Timeline for Isomorphic maximal commutative semi-simple sub algebras of complex matrices
Current License: CC BY-SA 3.0
7 events
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| Jun 8, 2016 at 17:06 | history | edited | Sean Lawton | CC BY-SA 3.0 |
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| Jul 24, 2013 at 11:44 | history | edited | user9072 |
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| Apr 15, 2013 at 6:31 | comment | added | user33117 | Thank you for your answers. Are the same arguments hold when replacing M_n(C) with twisted group algebra. Or maybe even for any semi-simple algebra? That is When giving A_1,A_2 two Isomorphic maximal commutative semi-simple sub algebras of a twisted group algebra A (or any semi-simple algebra). Are these algebras conjugate in A. | |
| Apr 14, 2013 at 14:47 | comment | added | user30379 | @ofir: Any such $A$ is a product of copies of $\mathbf{C}$, so a faithful representation $A \hookrightarrow {\rm{M}}_n(\mathbf{C})$ on $\mathbf{C}^n$ from a commutative semisimple $\mathbf{C}$-algebra $A$ sends the primitive idempotents to {\em distinct} pairwise orthogonal commuting nonzero idempotent linear operators on $\mathbf{C}^n$ whose sum is the identity operator. This is exactly a decomposition of $\mathbf{C}^n$ as a direct sum of nonzero subspaces. The "maximal" way to do this is with an ordered $n$-tuple of independent lines, and in a suitable basis all $n$-tuples look the same... | |
| Apr 14, 2013 at 12:09 | comment | added | ofir | Thank you for that answer. Where did you use the isomorphism condition? Are all the semi-simple commutative maximal sub-algebras of M_n(C) isomorphic? If I understand your answer, the only maximal commutative semi-simple sub-algebra of M_(C) is the diagonal algebra? | |
| Apr 14, 2013 at 11:21 | comment | added | Geoff Robinson | The answer is yes. Each corresponds to a decomposition of the identity as a sum of mutually orthogonal primitive idempotents, each of which has rank 1. | |
| Apr 14, 2013 at 10:44 | history | asked | ofir | CC BY-SA 3.0 |