Timeline for Analogue Bialgeras vs Lie bialgebras
Current License: CC BY-SA 2.5
9 events
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| Mar 20, 2011 at 17:40 | history | edited | Micah Miller | CC BY-SA 2.5 |
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| Mar 17, 2011 at 19:45 | history | edited | Micah Miller | CC BY-SA 2.5 |
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| Mar 15, 2011 at 20:04 | comment | added | Micah Miller | @Theo: that's a better way to put it. @Qiaochu: Let $A$ be an associative algebra and $[A]$ be the Lie algebra obtained by symmetrizing the product. Then $A \otimes A$ is an algebra, and $[A \otimes A]$ is a Lie algebra obtained by symmetrizing the product on $A \otimes A$. This is note the same as the bracket on $[A \otimes A]$, which won't satisfy Jacobi. I think this would cover your example. | |
| Mar 15, 2011 at 14:18 | comment | added | Theo Johnson-Freyd | @Qiaochu: One way to say Micah's point is that the category of Lie algebras does not have a functorial monoidal structure covering tensor product of underlying vector spaces. This is equivalent to saying that the operad Lie is not an operad of coalgebras. You can see this immediately: Lie(2) is one-dimensional, and the symmetric group acts by the sign representation. But the sign permutation does not have a nontrivial map to its tensor square. So just at the level of the anticommutativity you're hosed. | |
| Mar 15, 2011 at 2:43 | comment | added | Micah Miller | cont'd: I'm not sure how you would define another kind of bracket, if you start with two abstract Lie algebras. Perhaps, with certain examples like $gl(n)$, you can use extra knowledge about the Lie algebra to try something different. But if all you have is $L_i \otimes L_i \rightarrow L_i$, for $i=1,2$, then what else can you do with $(L_1 \otimes L_2)^{\otimes 2} \rightarrow L_1 \otimes L_2$ besides combining the brackets? | |
| Mar 15, 2011 at 2:37 | history | edited | Micah Miller | CC BY-SA 2.5 |
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| Mar 14, 2011 at 21:01 | comment | added | Micah Miller | You're right, there might be another bracket. But since the definition of a bialgebra used the fact that $A \otimes A$ is an associative algebra with naive combination of products, we would try to do the same thing with a Lie version of a bialgebra, if such a thing could exist. | |
| Mar 14, 2011 at 19:10 | comment | added | Qiaochu Yuan | It's not clear to me that there aren't other brackets you could try on the tensor product. If I wanted the tensor product of $\mathfrak{gl}(n)$ and $\mathfrak{gl}(m)$ to naturally be a subalgebra of $\mathfrak{gl}(nm)$, then this isn't the bracket I would use. | |
| Mar 14, 2011 at 18:05 | history | answered | Micah Miller | CC BY-SA 2.5 |