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Feb 4, 2015 at 16:39 vote accept hans
Feb 3, 2015 at 21:04 comment added hans @David: Maybe you could explain this with dual = $Hom(K_0(A),Z)$. When is it so and why (ok one explanation is the main answer).
Feb 3, 2015 at 14:01 answer added Ulrich Pennig timeline score: 2
S Feb 3, 2015 at 13:12 history suggested Rasmus CC BY-SA 3.0
replaced ? by Schochet
Feb 3, 2015 at 12:42 review Suggested edits
S Feb 3, 2015 at 13:12
Feb 3, 2015 at 3:46 comment added David Handelman And because $K_0(A)$ is free on countably many generators (if $A = C(X)$ where $X$ is Cantor), its dual is the product.
Feb 3, 2015 at 3:43 comment added David Handelman Isn't it just the dual group of K$_0 (A)$ (here $A = C(X)$, but could be any C*-algebra)? That is, Hom$(K_0(A),Z)$ (since K$_0 C = Z$). For $X$ a Cantor set, this is uncountable.
Feb 2, 2015 at 23:59 comment added Eric Wofsey I know almost nothing about KK-theory, but it seems to me that uncountability would follow from the fact that every infinite metrizable Stone space has $\mathbb{N}\cup\{\infty\}$ as a retract.
Feb 2, 2015 at 22:17 comment added Yemon Choi Have you got access to Higson and Roe's book? I haven't got my copy to hand, but that is the first place I would look.
Feb 2, 2015 at 22:00 history edited Eric Wofsey
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Feb 2, 2015 at 21:35 review First posts
Feb 2, 2015 at 21:45
Feb 2, 2015 at 21:30 history asked hans CC BY-SA 3.0