1 (15') 设 $R(X,Y):\ \calX(M)\to \calX(M)$ 为曲率, 求证:

(1)$R(X,Y)(fZ_1+gZ_2) =fR(X,Y)Z_1+gR(X,Y)Z_2$, $\forall\ X,Y,Z_1,Z_2\in \calX(M), f,h\in C^\infty (M)$;

(2)$R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$, $\forall\ X,Y,Z\in \calX(M)$.

证明:

(1) 回忆 $$\bex R(X,Y)=\n_X\n_Y-\n_Y\n_X-\n_{[X,Y]}, \eex$$ 我们有 $$\bex & &R(X,Y)(fZ)\\ & &=\n_X\n_Y(fZ)-\n_Y\n_X(fZ)-\n_{[X,Y]}(fZ)\\ & &=\n_X\sez{Yf\cdot Z+f\n_YZ} -\n_Y\sez{Xf\cdot Z+f\n_XZ} -\n_{[X,Y]}(fZ)\\ & &=\underline{XYf\cdot Z}+{\color{black}Yf\cdot\n_XZ} +{\color{red}Xf\cdot\n_YZ}+f\n_X\n_YZ\\ & &\quad-\underline{YXf\cdot Z}-{\color{red}Xf\cdot\n_YZ} -{\color{black}Yf\cdot\n_XZ}-f\n_Y\n_XZ\\ & &\quad-\underline{[X,Y]f\cdot Z}-f\n_{[X,Y]}Z\\ & &=fR(X,Y)Z, \eex$$ $$\bex & &R(X,Y)(Z_1+Z_2)\\ & &=\n_X\n_Y(Z_1+Z_2) -\n_Y\n_X(Z_1+Z_2) -\n_{[X,Y]}(Z_1+Z_2)\\ & &=R(X,Y)Z_1+R(X,Y)Z_2. \eex$$ 于是 $$\bex & &R(X,Y)(fZ_1+hZ_2)\\ & &=R(X,Y)(fZ_1)+R(X,Y)(hZ_2)\\ & &=fR(X,Y)Z_1+hR(X,Y)Z_2. \eex$$

(2) 由 $$\bex \ba{ccc} R(X,Y)Z=\n_X\n_YZ-\n_Y\n_XZ-\n_{[X,Y]}Z,\\ R(Y,Z)X=\n_Y\n_ZX-\n_Z\n_YX-\n_{[Y,Z]}X,\\ R(Z,X)Y=\n_Z\n_XY-\n_X\n_ZY-\n_{[Z,X]}Y \ea \eex$$ 即知 $$\bex & &R(X,Y)Z+R(Y,Z)X+R(Z,X)Y\\ & &=\n_X[Y,Z]-\n_{[Y,Z]}X +\n_Y[Z,X]-\n_{[Z,X]}Y\\ & &\quad+\n_Z[X,Y]-\n_{[X,Y]}Z\\ & &=[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\\ & &=0\quad\sex{Jacobi\mbox{ 恒等式}}. \eex$$  

2(10') 设 $V(t),\ J(t)$ 是沿最短测地线 $\gamma(t),\ t\in [0,1]$ 的向量场, 它们满足 $$\bex V(t)\perp \dot\gamma(t),\quad J(t)\perp \dot\gamma(t),\quad V(0)=J(0),\quad V(1)=J(1), \eex$$ 且 $J(t)$ 是 Jacobi 场, 求证: $$\bex I(J,J)\leq I(V,V), \eex$$ 其中 $I$ 为 $\gamma$ 上的指标形式.

证明: 由弧长第二变分公式知 $$\bex 0\leq I(V-J,V-J) =I(V,V)-2I(V,J)+I(J,J). \eex$$ 算出 $$\bex I(V,J) &=&\int_0^1\sez{\sef{\dot V,\dot J}-\sef{R(\dot \gamma,V)J,\dot \gamma}}\rd t\\ &=&\left.\sef{\dot J,V}\right|^1_0-\int_0^1 \sez{\sef{V,\ddot J}+\sef{R(J,\dot\gamma)\dot\gamma,V}}\rd t\\ &=&\left.\sef{\dot J,J}\right|^1_0 -\int_0^1 \sef{V,\ddot J+R(J,\dot \gamma)\dot\gamma}\rd t\\ &=&\left.\sef{\dot J,J}\right|^1_0; \eex$$ $$\bex I(J,J)&=&\int_0^1\sez{\sef{\dot J,\dot J}-\sef{R(\dot \gamma,J)J,\dot\gamma}}\rd t\\ &=&\left.\sef{\dot J,J}\right|^1_0-\int_0^1\sef{\ddot J+R(J,\dot\gamma)\dot\gamma,J}\rd t\\ &=&\left.\sef{\dot J,J}\right|^1_0. \eex$$ 我们得到 $$\bex 0\leq I(V,V)-2I(J,J)+2I(J,J) =I(V,V)-I(J,J), \eex$$ 而有结论.  

