Timeline for Homomorphism between exterior powers of a free module of finite rank
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2009 at 19:55 | vote | accept | Hideyuki Kabayakawa | ||
| Dec 10, 2009 at 14:58 | answer | added | Mariano Suárez-Álvarez | timeline score: 6 | |
| Dec 9, 2009 at 23:57 | comment | added | Alicia Garcia-Raboso | @a-fortiori: oops, I forgot the dual... Yes, your modification of the question has a canonical answer, while for the original one you need to choose an isomorphism of M with its dual. | |
| Dec 9, 2009 at 23:13 | vote | accept | Hideyuki Kabayakawa | ||
| Dec 9, 2009 at 23:13 | |||||
| Dec 8, 2009 at 21:15 | comment | added | user2035 | @Alberto: I was explaining my previous comment. You need to change the question in some way to get a canonical map, contraction does not work for m ∈ M and ω ∈ Λ^m M. (Please apologize my using the letter m twice.) | |
| Dec 8, 2009 at 20:03 | comment | added | Alicia Garcia-Raboso | @a-fortiori: the isomorphism you are giving takes (m-1)-forms to m-forms, not the other way around. I guess what Francisco wants would be m ↦ (ω ↦ m ∟ ω) (contraction). | |
| Dec 8, 2009 at 18:04 | comment | added | user2035 | The homomorphism is given by m ↦ (ω ↦ m ∧ ω). Using a basis you can see that it is an isomorphism if M is free of rank r. | |
| Dec 8, 2009 at 17:09 | vote | accept | Hideyuki Kabayakawa | ||
| Dec 9, 2009 at 23:13 | |||||
| Dec 8, 2009 at 16:17 | comment | added | Hideyuki Kabayakawa | How do you ask this question using exterior algebra? | |
| Dec 8, 2009 at 8:53 | comment | added | user2035 | It would feel more natural if you asked for the homomorphism M → Hom(Λ^{m-1}M, Λ^m M) which is induced by the graded algebra structure. | |
| Dec 8, 2009 at 1:00 | answer | added | Alicia Garcia-Raboso | timeline score: 6 | |
| Dec 8, 2009 at 0:46 | history | asked | Hideyuki Kabayakawa | CC BY-SA 2.5 |