# What are the generator for $K_1(C(\mathbb{T})\otimes\mathbb{K})$?

I've been working computing several K-groups associated to some $$C^*$$-algebras involved with my master's thesis, however I've just got stucked finding some generators for $$K_1(C(\mathbb{T})\otimes\mathbb{K})$$.

Just to elaborate my question let me explain the analogous problem associated to $$K_0(C(\mathbb{T})\otimes\mathbb{K})$$:

It is of course true that $$K_0(C(\mathbb{T})\otimes\mathbb{K})\cong K_0(C(\mathbb{T}))\cong \mathbb{Z}.$$ Also, it is a well known fact that $$$$ is a generator of $$K_0(C(\mathbb{T}))$$, so one would be tempted to find the generator of $$K_0(C(\mathbb{T})\otimes\mathbb{K})$$ using this information. Luckily, we have the next result:

The map from a $$C^*$$-algebra $$A$$ into $$A\otimes\mathbb{K}$$ given by $$a\mapsto a\otimes e_{11}$$, where $$e_{11}$$ is a rank one projection, induces an isomorphism between $$K_0(C(\mathbb{T})$$ and $$K_0(C(\mathbb{T})\otimes\mathbb{K})$$.

Finally, joining all the pieces it follows that a generator for $$K_0(C(\mathbb{T})\otimes\mathbb{K}$$ is $$[1\otimes e_{11}]$$.

Since the $$K_1$$-functor is also stable and finding generators for $$K_1(C(\mathbb{T}))$$ is not too hard, one would be tempted to do the same trick as above, however I have not found any analogous result for the $$K_1$$-groups. The main problem (I think) is that the proof (at least the one that I know) of the fact that the $$K_1$$-functor is stable is non-constructive, in the sense that the existence of the isomorphism follows from the continuity under direct limits of the $$K_1$$-functor.

With all this said, a generalized question is: Is there any known isomorphism between $$K_1(A)$$ and $$K_1(A\otimes\mathbb{K})$$? or more precisely (to my goals): Is there any way to find the generators of $$K_1(A\otimes\mathbb{K})$$ knowing the generators of $$K_1(A)$$?

• Edited, thanks! Apr 28, 2020 at 14:36

The morphism $$f : A \to A \otimes \mathbb K$$ which maps $$a$$ to $$a \otimes e_{11}$$ induce isomorphism $$K_1(f) : K_1(A) \to K_1(A \otimes \mathbb K)$$ on $$K_1$$ groups as well. In fact, we can construct very explicit inverse $$KK_0(A \otimes \mathbb K, A)$$ cocycle. Indeed, suppose $$\mathbb K$$ acts on separable Hilbert space $$\mathcal H$$, then $$A \otimes \mathcal H$$ is naturally Fredholm $$A \otimes \mathbb K, A$$ bimodule, which is naturally cocycle inside $$KK_0(A \otimes \mathbb K, A)$$. It's not so hard to see that Kasparov product with the morphism above is an identity.
• E.g. Wegge-Olsen "K-theory and C*-algebras" Lemma 7.1.8 (it's about the isomorphism $K_1(A) \to K_1(M_n(A))$, but after passing to inductive limit nothing change, by proposition 7.1.7) Apr 28, 2020 at 15:12