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$\newcommand{\sff}{II}$ $\newcommand{\mf}{\mathfrak}$ $\newcommand{\R}{\mathbf R}$ $\newcommand{\ab}[1]{\langle #1\rangle}$

Background

Let $M$ be a Riemannian submanifold of a Riemannian manifold $\tilde M$. Let $\tilde \nabla$ and $\nabla$ denote the Levi-Civita connections on $\tilde M$ and $M$ respectively. The $\tilde \nabla$ and $\nabla$ are related as follows: Let $X$ and $Y$ be smooth vector fields on $M$. Then $\nabla_XY$ at a point $p\in M$ is the orthogonal projection of $\tilde \nabla_XY$ on $T_pM$.

The second fundamental form of $M$ is defined as the map $\sff:\mf X(M)\times \mf X(M)\to \mf X(M)$ which takes a pair of vector fields $(X, Y)$ to be the orthogonal projection of $\tilde \nabla_XY$ onto $T_pM^\perp$. Thus $\tilde \nabla_X Y=\nabla_XY + \sff(X, Y)$.

Now suppose $M$ be a hypersurface in $\R^{n+1}$, and $N$ be a smooth unit normal vector field along $M$. Thus for any two smooth vector fields $X$ and $Y$ on $M$, we have $\sff(X, Y)=h(X, Y)N$, for some smooth map $h:\mf X(M)\times \mf X(M)\to \R$. Define a map $G: M\to S^n$ as $G(q)=N_q$ for all $q\in M$. The map $G$ is called the Gauss Map, and the Gaussian curvature of $M$ at $p$ is defined as $\det(dG_p)$.

Question

In the theorem below, the minus sign attached to $h$ is bugging me. For this means that the shape operator is the negative of the differential of the Gauss map. While the shape operator is supposed to be the same as the differential of the Gauss map. I am unable to find a mistake in my proof.

"Theorem." Let $M$ be a hypersurface in $\mathbf R^{n+1}$, and let $G:M\to S^n$ be the Gauss map. Let $N$ be a unit normal vector field along $M$. Then $\ab{dG_p u, v}= -h(u, v)$ for all $p\in M$ and $u, v\in T_pM$.

Proof. Let $\gamma:(-\varepsilon, \varepsilon)\to M$ be a smooth curve with $\gamma(0)=p$ and $\dot \gamma(0)=u$. Let $V:I\to TM$ be the parallel vector field along $\gamma$ with $V(0)=v$. Now we have $\ab{G\circ \gamma(t),\ V(t)}=0$ for all $t$. Taking the derivative at $t=0$, we have $$\ab{dG_p(\dot \gamma(0)),\ v} + \ab{G(\gamma(0)),\ \dot V(0)} = 0$$

Note that $\dot V$ is the covariant derivative of $V$ with respect to the Levi-Civita connection on $\R^{n+1}$. Thus $\dot V(0) = D_{t}V(0) + h(\dot \gamma(0), V(0)) N_{\gamma(0)}$. Substituting this in the equation above, we get

$$\ab{dG_p(\dot \gamma(0)),\ v} + \ab{G(\gamma(0)),\ h(\dot \gamma(0), V(0))N_{\gamma(0)}} = 0$$

Putting $\dot \gamma(0)=u$, $V(0)=v$, and $G(\gamma(0))=N_p$, we get $$\ab{dG_pu, v} + \ab{N_p,\ h(u, v) N_p} = \ab{dG_pu, v} + h(u, v) = 0$$ which proves the undesired result.

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  • $\begingroup$ Have you tried applying all this to the case of the unit circle in the plane? In that case, the Gauss map is very easy to compute. :) So is the shape operator. And there'll be a lot fewer fancy words. I suspect this'll all come down to the observation that when you look at a clock, the minute-hand moves clockwise over the face, but when you ride on the minute-hand of a clock, the clock's face appears to be moving counterclockwise beneath you. $\endgroup$ – John Hughes Dec 28 '17 at 13:36
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But your result and proof are both correct! If $M^n\subseteq \overline{M}^{n+k}$, for each normal field $\xi$ to $M$ you get a field $A_\xi$ of self-adjoint linear endomorphisms on the tangent spaces to $M$, satisfying $\langle A_\xi(X),Y\rangle =\langle {\rm II}(X,Y),\xi\rangle$. One proves that $A_\xi(X) = -(\overline{\nabla}_X\xi)^\top$ as follows: $$\langle A_\xi (X),Y\rangle =\langle {\rm II}(X,Y),\xi\rangle =\langle \overline{\nabla}_XY-\nabla_X Y,\xi\rangle =\langle \overline{\nabla}_X Y,\xi\rangle = -\langle \overline{\nabla}_X\xi, Y\rangle =\langle - (\overline{\nabla}_X\xi)^\top,Y\rangle,$$where in the penultimate step the minus sign appears from differentiating $\langle Y,\xi\rangle=0$ in the direction of $X$.

In the case $M^n\subseteq \Bbb R^{n+1}$, with $N$ an unit normal field along $M$, we have that $$A_N(X) = -(\overline{\nabla}_XN)^\top = -(dN(X))^\top = -dN(X), $$where the last step follows from differentiating $\langle N,N\rangle =1$ to obtain $\langle dN(X),N\rangle=0$. So the relation $\langle A_N(X),Y\rangle =\langle {\rm II}(X,Y),N\rangle$ now reads $$\langle- dN(X),Y\rangle =h(X,Y). $$

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