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.

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.

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
$$ \min_{x\in\mathbb{R}}\min_{i,j} |\lambda_i(x)-\lambda_j(x)| >0 $$ then the regularity of $a$ is that of $\sigma$, because the projectors are just fine. If not, trouble could occur, see that question.

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.