I have a little (possible dumb) technical question about local eigenvectors frame of a isometric immersion.

Let $N^n$ a smooth manifold and $(M^{n+1},g)$ a smooth Riemannian manifold. Consider $\phi: N\to M$ a isometric immersion and let $S$ be the shape operator of $N$.

Given $p\in N$, we can always assume that exists a local orthonormal frame $\{e_1,e_2,\cdots,e_n\}$ on a neighborhood of $p$, such that diagonalizes $S$? Or we need to put the condition that $p$ is not a umbilical point?


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    $\begingroup$ You mean $\{e_1,e_2,\dots,e_n\}$, of course. $\endgroup$ Commented Oct 24, 2018 at 16:50
  • $\begingroup$ @TedShifrin, yes, of course. $\endgroup$
    – Irddo
    Commented Oct 24, 2018 at 20:55

1 Answer 1


At each point $p$, of course, there's always an orthonormal basis for $T_pN$ diagonalizing $S_p$. You may likely have local smoothness issues whenever there are repeated eigenvalues. However, in dimension $n>2$, it's not good enough to say there are no umbilic points; you actually need to require distinct eigenvalues.

  • $\begingroup$ Is it really necessary to require all eigenvalues to be distinct? I think repeated eigenvalues don't automatically give issues. For instance: the $n$-sphere $S^n$ in $\mathbb{R}^{n+1}$ is totally umbilical and a frame in a point can be extended smoothly to a neighbourhood of that point. Or do I miss something? $\endgroup$
    – Ernie060
    Commented Oct 24, 2018 at 20:42
  • $\begingroup$ @Ernie060, of course there's no issue with all-umbilicity. I said “may likely” ... in the generic case with repeated roots, the usual discriminant issues arise. One can lift to branched double covers, etc., but there are issues. $\endgroup$ Commented Oct 24, 2018 at 20:56
  • $\begingroup$ Indeed, since you said "may likely" I assumed you were thinking about the generic case. The formulation "need to require distinct eigenvalues" however started to get me thinking. One question: if one assumes that in a neighbourhood the number of distinct eigenvalues of the shape operator is constant and that the dimension of the corresponding eigenspaces is constant, is it then possible to expand a frame in a point to a local frame field on the neighbourhood? I always thought this is possible. Or can issues still arise in this case? $\endgroup$
    – Ernie060
    Commented Oct 24, 2018 at 21:08
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    $\begingroup$ I think that case is manageable. Of course, this will not be "generic" behavior in most dimensions. This is, of course, one of the apparent weaknesses of the method of moving frames. They have difficulty handling local phenomena where rank fails to be constant. $\endgroup$ Commented Oct 24, 2018 at 21:19
  • $\begingroup$ I also believed this case is manageable. However, I only know this result as a "legend theorem": I tend to use it often, but never saw a fully detailed proof of it... Thank you very much for your help and clarifications, it's much appreciated. $\endgroup$
    – Ernie060
    Commented Oct 24, 2018 at 21:25

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