2

Consider a parameterization $r:U\rightarrow M$ for a regular surface $M$.Denote by $n$ the unit normal to the surface. I was given the following defition for the matrix $B$ that defines the second fundamental form: $$B= \begin{pmatrix} r''_{uu}\cdot n & r''_{uv}\cdot n \\ r''_{vu}\cdot n & r''_{vv}\cdot n \\ \end{pmatrix} $$ Also, I was given the following definition for the matrix $g$ that defines the Riemannian metric: $$g= \begin{pmatrix} r'_u\cdot r'_u & r'_u\cdot r'_v \\ r'_v\cdot r'_u & r'_v\cdot r'_v \\ \end{pmatrix} $$ The shape operator $S$ is $-dG$ (minus the differential of the Gauss map). It can be shown that $-dG=Bg^{-1}$.

What can be said about the eigenvalues of $B$? What can be said about the eigenvalues of $S$? Are they the same? Which of these eigenvalues are defined to be the principal curvatures of $M$? Is $\det B$ considered to be the Gauss curvature, or either $\det S$?

There is alot of information in Do Carmo's book and in other sources, but it is not consistent with the above definitions, which got me very cofused.

MasterJ
  • 1,036
  • 1
    Your equation for the shape operator is actually wrong. You should have $g^{-1}$ on the left. (I know of one published book that gets this, along with other things, definitely wrong.) The eigenvalues of $S$ are the principal curvatures, and $\det S$ is the Gaussian curvature. You might want to look at my text, which is a bit more readable than doCarmo. It's freely available from the link in my profile. – Ted Shifrin Jan 09 '18 at 00:48
  • So the only case that $B$ and $S$ are the same is when $r'_u,r'_v$ are orthonormal? Is there any way, given a parameterization $r$, to "force" them into being orthonormal? – MasterJ Jan 09 '18 at 09:09
  • 1
    @MasterJ: if your coordinate vectors are orthonormal then your surface is flat, so this cannot be done in general. What you can do is find an orthonormal frame $e_1(u,v),e_2(u,v)$ which spans the tangent space at each point, but this will typically not be $r_u,r_v$ for any parametrization $r$. – Anthony Carapetis Jan 09 '18 at 09:13

1 Answers1

3

I find the best way to resolve this kind of confusion is to take the perspective of abstract linear algebra: think of $B$ and $S$ as more than matrices. Really $B$ is a bilinear form on the tangent space $T_pM$, while $S$ is a linear operator $T_p M \to T_p M$. Without extra structure, it only makes sense to talk about eigenvalues and determinant of a linear operator; so the principal curvatures are the eigenvalues of $S$, and likewise $K = \det S.$

In a context where we have a preferred inner product, we often also talk about eigenvalues (or determinant, or...) of a bilinear form $b$; but what we mean by this is simply eigenvalues of the linear operator $L$ obtained from $b$ via the usual correspondence $b(X,Y) = g(L(X), Y).$ Since the second fundamental form and shape operator are related in this way (up to a sign), the eigenvalues will agree (up to sign).

Anthony Carapetis
  • 35,361
  • You claim that the eigenvalues of $B$ and $S$ will agree (up to sign). But if we take for example the torus $((2+\cos u)\cos v,(2+\cos u)\sin v,\sin u)$, then using the above we get that the eigenvalues of $B$ are ${1,(2+\cos u)\cos u }$ and for $S$ are ${1,\frac {\cos u}{2+\cos u}}$, and they are not the same. – MasterJ Jan 09 '18 at 09:05
  • @MasterJ: When I say the eigenvalues of $B$, I'm talking about the eigenvalues of the bilinear form $B$ with respect to the matrix $g$. These are the eigenvalues of the matrix $g^{-1}B,$ not the matrix $B$. (The latter depend on the choice of coordinates, so they're geometrically meaningless.) – Anthony Carapetis Jan 09 '18 at 09:10