1

I am self-studying KK-theory. I came up with the following lemma: Let A, B and C be graded $C^{*}$-algebras, and $\phi: B \rightarrow C $ be an even *-homomorphism, and $X=(E, \pi , T)\in \mathbb{E}(A,B)$, the class of all Kasparov A-B-module. Then $(E\otimes_{\phi}C , \pi \otimes id_{c}, T \otimes id_{c})$ is an element of $\mathbb{E}(A,C)$. The tensor here is the tensor produt on the graded $C^{*}$-algebras.

(1) My first question is what does the tensor $E\otimes_{\phi}C$? (I think it means the scalars come from $\phi(B)$). (2) The reason that the homomorphism $\phi$ has to be even is that in the definition of Kasparov A-B-modules $T$ is even?

Emilly
  • 65

1 Answers1

1

This is just the tensor product as outlined (for non-graded case) in section 13.5 of Blacakdar's K-Theory for Operator Algebras. In this case, what we do is with $\phi$ we identify $C$ as a left $B$-module, and form the algebraic tensor product $E\odot_BC$. This is a right $C$-module, and has a $C$-valued inner product given by $$\langle e_1\otimes c_1,e_2\otimes c_2\rangle=c_1^*\phi(\langle e_1,e_2\rangle)c_2.$$ Now quotient $E\odot_BC$ by vectors of length zero, and take the completion to obtain the Hilbert $C$-module $E\otimes_\phi C$.

Having $\phi$ even makes $E\otimes_\phi C$ a graded Hilbert $C$-module, and eventually (after checking lots of details) makes $(E\otimes_{\phi}C , \pi \otimes id_{C}, T \otimes id_{C})$ a (graded) Kasparov module.

Aweygan
  • 23,232