In Exercise 9.8.9 of the book "Analytic K-Homology" by Higson and Roe one has to construct a cap product $K_p(A) \otimes K^q(A) \to K^{q-p}(A)$, if A is commutative.

Is the commutativity assumption on A really needed for the construction of this cap product?

I mean ... isn't the cap product just induced by the map $A \otimes \mathfrak{D}(A) \to \mathfrak{D}(A) / \mathfrak{K}$, $a \otimes T \mapsto [aT]$? (Or a slight variation of it accounting for the fact that we have to take the unitization of A somewhere.) This map is a $^\ast$-homomorphism regardless of the commutativity of $A$.

  • $\begingroup$ It would be easier to answer this question if it were more self-contained, so that we didn't have to have this book in order to answer it. For example, what is $D(A)$, and what is $K$? $\endgroup$ Dec 5, 2014 at 8:23
  • $\begingroup$ Given a representation $\rho\colon A \to \mathbb{B}(H)$ of the $C^\ast$-algebra $A$ on a separable Hilbert space $H$, one defines the dual algebra $\mathfrak{D}(A)$ of $A$ as $\mathfrak{D}(A) := \{ T \in \mathbb{B}(H) \colon [T,\rho(a)] \in \mathfrak{K} \text{ for all }a \in A\}$. Here $\mathfrak{K}$ are the compact operators. The $K$-homology of $A$ is then defined as $K^p(A) := K_{1-p}(\mathfrak{D}(A^+))$, where $A^+$ is the unitization of $A$ and we use an ample representation to form the dual algebra. $\endgroup$
    – AlexE
    Dec 5, 2014 at 8:49
  • $\begingroup$ @NajibIdrissi: why did you delete the $K$-homology tag? $\endgroup$
    – AlexE
    Dec 5, 2014 at 8:51
  • $\begingroup$ @AlexE There were only two questions tagged "K homology", both by you. I think the tag is too specific: as far as I understand, K-homology is simply the homology theory associated to the spectra that comes from K-theory. I don't think this deserves its own tag different from k-theory... It would be like having two tags "left-derived-functors" and "right-derived-functors", or "lie-algebra-homology" and "lie-algebra-cohomology", that's too much. $\endgroup$ Dec 5, 2014 at 9:01
  • $\begingroup$ @QiaochuYuan: Oh, yes, you are right ... $aT$ is an element of $\mathfrak{D}(A)$ in general only if $A$ is commutative. I completely forgot to check this, because I was somehow fixed only on checking if the map is a $^\ast$-homomorphism. Thank you! You can post it as an answer so that I can accept it. $\endgroup$
    – AlexE
    Dec 5, 2014 at 9:52

1 Answer 1


Reposting my comment: in order for the map you describe to be defined it seems like you at least need $D(A)$ to contain $A$ (I don't see the point of the fraktur here; most of the time it just makes things harder to read). I don't see why this needs to hold if $A$ is noncommutative.


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