Timeline for Coordinate free way to construct inner product on exterior powers
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2023 at 17:40 | comment | added | anon | For pure elements, I believe $\langle u_1\wedge\cdots\wedge u_k,v_1\wedge\cdots\wedge v_k\rangle=\det[\omega(u_i,v_j)]$, so basically the Gramian determinant but polarized. Also see this answer of mine for a geometric interpretation. tldr: it generalizes $u\cdot v=\|u\|\|v\|\cos\theta$ to angles between subspaces. | |
| Mar 3, 2023 at 16:36 | vote | accept | Cindy | ||
| Mar 1, 2023 at 18:27 | answer | added | Cindy | timeline score: 3 | |
| Mar 1, 2023 at 9:39 | comment | added | Igor Khavkine | For uniqueness, you need to bring in a bit of representation theory. Let's say $\jmath\colon \wedge^k V \to (\wedge^k V)^* \cong \wedge^k V$ represents a different $O(\omega,V)$ invariant inner product, while the maps and isomorphism should be read as those of $O(\omega,V)$ representations. Since $\wedge^k V$ is an irreducible representation, by Schur's lemma, $\jmath$ must be proportional to the identity, which represents your standard inner product. Uniqueness fails for example for $S^k V$, because it is not an irreducible representation (exercise). | |
| Mar 1, 2023 at 1:55 | comment | added | Willie Wong | Re: it is not obvious that it is positive-definite. Of course not. The procedure you described works for ANY non-degenerate bilinear form $\omega$, with arbitrary signature. (Any such form provides an isomorphism $V\to V^*$ and the procedure you outlined works.) The fact that you need to do a computation to show that pos def $\omega$ generates a pos def form on $\wedge^k V$ is probably a feature, not a bug, of your argument. | |
| Feb 28, 2023 at 23:54 | comment | added | Tom Goodwillie | I would say that the calculation you have to do to show that your inner product is positive-definite is in fact easier than the one you want to avoid. Your definition quickly implies that the $e_{i_1}\wedge\dots e_{i_k}$ are orthonormal. | |
| Feb 28, 2023 at 20:33 | review | Close votes | |||
| Mar 4, 2023 at 17:36 | |||||
| S Feb 28, 2023 at 18:38 | review | First questions | |||
| Feb 28, 2023 at 19:35 | |||||
| S Feb 28, 2023 at 18:38 | history | asked | Cindy | CC BY-SA 4.0 |