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Apr 26, 2020 at 21:20 vote accept Libli
Apr 23, 2020 at 20:46 comment added dhy @Libli: Now that Friedrich's Knop's answer has appeared, feel free to unaccept mine and accept his instead. Interestingly, the Andreev-Vinberg-Èlašvili argument also does this split into semi-simple and unipotent, but then they cleverly (almost magically) apply the classification of invariant inner forms on $\mathfrak{g}$... it seems plausible that my long calculations are somehow encoding their more conceptual Lie-theoretic argument.
Apr 23, 2020 at 10:01 vote accept Libli
Apr 26, 2020 at 21:18
Apr 22, 2020 at 21:08 comment added dhy @RobertBryant: Upon closer inspection, it appears that the paper may be assuming $m\geq 4$. In any case, the $m=3$ case is not needed for Voisin's argument.
Apr 22, 2020 at 19:00 comment added Libli @RobertBryant : still interested by your proof. And it is included in my question (I wrote bigger meaning bigger or equal).
Apr 22, 2020 at 18:58 comment added Libli @RobertBryant : Well, the proof of the lemma which uses this result is done in a rather offhand way, to say the least. It's not clear at all what cases she needs precisely. At least, she seems to claim that it must be true for any $r \geq 3$. You can have a look at the paper here : arxiv.org/pdf/2004.09310.pdf page 11, proof of lemma 2.2.
Apr 22, 2020 at 17:27 comment added Robert Bryant @Libli: I have a proof, but since you omitted this case in your question, I am guessing that this may already be known. On the other hand, in dhy's answer, it was stated that the case m=r=3 was claimed in the paper. Can you tell me whether this is so? (I don't have a copy of the paper to look at myself.)
Apr 22, 2020 at 15:30 comment added Libli @RobertBryant : Very interesting! Do you have a proof or a reference for this?
Apr 22, 2020 at 15:15 comment added Robert Bryant I don't know whether the case $m=r=3$ is claimed in the paper, but if it is, this is an error. The $\mathrm{SL}_6(\mathbb{C})$-stabilizer of a generic 3-plane in $\Lambda^2(\mathbb{C}^6)$ has dimension $1$.
Apr 22, 2020 at 6:19 comment added Libli @dhy : thanks for the answer. I think I can follow the argument in the semi-simple case, the unipotent case is still a bit difficult for me. But at least, I understood that the general idea of decomposing into semi-simple and unipotent will provide bounds on the possible exceptions. That's a nice idea. I am just waiting a day or two to see if someone comes up with a simpler proof or a reference to a simpler proof.
Apr 21, 2020 at 22:05 comment added Sam Hopkins Are "worst possible proofs" collected in a book kept by the devil?
Apr 21, 2020 at 22:03 history answered dhy CC BY-SA 4.0