Conditions for existence of $m$-th differentiable root of a non-negative definite matrix

Question

In M.I. Friedlin's famous paper "On the Factorization of Non-Negative Definite Matrices", he shows that if a non-negative definite symmetric matrix $a(x)=\{a^{ij}(x)\}_{i,j=1}^n$ is in $C^2(\mathbb{R}^n)$, then there exists a symmetric Lipschitz continuous matrix $\sigma(x)$ such that $\sigma^2(x)=a(x)$.

Let $m>0$ be an integer. Are there any reasonable conditions on $a(x)$ that ensure there exists a symmetric matrix $\sigma(x)$ such that $\sigma^2(x)=a(x)$ and $\sigma\in C^m(\mathbb{R}^d)$?

Answer 1

The problem is multiple eigenvalues. If $a$ has "truly" isolated eigenvalues for all $x$, that is, if $\lambda_i(x)$ is the i-th eigenvalue of $a(x)$ it is true that
$$ \inf_{x\in\mathbb{R}}\min_{i,j} |\lambda_i(x)-\lambda_j(x)| >0 $$ then the regularity of $a$ is that of $\sigma$, because the projectors are just fine. If not, trouble could occur, see that question.