Bott and Duistermaat-Herckman Formulas

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 $A$. 1. Berline-Vergne Localization Formula 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 $a,b\in S^1$, $P\in M$

  1. $P1=\phi(P,1)=P$;
  2. $(Pa)b=P(ab)$.
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
\[
\tilde{g}^{TM}=\int_{a\in S^1}a^*g^{TM}.
\]
练习 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 $f\in C^{\infty}(M)$,
\[
K_{_P}f=\left.\frac{\rd}{\rd \epsilon}f\left(P\exp(\epsilon t)\right)\right|_{\epsilon=0}, \quad\text{ for any }P\in M.
\]

练习 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 Cartan homotopy formula 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 $\mathcal{L}_K$-invariant form. 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 $S^1$ equivariant cohomology of $M$.

Now, consider any element $\omega\in \Omega^*(M)$. We say $\omega$ is $\rd_K$-closed 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 $K$.

练习 3. Prove that
\[
\mathrm{zero}\{K\}=\emptyset\Leftrightarrow\text{ the fixed point set of }M\text{ under }S^1\text{-action is empty}. \]

命题 1. If $K$ has no zeros on $M$, then for any $\omega\in \Omega^*(M)$ which is $\rd_K$-closed, one has $\int_M\omega=0$.

证明 . 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.

定理 2. If $\mathrm{zero}(K)$ is discrete, then for any $\rd _K$-closed form $\omega\in\Omega^*(M)$, one has
\[
\int_M\omega=(2\pi)^l\sum_{p\in\mathrm{zero}(K)}\frac{\omega^{[0]}(p)}{\lambda(p)}.
\]

证明 . 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.

One Comment

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

发表评论