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...