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.

  • 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? – Andreas Cap May 27 '17 at 12:59
  • 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. – Misha May 27 '17 at 13:11
  • 1
    @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 $2^n$. – Robert Bryant May 27 '17 at 13:18
  • 1
    @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. – Robert Bryant May 27 '17 at 13:25
  • 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). – Robert Bryant May 27 '17 at 13:38
up vote 7 down vote accepted

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 $a$.

  • 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? – Anthony Carapetis May 28 '17 at 13:47
  • 1
    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. – Friedrich Knop May 28 '17 at 14:29
  • 1
    @FriedrichKnop Old trick also known as Hadamard's lemma en.wikipedia.org/wiki/Hadamard%27s_lemma – Vít Tuček Jun 2 '17 at 10:54

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.