Timeline for Structure of the adjoint representation of a (finite) group (Hopf algebra) ?
Current License: CC BY-SA 3.0
6 events
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| Jan 5, 2014 at 15:18 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jan 5, 2014 at 15:17 | comment | added | Will Sawin | Initially this tensor product is only equal to the matrix algebra as an $A \otimes A$-module, with one $A$ acting on the left and the other $A$ acting on the right. The Hopf algebra structure gives us a map $A \to A \otimes A$, allowing us to turn this into an $A$-module. This is exactly the same structure we use to take the left action and the right action and combine them to a single action, forming the adjoint representation. So the adjoint representation is equivalent to the sum of tensor products. | |
| Jan 5, 2014 at 15:16 | comment | added | Alexander Chervov | It seems "Hopf property" is not mentioned explicitly in your argument ? | |
| Jan 5, 2014 at 15:15 | comment | added | Will Sawin | Well I meant to assume semi simplicity. So we can view it as a sum of matrix algebras over division algebras. Each $n \times n$ matrix algebra over a division algebra corresponds to a simple representation, which is just an $n$-dimensional vector space over that division algebra. If we tensor this with its dual (I guess we have to tensor it over the division algebra = the automorphism group of the module, not over the base field), then we get the matrix algebra. | |
| Jan 5, 2014 at 9:43 | comment | added | Alexander Chervov | Thank you. Can you give a comment/refrence for a "Hopf algebra, as a diagonal representation of itself, is equal to a sum of the tensor product of each simple representation with its dual". | |
| Jan 4, 2014 at 23:08 | history | answered | Will Sawin | CC BY-SA 3.0 |