3
$\begingroup$

Terminology note

In the following, a scalar product will be a symmetric bilinear form, and a euclidean scalar product will be a positive definite scalar product. This is the terminology used by my Differential Geometry teacher.

Background

Let me sum up the operators coming in play in the question. To start right from the start, we have, in as general a context as possible, a Riemannian manifold $(M,g)$. The metric induces a single Levi-Civita connection, which I will denote $\nabla_XY$, where $X,Y$ are fields and $\nabla_XY$ is the covariant derivative of $Y$ along $X$, as it is also called. The curvature tensor is then defined as:

$$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z.$$

This is the type $(1,3)$ curvature tensor, whereas the type $(0,4)$ tensor is:

$$R(X,Y,Z,W)=g(R(X,Y)Z,W).$$

This can be proven to be antisimmetric in the first two and second two entries, and symmetric w.r.t. swapping the two couples (i.e. $R(X,Y,Z,W)=R(Z,W,X,Y)$). This means it induces a scalar product on the second exterior power, which we define as:

$$\mathcal{R}(X\wedge Y,Z\wedge W)=R(X,Y,W,Z).$$

$g$ also induces a scalar product on the second exterior power:

$$g'(v\wedge w,u\wedge t)=g(v,u)g(w,t)-g(v,t)g(w,u).$$

This can be shown to be positive definite -- in fact, if $\{v_j\}$ is orthonormal w.r.t $g$< $\{v_i\wedge v_j\}_{i<j}$ is orthonormal w.r.t. $g'$. Fixing this scalar product as a "reference", since $\mathcal{R}$ is also a scalar product, we get a self-adjoint operator, which we still denote $\mathcal{R}$, by setting:

$$g'(\mathcal{R}(v\wedge w),u\wedge t)=\mathcal{R}(v\wedge w,u\wedge t).$$

Oh I'm writing all of this with decomposable vectors but this is extended by additivity to non-decomposable ones of course.

The last operator we need is the shape operator. So we consider a hypersurface $N\subseteq M$. Locally, we can always find a normal unitary vector field $\mathcal{N}$ for $N$. Modulo a sign, this induces an operator, called shape operator, which is defined by:

$$S(v)=\nabla_X\mathcal{N},$$

where $X$ is any field which evalues to $v$ at the tangency point $p$ of $v$ to $M$. It can be shown that this is a $g$-self-adjoint endomorphism of $T_pN$, and that $S(\mathcal{N}(p))=0$.

Question

So here comes, at last, the question. I have been given the following exercise.

Let $M$ be a hypersurface of $\mathbb{R}^n$. Suppose $M$ is compact, connected, orientable (i.e. has a global unitary normal field). Assume $\{v_j\}$ is an orthonormal basis for the tangent space $T_pM$ consisting of eigenvalues of the shape operator $S$. Prove that $\{v_i\wedge v_j\}$ is an o.n.b. of $\Lambda^2T_pM$ composed of eigenvalues of $\mathcal{R}$, and more precisely $\mathcal{R}(v_i\wedge v_j)=\lambda_i\lambda_jv_i\wedge v_j$, where $S(v_i)=\lambda_iv_i$.

Now the fact that those wedges provide an orthonormal basis is easy, since we know (by earlier proof in the course) that it is a basis, and orthonormality follows from a simple computation:

$$g'(v_i\wedge v_j,v_k\wedge v_\ell)=g(v_i,v_k)g(v_j,v_\ell)-g(v_i,v_\ell)g(v_j,v_k)=\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk},$$

we need $i<j$ and $k<\ell$ by definition of the basis, if the two are equal the first term is $1\cdot1$ and the second one $0\cdot0$, so it is a normal basis, and if they are distinct then the first term is 0, and if $i\neq k$ it could be that $i=\ell$ but the $j\neq k$ since $j>i=\ell>k$, so the second term is also zero, whereas if $j\neq\ell$ it could be $j=k$, but $i\neq\ell$ by a similar argument, giving again that both terms are zero. As for the fact those vectors are eigenvalues of $\mathcal{R}$, I am pretty much at a loss: what is the link between $\mathcal{R}$ and $S$? I must be missing something fundamental, but I just don't see a way to link them.

PS

I also know that the hypotheses given imply $S$ is everywhere positive definite or everywhere negative definite, and that $M$ is diffeomorphic to $S^d$. That is (or should be) called Hadamard's theorem. "or should be" because the statement was given after the proof of the theorem and along the lines of «This set of considerations are called Hadamard's theorem». At this point in the course we have not yet covered geodesics, and we never saw anything involving both geodesics and $S$. I remark this because I have seen this question. Also, principal curvature is an unknown term for me. Seeing this question makes me remember the Theorema Egregium, which we proved, and seems to be potentially useful here. However, T.E. works in $\mathbb{R}^3$, not $\mathbb{R}^{d+1}$, so it would need a generalization. Obviously, $d=2$ is solved by T.E., because if the sectional curvature (which equates to $\mathcal{R}(v_i\wedge v_j,v_i\wedge v_j)$ since the denominator is 1) is the determinant of the shape operator, then it is the product of the eigenvalues, which are precisely the $\lambda_i$.

$\endgroup$
2
$\begingroup$

By the Gauss (or is it Codazzi-Mainardi?) equation (which I had completely forgotten about), we have:

$$R(X,Y)Z=A(S(X),S(Y))Z=\lambda_i\lambda_jA(X,Y)Z=\lambda_i\lambda_j\langle Y,Z\rangle X-\langle X,Z\rangle Y,$$

so: $\newcommand{\ang}[1]{\langle #1 \rangle}$

\begin{align*} \mathcal{R}(X\wedge Y,Z\wedge T)={}&R(X,Y,T,Z)=\ang{R(X,Y)T,Z}=\lambda_i\lambda_j\ang{\ang{Y,T}X-\ang{X,T}Y,Z}={} \\ {}={}&\lambda_i\lambda_j\ang{X,Z}\ang{Y,T}-\lambda_i\lambda_j\ang{X,T}\ang{Z,Y}. \end{align*}

By definition of the operator $\mathcal{R}$:

$$\ang{\mathcal{R}(X\wedge Y),Z\wedge T}=\mathcal{R}(X\wedge Y,Z\wedge T)=\lambda_i\lambda_j\ang{X,Z}\ang{Y,T}-\lambda_i\lambda_j\ang{X,T}\ang{Z,Y},$$

but $X,Y,Z,T$ are part of an orthonormal system (the one chosen on $N$), so if $X\neq Z$ or $Y\neq T$ this is 0, and otherwise this is $\lambda_i\lambda_j$, proving this is indeed an eigenvalue.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.