Pullback of $2$-sphere volume form via Gauss map

Question

This is problem 17.3 from Tu's Differential Geometry text. This post appears to answer my question, but I don't follow many of the explanations, including the construction and usage of $det$.

Problem

Let $M$ be a smooth, compact, oriented surface in $\mathbb{R^3}$ with $K$ the Gaussian curvature on $M$. If $v : M \to S^2 $ is the Gauss map, prove that $$ v^* (vol_{S^2}) = K vol_M $$

Corrected Proof

Lemma 1. If $N$ is an $n$-dimensional smooth, oriented manifold, $(U, x^1, \ldots, x^n)$ a chart a $p \in N$ with onb $\{e_1, \ldots e_n \} =: e $ , then $$ vol_N = det_e $$

Proof. If $\theta^1, \ldots \theta^n$ is dual to $e_1, \ldots e_n$, then for any $X_1, \ldots X_n$ in $T_p N$ with $X_j = \sum a^i_j e^i$ \begin{align} vol_N (X_1, \ldots X_n) &= \theta^n \wedge \cdots \wedge \theta^n (X_1, \ldots X_n) \\ &= \sum_{\sigma \in S_n} sgn(\sigma) a^i_{\sigma(1)} \cdots a^i_{\sigma(n)} \\ &= det_e [ a^i_j ] \end{align} $\square$

Lemma 2. There exists $\{ e_1, e_2 \} =: e$ principal vectors at $p$ in $M$ such that $e$ is an onb for $M$, and $ \{ \kappa_1 e_1, \kappa_2 e_2 \} = \{ dv_p(e_1), dv_p(e_2) \} =: e'$ is an orthogonal basis for $S^2$.

We will use these facts: $dv_p = -L$ (the shape operator at $p$) (Problem 5.3), the $L$ has the principal vectors as eigenvectors and principal curvatures as eigenvalues (Prop 5.6), and $L$ is self-adjoint.

Note that if $\kappa_1 = \kappa_2$, all vectors in $T_p M$ are principal, so we can freely choose an onb. So assume they're unequal.

Then \begin{align} \kappa_1^2 \langle e_1, e_2 \rangle &= \langle -L (e_1), -L (e_2) \rangle \\ &= \langle dv_p (e_1), dv_p (e_2) \rangle \\ &= \kappa_2^2 \langle e_1, e_2 \rangle \end{align}

Because the principal curvatures are not equal, we must have $$\langle e_1, e_2 \rangle = 0 = \langle dv_p (e_1), dv_p (e_2) \rangle = \langle \kappa_1 e_1, \kappa_2 e_2 \rangle.$$

$\square$

Note that Lemma 2 implies that $e$ is an onb for both $T_p M$ and $T_p S^2$.

Now if $K(p) = det(J_v(p)) = 0$ then the claim obviously holds because $v^* vol_{S^2} = 0 = K(p) vol_M$. So assume $K(p) = det(J_V(p)) \neq 0$.

Applying Lemma 2, Choose onb $e$ for $M$ (which is also onb for $S^2$). Let $\Theta^1 \wedge \Theta^2$ be the volume form for $S^2$ (with $\Theta^i$ dual to $e_i$). Then for any $X_1, X_2$ vectors in $S^2$ with $X_j = \sum x^{i}_j e_i$.

\begin{align} v^* (vol_{S^2})(X_1, X_2) &= \Theta^1 \wedge \Theta^2 (dv_p X_1, dv_p X_2) \\ &= \Theta^1(dv_p X_1) \Theta^2(dv_p X_2) - \Theta^1(dv_p X_2) \Theta^2(dv_p X_1) \\ \text{(Lemma 2)} &= (\kappa_1 x^1_1) (\kappa_2 x^2_2) - (\kappa_1 x^1_2) (\kappa_2 x^2_1) \\ &= K det_e(X_1, X_2) \\ \text{(Lemma 1)} &= K vol_M \end{align} $\square$

Update

Special thanks to @cbishop for providing the key insight (Prop 8.2.1) that helped me solve the problem!

Answer 1

Corollary 8.2.2 of Pressley's book "Elementary Differential Geometry" states that there is an orthonormal basis of the tangent plane of a surface consisting of the principal vectors. I believe that this fact plus multi-linearity should complete your proof.

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I had the wrong author above, that is now fixed. A screenshot of the proof of the corollary is here:

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Here is the statement of A.0.3 from Pressley:

So since the Weingarten map is self-adjoint, it is diagonalizable and the eigenvalues are the principal curvatures. It appears that this is covered in Tu's book in Section 5.3 up through Corollary 5.7.

