If an equivariant map is smooth on diagonal matrices, is it smooth everywhere? This is a followup from a question I asked on math.SE, which received a helpful answer but unfortunately not a complete one. $\def\Sym{\mathrm{Sym}_{n\times n}}$
$\def\s{\mathrm{Sym}}\def\sp{\s^+}$Let $\sp \subset GL(n,\mathbb R)$ denote the space of symmetric positive-definite $n \times n$ matrices. (It might be more fruitful to think of the full subspace of symmetric matrices, since it turns all the group actions below in to representations.)
I am interested in functions $A : \sp \to \sp$ that are equivariant under the natural conjugation action of $O(n)$; i.e. such that$$A(R^T X R) = R^T A(X) R$$ for all $X \in \sp, R \in O(n,\mathbb R)$.
Since we can diagonalize any $X$, we know that such an $A$ is determined by its restriction to diagonal matrices, which gives a function $a : (0,\infty)^n \to (0,\infty)^n$ which is equivariant under the natural permutation action of the symmetric group $S_n$. Conversely, any such $a$ can be extended uniquely to an equivariant $A$. Thus we can specify an $A$ by just declaring what it does to eigenvalues.

Question: If we know $a$ is smooth, can we conclude $A$ is smooth?

In the analogous problem for $O(n)$-invariant scalars $F : \sp \to \mathbb R$ (which reduce to symmetric functions $f : (0,\infty)^n \to \mathbb R$ of the eigenvalues), we can solve this problem using Glaeser's "differentiable Newton's theorem" - we get that a smooth symmetric function of the eigenvalues is a smooth function of the symmetric matrix invariants, which are in turn smooth functions of the matrix itself. The key is that $S_n$-invariant polynomials of the eigenvalues and $O(n)$-invariant polynomials of the matrices are the same thing.
Since I couldn't find any similar work on equivariant maps (please relieve me of my ignorance!), my only thought was to use something like this theorem of Schwarz to study the scalar $\tilde A : \sp \times \sp\to \mathbb R$ defined by $$\tilde A(X,Y) = \langle A(X), Y\rangle,$$ which is invariant under the action $\rho_R(X,Y) = (R^T X R, R^T Y R)$. We can define an $S_n$-invariant $\tilde a : (0,\infty)^n \times (0,\infty)^n \to \mathbb R$ similarly; but unfortunately it seems to me that there is no obvious relation between $\tilde a$ and $\tilde A$ - we only have $a(\lambda(X)) \cdot \lambda(Y) = \langle A(X), Y \rangle$ when $X,Y$ have the same eigenvectors.
Any pointers would be great - this is a tangent from my usual research, so there is probably a whole body of relevant work I'm unaware of.
 A: I think, one can argue as follows.


*

*Let $D\subseteq\text{Sym}$ be the diagonal matrices. Since $\exp:D\to D^+$ and $\exp:\text{Sym}\to\text{Sym}^+$ are compatible diffeomorphisms it suffices to answer the analogous problem for $D\subseteq\text{Sym}$. 

*For $m=0,\ldots,n-1$ let $c_m:D\to D:(x_i)\mapsto(x_i^m)$. These are certainly $S_n$-covariants. It is well known that they freely generate the space of all polynomial $S_n$-covariants as a module over the ring of all polynomial invariants. I claim that this also holds for smooth functions. In other words, for every smooth covariant $a$ there are smooth invariants $f_0,\ldots,f_{n-1}$ with
$$
a=\sum_{m=0}^{n-1}f_mc_m
$$
There is probably a general theorem but here is an ad hoc argument which independently shows that the $c_m$ form a basis. Let $a=(a_1,\ldots,a_n)$ be the components. Then we have to solve
$$
a_i=\sum_{m=0}^{n-1}f_m x_i^m
$$
This is a linear system of equations for the $f_m$ with the Vandermonde matrix as coefficients. So we can uniquely solve it. One gets $f_m=\frac{\tilde f_m}{V}$ where $\tilde f_m$ is smooth and $V$ is the Vandermonde determinant. The equivariance of $a$ implies that $\tilde f_m$ vanishes where $V$ vanishes so $f_m$ is a smooth function.

*Glaeser's theorem shows that $f_i(x)$ can be extended to a smooth $O(n)$-invariant $F_i(X)$ on $\text{Sym}$. Thus
$$
A(X)=\sum_{m=0}^{n-1}F_m(X)X^m
$$
is a smooth extension of $a$.
