Timeline for Optimal reference for tensor product of symmetric bilinear forms?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2011 at 17:26 | vote | accept | Jim Humphreys | ||
| May 25, 2011 at 2:36 | comment | added | Allen Knutson | I really don't see the risk here. From an inner product $\langle,\rangle$, define $\phi:V\to V^*$ by $\vec v \mapsto \langle \vec v,\bullet \rangle$. In general (and without Choice), I am ready to believe that e.g. $V^*$ might be just $0$, but then $V$ isn't going to have an inner product, either. I'm not using that the natural map $V\to (V^*)^*$ is 1:1 anywhere. | |
| May 24, 2011 at 17:18 | comment | added | Jim Humphreys | @Marty: All of this is in principle purely algebraic (from my point of view). | |
| May 24, 2011 at 16:16 | comment | added | Marty | In the infinite-dimensional case, are you working with topological vector spaces and continuous (in some topology) bilinear forms? Or just purely algebraic vector spaces? | |
| May 24, 2011 at 14:40 | comment | added | Jim Humphreys | @Allen: This makes good sense, up to a point, but I'm especially concerned with treatment of infinite dimensional vector spaces where it gets risky to bring in dual spaces this way (?) | |
| May 24, 2011 at 12:33 | answer | added | Francesco Polizzi | timeline score: 2 | |
| May 24, 2011 at 0:43 | comment | added | Allen Knutson | [An index-free argument.] Lemma: if $A\to B$ is injective, then $A\otimes C \to B\otimes C$ is too. Think of a bilinear form on $V$ as a map $V \to V^\ast$. Proof of desired theorem: compose $V \otimes W \to V^\ast \otimes W \to V^\ast \otimes W^\ast$ to get an injective map $V\otimes W \to V^\ast \otimes W^\ast$, the desired bilinear forms. The only space I see finite-dimensionality being relevant is to know the 1:1 maps are also onto ("strong nondegeneracy"). | |
| May 23, 2011 at 22:43 | history | asked | Jim Humphreys | CC BY-SA 3.0 |