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I suppose this is a question with a well known answer. Suppose $A$ and $B$ are two algebras over some field and there is a map $$ f: \operatorname{K_0}(A) \to \operatorname{K_0}(B) $$ is it necessarily induced by a tensor product with some bimodule? If not in general, is it true for some reasonable $A$, $B$ and $f$?

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up vote 10 down vote accepted

I'd say no to both questions.

1) If $k$ is a field then $K_0(k)= \mathbb Z$ generated by $[k]$ and the class of any $k$-module is positive, so $-n\colon K_0(k)\rightarrow K_0(k)$ cannot be induced by a bimodule, $n>0$.

2) If $k$ is a field of positive characteristic and $k'$ is a field of characteristic $0$ then the only map $K_0(k)\rightarrow K_0(k')$ induced by a bimodule is the trivial map, since the only $k$-$k'$-bimodule is the trivial one. The same the other way round.

EDIT: Answering Sasha's comment below. If $A=\mathbb{C}[\epsilon]/(\epsilon^2)$, any left and right projective $k$-$A$-bimodule is even-dimensional over $\mathbb C$, hence all induced homomorphisms $K_0(A)=\mathbb Z\rightarrow K_0(\mathbb C) =\mathbb Z$ are multiples of $2$, in particular the identity fails to be induced.

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The second answer might need a little adjustment in order for $k$ and $k'$ to be algebras over the same field. – S. Carnahan Mar 1 '13 at 7:56
Even if you ask for the homomorphism to respect some sort of positivity (which your first example would violate), there are lots of counterexamples. For instance, if $K$ and $L$ are two fields extending the base field $k$ such that there is no embedding of $K$ into $L$, then $K_0(K)=K_0(L)=\mathbb{Z}$ but the identity map cannot be realized by a bimodule. – Eric Wofsey Mar 1 '13 at 8:16
Thanks, answer and comments clarifies the situation. But what if we restrict the question to the algebras over complex numbers and morphisms preserving positivity is the statement still incorrect? – Sasha Pavlov Mar 1 '13 at 12:12

In the world of $C^*$-algebras, there are cases, where a homomorphism $f \colon K_0(A) \to K_0(B)$ is induced not only by a bimodule, but by an honest $*$-homomorphism. For example, if $A$ is a separable, unital, nuclear simple $C^*$-algebra and $B$ is separable and nuclear and if the universal coefficient theorem holds for the pair $A,B$, then $f$ is the image of some element in $KK(A,B)$ with respect to the map $KK_*(A,B) \to Hom(K_*(A),K_*(B))$ (as defined for example in the book by Blackadar) and any such element comes from a full homomorphism $\varphi \colon A \otimes \mathcal{O}_{\infty} \otimes \mathbb{K} \to B \otimes \mathcal{O}_{\infty} \otimes \mathbb{K}$ by Phillips' classification theorem (Theorem 4.1.3 in that paper. Here $\mathbb{K}$ denotes the compact operators on a separable infinite dimensional Hilbert space and $\mathcal{O}_{\infty}$ is the infinite Cuntz algebra.).

In particular, if $B \cong B \otimes \mathcal{O}_{\infty}$, $A \cong A \otimes \mathcal{O}_{\infty}$ and if you can lift $f$ to an invertible element in $KK(A,B)$, then $f$ comes from a stable isomorphism $\varphi \colon A \otimes \mathbb{K} \to B \otimes \mathbb{K}$, which in turn corresponds to a Morita equivalence between $A$ and $B$ by a theorem of Brown, Green and Rieffel. The conditions $A \cong A \otimes \mathcal{O}_{\infty}$ and $B \cong B \otimes \mathcal{O}_{\infty}$ are for example satisfied if $A$ and $B$ are simple and purely infinite $C^*$-algebras.

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Considering the tags to your question, this is probably a little off-topic, but nevertheless. – Ulrich Pennig Mar 1 '13 at 10:02
One might add that this works because every element in $K_0(A)$ is positive if $A$ is purely infinite simple, so that the obstruction used in Fernando's first counterexample disappears. – Rasmus Bentmann Mar 1 '13 at 10:54

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