Generalized Gauss map, giving rise to second fundamental form

Question

I know that the tangent bundle of $G_n(\mathbb{R}^{n+k})$ is isomorphic to $\text{Hom}(\gamma^n(\mathbb{R}^{n+k}), \gamma^\perp)$, where $\gamma^\perp$ denotes the orthogonal complement of $\gamma^n(\mathbb{R}^{n+k})$ in $\epsilon^{n+k}$. Now, consider a smooth manifold $M \subset \mathbb{R}^{n+k}$. If $\overline{g}: M \to G_n(\mathbb{R}^{n+k})$ denotes the generalized Gauss map, how do I see that $$D\overline{g}: DM \to DG_n(\mathbb{R}^{n+k})$$gives rise to a cross-section of the bundle$$\text{Hom}(\tau_M, \text{Hom}(\tau_M, \nu)) \cong \text{Hom}(\tau_M \otimes \tau_m, \nu)?$$

Answer 1

Examining the Gauss map, we see that $\overline{g}$ is covered by the bundle map$$g \oplus g^\perp: \tau \oplus \nu \to \gamma_n \oplus \gamma^\perp$$and the differential$$D\overline{g}: \tau M \to \tau G_n(\mathbb{R}^{n+k}) \simeq \text{Hom}(\gamma^n, \gamma^\perp).$$This thus yields a fiber-linear morphism$$\sigma: \tau M \to \text{Hom}(\tau M, \nu)$$via$$(p, v) \mapsto (g^\perp)_{\overline{g}(p)}^{-1} \circ D\overline{g}_p(v) \circ g_p.$$This is precisely a smooth section of the desired bundle.

Answer 2

What is $\nu$? What is $\gamma^n(\mathbb R^{n+k})$?

The derivative of a map $f:X\to Y$ is a section of the bundle $\mathrm{Hom}(TX,f^*TY)$ over $X$, where $f^*TY$ is the pullback of the vector bundle $TY$ to $X$ via $f$.

Let's work this out for the example you've given. The general case is exactly the same. Fix a point $p$ of $M$. Then $D\bar g_p$ is a linear map $T_pM\to T_{\bar g(p)}G_n(\mathbb R^{n+k})$, i.e. an element of $\mathrm{Hom}(T_pM,T_{\bar g(p)}G_n(\mathbb R^{n+k}))=\mathrm{Hom}(T_pM, (\bar g^*TG_n(\mathbb R^{n+k}))_p)$. Thus, $D\bar g$ is a section of the bundle $\mathrm{Hom}(TM,\bar g^*TG_n(\mathbb R^{n+k}))$ over $M$.

Now, it's just a matter of computing what $\bar g^*TG_n(\mathbb R^{n+k})$ is. This is the space of all pairs $(p,v)\in M\times TG_n(\mathbb R^{n+k})$ such that $v$ lies in the tangent space of $\bar g(p)$.

I will try to proceed once I figure out what $\nu$ is.

Answer 3

The last equality is basic linear algebra, of course. As far as I can tell, all you're missing is the tautological facts that $\bar g^*\gamma^n = \tau_M$ and $\bar g^*\gamma^\perp = \nu$. These follow immediately from the definitions of $\bar g$, $\gamma$, $\gamma^{\perp}$, $\tau_M$, and $\nu$.