Cross Product of Image of Self-Adjoint Operator

Question

This has been asked before, but I'm trying to prove that if $N: S \to S^2$ is a normal Gauss map of a regular surface $S$, then $$dN_p(v) \wedge dN_p(w) = K(p)(v \wedge w), $$ in which $K(p)$ denotes de Gaussian curvature of $S$ at $p$. I know that I'm supposed to use properties of cross product, and remind that $det(-dN_p)=K(p)$, as indicated on Question about cross product of images of linear transformation. But I can't see how, following the link's answers, $dN_p$ being self-adjoint implies that $((dN_p)^T)^{-1}=I$.

Answer 1

Because $dN_p$ is a self-adjoint map on the two-dimensional tangent space of the surface, we know there is an orthonormal basis for the tangent space consisting of eigenvectors, and, moreover, the eigenvalues (the principal curvatures) are necessarily real. Let's write $v_1,v_2$ for this orthonormal basis, and so $$dN_p(v_1) = k_1v_1 \quad\text{and}\quad dN_p(v_2) = k_2v_2.$$ Then it's clear that \begin{align*} dN_p(v_1)\wedge dN_p(v_2) &= (k_1v_1)\wedge (k_2v_2) = (k_1k_2)(v_1\wedge v_2) \\ &= K(p) v_1\wedge v_2. \end{align*} Now you can just verify (using properties of $\wedge$) that for any tangent vectors $v,w$, the same property holds. Write $v=a_{11}v_1+a_{12}v_2$, $w=a_{21}v_1+a_{22}v_2$, and both sides of the equation will pick up a factor of $a_{11}a_{22}-a_{12}a_{21}$. Thus, the equation holds.

The attempted solution in the link you gave is just not going to work. Here is what is correct. If you have a $3\times 3$ matrix $A$, i.e., a linear map on all of $\Bbb R^3$, then (continuing to write $\wedge$, as doCarmo does, for the cross product) $$Av\wedge Aw = (\det A)(A^\top)^{-1}(v\wedge w). \tag{$\star$}$$ It follows that if $A$ is an orthogonal $3\times 3$ matrix, then we have $Av\wedge Aw = (\det A) A(v\wedge w)$, since $AA^\top = I$ implies $(A^\top)^{-1} = A$. As you can see, this is totally removed from the topic at hand. If you're curious, the formula ($\star$) follows from the classic formula $$A^{-1} = \frac1{\det A}(\text{cof }A)^\top,$$ where $\text{cof }A$ is the matrix of cofactors. This means that $$\text{cof }A = (\det A)(A^{-1})^\top = (\det A)(A^\top)^{-1}.$$ If you write out $v$ and $w$ in terms of the standard basis, these cofactors are precisely what appear in the coefficients of $Av\wedge Aw$. (If you know some exterior algebra, what is going on here is that the standard matrix representation of $\Lambda^2 A$ is the cofactor matrix of $A$.) Self-adjointness is irrelevant here, as is the whole formula, since we're talking about a linear map defined just on the tangent space of the surface, as you observed in your query.