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Let $(\cal A,\|\cdot\|)$ be a finite dimensional norm algebra (Banach Algebra). Can we say any thing about the amenability of $\cal A$. What if we impose some extra conditions on $\cal A$, say commutativity or the existence of a unit.

Thank you very much.

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Every finite-dimensional Banach algebra should be amenable. You can probably find this in the sources link on wikipedia. –  Rasmus Oct 6 '12 at 7:27
    
@ Rasmus: If it is really easy to show could you tell me how should I start a proof. –  User3060 Oct 7 '12 at 5:02
    
@Rasmus: I guess the answer should be more complicated, as for a finite normed space $X$, one may make a Banach algebra by defining $ab=0$ for every $a,b\in X$ and it is not amenable. –  User3060 Jun 10 at 2:17
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Yes, you are right. It's nice to see that, apparently, you've found the answer to your question. –  Rasmus Jun 10 at 7:26

1 Answer 1

up vote 2 down vote accepted

I guess the answer to my question is as follows.

If $A$ is a finite dimensional Banach algebra with a trivial Jacobson radical, then it is amenable. The answer is easily based on this fact that this class of finite dimensional algebras are isomorphic to a product of finitely many $n_i\times n_i$ matrix rings over division rings $D_i$, by Artin–Wedderburn theorem.

Recall that each of this matrix algebras is amenable and a simple argument regarding adjoining the (virtual) diagonals together implies that the product is amenable.

The converse is also true. If $A$ is a finite dimensional Banach algebra with a non-trivial radical, it cannot admit any (virtual) diagonal; hence, it cannot be amenable.

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