Derivative of Gauss map is the second fundamental form

Question

I have been messing around with Grassmannians lately.

Let $M^k\subseteq \Bbb R^n$ be an embedded submanifold equipped with the induced Riemannian metric, and consider the Gauss map $G\colon M \to {\rm Gr}_k(\Bbb R^n)$ given by $G(p) = T_pM$, taking values in the Grassmannian of $k$-planes of $\Bbb R^n$. Then the derivative is a map ${\rm d}G_p\colon T_pM \to {\rm Hom}(T_pM, T_pM^\perp)$, which by currying may be seen as a bilinear map ${\rm d}G_p\colon T_pM \times T_pM \to T_pM^\perp$.

The only reasonable guess is that ${\rm d}G_p$ is the second fundamental form ${\rm II}_p$, but I'm not sure how to even start this. Can someone show me the calculation or give a reference? Thanks.

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Note: I can see that $${\rm d}G_p(v,w) = \frac{\partial \alpha}{\partial s}(0,0)$$where $\alpha \colon (-\epsilon,\epsilon)^2 \to M$ is given by $\alpha(t,s) = \exp_{\gamma(t)}(sw(t))$, where $t \mapsto \gamma(t)$ is any curve in $M$ with $\gamma(0) = p$ and $\gamma'(0) = v$, while $t \mapsto w(t)$ is a curve of vectors such that $w(0) = w$ and $w(t) \in T_{\gamma(t)}M$ for all $t$. But it is not clear to me how to make $\nabla$ and ${\rm D}$ appear here.

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Nevermind, it is trivial. We may assume that $t \mapsto w(t)$ is a parallel field along $\gamma$, so $$w'(t) = \frac{{\rm D}w}{{\rm d}t}(t) + {\rm II}_{\gamma(t)}(\gamma'(t),w(t))\implies w'(0) = {\rm II}_p(v,w)$$as wanted. I'll leave this open in case someone realizes I'm royally screwing the pooch here...

Answer 1

I don't see why ${\rm d}G_p(v,w) = \frac{\partial \alpha}{\partial s}(0,0)$, but there is a direct calculation.

There is a local paramatrization $\phi : {\rm Hom}(T_pM,T_pM^\perp)\to{\rm Gr}_k(\Bbb R^n)$ given by sending a linear transformation to its graph. Let $G_\phi=\phi^{-1}\circ G$.

Now choose $u,v\in T_pM$, we have $${\rm d}G_p(v,u)=v\cdot(G_\phi(\cdot)(u))$$ Here, $G_\phi(\cdot)(u):N\to T_pM^\perp$ is a function defined in a neighbourhood $N$ of $p$ in $M$, and $v\cdot$ means taking the directional derivative w.r.t $v$.

On the other hand, the tangent vector field $$U(q)=u+G_\phi(q)(u)$$ is tangent to $M$ in a neighbourhood of $p$. By definition, the second fundamental form is calculated as:

\begin{align}\Pi_p(u,v)&=(v\cdot(U))^{T_pM^\perp}\\ &=(v\cdot u+v\cdot G_\phi(\cdot)(u))^{T_pM^\perp}\\ &=(v\cdot G_\phi(\cdot)(u))^{T_pM^\perp}\\ &=v\cdot G_\phi(\cdot)(u) \end{align}

Answer 2

$\newcommand{\R}{\mathbb{R}}$ Here's a sketch of a relatively low tech way to see this.

First, a general observation: If $\Phi: M \rightarrow \R^n$ is an isometric embedding, then the second fundamental form is the tensor $H = H_{ij}\,dx^i\,dx^j \in S^2T^*\otimes T^\perp$ given by $$ H_{ij} = \text{ normal component of }\partial^2_{ij}\Phi. $$

Given a point $p \in M$, you can always position $M$ so that $p = 0$ and $T_pM = \{ x^{k+1} = \cdots = x^n = 0 \}$. Locally, $M$ is now the graph of a map $u: \R^k \rightarrow \R^{n-k}$: $$ M = \{ (x, u(x)) \}.$$ On one hand, as you've observed, the Gauss map is the map $$ G(p) = T_pM = \{ (v, L_p(v))\ :\ v \in \R^k \}, $$ where $L_p = \partial u: \R^k \rightarrow \R^{n-k}$. Therefore, its derivative at $0$ is given by $\partial^2u(0): \R^k \rightarrow \mathrm{Hom}(\R^k,\R^{n-k})$.

On the other hand, by the observation at the start, the second fundamental form at $0$ is also $$\partial^2u(0) \in S^2(\R^k)^*\otimes \R^{n-k}$$