Timeline for Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 22, 2016 at 20:26 | answer | added | Abdelmalek Abdesselam | timeline score: 2 | |
| Dec 21, 2016 at 14:01 | answer | added | Robert Bryant | timeline score: 7 | |
| Dec 21, 2016 at 9:24 | vote | accept | evgeny | ||
| Dec 21, 2016 at 8:43 | answer | added | Friedrich Knop | timeline score: 10 | |
| Dec 21, 2016 at 7:47 | comment | added | evgeny | @YCor, edited: I hope that now $\mathbb C[W]^{\mathbb C^* \rtimes \mathbb Z/2\mathbb Z}$ is right? | |
| Dec 21, 2016 at 7:45 | history | edited | evgeny | CC BY-SA 3.0 |
corrected about degree 4, not 2
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| Dec 21, 2016 at 7:29 | comment | added | YCor | @evgeny, you should edit further, since all expectations about $C[sq]$ make little sense now. | |
| Dec 21, 2016 at 6:21 | history | edited | evgeny | CC BY-SA 3.0 |
corrected, thanks!
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| Dec 20, 2016 at 21:49 | comment | added | Robert Bryant | @evgeny I hate to have to point this out, but if $\omega$ is a $3$-form then $\omega^2 \equiv 0$ (after all, $\omega\wedge\eta = -\eta\wedge\omega$ when $\omega,\eta\in\Lambda^3$), so your attempt to construct an invariant this way fails. In fact, there is a nonzero invariant polynomial under $\mathrm{SL}(6,\mathbb{C})$ and it does generate the ring of invariants, but it is irreducible of degree $4$, not degree $2$. | |
| Dec 20, 2016 at 21:11 | comment | added | Sasha | Vinberg-Popov, Invariant theory (VINITI, Algebraic geometry - IV), and references therein. | |
| Dec 20, 2016 at 20:57 | comment | added | evgeny | @Sasha, could you give me a reference or a name for this result? | |
| Dec 20, 2016 at 20:45 | comment | added | Sasha | It does. This is a well-known result in invariant theory. | |
| Dec 20, 2016 at 20:41 | history | asked | evgeny | CC BY-SA 3.0 |