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Let $I$ be a filtered poset, which you should think of as being huge. Let $A_i$ be an $I$-diagram of $C^{\star}$-algebras and let $A$ be the colimit of this diagram; if necessary, we can also assume that all structure maps $A_i \rightarrow A_j$ are inclusions. Given a nuclear $C^{\star}$-algebra $N$, is it then true that tensoring with $N$ commutes with the colimit, i.e. that $A \otimes N$ is the colimit of $A_i \otimes N$? In particular, I am interested in the case $N = C(X)$ for some compact space $X$.

I expect this to be true (and seem to have a proof for C(X) at least), but have not been able to find anything in the literature, so a reference would be welcome.

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    $\begingroup$ In the category of von Neumann algebras, the functor $- \otimes N$ preserves coequalizers [Guichardet, Bull Sci Math 90:41-64, 1966, Prop 8.3]. I would hope the same holds for C*-algebras, so that it would suffice to concentrate on coproducts. But also, in the category of von Neumann algebras, colimits do not preserve flatness [Guichardet, Remark 8.2], which is a bad sign. $\endgroup$ Nov 28, 2011 at 16:15
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    $\begingroup$ By the way, preserving filtered colimits is the same as preserving colimits of chains [Adamek&Rosicky, Cambridge Univ Press, 1994, Cors 1.5 & 1.7]. So you may assume that I is a total order. If all the maps $A_i \to A_j$ are inclusions, doesn't that make the colimit just the closure of the union of all the $A_i$, leading to a trivial proof? $\endgroup$ Nov 28, 2011 at 16:43
  • $\begingroup$ You definitely want $N$ to be an exact $C^*$-algebra, not just a nuclear one, for something like that to be true. $\endgroup$ Nov 28, 2011 at 21:11
  • $\begingroup$ Chris, what tensor product are you/Guichardet using? I seem to recall that Guichardet defines a tensor which gives decent SMC properties but isn't the spatial t.p. And at C*-level the whole issue of tensor products is MUCH more complicated $\endgroup$
    – Yemon Choi
    Nov 28, 2011 at 21:31
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    $\begingroup$ @Ulrich Pennig: all nuclear C*-algebras are exact... $\endgroup$
    – Yemon Choi
    Nov 28, 2011 at 21:32

1 Answer 1

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If $N$ is exact and the tensor products are minimal then $A\otimes N$ is the colimit of the $A_i\otimes N$'s. Say the connecting maps are $\phi_{i,j}$. Then to check that $A\otimes N$ is the colimit of $\{A_i\otimes N,\phi_{i,j}\otimes\mathrm{id}_N\}$ two properties must be verified:

(1) the union of the ranges of the maps $\phi_{i,\infty}\otimes \mathrm{id}_N$ is dense in $A\otimes N$,

(2) for each $i$, $\ker (\phi_{i,\infty}\otimes \mathrm{id}_N)=\overline{\bigcup_{j>i} \ker({\phi_{i,j}\otimes \mathrm{id}_N})}$.

The first property is straightforward, since the span of the elementary tensors $\phi_{i,\infty}(a)\otimes n$ is dense. The second property follows from the fact that $\ker(\phi_{i,\infty}\otimes \mathrm{id}_N)=\ker(\phi_{i,\infty})\otimes N$ and the correspoding property (2) for the colimit of the $A_i$'s.

That $\ker(\phi\otimes \mathrm{id}_N)=\ker(\phi)\otimes N$ follows, if $\phi$ is surjective, by exactness of $N$, if $\phi$ is injective, by the minimality of $\otimes$ (the minimal tensor product behaves well with respect to inclusions), and the general case is a composite of those two.

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