Timeline for What are the conditions for the dual of the exterior algebra to be isomorphic to the exterior algebra of the dual?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 25 at 8:24 | comment | added | Cameron | @LSpice the map is usually constructed as a perfect pairing $(\Lambda^*_kM)\times(\Lambda^*_kM^\vee)$. See Theorem 3.1 in Conrad's virtualmath1.stanford.edu/~conrad/diffgeomPage/handouts/… | |
| Sep 24 at 16:15 | comment | added | Tom Goodwillie | You certainly need finite generation, even in the field case. Or at least, if $M$ is graded, finite generation in each degree. The canonical map $M^\vee\otimes M^\vee\to (M\otimes M)^\vee$ will not be surjective otherwise. | |
| Sep 24 at 14:56 | comment | added | LSpice | When you refer to several proofs constructing a homomorphism, do they all go the same way? Which way? (I assume right to left, i.e., $\Lambda_k^*(M^\vee) \to (\Lambda_k^*(M))^\vee$.) | |
| Sep 24 at 14:54 | history | edited | LSpice | CC BY-SA 4.0 |
DOI'd link; titles of links
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| Sep 24 at 14:29 | comment | added | Antonius | I think the field case is ok, you can work with a Hamel basis. | |
| Sep 24 at 14:27 | comment | added | Z. M | I think that the linked MO question says that this also holds for finite projective $k$-modules for any commutative ring $k$. | |
| Sep 24 at 14:11 | history | asked | Cameron | CC BY-SA 4.0 |