Bott Residue Formula
We make the same assumptions as in previous section. Let $i_1,\cdots,i_k$ be $k$ positive even integers. For any $p\in\mathrm{zero}(K)$ and $1\leq j \leq k$, set
\[
\lambda^{i_j}(p)=\lambda_1^{i_j}+\cdots+\lambda_l^{i_j}.
\]By following theorem, we reduce the computation of characteristic numbers of $TM$ to quantities on $\mathrm{zero}(K)$.
定理 1. If $i_1+\cdots+i_k=l$, then
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]
=(2\pi\sqrt{-1})^l\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)},
\]where $R^{TM}$ is the curvature of the Levi-Civita connection $\nabla^{TM}$. And, if $i_1+\cdots+i_k<l$, then
\[
\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)}=0.
\]
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]
=(2\pi\sqrt{-1})^l\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)},
\]where $R^{TM}$ is the curvature of the Levi-Civita connection $\nabla^{TM}$. And, if $i_1+\cdots+i_k<l$, then
\[
\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)}=0.
\]
证明 . We define a operator
\[
L_K=\nabla^{TM}_K-\left.\mathcal{L}_K\right|_{\Gamma(TM)}.
\]Obviously, for any $Y\in \Gamma(TM)$
\[
L_KY=\nabla^{TM}_KY-[K,L]=\left(\nabla^{TM}K\right)(Y).
\]i.e., $L_K=\left(\nabla^{TM}K\right)\in\Gamma(\mathrm{End}(TM))=\Omega^0(M,\mathrm{End}(TM))$. We need to find a $\omega\in\Omega^*(M)$ satisfied $\rd_K\omega=0$ and
\[
\int_{M}\omega=\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right].
\]
Of course, it is necessary that we can calculate $\int_{M}\omega$.
For any integer $h$, if we have $\rd_K\tr\left[(R^{TM}+L_K)^h\right]=0$. Since
\[
\left[\tr\left[\left(R^{TM}+L_K\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}+L_K\right)^{i_k}\right]\right]^{[\mathrm{top}]}=\tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right],
\]it means
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]=\int_{M}\tr\left[\left(R^{TM}+L_K\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}+L_K\right].
\]
By the Berline-Vergne localization formula, one can easily obtain
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]=(2\pi)^l\sum_{p\in\mathrm{zero}(K)}\frac{\tr\left[\left(L_K(p)\right)^{i_1}\right]\cdots\tr \left[\left(L_K(p)\right)^{i_k}\right]}{\lambda(p)}
\]On the local coordinate system $\left(U_p,(x^1,\cdots,x^{2l})\right)$ from the previous section,
\[
K=\sum_{i=1}^{l}\lambda_i\left(x^{2i}\frac{\partial}{\partial x^{2i-1}}-x^{2i-1}\frac{\partial}{\partial x^{2i}}\right).
\]And, we know $L_K=\nabla^{TM}K$ is a anti-symmetry operator. So on $U_p$,
\[
L_K(p)=\begin{pmatrix}
0 & \lambda_1 & & &\\
-\lambda_1 & 0 & & &\\
& & \ddots & &\\
& & & 0 & \lambda_l\\
& & & -\lambda_l & 0
\end{pmatrix},
\]then $\left(L_K(p)\right)^2=-\mathrm{diag}\left\{\lambda_1^2,\lambda_1^2,\cdots,\lambda_l^2,\lambda_l^2\right\}$.
Thus, for each $1\leq j \leq k$, $\tr \left[\left(L_K(p)\right)^{i_j}\right]=2(-1)^{i_j/2}\lambda^{i_j}(p)$. From this result,
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]
=(2\pi\sqrt{-1})^l\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)}.
\]
Otherwise, we have
\begin{align*}
(\rd+i_K)\tr\left[\left(R^{TM}+L_K\right)^h\right]=&\rd\tr\left[\left(R^{TM}+L_K\right)^h\right]+i_K\tr\left[\left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM},\left(R^{TM}+L_K\right)^h\right]+\tr\left[i_K\left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM},\left(R^{TM}+L_K\right)^h\right]+\tr\left[i_K, \left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM}+i_K,\left(R^{TM}+L_K\right)^h\right];
\end{align*}
\begin{align*}
\left(\nabla^{TM}+i_K\right)^2=&\left(\nabla\right)^2+(i_K)^2+\nabla^{TM}\circ i_K+ i_K\circ \nabla^{TM}\\
=&R^{TM}+0+\nabla^{TM}\circ i_K+ i_K\circ \nabla^{TM}\\
=&R^{TM}+0+\nabla^{TM}\circ i_K+ i_K\left(\nabla^{TM}\right)-\nabla^{TM}\circ i_K\\
=&R^{TM}+\nabla^{TM}_K\\
=&R^{TM}+L_K+\mathcal{L}_K;
\end{align*}
i.e.,$R^{TM}+L_K=\left(\nabla^{TM}+i_K\right)^2-\mathcal{L}_K$. By Bianchi identity, one can obtain
\begin{align*}
\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=&\left[\nabla^{TM}+i_K,\left(\nabla^{TM}+i_K\right)^2-\mathcal{L}_K\right]\\
=&-\left[\nabla^{TM}+i_K,L_K\right].
