Let $S$ be a symmetric $(0, 2)$ tensor on a Riemannian manifold $M$. Define $E\_S : M \to \mathbb{Z}$ by $E\_S(x) = \left(\text{the number of distinct eigenvalues of } S\_x\right)$. I've seen the following claims in several papers (also see 16.10 in Besse's Einstein Manifolds):



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> * $M\_S \doteq \left\{x \in M \ \vert \ E\_S \text{ is constant in a neighbourhood of } x \right\}$ is an open and dense subset of $M$
> * The eigenvalues of $S$ are distinct and smooth in each connected component $U$ of $M\_S$.
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I'm a having a hard time proving these facts in a rigorous enough manner (openness is clear, but I'm having a hard time with denseness). It "feels" true since in some sense the eigenvalues should be smooth, but I can't see how to formalize this precisely (smooth from where to where? how to prove smoothness?). I'd appreciate any help.



Why is this subset associated to a $2$-tensor dense?
I claim that the function $E\_S$ is lower semi-continuous, meaning that for any sequence $x\_k \in M$ converging to some $x \in M$, $\lim\_{k \to \infty} E\_S(x\_k) \geq E\_S(x)$.


We want to show that the complement of $M\_S$ has empty interior. Points $x$ in this complement are characterized by the fact that there exists a sequence of points $(x\_k)\_k$ converging to $x$ which is not eventually constant.


Assume by contradiction that there exists an open subset $U$ contained in $\overline{M\_S}$.


Let $n = \max\_{y \in U} E\_S(y)$ (this max is finite because $E\_S(y) \leq \dim(M)$. As $E\_S$ is lower semi-continuous, the set $V = E\_S^{-1}(n) \cap U = E\_S^{-1}(n-1, \infty) \cap U$ is open and non-empty.


But all points in $V$ belong to $M\_S$ as the number of distinct eigenvalues of $S$ is constant on $V$ and $V$ is open. This contradicts the fact that $\overline{M\_S}$ has non-empty interior.


Let us now return to the lower semi-continuity of $E\_S$. I guess this is more or less standard stuff. A nice way to see it is via Rouché's theorem.


Let me work in a coordinate chart.


Assume given a point $x\_0 \in M$ and let $n\_0 = E\_S(x\_0)$. Let $\lambda\_1, \ldots, \lambda\_{n\_0}$ be the distinct eigenvalues of $S$ at $x\_0$. You can consider small disjoint intervals $I\_k$ centered at the eigenvalues and view them as the restriction to $\mathbb{R}$ of balls $B\_1, \ldots B\_k$ in the complex plane.


If $x$ is close enough to $x\_0$, the characteristic polynomial $\chi\_{S(x)}$ of $S$ at $x$ will be close (in the sup-norm over some large compact set encompassing all the balls) to $\chi\_{S(x\_0)}$. Hence, the number of eigenvalues in each ball $B\_i$ fox $S(x)$ (counted with multiplicity) will be the same as for $S(x\_0)$.


This means that locally, distinct eigenvalues cannot merge, i.e. $E\_S(x) \geq E\_S(x\_0)$ for $x$ close enough to $x\_0$.


Rouché's theorem is a very nice tool to prove continuity of the eigenvalues even assuming multiplicity. What you also see is that multiplicity of each of the eigenvalues are locally constant on $M\_S$.


To prove differentiability of an eigenvalue $\lambda = \lambda(x)$ with multiplicity $k \geq 1$, the easiest way is to remark that the derivatives of the characteristic polynomial satisfy $\chi\_{S(x)}^{(k-1)}(\lambda) = 0$ while $\chi\_{S(x)}^{(k)}(\lambda) \neq 0$. So you can just apply the implicit function theorem to $(x, \lambda) \mapsto \chi\_{S(x)}^{(k-1)}(\lambda)$.


Let me know if you need a more detailed answer.


