat{\omega}_j\wedge\cdots\wedge\omega_k .$ Let $\Omega^*_K(M)=\left\{\omega\in \Omega^*_K(M):\mathcal{L}_K\omega=0 \right\}$ be the subspace of . Set
$\rd _K=\rd +i_{_k}: \Omega^*(M)\rightarrow \Omega^*(M).$ Then, by $\rd^2=0$ and $i_{_K}^2=0$ one has
$\rd^2_{K}=\rd i_{_K}+i_{_K}\rd=\mathcal{L}_K.$ Thus $\rd_K$ preserves $\Omega^*_K(M)$ and $\left.\rd_K^2\right|_{\Omega^*_K(M)}=0$ . The corresponding cohomology group
$H^*_K(M)=\frac{\ker \rd_{K}|_{\Omega^*_K(M)}}{\mathrm{Im} \rd_{K}|_{\Omega^*_K(M)}}$ is called the of $M$ .

Now, consider any element $\omega\in \Omega^*(M)$ . We say $\omega$ is if $\rd_K\omega=0$ . The equivariant localization formula duo to Berline-Vergne(or Atiyah-Bott) shows that the integration of a $\rd_K$ -closed differential form over $M$ can be localized to the zero set of the Killing vector field $K$ .

Exercise 3. Prove that

$\mathrm{zero}\{K\}=\emptyset\Leftrightarrow\text{ the fixed point set of }M\text{ under }S^1\text{-action is empty}.$

Proposition 1. If

$K$ has no zeros on

$M$ , then for any

$\omega\in \Omega^*(M)$ which is

$\rd_K$ -closed, one has

$\int_M\omega=0$ .

Proof . Let

$\theta\in\Omega^1(M)$ be the one form on

$M$ such that for any

$X\in\Gamma(TM)$ ,

$i_{_K}\theta=\langle X,K\rangle$ . By

$\mathcal{L}_K$ preserves

$g^{TM}$ , one has

$\mathcal{L}_K\big(\theta(X)\big)=\mathcal{L}_K\big(i_{_X}\theta\big)=\mathcal{L}_K\langle X,K\rangle.$ i.e.,

$\begin{align*} \left(\mathcal{L}_K\theta\right)(X)+\theta\left(\mathcal{L}_KX\right)=&\langle \mathcal{L}_KK,X \rangle+\langle \mathcal{L}_KX,K \rangle\\ =&i_{\mathcal{L}_KX}\theta=\theta\left(\mathcal{L}_KX\right). \end{align*}$
Thus

$\mathcal{L}_K\theta=0$ . one then sees that

$\rd_K\theta$ is

$\rd_K$ -closed.
By

$\omega$ is

$\rd_K$ -closed, for any

$T\geq0$ one has

$\begin{align*} \int_M\omega\exp(-T\rd_K\theta)=&\int_M\omega\left[1+\sum_{i=1}^{\infty}\frac{(-1)^i}{i!}T^i(\rd_K\theta)^i\right]\\ =&\int_M\omega + \int_M\omega\wedge\rd_K\left[\sum_{i=1}^{\infty}\frac{(-1)^i}{i!}T^i\theta\wedge(\rd_K\theta)^{i-1}\right]\\ =&\int_M\omega +(-1)^{\mathrm{deg}(\omega)} \int_M\rd_K\left[\omega\wedge\sum_{i=1}^{\infty}\frac{(-1)^i}{i!}T^i\theta\wedge(\rd_K\theta)^{i-1}\right]\\ =&\int_M\omega. \end{align*}$
Otherwise,

$\begin{align*} \int_M\omega\exp(-T\rd_k\theta)=&\int_M\omega\exp\left[-T(\rd\theta+i_K\theta)\right]\\ =&\int_M\omega\exp\left[-T(\rd\theta+|k|^2)\right]\\ =&\int_M\Big(\omega\exp(-T|K|^2)\Big)\left[\sum_{i=1}^{\dim M/2}\frac{(-1)^i}{i!}T^i(\rd\theta)^i \right]. \end{align*}$
And, as

$K$ has no zeros on

$M$ ,

$|K|$ has a positive lower bound

$\delta>0$ on

$M$ . By

$M$ is closed, one sees easily that

$\int_M\Big(\omega\exp(-T|K|^2)\Big)\left[\sum_{i=1}^{\dim M/2}\frac{(-1)^i}{i!}T^i(\rd\theta)^i\right]\rightarrow 0, \quad\text{ as }T\rightarrow0.$ Thus

$\int_M\omega=0.$

In the previous discussion we considered the case of

$\mathrm{zero}(K)=\emptyset$ . Now we assume the zero set of

$K$ is discrete.

For every $p\in \mathrm{zero}(K)$ , there is a small open neighborhood $U_p$ of $p$ and an oriented coordinate system $(x^1,\cdots,x^{2l})$ with $2l=\dim M$ such that we have
$\left.g^{TM}\right|_{U_p}=\left(\rd x^1\right)^2+\cdots+\left(\rd x^{2l}\right)^2$ and
$K|_{U_p}=\sum_{i=1}^{l}\lambda_i\left(x^{2i}\frac{\partial}{\partial x^{2i-1}}-x^{2i-1}\frac{\partial}{\partial x^{2i}}\right)$ with each $\lambda_i\neq0$ for $1\leq i\leq l$ .