3(10') 设 $\gamma(t):\ (-\infty,+\infty)\to M$ 为一条测地直线, 相应地记 $$\bex \gamma_+=\gamma|_{[0,+\infty)},\quad \gamma_-=\gamma|_{(-\infty,0]} \eex$$ 及两 Busemann 函数 $$\bex B_{\gamma_+}(x)=\lim_{t\to+\infty}\sez{d(x,\gamma(t))-t}; \eex$$ $$\bex B_{\gamma_-}(x)=\lim_{t\to-\infty}\sez{d(x,\gamma(t))+t}. \eex$$ 求证: $$\bex B_{\gamma_+}+B_{\gamma_-}=0,\quad\mbox{在 } \gamma \mbox{ 上}; \eex$$ $$\bex B_{\gamma_+}+B_{\gamma_-}\geq 0,\quad\mbox{在 } M \mbox{ 上}. \eex$$

证明:

(1)设 $x\in\gamma$, 则 $x=\gamma(t_0)$, 对某个 $t_0$. 而当 $t$ 充分大时, $$\bex \ba{cc} d(x,\gamma(t))-t =(t-t_0)-t=-t_0,\\ d(x,\gamma(-t))-t =(t_0-(-t))-t =t_0. \ea \eex$$ 于是 $$\bex B_{\gamma_+}(x) +B_{\gamma_-}(x) &=&\lim_{t\to+\infty} \sed{\sez{d(x,\gamma(t))-t} +\sez{d(x,\gamma(-t))-t}}\\ &=&-t_0+t_0\\ &=&0. \eex$$

(2)当 $x\in M$, 则于 $$\bex & &\sez{d(x,\gamma(t))-t} +\sez{d(x,\gamma(-t))-t}\\ & &=\sez{d(x,\gamma(t))+d(x,\gamma(-t))}-2t\\ & &=d(\gamma(t),\gamma(-t))-2t\\ & &\geq 2t-2t\\ & &=0 \eex$$ 两端令 $t\to+\infty$, 有 $$\bex B_{\gamma_+}+B_{\gamma_-}\geq 0. \eex$$  

4(15') 设 $M$ 为紧流形, 再设 $g_{ij}(t)$ 满足 Ricci 流, 且 $f(t),\tau(t)$ 满足 $$\bex \frac{\p }{\p t}f=-\lap f+\sev{\n f}^2-R+\frac{n}{2\tau},\quad \frac{\p }{\p t}\tau =-1. \eex$$ 求证:

(1) $$\bex \frac{\rd}{\rd t}\int_M \sex{4\pi \tau}^{-\frac{n}{2}} e^{-f}\, \rd vol_{g_{ij}}=0; \eex$$

(2) $$\bex & &\frac{\rd }{\rd x}\int_M \sez{\tau \sex{R+\sev{\n f}^2} +f-n}(4\pi^\tau)^{-\frac{n}{2}} e^{-f}\,\rd vol_{g_{ij}}\\ & &=\int_M 2\tau \sev{R_{ij}+\n_i\n_j f-\frac{1}{2\tau}g_{ij}}^2 (4\pi \tau)^{-\frac{n}{2}} e^{-f}\,\rd vol_{g_{ij}}. \eex$$

证明:

(1)由 $$\bex \frac{\p}{\p t}g &=&\frac{\p}{\p t}\sev{\ba{ccc} g_{11}&\cdots&g_{1n}\\ \vdots&\ddots&\vdots\\ g_{n1}&\cdots&g_{nn} \ea}\\ &=&\sum_{j=1}^n \sev{\ba{ccccc} g_{11}&\cdots&\frac{\p}{\p t}g_{1j}&\cdots&g_{1n}\\ \vdots&&\vdots&&\vdots\\ g_{n1}&\cdots&\frac{\p}{\p t}g_{nj}&\cdots&g_{nn} \ea}\\ &=&\sum_{j=1}^n \sev{\ba{ccccc} g_{11}&\cdots&-2R_{1j}&\cdots&g_{1n}\\ \vdots&&\vdots&&\vdots\\ g_{n1}&\cdots&-2R_{nj}&\cdots&g_{nn} \ea}\\ &=&-2\sum_{j=1}^n\sez{\sum_{i=1}^n (gg^{ij})R_{ij}}\quad \sex{g^{ij}g_{jk}=\delta_{ik}}\\ &=&-2gR \eex$$ 及单位分解知 $$\bex & &\frac{\rd}{\rd t}\int_M \sex{4\pi \tau}^{-\frac{n}{2}} e^{-f}\, \rd vol_{g_{ij}}\\ &=&\int_M (4\pi)^{-\frac{n}{2}} \left[-\frac{n}{2}\tau^{-\frac{n}{2}-1}\frac{\p \tau}{\p t} e^{-f}\sqrt{g} +\tau^{-\frac{n}{2}}e^{-f}\sex{-\frac{\p f}{\p t}}\sqrt{g}\right.\\ & &\left. +\tau^{-\frac{n}{2}}e^{-f}\frac{1}{2}g^{-\frac{1}{2}}\frac{\p g}{\p t} \right]\,\rd x\\ &=&\int_M (4\pi)^{-\frac{n}{2}} \left[ \frac{n}{2}\tau^{-\frac{n}{2}-1} e^{-f}\sqrt{g} \right.\\ & &\left. +\tau^{-\frac{n}{2}} e^{-f}\sex{\lap f-\sev{\n f}^2+R-\frac{n}{2\tau}}\sqrt{g} -\tau^{-\frac{n}{2}} e^{-f}R\sqrt{g} \right]\,\rd x\\ & &=\int_M \sex{4\pi\tau}^{-\frac{n}{2}} e^{-f}\sex{\lap f-\sev{\n f}^2}\sqrt{g}\,\rd x\\ &=&-\int_M\sex{4\pi\tau}^{-\frac{n}{2}} \lap\sex{e^{-f}}\,\rd vol_{g_{ij}}\\ &=&0\quad\sex{\tau\mbox{ 为参数, 由 } Stokes\mbox{ 公式}}. \eex$$

(2)请参考 [H.D. Cao, X.P. Zhu, A complete proof of the Poincar\'e and geometrization conjectures --- application of the Hamilton -- Perelman theory of the Ricci flow, Asian J. Math., 10 (2006), no. 2, 165—492] 第206 页.