MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When giving $A_1,A_2$ two isomorphic maximal commutative semi-simple sub algebras of $M_n(\mathbb{C})$, are these algebras conjugate in $M_n(\mathbb{C})$? Namely, does there exists a regular matrix $P$ such that $P^{-1}A_1P=A_2$?

share|cite|improve this question
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. – Geoff Robinson Apr 14 '13 at 11:21
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? – ofir Apr 14 '13 at 12:09
@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... – user30379 Apr 14 '13 at 14:47
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. – user33117 Apr 15 '13 at 6:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.