Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A$.

share|improve this question
    
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 $K$? –  Qiaochu Yuan Dec 5 at 8:23
    
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. –  AlexE Dec 5 at 8:49
    
@NajibIdrissi: why did you delete the $K$-homology tag? –  AlexE Dec 5 at 8:51
    
@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. –  Najib Idrissi Dec 5 at 9:01
    
@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. –  AlexE Dec 5 at 9:52

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

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.