Set
$\lambda(p)=\lambda_1\cdots\lambda_l.$
In fact, the existence of $U_p$ is not trivial. But, I am very sorry about that I don’t understand about the existence.

Theorem 2. If

$\mathrm{zero}(K)$ is discrete, then for any

$\rd _K$ -closed form

$\omega\in\Omega^*(M)$ , one has

$\int_M\omega=(2\pi)^l\sum_{p\in\mathrm{zero}(K)}\frac{\omega^{[0]}(p)}{\lambda(p)}.$

Proof . Since

$M\setminus \cup_{p\in\mathrm{zero}(K)}U_p$ is closed manifold and

$K$ has no zeros on

$M\setminus \cup_{p\in\mathrm{zero}(K)}U_p$ , using the proposition 1 we have

$\int_{M\setminus \cup_{p\in\mathrm{zero}(K)}U_p}\omega=\int_{M\setminus \cup_{p\in\mathrm{zero}(K)}U_p}\omega\exp\left(-T\rd_K\theta\right)=0.$ Hence

$\begin{align*} \int_M\omega=&\int_{M\setminus \cup_{p\in\mathrm{zero}(K)}U_p}\omega\exp\left(-T\rd_K\theta\right) +\sum_{p\in\mathrm{zero}(K)}\int_{U_p}\omega\exp\left(-T\rd_K\theta\right)\\ =&\sum_{p\in\mathrm{zero}(K)}\int_{U_p}\omega\exp\left(-T\rd_K\theta\right). \end{align*}$
On

$U_p$ we have

$|K|^2=\sum_{i=1}^{l}\lambda_i^2\left[(x^{2i})^2+(x^{2i-1})^2\right],\quad \theta=\sum_{i=1}^{l}\lambda_i\left(x^{2i}\rd x^{2i-1}-x^{2i-1}\rd x^{2i}\right).$ Then,

$\begin{align*} &\int_{U_p}\omega\exp\left(-T\rd_K\theta\right)\\ &=\int_{U_p}\omega\exp\left(-T|K|^2-T\rd\theta\right)\\ &=\int_{U_p}\omega\exp\left[-T\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] \exp\left(2T\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)\\ &=\int_{U_p}\exp\left[-T\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] \omega\sum_{k=0}^{\infty}\left(2T\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)^k\\ &=\int_{U_p}\exp\left[-T\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] \sum_{j=0}^{l}\omega^{[2j]}\left(2T\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)^{l-j}\\ \end{align*}$
Make the change of the coordinate system

$(x^1,\cdots,x^{2l})\rightarrow\sqrt{T}(x^1,\cdots,x^{2l})$ . Above integral is rewritten

$\begin{align*} \int_{\sqrt{T}U_p}&\exp\left[-T\sum_{i=1}^{l}\lambda_i^2\frac{1}{T}\left((x^{2i-1})^2+(x^{2i})^2\right)\right]\\ &\sum_{j=0}^{l}T^{-j}\omega^{[2j]}\left(\frac{x}{\sqrt{T}}\right)2^{l-j}\left(T\sum_{i=1}^l\lambda_iT^{-1}\rd x^{2i-1}\wedge\rd x^{2i}\right)^{l-j}\\ =\int_{\sqrt{T}U_p}&\exp\left[-\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right]\\ &\sum_{j=0}^{l}T^{-j}\omega^{[2j]}\left(\frac{x}{\sqrt{T}}\right)2^{l-j}\left(\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)^{l-j} \end{align*}$
When

$0<j\leq l$ , using

$\int_{R}\exp{(-x^2)}\rd x=\sqrt{\pi}$ one can easily find that

$\begin{align*} \int_{\sqrt{T}U_p}&\exp\left[-\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] T^{-j}\omega^{[2j]}\left(\frac{x}{\sqrt{T}}\right)2^{l-j}\left(\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)^{l-j}\\ &\rightarrow 0 \quad as\quad T\rightarrow+\infty. \end{align*}$
Thus,

$\begin{align*} &\int_{U_p}\omega\exp\left(-T\rd_K\theta\right)\\ &=\int_{\sqrt{T}U_p}\exp\left[-\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] \omega^{[0]}\left(\frac{x}{\sqrt{T}}\right)2^{l}\left(\sum_{i=1}^l\lambda_i\rd x^{2i-1}\wedge\rd x^{2i}\right)^{l}\\ &=\int_{\sqrt{T}U_p}\exp\left[-\sum_{i=1}^{l}\lambda_i^2\left((x^{2i-1})^2+(x^{2i})^2\right)\right] \omega^{[0]}\left(\frac{x}{\sqrt{T}}\right)2^{l} \lambda(p)^{-1}\rd\lambda_1x^1\rd\lambda_1x^2\cdots\rd\lambda_lx^{2l}\\ &\rightarrow(2\pi)^l\lambda(p)^{-1}\omega^{[0]}(0)\quad as\quad T\rightarrow+\infty. \end{align*}$
this completes the proof.

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《“Bott and Duistermaat-Herckman Formulas”》有 1 条评论

游神

12/02/2012

建议你把Bismut的文章Localization formulas, superconnections, and the index theorem for families拿出来写写。

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