Timeline for Ring of invariants for $n$-tuples of Lie algebras
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2021 at 8:42 | history | edited | Sean Lawton | CC BY-SA 4.0 |
Added some hopefully clarifying remarks.
|
| Apr 26, 2021 at 17:06 | vote | accept | skeptic | ||
| Apr 26, 2021 at 17:06 | comment | added | skeptic | @ Sean Lawton, thanks once again | |
| Apr 26, 2021 at 17:03 | comment | added | Sean Lawton | Depending on what you want, there are many open questions. But some things are known. For example, Gerald Schwarz has studied the case of $G_2$. To see what I outlined above played out in detail for $\mathrm{SL}(n,\mathbb{C})$ see page 11 of my paper: arxiv.org/pdf/0709.4403.pdf. | |
| Apr 26, 2021 at 16:57 | comment | added | skeptic | thanks a lot, I need to think over your suggestions before I can make any more comments ... but final question if I may, just a curiosity .. what happens for exceptional groups ? has it been studied? | |
| Apr 26, 2021 at 16:50 | comment | added | Sean Lawton | To go the other way around, you need to use a "variation function" $F:G\to \mathfrak{g}$ which is conjugation invariant. This will turn $G$-invariants of $\mathfrak{g}^k$ into $G$-invariants of $G^k$. This is essentially "integration". Again, a simple example is $\mathrm{SL}(n,\mathbb{C})$ which is $X\mapsto X-(tr(X)/n)I$. In general, you take an orthogonal structure $B$ on $\mathfrak{g}\times \mathfrak{g}$ which always exists for reductive $G$ (killing form if semisimple) and solve: $B(F(A),X)=(d/dt)|_{t=0}f(AexptX)$ for any $G$-invariant $f:G\to\mathbb{C}$. | |
| Apr 26, 2021 at 16:44 | comment | added | Sean Lawton | There is something to do here. First understand the simple case of $\mathrm{SL}(n,\mathbb{C})$. In that case the invariants you care about are those of $nxn$ traceless matrices. So there are two steps to related the two invariant rings. Step 1: add in the traces of single generic matrices; and Step 2: quotient by $det-1$ for each generic matrix. In general, the relationship is in one direction you take the tangent space of the identity in the $G$-character variety of $F_k$ to obtain $\mathfrak{g}^k//G$ (and invariants of the former give invariants of the latter). | |
| Apr 26, 2021 at 16:34 | comment | added | skeptic | I see .. thanks .. this is nice .. if I may ask .. is it very obvious how to pass from $G$ to $\mathfrak{g}$ (or the other way) ? | |
| Apr 26, 2021 at 16:21 | comment | added | Sean Lawton | Yes. You are comparing the $G$-invariants of $\mathrm{Hom}(F_k,G)\cong G^k$ versus $\mathfrak{g}^k$. The invariant rings are "morally'' the same. | |
| Apr 26, 2021 at 16:10 | comment | added | skeptic | Thank you for taking interest in my question, but how are you connecting $k >2$ case with character variety ? are we considering $Hom(F_k, G)$ where $F_k$ is the free group on $k$-generators? | |
| Apr 26, 2021 at 16:02 | history | edited | Sean Lawton | CC BY-SA 4.0 |
Added a remark to address a more general question, or maybe the one the OP actually wanted (not sure).
|
| Apr 26, 2021 at 12:57 | history | edited | Sean Lawton | CC BY-SA 4.0 |
added 33 characters in body
|
| Apr 26, 2021 at 12:47 | history | answered | Sean Lawton | CC BY-SA 4.0 |