\end{align*}
Since both $\nabla^{TM}$ and $K$ are $S^1$-invariant, one has
\[
\left[\nabla^{TM},\mathcal{L}_K\right]=0,\quad \left[i_K,\mathcal{L}_K\right]=0.
\]Hence $\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=0$. It shows
\[
\rd_K\tr\left[\left( R^{TM}+L_K\right)^h\right]=\tr\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=0.
\]
\[
L_K=\nabla^{TM}_K-\left.\mathcal{L}_K\right|_{\Gamma(TM)}.
\]Obviously, for any $Y\in \Gamma(TM)$
\[
L_KY=\nabla^{TM}_KY-[K,L]=\left(\nabla^{TM}K\right)(Y).
\]i.e., $L_K=\left(\nabla^{TM}K\right)\in\Gamma(\mathrm{End}(TM))=\Omega^0(M,\mathrm{End}(TM))$. We need to find a $\omega\in\Omega^*(M)$ satisfied $\rd_K\omega=0$ and
\[
\int_{M}\omega=\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right].
\]
Of course, it is necessary that we can calculate $\int_{M}\omega$.
For any integer $h$, if we have $\rd_K\tr\left[(R^{TM}+L_K)^h\right]=0$. Since
\[
\left[\tr\left[\left(R^{TM}+L_K\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}+L_K\right)^{i_k}\right]\right]^{[\mathrm{top}]}=\tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right],
\]it means
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]=\int_{M}\tr\left[\left(R^{TM}+L_K\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}+L_K\right].
\]
By the Berline-Vergne localization formula, one can easily obtain
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]=(2\pi)^l\sum_{p\in\mathrm{zero}(K)}\frac{\tr\left[\left(L_K(p)\right)^{i_1}\right]\cdots\tr \left[\left(L_K(p)\right)^{i_k}\right]}{\lambda(p)}
\]On the local coordinate system $\left(U_p,(x^1,\cdots,x^{2l})\right)$ from the previous section,
\[
K=\sum_{i=1}^{l}\lambda_i\left(x^{2i}\frac{\partial}{\partial x^{2i-1}}-x^{2i-1}\frac{\partial}{\partial x^{2i}}\right).
\]And, we know $L_K=\nabla^{TM}K$ is a anti-symmetry operator. So on $U_p$,
\[
L_K(p)=\begin{pmatrix}
0 & \lambda_1 & & &\\
-\lambda_1 & 0 & & &\\
& & \ddots & &\\
& & & 0 & \lambda_l\\
& & & -\lambda_l & 0
\end{pmatrix},
\]then $\left(L_K(p)\right)^2=-\mathrm{diag}\left\{\lambda_1^2,\lambda_1^2,\cdots,\lambda_l^2,\lambda_l^2\right\}$.
Thus, for each $1\leq j \leq k$, $\tr \left[\left(L_K(p)\right)^{i_j}\right]=2(-1)^{i_j/2}\lambda^{i_j}(p)$. From this result,
\[
\int_M \tr\left[\left(R^{TM}\right)^{i_1}\right]\cdots\tr \left[\left(R^{TM}\right)^{i_k}\right]
=(2\pi\sqrt{-1})^l\sum_{p\in\mathrm{zero}(K)}\frac{2^k\lambda^{i_1}(p)\cdots\lambda^{i_k}(p)}{\lambda(p)}.
\]
Otherwise, we have
\begin{align*}
(\rd+i_K)\tr\left[\left(R^{TM}+L_K\right)^h\right]=&\rd\tr\left[\left(R^{TM}+L_K\right)^h\right]+i_K\tr\left[\left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM},\left(R^{TM}+L_K\right)^h\right]+\tr\left[i_K\left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM},\left(R^{TM}+L_K\right)^h\right]+\tr\left[i_K, \left(R^{TM}+L_K\right)^h\right]\\
=&\tr\left[\nabla^{TM}+i_K,\left(R^{TM}+L_K\right)^h\right];
\end{align*}
\begin{align*}
\left(\nabla^{TM}+i_K\right)^2=&\left(\nabla\right)^2+(i_K)^2+\nabla^{TM}\circ i_K+ i_K\circ \nabla^{TM}\\
=&R^{TM}+0+\nabla^{TM}\circ i_K+ i_K\circ \nabla^{TM}\\
=&R^{TM}+0+\nabla^{TM}\circ i_K+ i_K\left(\nabla^{TM}\right)-\nabla^{TM}\circ i_K\\
=&R^{TM}+\nabla^{TM}_K\\
=&R^{TM}+L_K+\mathcal{L}_K;
\end{align*}
i.e.,$R^{TM}+L_K=\left(\nabla^{TM}+i_K\right)^2-\mathcal{L}_K$. By Bianchi identity, one can obtain
\begin{align*}
\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=&\left[\nabla^{TM}+i_K,\left(\nabla^{TM}+i_K\right)^2-\mathcal{L}_K\right]\\
=&-\left[\nabla^{TM}+i_K,L_K\right].
\end{align*}
Since both $\nabla^{TM}$ and $K$ are $S^1$-invariant, one has
\[
\left[\nabla^{TM},\mathcal{L}_K\right]=0,\quad \left[i_K,\mathcal{L}_K\right]=0.
\]Hence $\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=0$. It shows
\[
\rd_K\tr\left[\left( R^{TM}+L_K\right)^h\right]=\tr\left[\nabla^{TM}+i_K, R^{TM}+L_K \right]=0.
\]
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