[家里蹲大学数学杂志]第050期2011年广州偏微分方程暑期班试题

[家里蹲大学数学杂志]第050期2011年广州偏微分方程暑期班试题

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

(1)$R(X,Y)(fZ_1+gZ_2) =fR(X,Y)Z_1+gR(X,Y)Z_2$, $forall X,Y,Z_1,Z_2in calX(M), f,hin C^infty (M)$;

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

证明:

(1) 回忆 $$bex R(X,Y)=n_Xn_Y-n_Yn_X-n_{[X,Y]}, eex$$ 我们有 $$bex & &R(X,Y)(fZ)\ & &=n_Xn_Y(fZ)-n_Yn_X(fZ)-n_{[X,Y]}(fZ)\ & &=n_Xsez{Yfcdot Z+fn_YZ} -n_Ysez{Xfcdot Z+fn_XZ} -n_{[X,Y]}(fZ)\ & &=underline{XYfcdot Z}+{color{black}Yfcdotn_XZ} +{color{red}Xfcdotn_YZ}+fn_Xn_YZ\ & &quad-underline{YXfcdot Z}-{color{red}Xfcdotn_YZ} -{color{black}Yfcdotn_XZ}-fn_Yn_XZ\ & &quad-underline{[X,Y]fcdot Z}-fn_{[X,Y]}Z\ & &=fR(X,Y)Z, eex$$ $$bex & &R(X,Y)(Z_1+Z_2)\ & &=n_Xn_Y(Z_1+Z_2) -n_Yn_X(Z_1+Z_2) -n_{[X,Y]}(Z_1+Z_2)\ & &=R(X,Y)Z_1+R(X,Y)Z_2. eex$$ 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 $$bex ba{ccc} R(X,Y)Z=n_Xn_YZ-n_Yn_XZ-n_{[X,Y]}Z,\ R(Y,Z)X=n_Yn_ZX-n_Zn_YX-n_{[Y,Z]}X,\ R(Z,X)Y=n_Zn_XY-n_Xn_ZY-n_{[Z,X]}Y ea eex$$ 即知 $$bex & &R(X,Y)Z+R(Y,Z)X+R(Z,X)Y\ & &=n_X[Y,Z]-n_{[Y,Z]}X +n_Y[Z,X]-n_{[Z,X]}Y\ & &quad+n_Z[X,Y]-n_{[X,Y]}Z\ & &=[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\ & &=0quadsex{Jacobimbox{ 恒等式}}. eex$$  

2(10') 设 $V(t), J(t)$ 是沿最短测地线 $gamma(t), tin [0,1]$ 的向量场, 它们满足 $$bex V(t)perp dotgamma(t),quad J(t)perp dotgamma(t),quad V(0)=J(0),quad V(1)=J(1), eex$$ 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 $$bex 0leq I(V-J,V-J) =I(V,V)-2I(V,J)+I(J,J). eex$$ 算出 $$bex I(V,J) &=&int_0^1sez{sef{dot V,dot J}-sef{R(dot gamma,V)J,dot gamma}}rd t\ &=&left.sef{dot J,V}right|^1_0-int_0^1 sez{sef{V,ddot J}+sef{R(J,dotgamma)dotgamma,V}}rd t\ &=&left.sef{dot J,J}right|^1_0 -int_0^1 sef{V,ddot J+R(J,dot gamma)dotgamma}rd t\ &=&left.sef{dot J,J}right|^1_0; eex$$ $$bex I(J,J)&=&int_0^1sez{sef{dot J,dot J}-sef{R(dot gamma,J)J,dotgamma}}rd t\ &=&left.sef{dot J,J}right|^1_0-int_0^1sef{ddot J+R(J,dotgamma)dotgamma,J}rd t\ &=&left.sef{dot J,J}right|^1_0. eex$$ 我们得到 $$bex 0leq I(V,V)-2I(J,J)+2I(J,J) =I(V,V)-I(J,J), eex$$ 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 $$bex B_{gamma_+}(x)=lim_{tto+infty}sez{d(x,gamma(t))-t}; eex$$ $$bex B_{gamma_-}(x)=lim_{tto-infty}sez{d(x,gamma(t))+t}. eex$$ 求证: $$bex B_{gamma_+}+B_{gamma_-}=0,quadmbox{在 } gamma mbox{ 上}; eex$$ $$bex B_{gamma_+}+B_{gamma_-}geq 0,quadmbox{在 } M mbox{ 上}. eex$$

证明:

(1)设 $xingamma$, 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 $$bex B_{gamma_+}(x) +B_{gamma_-}(x) &=&lim_{tto+infty} sed{sez{d(x,gamma(t))-t} +sez{d(x,gamma(-t))-t}}\ &=&-t_0+t_0\ &=&0. eex$$

(2)当 $xin M$, 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 $tto+infty$, 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 $$bex frac{p }{p t}f=-lap f+sev{n f}^2-R+frac{n}{2tau},quad frac{p }{p t}tau =-1. eex$$ 求证:

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

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

证明:

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

(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.

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