Timeline for Schubert calculus expressed in terms of the cotangent space of the Grassmannians
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2017 at 18:23 | comment | added | Anton Mellit | Sorry, I should have made it more explicit. It is known that irreducible modules of $U(n)$ are precisely the Schur functors applied to the standard representation. That's why these are irreducible modules. | |
| Aug 14, 2017 at 9:54 | comment | added | Han Jin Ma | Thanks for the great answer! Just to be sure I understand what's happening, in the last displayed equation, is it clear that each of the summands are irreducible modules? If they're not irreducible then I don't understand why Taking invariants we see that only terms with $\lambda = \mu$ survive, | |
| Aug 12, 2017 at 16:23 | vote | accept | Han Jin Ma | ||
| Aug 12, 2017 at 16:23 | history | bounty ended | Han Jin Ma | ||
| Aug 12, 2017 at 16:23 | vote | accept | Han Jin Ma | ||
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| Aug 12, 2017 at 16:23 | vote | accept | Han Jin Ma | ||
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| Aug 12, 2017 at 16:21 | vote | accept | Han Jin Ma | ||
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| Aug 12, 2017 at 16:21 | vote | accept | Han Jin Ma | ||
| Aug 12, 2017 at 16:21 | |||||
| Aug 11, 2017 at 15:26 | history | answered | Anton Mellit | CC BY-SA 3.0 |