Is there a finite dimensional Banach algebra $A$ for which $K_{0}(A)$ is a finite group?
I asked this question in MSE but I received no answer
https://math.stackexchange.com/questions/1624250/k-theory-of-finite-dimenional-banach-algebras
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3$\begingroup$ No, not even removing the word Banach, taking dimension over the ground field defines a surjection onto an infinite cyclic group. $\endgroup$– Fernando MuroCommented Jan 29, 2016 at 7:41
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3$\begingroup$ I'm using that the algebra has finite dimension so that any f.g. projective module too, over the ground field. $\endgroup$– Fernando MuroCommented Jan 29, 2016 at 13:44
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1$\begingroup$ @Fernando: I think you are implicitly assuming that your algebra has a unit. $\endgroup$– Alain ValetteCommented Jan 29, 2016 at 16:11
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2$\begingroup$ @Fermando: Because, for non-unital algebras $K_0(A)$ is defined as $\ker[K_0(\tilde{A})\rightarrow K_0(k)=\mathbb{Z}]$, where $\tilde{A}$ is the unitization of $A$, so you don't deal with f.g. projective modules, but with differences of modules having the same dimension. $\endgroup$– Alain ValetteCommented Jan 29, 2016 at 21:55
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1$\begingroup$ If you allow non-unital algebras, it is easy to find a finite-dimensional algebra $A$ such that $K_0(A)$ is trivial (for instance, let $A=\mathbb{C}$ with the zero multiplication; then over $\tilde{A}=\mathbb{C}[x]/(x^2)$ all projective modules are free). Whether $K_0(A)$ can be finite but nontrivial seems a lot harder. $\endgroup$– Eric WofseyCommented Jan 30, 2016 at 10:17
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1 Answer
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The definition of K$_0 (A)$ (where $A$ is a finite dimensional Banach algebra), that it be the kernel of $K_0(\tilde A) \to K_0(C) \cong Z$ leads immediately to it being a free abelian group, possibly zero (free on no generators); the latter occurs iff $A$ is nilpotent; $\tilde A$ denotes the unitification.
Since $\tilde A$ is a finite dimensional unital algebra, $\overline{A}:= \tilde A/rad(\tilde A)$ is finite-dimensional semisimple, hence $K_0(\overline A) $ is free finitely generated; since the radical of a finite dimensional algebra is nilpotent, $K_0(\tilde A) \cong K_0(\overline A) \cong Z^d$ for some $d$.
As $K_0(\tilde A) \to K_0(C) = Z$ is onto, $K_0(A)$ is free on $d-1$ generators. So $K_0(A)$ is either zero or torsion free. If $A$ is nilpotent, then $\tilde A$ has a unique maximal ideal, and then $d=1$. Conversely, if $d=1$, then $\overline A = C$, forcing $\tilde A$ again to have a unique maximal (two-sided) ideal. For finite dimensional algebras (over any field), $\tilde A$ having unique maximal ideal is equivalent to $A$ being nilpotent (otherwise, $A$ would have a nontrivial idempotent, etc).
We only need that $A$ be a finite dimensional algebra over some field for this to work, not a Banach algebra. The outcome is that $K_0(A)$ is finite iff it is zero iff $A$ is nilpotent.
answered Feb 14, 2016 at 6:42
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$\begingroup$ Prof. Handelman Thank you very much for your answer. $\endgroup$ Commented Feb 14, 2016 at 20:56