Timeline for Subgroup of $\mathrm{GL}_n$ stabilizing linear subspace skew-symmetric matrices
Current License: CC BY-SA 4.0
22 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2020 at 21:20 | vote | accept | Libli | ||
| Apr 23, 2020 at 10:01 | vote | accept | Libli | ||
| Apr 26, 2020 at 21:18 | |||||
| Apr 23, 2020 at 9:25 | answer | added | Friedrich Knop | timeline score: 5 | |
| Apr 22, 2020 at 21:00 | answer | added | Robert Bryant | timeline score: 6 | |
| Apr 22, 2020 at 6:15 | history | edited | Libli | CC BY-SA 4.0 |
edited body
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| Apr 22, 2020 at 6:14 | comment | added | Libli | @RobertBryant : you're obviously right. I will edit that. | |
| Apr 21, 2020 at 23:05 | comment | added | Robert Bryant | I'm sure that I'm not the first person to point this out, but I didn't see it in the previous comments: The statement as you have made it is false. Clearly the stabilizer subgroup contains all the multiples of the identity in $\mathrm{GL}_{2m}(\mathbb{C})$, so it's never of dimension $0$. Maybe you meant to write $\mathrm{SL}_{2m}(\mathbb{C})$ or $\mathrm{PGL}_{2m}(\mathbb{C})$? | |
| Apr 21, 2020 at 22:03 | answer | added | dhy | timeline score: 3 | |
| Apr 21, 2020 at 20:05 | comment | added | Libli | @paulgarrett : As I told you in my first comment, we know the action and the linearization is obtained easily. But computing the isotropy subalgebra at a generic point is not trivial. Otherwise, why would Kimura and Sato have had witten a 156-pages long paper on prehomogeneous vector spaces? | |
| Apr 21, 2020 at 19:52 | comment | added | paul garrett | It just occurred to me: linearizing to Lie algebras, is a highest-weight computation feasible? | |
| Apr 21, 2020 at 18:14 | comment | added | Libli | Well, I might be mistaken, but I think that the sentence is badly written and "this space is zero" refers to the stabilizer being finite. Furthermore, it is claimed at the beginning of the proof that the space of infinitesimal automorphisms of a linear section of the Pfaffian will be proven to vanish. So following her proof, I think that she deduces that from the fact that the tangent space to the stabilizer is zero. | |
| Apr 21, 2020 at 18:05 | comment | added | LSpice | @Libli, thanks. I see conditions there that look like yours, but I don't see the claim that the group is finite. I do see a lot of notation I don't recognise, so maybe you've just translated into concrete language …. | |
| Apr 21, 2020 at 18:03 | comment | added | LSpice | @dhy, it seems to me that the entire subspace is generic in the Zariski sense among subspaces of its dimension …. Are you saying that the condition of dimension bigger than 3 should be codimension bigger than 3? | |
| Apr 21, 2020 at 18:02 | comment | added | Libli | @LSpice : end of page 11, proof of lemma 2.2 | |
| Apr 21, 2020 at 18:02 | comment | added | dhy | @LSpice It is assumed in the paper that the codimension of this subspace is more than $2(2m-2).$ (By duality really the only relevant condition is that the codimension is $\geq 3$ (speaking of, I believe in this question it should be dimension $\geq 3$, not $>3$, no?)). Generic here means generic (in the Zariski sense) among subspaces of that fixed dimension. | |
| Apr 21, 2020 at 17:45 | comment | added | LSpice | What does "generic" mean? The condition is very surprising to me, because it seems easier to preserve a bigger subspace. For example, the preserver of $\bigwedge^2\mathbb C^{2m}$ is all of $\operatorname{GL}_{2m}$ …. Could you point to the precise statement? | |
| Apr 21, 2020 at 17:44 | history | edited | LSpice | CC BY-SA 4.0 |
Name of paper; PDF -> abs
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| Apr 21, 2020 at 17:37 | comment | added | Libli | @paulgarrett : probably, but that means finding a "generic" $A \in W \otimes \bigwedge^2 \mathbb{C}^n$ and computing the isotropy Lie sub-algebra of $\mathfrak{sl}_{W} \oplus \mathfrak{gl}_{2m}$. I am not sure how this can be easily checked. | |
| Apr 21, 2020 at 17:36 | history | edited | Libli | CC BY-SA 4.0 |
references to the paper added
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| Apr 21, 2020 at 16:59 | comment | added | paul garrett | Presumably easier to check at the Lie algebra level, in any case. | |
| Apr 21, 2020 at 13:21 | history | edited | Libli | CC BY-SA 4.0 |
dimension of the ambiant space is even
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| Apr 21, 2020 at 8:20 | history | asked | Libli | CC BY-SA 4.0 |