Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $f$?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
1  
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.