Timeline for Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 23, 2016 at 8:12 | history | suggested | evgeny | CC BY-SA 3.0 |
added a link on an article with the details of the construction
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| Dec 23, 2016 at 6:56 | review | Suggested edits | |||
| S Dec 23, 2016 at 8:12 | |||||
| Dec 22, 2016 at 21:05 | comment | added | Robert Bryant | @Sasha I guess the difference in point of view for me is that Friedrich's construction of $\alpha$ is a construction that produces a quartic invariant for any symplectic representation $V$ of any semisimple Lie group $G$ (via its Killing form) (of course, it might vanish identically in some situations), while, from my point of view, the space $S$ for which $V=\Lambda^3(S^*)$ is the fundamental object, and interpreting $J_\phi$ as an element of $\mathrm{End}(S)$ (so that it can be squared) is something that you wouldn't see if all you had was $V$. | |
| Dec 21, 2016 at 16:47 | comment | added | Sasha | Honestly, I don't see any difference (except for the normalization). But definitely, you wrote the map in a more explicit way and explained more of its properties. | |
| Dec 21, 2016 at 15:58 | comment | added | Robert Bryant | @Sasha I agree that it's similar (though the normalization of $\alpha$ is different), but thinking of it this way yields a bit more information: The fact that $(J_\phi)^2$ is a multiple of the identity map on $S$ turns out to be very useful in deriving the normal forms of the orbits, even the degenerate ones. | |
| Dec 21, 2016 at 14:28 | comment | added | Sasha | This actually is the same construction as in Friedrich's answer. The map $\phi \mapsto J_\phi$ is just the moment map, and the trace of the square is just the Killing form. | |
| Dec 21, 2016 at 14:01 | history | answered | Robert Bryant | CC BY-SA 3.0 |