Answer 2

$\newcommand{\R}{\mathbb{R}}$ Following Tu, section 5 chapter 1, consider a hypersurface $M$ that is the zero set of a function $f:\R^n\mapsto \R$. Let $N(x) = grad_f(x)$ for $x\in \R^n$, the gradient of $f$ considered as a column vector. It is normal to the hypersurface.

• The following concepts are defined in the book:
  1. The shape operator: $L: X\mapsto -D_X N_{unit}$. (section 5.2). Note, although Tu uses the same notation $N$ as the gradient above, he says $N$ is unit normal (instead of being just gradient). I add the subscript unit to be clear.
  2. The Gaussian curvature is the determinant of the Shape operator (Corollary 5.7).
  3. The Gauss map maps $x\in M$ to the unit vector in the normal direction, $x\mapsto N_{unit}(x) = \frac{1}{|N(x)|}N(x)=\frac{1}{(N(x)^TN(x))^{1/2}}N(x)$ (problem 5.3)

Items 1 and 3 show the differential of the Gauss map is $-L$, which Tu states in problem 5.3. Problem 17.3 follows from the fact that the pullback of the volume form requires multiplying by the determinant of the differential, (the manifold version of the change of variable formula for multiple integrals in vector/multivariable calculus.)

Here is an explicit description of the shape operator, by differentiating the formula of the Gauss map in 3. Let $H$ be the Hessian of the function $f$ defining the hypersurface $M$. The Jacobian of the Gauss map is given by $$D_XN_{unit} = (-\frac{1}{(N^TN)^{3/2}}NN^TH + \frac{1}{(N^TN)^{1/2}}H)X $$ where we use $D_X(N^TN) = 2N^THX$ when differentating $\frac{1}{(N^TN)^{1/2}}$. Equivalently, the Jacobian of the Gauss map is given by the matrix $-L_n = -\frac{1}{(N^TN)^{3/2}}NN^TH + \frac{1}{(N^TN)^{1/2}}H$. We can check easily $$N^TD_XN_{unit} = -\frac{1}{(N^TN)^{3/2}}N^TNN^TH + \frac{1}{(N^TN)^{1/2}}N^TH = -\frac{1}{(N^TN)^{1/2}}N^TH + \frac{1}{(N^TN)^{1/2}}N^TH=0 $$ confirming the fact that the shape operator maps to $T_xM$. Note we use global coordinates so the $L_n$ is an $n\times n$ matrix. To compute the Gaussian curvature, we need to restrict it to $T_xM$ (on $\R^n$, since $-L_n$ is of rank n-1, its determinant is zero)

As an example, for the sphere $S^{n-1}$ defined by $x^Tx = 1$, where $x$ is considered as a column vector, $N(x) = 2x$, $H(x) = 2I$, the shape operator is $$L_n = \frac{1}{(N^TN)^{3/2}}NN^TH - \frac{1}{(N^TN)^{1/2}}H = xx^T - I, $$ which is minus the identity map when restricted to $T_xS^{n-1}$, as it sends $v$ satisfying $x^Tv = 0$ to $(xx^T - I)v = -v$. Thus, in this case, the determinant is $(-1)^{n-1}$. In general, we need a basis of $T_xM$ to compute the determinant.

Answer 3

Here's a quick sketch: $\newcommand\R{\mathbb{R}}$ Let $\gamma: M \rightarrow S^2 \subset \R^3$ be the Gauss map. As @TedShifrin said, for each $p \in M$, $$ T_pM = T_{\gamma(p)}S^2 = H, $$ where $$ H = \{ v\in\R^3\ :\ v\cdot\gamma(p) = 0 \}. $$ Therefore, if you choose an orthonormal basis $(e_1, e_2)$ of $H$ and dual basis $(\omega^1,\omega^2)$, then $$\omega^1\wedge\omega^2 = d\mathrm{vol}_M(p) = d\mathrm{vol}_{S^2}(\gamma(p)).$$ The differential of the Gauss map at $p$ is a linear map $$\gamma_*: H \rightarrow H. $$ This is the shape operator. The matrix $A$ given by $$ A_{ij} = e_i\cdot \gamma_*e_j $$ is the second fundamental form written with respect to the orthonormal frame and its determinant is the Gauss curvature. You can now verify that $$\gamma^*(\omega^1\wedge\omega^2) = K\omega^1\wedge\omega^2. $$