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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!

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  • $\begingroup$ It's very hard to figure out what your questions really are. Remember that $T_pM = T_{\nu(p)}S^2$ and that are using the same basis, $e$, for both those vector spaces. $\endgroup$ Commented Jul 24, 2023 at 17:16
  • $\begingroup$ @TedShifrin could you explain your 2 claims? If the tangent spaces are equal (or did you mean isomorphic?), with the same onb's, then the solution becomes very clear. $\endgroup$
    – IsaacR24
    Commented Jul 24, 2023 at 17:34
  • $\begingroup$ No, I meant equal. That’s crucial and is one of the main distinguishing features of the Gauss map. Remember that the tangent space of the unit sphere at $q$ is the plane orthogonal to $q$. $\endgroup$ Commented Jul 24, 2023 at 17:58
  • $\begingroup$ Then why is the tangent space at p in M the plane orthogonal to p? It's not clear to me this would be a plane. $\endgroup$
    – IsaacR24
    Commented Jul 24, 2023 at 21:00
  • $\begingroup$ No, the tangent space at $p\in M$ is orthogonal to $\nu(p)$. Put the pieces together carefully. $\endgroup$ Commented Jul 24, 2023 at 21:29

3 Answers 3

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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.


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

enter image description here



Here is the statement of A.0.3 from Pressley:

enter image description here

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.

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  • $\begingroup$ I can't readily access this book, could you provide a snapshot of the statement of proof of that corrolary? $\endgroup$
    – IsaacR24
    Commented Aug 1, 2023 at 23:11
  • $\begingroup$ I have updated my post with a snapshot and corrected the author (forgot to double check that) $\endgroup$
    – cbishop
    Commented Aug 2, 2023 at 0:59
  • $\begingroup$ I was able to look at the 2nd Ed of Presley's book, but I don't see corollary 8.2.2. Can you also provide a snapshot of Theorem A.0.3? I'm just trying to see a complete solution to my problem at once. I think this would close the loop. $\endgroup$
    – IsaacR24
    Commented Aug 2, 2023 at 14:26
  • $\begingroup$ Hopefully the second screenshot + my comment clears things up. You can read 8.2.2 directly from the first screenshot as well $\endgroup$
    – cbishop
    Commented Aug 2, 2023 at 20:43
  • $\begingroup$ I'm very close I think -- see my updated proof in the original post. I actually attempted to prove part of theorem A.0.3 in Lemma 2. $\endgroup$
    – IsaacR24
    Commented Aug 2, 2023 at 20:59
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$\newcommand{\R}{\mathbb{R}}$ 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.

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  • $\begingroup$ I really don't understand most of these computations and it doesn't seem like Tu would directly prepare me for them. For example, I'm not sure how the Hessian enters into this (Tu never mentions this, but I looked up what it means). You're also solving this at a greater level of generality than the problem calls for. $\endgroup$
    – IsaacR24
    Commented Jul 30, 2023 at 18:09
  • $\begingroup$ Is there a relatively easy fix to my current proof that would make it correct? Based on Professor's Shifrin's comments, it seemed like I was very close. For example, if the push forward $dv_p$ of an onb in $T_p M$ is an onb in $S^2$, I think this would close the gap. $\endgroup$
    – IsaacR24
    Commented Jul 30, 2023 at 18:11
  • $\begingroup$ I apologize for stepping in this discussion with Professor Shifrin, but I hope the following are helpful. If $N(p)$ is the gradient of the constraint of the manifold at $p$, what are the equations for the tangent spaces $T_pM$ and $T_{\nu(p)}S^2$ ? Note, $\nu(p) =\frac{1}{|N(p)|}N(p)$. For a concrete example, if $M$ is defined by $x^4 + y^2 +z^2=1$, if $p= (a, b, c)$, what is $N(p)$ and what are the two tangent spaces in questions ? $\endgroup$ Commented Jul 30, 2023 at 20:29
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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 $$\gamma^*(\omega^1\wedge\omega^2) = K\omega^1\wedge\omega^2. $$

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  • $\begingroup$ Professor Shifrin said that the tangent spaces were not "literally" equal. As best I can tell, they are at least isomorphisms because $\gamma$ is a local diffeomorphism. Even assuming this is true, how do you justify the final equality? I don't see how the matrix of the second fundamental form justifies this. $\endgroup$
    – IsaacR24
    Commented Aug 3, 2023 at 10:55
  • $\begingroup$ The tangent space of a surface in $\mathbb{R}^3$ is a subspace of $\mathbb{R}^3$. You’ll also have to review carefully the definition of the second fundamental form. You should try to figure out why both what he said and what I said are correct. $\endgroup$
    – Deane
    Commented Aug 3, 2023 at 11:41
  • $\begingroup$ By the way, the Gauss map is not necessarily a diffeomorphism. $\endgroup$
    – Deane
    Commented Aug 3, 2023 at 11:46

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