数学代写|MATH3405 Differential geometry

A Riemannian submanifold $M \subset(\bar{M}, \bar{g})$ with the induced metric $g=\left.\bar{g}\right|_{T M}$ is called totally geodesic if for every $p \in M, v \in T_p M$, the $\bar{g}$-geodesic $\gamma$ with initial conditions $\gamma(0)=p, \gamma^{\prime}(0)=v$ lies in $M$. Show the following are equivalent:

$M$ is totally geodesic.

Every $g$-geodesic in $M$ is also a $\bar{g}$-geodesic in $\bar{M}$.

The second fundamental form of $M$ vanishes.

Solution:

Let $\bar{\nabla}$ be the connection on $\bar{M}$ and let $\nabla$ be the induced connection on $M$. Take $p \in M, v \in$ $T_p M$ and let $\gamma$ be a $g$-geodesic with $\gamma(0)=p, \gamma^{\prime}(0)=v$ and let $\bar{\gamma}$ be a $\bar{g}$-geodesic with $\bar{\gamma}(0)=\gamma(0)$ and $\bar{\gamma}^{\prime}(0)=\gamma^{\prime}(0)$. First we will show that statement 1 implies statement 2 . If $M$ is totally geodesic, then $\bar{\gamma}$ lies in $M$, so $\bar{\gamma}^{\prime}(t) \in T_{\bar{\gamma}(t)} M$ and hence $\nabla_{\bar{\gamma}^{\prime}(t)} \bar{\gamma}^{\prime}(t)=\bar{\nabla}{\bar{\gamma}^{\prime}(t)} \bar{\gamma}^{\prime}(t)=0$ for any $t$, so $\bar{\gamma}=\gamma$ by the uniqueness of solutions for ODE’s. Hence, every $g$-geodesic is a $\bar{g}$-geodesic. Now, assuming statement 2 holds, we have $\nabla{\gamma^{\prime}(0)} \gamma^{\prime}(0)=\bar{\nabla}{\bar{\gamma}^{\prime}(0)} \bar{\gamma}^{\prime}(0)$, it follows that $$ \left.I I\right|_p(v, v)=-\nabla{\gamma^{\prime}(0)} \gamma^{\prime}(0)+\bar{\nabla}{\bar{\gamma}^{\prime}(0)} \bar{\gamma}^{\prime}(0)=0 $$ so $I I=0$ for any $p, v$, which is statement 3. Finally, if $I I \equiv 0$, then $\bar{\nabla}{\bar{\gamma}^{\prime}} \bar{\gamma}^{\prime}=\nabla_{\gamma^{\prime}} \gamma^{\prime}+I I\left(\gamma^{\prime}, \gamma^{\prime}\right)=0$, so $\bar{\gamma}$ lies in $M$ and so $M$ is totally geodesic.

Let $M \subset \mathbb{R}^3$ be the catenoid, which is the surface of revolution obtained by revolving the curve $x=\cosh z$ around the $z$-axis. Show that $M$ has zero mean curvature.

Solution:

We can parametrize the catenoid by
$$
F(\theta, z)=(\cosh (z) \cos \theta, \cosh (z) \sin \theta, z)
$$
for $\theta \in[0,2 \pi)$ and $z \in \mathbb{R}$. Now, we calculate
$$
\begin{gathered}
\partial_\theta F=(-\cosh (z) \sin \theta, \cosh (z) \cos \theta, 0) \
\partial_z F=(\sinh (z) \cos \theta, \sinh (z) \sin \theta, 1)
\end{gathered}
$$
and so we see that $g_{i j}=\cosh ^2(z) \delta_{i j}$. Choose an outward unit normal $\nu=\left(1+\sinh ^2(z)\right)^{-1 / 2}(-\cos \theta,-\sin \theta, \sinh (z))$. In coordinates, the scalar second fundamental form then becomes $h_{i j}=\partial_{i j}^2 F \cdot \nu$, so we have
$$
\begin{gathered}
h_{\theta \theta}=\cosh (z)\left(1+\sinh ^2(z)\right)^{-1 / 2}=1 \
h_{\theta z}=h_{z \theta}=0 \
h_{z z}=-\cosh (z)\left(1+\sinh ^2(z)\right)^{-1 / 2}=-1
\end{gathered}
$$
and so
$$
h_j^i=g^{i p} h_{p j}=\frac{1}{\cosh ^2(z)}\left(\begin{array}{cc}
1 & 0 \
0 & -1
\end{array}\right)
$$

Since the mean curvature is the trace of $h_j^i$, we see that the mean curvature must vanish.

A geodesic triangle in a Riemannian 2-manifold $\left(M^2, g\right)$ is a domain $\Omega$ with piecewise-smooth boundary $\partial \Omega$ consisting of three geodesics meeting at three vertices. If $M$ has constant Gauss curvature $K$, show that the sum of interior angles of any geodesic triangle is $\pi+K A$ where $A$ is the area of $\Omega$.

Solution:

Recall that by the Gauss-Bonnet theorem, if $\Omega$ is a curved polygon in $\left(M^2, g\right)$, then
$$
\int_{\Omega} K+\int_{\partial \Omega} k+\sum \varepsilon_i=2 \pi \chi(\Omega)
$$
where $k$ is the geodesic curvature of $\partial \Omega, \varepsilon_i$ are the exterior angles of $\Omega$, and $\chi(\Omega)$ is the Euler characteristic of $\Omega$. If $\Omega$ is a geodesic triangle, then $\chi(\Omega)=1, \partial \Omega$ has zero geodesic curvature, and $\sum_{i=1}^3 \varepsilon_i=3 \pi-\sum_{i=1}^3 \theta_i$, where $\theta_i$ are the interior angles. Hence, if $K$ is constant,
$$
K A+3 \pi-\sum_{i=1}^3 \theta_i=2 \pi
$$
or, equivalently
$$
\sum_{i=1}^3 \theta_i=\pi+K A
$$