Timeline for If tensor product of representations is a representation, must we have a bialgebra?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2014 at 2:37 | answer | added | gguest | timeline score: 2 | |
| Oct 6, 2013 at 19:17 | answer | added | Evan Jenkins | timeline score: 1 | |
| Oct 6, 2013 at 18:41 | comment | added | Joshua Grochow | @AllenKnutson: I thought of that, but couldn't quite get it to work. I think I agree with Peter Samuelson's guess, that all this leads to is an algebra $A$ such that $A \otimes A$ is an $A - A \otimes A$-bimodule that satisfies an associativity condition... | |
| Oct 6, 2013 at 17:00 | answer | added | Peter Samuelson | timeline score: 5 | |
| Oct 6, 2013 at 16:58 | comment | added | Qiaochu Yuan | @Allen: I suspect the symmetric monoidal axioms are only going to give you weak compatibility; see en.wikipedia.org/wiki/Weak_Hopf_algebra . | |
| Oct 6, 2013 at 16:48 | answer | added | Qiaochu Yuan | timeline score: 7 | |
| Oct 6, 2013 at 16:29 | comment | added | Qiaochu Yuan | You need to be careful what you mean by "the tensor product of vector spaces gives a symmetric monoidal structure." What you want to say is that there exists a symmetric monoidal structure which, after being hit with the forgetful functor, is the tensor product of vector spaces. | |
| Oct 6, 2013 at 16:19 | comment | added | Allen Knutson | Take $V=W=A$, so you get the map $A \to A\otimes A$, $a \mapsto a(1\otimes 1)$. I'm guessing the symmetric monoidal axioms are going to tell you that's a ring homomorphism (but haven't checked at all). | |
| Oct 6, 2013 at 15:40 | history | asked | Joshua Grochow | CC BY-SA 3.0 |