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
\n\\[
\n\\lambda^{i_j}(p)=\\lambda_1^{i_j}+\\cdots+\\lambda_l^{i_j}.
\n\\]By following theorem, we reduce the computation of characteristic numbers of $TM$ to quantities on $\\mathrm{zero}(K)$.
\n
\n\\begin{thm}
\nIf $i_1+\\cdots+i_k=l$, then
\n\\[
\n\\int_M \\tr\\left[\\left(R^{TM}\\right)^{i_1}\\right]\\cdots\\tr \\left[\\left(R^{TM}\\right)^{i_k}\\right]
\n=(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)},
\n\\]where $R^{TM}$ is the curvature of the Levi-Civita connection $\\nabla^{TM}$. And, if $i_1+\\cdots+i_k<l$, then
\n\\[
\n\\sum_{p\\in\\mathrm{zero}(K)}\\frac{2^k\\lambda^{i_1}(p)\\cdots\\lambda^{i_k}(p)}{\\lambda(p)}=0.
\n\\]
\n\\end{thm}
\n\\begin{proof}
\nWe define a operator
\n\\[
\nL_K=\\nabla^{TM}_K-\\left.\\mathcal{L}_K\\right|_{\\Gamma(TM)}.
\n\\]Obviously, for any $Y\\in \\Gamma(TM)$
\n\\[
\nL_KY=\\nabla^{TM}_KY-[K,L]=\\left(\\nabla^{TM}K\\right)(Y).
\n\\]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
\n\\[
\n\\int_{M}\\omega=\\int_M \\tr\\left[\\left(R^{TM}\\right)^{i_1}\\right]\\cdots\\tr \\left[\\left(R^{TM}\\right)^{i_k}\\right].
\n\\]
\nOf course, it is necessary that we can calculate $\\int_{M}\\omega$.
\nFor any integer $h$, if we have $\\rd_K\\tr\\left[(R^{TM}+L_K)^h\\right]=0$. Since
\n\\[
\n\\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],
\n\\]it means
\n\\[
\n\\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].
\n\\]
\nBy the Berline-Vergne localization formula, one can easily obtain
\n\\[
\n\\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)}
\n\\]On the local coordinate system $\\left(U_p,(x^1,\\cdots,x^{2l})\\right)$ from the previous section,
\n\\[
\nK=\\sum_{i=1}^{l}\\lambda_i\\left(x^{2i}\\frac{\\partial}{\\partial x^{2i-1}}-x^{2i-1}\\frac{\\partial}{\\partial x^{2i}}\\right).
\n\\]And, we know $L_K=\\nabla^{TM}K$ is a anti-symmetry operator. So on $U_p$,
\n\\[
\nL_K(p)=\\begin{pmatrix}
\n 0 & \\lambda_1 & & &\\\\
\n -\\lambda_1 & 0 & & &\\\\
\n & & \\ddots & &\\\\
\n & & & 0 & \\lambda_l\\\\
\n & & & -\\lambda_l & 0
\n \\end{pmatrix},
\n\\]then $\\left(L_K(p)\\right)^2=-\\mathrm{diag}\\left\\{\\lambda_1^2,\\lambda_1^2,\\cdots,\\lambda_l^2,\\lambda_l^2\\right\\}$.
\nThus, 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,
\n\\[
\n\\int_M \\tr\\left[\\left(R^{TM}\\right)^{i_1}\\right]\\cdots\\tr \\left[\\left(R^{TM}\\right)^{i_k}\\right]
\n=(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)}.
\n\\]
\nOtherwise, we have
\n\\begin{align*}
\n(\\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]\\\\
\n=&\\tr\\left[\\nabla^{TM},\\left(R^{TM}+L_K\\right)^h\\right]+\\tr\\left[i_K\\left(R^{TM}+L_K\\right)^h\\right]\\\\
\n=&\\tr\\left[\\nabla^{TM},\\left(R^{TM}+L_K\\right)^h\\right]+\\tr\\left[i_K, \\left(R^{TM}+L_K\\right)^h\\right]\\\\
\n=&\\tr\\left[\\nabla^{TM}+i_K,\\left(R^{TM}+L_K\\right)^h\\right];
\n\\end{align*}
\n\\begin{align*}
\n\\left(\\nabla^{TM}+i_K\\right)^2=&\\left(\\nabla\\right)^2+(i_K)^2+\\nabla^{TM}\\circ i_K+ i_K\\circ \\nabla^{TM}\\\\
\n=&R^{TM}+0+\\nabla^{TM}\\circ i_K+ i_K\\circ \\nabla^{TM}\\\\
\n=&R^{TM}+0+\\nabla^{TM}\\circ i_K+ i_K\\left(\\nabla^{TM}\\right)-\\nabla^{TM}\\circ i_K\\\\
\n=&R^{TM}+\\nabla^{TM}_K\\\\
\n=&R^{TM}+L_K+\\mathcal{L}_K;
\n\\end{align*}
\ni.e.,$R^{TM}+L_K=\\left(\\nabla^{TM}+i_K\\right)^2-\\mathcal{L}_K$. By Bianchi identity, one can obtain
\n\\begin{align*}
\n\\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]\\\\
\n=&-\\left[\\nabla^{TM}+i_K,L_K\\right].
\n\\end{align*}
\nSince both $\\nabla^{TM}$ and $K$ are $S^1$-invariant, one has
\n\\[
\n\\left[\\nabla^{TM},\\mathcal{L}_K\\right]=0,\\quad \\left[i_K,\\mathcal{L}_K\\right]=0.
\n\\]Hence $\\left[\\nabla^{TM}+i_K, R^{TM}+L_K \\right]=0$. It shows
\n\\[
\n\\rd_K\\tr\\left[\\left( R^{TM}+L_K\\right)^h\\right]=\\tr\\left[\\nabla^{TM}+i_K, R^{TM}+L_K \\right]=0.
\n\\]
\n\\end{proof}<\/p>\n","protected":false},"excerpt":{"rendered":"
We make the same assumptions as in previ …<\/p>\n