题号:14628    题型:解答题    来源:2024年丘成桐大学生数学竞赛(几何与拓扑类)-无答案
Let $M$ be a closed oriented Riemannian manifold, where $g_t$ is a family of smooth Riemannian metrics smoothly depending on $t \in(-\varepsilon, \varepsilon)$. Suppose there exists a family of eigenfunctio ns $f_t$ and eigenvalues $\lambda_t$ smoothly depending on $t$ such that
$$
\Delta_{g_t} f_t=\lambda_t f_t,
$$
where $\Delta_{g_t}$ is the Laplace-Beltrami operator defined using the Riemannian metric $g_t$. Additionally, assume that $f_0$ is not a co nstant function. We define $\dot{\lambda}:=\left.\frac{d}{d t}\right|_{t=0} \lambda_t$ and $\dot{\Delta}:=\left.\frac{d}{d t}\right|_{t=0} \Delta_{g_t}$. Prove the following:
(i) As $\lambda_0$ is an eigenvalue of $\Delta_{g_0}$, let $V_{\lambda_0}:=\operatorname{Ker}\left(\Delta_{g_0}-\lambda_0\right)$ be the eigenspace of $\lambda_0$, and let $\Pi: L^2\left(M, g_0\right) \rightarrow V_{\lambda_0}$ be th e orthogonal projection onto the eigenspace. Prove that $\dot{\lambda}$ is an eigenvalue of the operator $\Pi \circ \Delta^{\prime}: V_{\lambda_0} \rightarrow V_{\lambda_0}$.
(ii) Let $\varphi_t: M \rightarrow M$ be a 1-parameter family of diffeomorph isms of $M$ and assume $g_t=\varphi_t^* g_0$. Prove that $\dot{\lambda}=0$.
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