Let $M \subset \mathbb{R}^{n+1}$ be a Riemannian hypersurface, and let $N$ be a smooth unit normal vector field along $M$. Denote by $\nu : M \to \mathbb{S}^n$ the Gauss map associated to $N$. Show that $$\nu^*\mathrm{vol}_{\mathbb{S}^n} = (-1)^n K \mathrm{vol}_M, $$ where $K$ is the Gaussian curvature of $M$, $\mathrm{vol}_{\mathbb{S}^n}$ is the standard volume form of $\mathbb{S}^n$, and $\mathrm{vol}_M$ is the volume form of $M$.
I am not particularly sure how to prove this statement. I tried proving this locally, in a coordinate chart, but the computations get messy. More precisely, I am not sure how to relate the volume form on $\mathbb{S}^n$ with the Gaussian curvature of $M$ and its volume form.
Best Answer
Recall that for an $n$-dimensional real vectorspace $V$ there is the determinant $\det:\mathrm{End}V\to \mathbb R$. On the other hand if $V$ is equipped with an inner product and an orientation one can define $\det_V:V^n\to\mathbb R$ as follows: For $v_1,...,v_n\in V$ choose any positively oriented onb $e_1,...,e_n\in V$ and set $\det_V(v_1,...,v_n)=\det (A)$, where $A$ is the linear map $e_i\mapsto v_i$. In particular ${\det}_{\mathbb R^{n+1}}$ is the volume form on $\mathbb R^{n+1}$.
Now back to the question. Since $\dim M=n$ it suffices to show the equality of $n$-forms at each $p\in M$ applied to a basis. So let $X_1,...,X_n$ be a positively oriented onb of $T_pM$, then
$$\nu^*{\mathrm{vol}_{\mathbb{S}^n}} (X_1,...,X_n)= {\det}_{\mathbb R^{n+1}}(\nu_p,d_p\nu (X_1),...,d_p\nu(X_n))$$
Now we convert this expression into a determinant of a linear map: Since $\nu_p, X_1,...,X_n$ is a positively oriented onb of $\mathbb R^{n+1}=\langle\nu_p\rangle\oplus T_pM$ we set $A=\mathrm{id}\oplus d_p\nu$. Using $\det (A_2\oplus A_1)=\det A_2\cdot\det A_1$ (see here) the expression above is equal to
$$\det(A)=1\cdot\det(d_p\nu)=(-1)^n\det (d_p-\nu)=(-1)^nK_p\\ =(-1)^nK_p{\mathrm{vol}_M}(X_1,...,X_n)$$
and this is what we wanted to show.