Let $A$ and $B$ be unital $C^*$-algebras. Let $u\in M_n(A)$ be a unitary representing an element $[u]\in K_1(A)$ and $p\in M_m(B)$ be a projection representing an element $[p]\in K_0(B)$. Then the unitary $$u\otimes p+1_A\otimes (1_B-p)$$ represets the product $[u]\times [p]\in K_1(A\otimes B)$ (see in Nigel Higson's and John Roes book "Analytic K-homology", proposition 4.8.3). This gives us a description of a product map $K_1(A)\times K_0(B)\to K_1(A\otimes B)$.

Now, let $A$ non-unital and $B$ unital, but $u$ and $p$ as above, where $u\in K_1(A)\cong K_1(A^+)$ with $A^+$ the unitization of $A$.

My question is: Is it possible to give an explicit description for $K_1(A)\times K_0(B)\to K_1(A\otimes B)$ as above for this case to map $[u]$ and $[p]$ to $K_1(A\otimes B)$ naturally?

I guess that this is not possible. My first try was to consider $$u\otimes p+1_{A^+}\otimes (1_B-p),$$ which represents an element in $K_1(A^+\otimes B)$. However, since $A\otimes B$ is non-unital, the product map should map to $K_1(A\otimes B)\cong K_1((A\otimes B)^+)$, but $ K_1((A\otimes B)^+)$ is not isomorphic to $K_1(A^+\otimes B)$. For $A$ nonunital and $B$ unital and nuclear, there is the following relation: there is a commutative diagram

$$\require{AMScd}\def\Z{\mathbb{Z}} \begin{CD} 0 @>>> K_1(A)\otimes K_0(B) @>>> K_1(A^+)\otimes K_0(B) @>>>> K_1(\mathbb{C})\otimes K_0(B) @>>> 0 \\ @. @VV V @VV\alpha V @VV\alpha V @. \\ 0 @>>>K_1(A\otimes B) @>>> K_1(A^+\otimes B) @>>> K_1(\mathbb{C}\otimes B) @>>> 0 \\ \end{CD} $$ where $K_1(\mathbb{C})=0$, $K_1(\mathbb{C}\otimes B)\cong K_1(B)$ and $K_1(A)\otimes K_0(B) \to K_1(A^+)\otimes K_0(B)$ is an isomorphism. If $B$ is nonnuclear, then the sequence in this diagram is not exact at $K_1(A\otimes B)$.


1 Answer 1


I believe the correct formula is $$(u-1_{A^+}) \otimes p + 1_{(A \otimes B)^+}$$ which makes sense if you work this out in the following special case. Let $A$ be $C_0(0,1)$ and let $B$ be $C(\{0,1\})$, and take $u(s)=e^{2\pi is}$ and $p(t)=t$. In this case $$((u-1_{A^+}) \otimes p + 1_{(A \otimes B)^+})(s,t)= (e^{2\pi i s}-1)t + 1$$ which is constantly one on $(0,1)\times\{0\}$ and loops around the unit circle on $(0,1)\times\{1\}$. Check that this is plus-or-minus the correct answer for this special case. Now use the universal properties of $C_0(0,1)$ and $C(\{0,1\})$ to get maps from these to a generic $A_1$ and $B_1$ and use the fact that the product map is natural to show this is the correct formula in general.

  • 1
    $\begingroup$ why is $p$ a projection, and why is $1_{A^+}$ in $1_{(A \otimes B)^+}$? $\endgroup$
    – hänsel
    Aug 20, 2017 at 8:21
  • 1
    $\begingroup$ Now B and so p should make more sense. $\endgroup$ Aug 20, 2017 at 13:28

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