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Cross Product of Image of Self-Adjoint Operator

Mathematics Asked by José Victor Gomes on September 30, 2020

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 $((dN_p)^T)^{-1}=I$.

One Answer

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 $$dN_p(v_1) = k_1v_1 quadtext{and}quad dN_p(v_2) = k_2v_2.$$ 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_1wedge v_2) \ &= K(p) v_1wedge 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 $3times 3$ 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) $$Avwedge Aw = (det A)(A^top)^{-1}(vwedge w). tag{$star$}$$ It follows that if $A$ is an orthogonal $3times 3$ matrix, then we have $Avwedge Aw = (det A) A(vwedge w)$, 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 $Avwedge Aw$. (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.

Correct answer by Ted Shifrin on September 30, 2020

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