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Consider the associative algebra A with generators $T_i$ and rule $T_i*T_j=\Sigma_kC^{ij}_k*T_k$. Even if it makes no sense for a fusion ring (my momentary pet :-) to change basis it is still possible to linear transform the basis (via $T_j=M_{ij}E_i$) into diagonal form: $E_i*E_j=\delta_{ij}*E_i$. (Or so I think.)
1. Is commutativity (which I always assume tacitly) necessary? Is associativity necessary and/or sufficient? (For that A in fact can be "diagonalized" with some $M_{ij}$)
2. Has this M a name then? I compute-guessed a few cute relations of M and (assuming A is a fusion ring) the Verlinde matrix S of A (which then surely are known).

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  • $\begingroup$ I considered this question years ago and found that associativity is necessary. If I recall correctly I was not able to decide if it is also sufficient (in the case of finite dimensional algebras over a field). I will try to look up my old notes when I get home. The counterexample was some quite ugly non-associative deformation of the quaternions which someone else on the internet used as a counterexample to something else, but maybe more elegant counterexamples are possible. Details will follow. $\endgroup$
    – Vincent
    Apr 22, 2015 at 11:07
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    $\begingroup$ How does this diagonalization work for the two by two matrices over the complex numbers? $\endgroup$
    – Yemon Choi
    Apr 22, 2015 at 11:11

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I misread your question. My comment alluded to the question whether you can have a basis such that $E_i * E_j = $ some single $E_k$, (as opposed to a linear combination of all of them) but what you ask is much stronger: does every commutative associative algebra have a basis of orthogonal idempotents. This implies commutativity: it is true if and only if your algebra is (as an algebra) isomorphic to a direct sum of copies of the ground field.

From here we find another necessary condition: no nilpotent elements. On the other hand we know from the Artin-Wedderburn structure theorem that if your algebra is commutative and semi-simple it must have this form.

I'm not sure if commutative and without nilpotent elements are also sufficient conditions in the infinite dimensional case. In the finite dimensional case I believe they are: finite dimensional algebras are Artinian, which means that the Jacobson radical coincides with the nilradical, hence if the latter is zero, so is the former, which in turn implies semi-simplicity and we can apply Artin-Wedderburn. (I hope I got this right, otherwise please correct.)

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    $\begingroup$ You need more than commutative and semisimple in the finite dimensional case: you need that the algebra splits over the ground field. If the ground field is not algebraically closed, you can get commutative semisimple algebras which are not diagonalized by just taking an finite extension. $\endgroup$ Apr 22, 2015 at 14:15
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    $\begingroup$ Regarding the infinite dimensional case: $k[x]$ does not have a basis of orthogonal idempotents: The only idempotent elements are $0$ and $1$. $\endgroup$ Apr 22, 2015 at 17:54
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As Benjamin said: an associative commutative finite-dimensional algebra is diagonalizable iff its of the form $K^n$. In any commutative associative algebra the diagonalizable elements form a subalgebra: its the largest diagonalizable subalgebra of the entire algebra. Maybe this helps you, too.

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