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.

  • $\begingroup$ Is it clear that the image of a diagonal matrix under an equivariant map must be diagonal, too, or is this an assumption that you are willing to make? $\endgroup$ May 27, 2017 at 12:59
  • $\begingroup$ The map $a: R_+^n\to R_+^n$ (recording the eigenvalues of the matrix $A(X)$, $X$ is diagonal) is insufficient to recover $A$ (already in the case $n=2$), so smoothness of $a$ is not enough. What you want to assume is smoothness of the restriction of $A$ to the subspace of diagonal matrices. $\endgroup$
    – Misha
    May 27, 2017 at 13:11
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    $\begingroup$ @AndreasCap: Yes, $A(X)$ must be diagonal if $X$ is diagonal. This is because the diagonal matrices are the fixed subspace under conjugation by the subgroup of all diagonal orthogonal matrices, a group of order $2^n$. $\endgroup$ May 27, 2017 at 13:18
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    $\begingroup$ @Misha: Because of equivariance, the map $A$ on diagonal matrices must be of the form $$A\bigl(\mathrm{diag}(\lambda_1,\ldots,\lambda_n)\bigr) = \mathrm{diag}\bigl(f(\sigma_1,\ldots,\sigma_{n-1},\lambda_1),\ldots,f(\sigma_1,\ldots,\sigma_{n-1},\lambda_n)\bigr)$$ for some function $f$, where $\sigma_i$ are the elementary symmetric functions of $(\lambda_1,\ldots,\lambda_n)$. Clearly, the function $f$ is sufficient to determine $A$ completely. $\endgroup$ May 27, 2017 at 13:25
  • $\begingroup$ If $f(\sigma_1,\ldots,\sigma_{n-1},\lambda)$ is polynomial in its last argument, then clearly $A$ will extend smoothly to all of $\mathrm{Sym}^+$. I believe that this will suffice (by Taylor expansion) to prove that $A$ will be smooth whenever $f$ is smooth (on the natural domain in $(\mathbb{R}^+)^n$ needed to cover the all-eigenvalues-positive assumption). $\endgroup$ May 27, 2017 at 13:38

1 Answer 1


I think, one can argue as follows.

  1. 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 $D\subseteq\text{Sym}$.

  2. 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.

  3. 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 $a$.

  • $\begingroup$ Thanks for your answer. I see why $\tilde f_i$ vanishes wherever $V$ does, but I'm not sure how to get smoothness of their quotient - the general expression for the Vandermonde inverse is quite intimidating. Is there some easy trick I'm missing? $\endgroup$ May 28, 2017 at 13:47
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    $\begingroup$ There are very general criteria for the divisibility of a smooth function by a polynomial or a real analytic function. See Malgrange's "Ideals of differentiable functions". In the given case, it is elementary, though, since the Vandermonde is a product of linear functions. Claim: The linear function $\ell$ divides $f$ iff $f$ vanishes in $\{\ell=0\}$. Proof: W.l.o.g. $\ell=x_1$ such that $f(0,x_2,\ldots,x_n)=0$. Then $f=x_1g$ with $g=\int_0^1 f_{x_1}(tx_1,x_2,\ldots,x_n)dt$ smooth. This trick is old. $\endgroup$ May 28, 2017 at 14:29
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    $\begingroup$ @FriedrichKnop Old trick also known as Hadamard's lemma en.wikipedia.org/wiki/Hadamard%27s_lemma $\endgroup$ Jun 2, 2017 at 10:54

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