Let $A$ be a unital C*-algebra.

Let me define its "$K$-theory space" to be the image of its $K$-theory spectrum under the functor $\Omega^\infty:$ Spectra $\to$ Spaces. I denote the $K$-theory space of $A$ by $K(A)$.

Let $\mathcal K$ be the C*-algebra of compact operators on an infinite dimensional Hilbert space (usually taken to be separable).

Write $Pr(A \otimes \mathcal K)$ for the space of projections in $A \otimes \mathcal K$. This is an $E_\infty$-space with respect to the operation of direct sum (the latter involves reshuffling the coordinates on $\mathcal K$)

Question: Is $K(A)$ equivalent to the group completion of $Pr(A \otimes \mathcal K)$?

I'm being told by Ulrich Pennig that the above statement is false for non-unital algebras, as there exist non-unital C*-algebras which are stably projectionless. They contain no projections, even after forming the tensor product with $\mathcal K$. ($C_0(\mathbb R)$ is such an algebra.)

  • $\begingroup$ Do you have a "natural" map $Pr(A\otimes\mathcal{K})\rightarrow K(A)$? $\endgroup$
    – mathphys
    Mar 22, 2017 at 22:31
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    $\begingroup$ I am puzzled by the down-vote... Does someone think my question is too easy? $\endgroup$ Mar 24, 2017 at 8:08
  • 1
    $\begingroup$ I don't understand it either. I think this is a perfectly fine question. $\endgroup$ Mar 24, 2017 at 12:07

1 Answer 1


The fundamental observation is that $\pi_1$ of the $E_\infty$-space $Pr(A\otimes \mathcal K)$ is stably abelian.

(Here "stably abelian" means that for any connected component $Y\subset Pr(A\otimes \mathcal K)$, and any pair of loops $a,b\in\pi_1(Y)$, there exists an element $p\in Pr(A\otimes \mathcal K)$ such that $p\oplus a$ and $p\oplus b$ commute in $\pi_1(p\oplus Y)$.)

Given the above fact, the group completion of $Pr(A\otimes \mathcal K)$ can be computed as an infinite telescope: $$ Pr(A\otimes \mathcal K)^{gr}=\mathrm{hocolim} Pr(A\otimes \mathcal K) $$ with respect to the self-map $1_A\oplus- :Pr(A\otimes \mathcal K)\to Pr(A\otimes \mathcal K)$.

(For more general $E_\infty$-spaces, one also needs to perform a Quillen's plus-construction, after having done the infinite telescope.)

Given a compact topological space $X$, mapping out of $X$ commutes with the formation of filtered homotopy colimits. We therefore have: $$ [X,Pr(A\otimes \mathcal K)^{gr}]= [X,\mathrm{hocolim} Pr(A\otimes \mathcal K)]= \mathrm{colim} [X,Pr(A\otimes \mathcal K)]= [X,Pr(A\otimes \mathcal K)]^{gr}. $$

Now, we may identify the space of maps from $X$ to $Pr(A\otimes \mathcal K)$ with the space of projections in the C*-algebra $C(X,A\otimes \mathcal K)=C(X,A)\otimes \mathcal K$. We therefore have $$ [X,Pr(A\otimes \mathcal K)]= \pi_0(\mathrm{Map}(X,Pr(A\otimes \mathcal K)))= \pi_0(Pr(C(X,A)\otimes \mathcal K)), $$ and so $$ [X,Pr(A\otimes \mathcal K)^{gr}]= [X,Pr(A\otimes \mathcal K)]^{gr}= \pi_0(Pr(C(X,A)\otimes \mathcal K))^{gr}. $$ The latter is, by definition, $K_0(C(X,A))=[X,K(A)]$.

At last, by the Yoneda lemma, the equation $$ [X,Pr(A\otimes \mathcal K)^{gr}]=[X,K(A)] $$ implies the desired homotopy equivalence $Pr(A\otimes \mathcal K)^{gr}\approx K(A)$.

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    $\begingroup$ The application of Yoneda is dubious: you only established the bijection for compact X. $\endgroup$ Apr 3, 2017 at 11:06
  • $\begingroup$ @Oscar. You're right. But I feel that the argument is fixable. The infinity-category of spaces is the ind-category of the infinity-category of compact spaces. So the homotopy type of a space is completely remembered by the presheaf it induces on compact spaces. Now, the map $Ho([$compact spaces, spaces$])\to [Ho($compact spaces$),Ho($spaces$)]$ is not an equivalence of categories, but I seem to remember that it is full (the kernel consisting of phantom maps). $\endgroup$ Apr 3, 2017 at 13:50
  • $\begingroup$ @OscarRandal-Williams : are there spaces $Y_1$ and $Y_2$ which are not homotopy equivalent, but which induce isomorphic presheaves on $Ho($compact spaces$)$? I.e. such that $[X,Y_1]\cong [X,Y_2]$, naturally in $X$? $\endgroup$ Apr 3, 2017 at 20:28
  • $\begingroup$ I can't see how to build any, but I can't see how to rule it out either. $\endgroup$ Apr 3, 2017 at 21:53

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