Timeline for If an equivariant map is smooth on diagonal matrices, is it smooth everywhere?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2017 at 11:51 | vote | accept | Anthony Carapetis | ||
| May 27, 2017 at 17:40 | answer | added | Friedrich Knop | timeline score: 9 | |
| May 27, 2017 at 13:59 | comment | added | Misha | @RobertBryant: I see, I forgot about this finite subgroup. | |
| May 27, 2017 at 13:38 | comment | added | Robert Bryant | 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). | |
| May 27, 2017 at 13:25 | comment | added | Robert Bryant | @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. | |
| May 27, 2017 at 13:18 | comment | added | Robert Bryant | @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 $2^n$. | |
| May 27, 2017 at 13:11 | comment | added | Misha | 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. | |
| May 27, 2017 at 12:59 | comment | added | Andreas Cap | 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? | |
| May 27, 2017 at 10:04 | review | First posts | |||
| May 27, 2017 at 10:12 | |||||
| May 27, 2017 at 9:59 | history | asked | Anthony Carapetis | CC BY-SA 3.0 |