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In noncommutative geometry when one want to go to the differentiable level, one is forced to work with algebras which are no longer $C^*$. It is nice if we don't loose much information by the replacement of $C^*$-algebras by more general algebra. One possible choice is the so called pre $C^*$-algebra which is by definition a subalgebra $\mathcal{A}$ of a $C^*$-algebra $A$ of which is complete with respect to some locally convex topology finer that the topology of $A$ and is closed under the holomorphic functional calculus. Suppose that we have such $\mathcal{A}$ which is moreover a Fréchet algebra and assume that $\mathcal{A}$ is dense in $A$ (in the norm topology of $A$). The claim is that:

The inclusion $i:\mathcal{A} \to A$ induces an isomorphism in $K_0$ groups.

I found the proof of this fact in the book "Elements of Noncommutative Geometry" by Várilly, Gracia-Bondía and Figueroa. There are two moments which are not clear for me:
1. The first is that authors state that it is enough to prove this fact for unital algebras. I believe that it is indeed the case however I'm not sure whether there are no technical problems.
2. The second thing is the end of the proof: let us recall that $K$-theory may be defined in terms of idempotents (not projections, i.e. our idempotents need not to be self-adjoint) in matrix algebras $Q_n(A)=\{e \in M_n(A): e^2=e\}$ (the same for $\mathcal{A}$: one checks that matrix algebras for $\mathcal{A}$ are again Fréchet pre $C^*$-algebras). Of this the quotient space is made by equivalence relation where $e \sim f$ iff there is some invertible matrix $g$ (possibly bigger that $e$ and $f$) such that $(e \oplus 0)g=g(f \oplus 0)$. At the end of the proof authors arrived to the conclusion that the inclusions $Q_n(\mathcal{A} \to Q_n(A)$ are all homotopy equivalences and they claim that this finishes the proof provided we are able to prove that two idempotents which are homotopic are equivalent in the above sense. I don't quite see two things:

Why it is enough to prove this to conclude that the inclusion induces iso on $K_0$ groups?

And also

How to prove that being homotopic implies being equivalent (in the context of general idempotents and invertibles)?

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1) First question: unital algebras are sufficient.

If $i:K_0(\tilde {\mathcal A}) \rightarrow K_0(\tilde A)$ is an isomorphism, and since $id:K_0(C) \rightarrow K_0(C)$ is an isomorphism, you make a diagram chase by doubling the exact sequence $$0 \rightarrow K_0(A) \rightarrow K_0(\tilde A) \rightarrow^\rho K_0(C) \rightarrow 0$$ (second case with ${\mathcal A}$) and connect both lines with $i$ to see that $K_0({\mathcal A}) \rightarrow K_0(A)$ is an isomorphism as well. Note that $K_0(A)$ is the kernel of $\rho$ by definition!

3) "How to prove that being homotopic implies being equivalent (in the context of general idempotents and invertibles)?"

Note that ${\mathcal A}$ is a subalgebra of $A$, so you have all the notions from $C^*$-algebras like projections in ${\mathcal A}$ as well. I think the elementary standard proofs from $C^*$-algebras work in ${\mathcal A}$ as well. I think you must go by yourselves through the proofs.

See Wegge-Olsen Proposition 5.2.10 & 5.2.12 that homotopy-equivalence and unitary equivalence (or invertible equivalence (Proposition 4.2.6)) is the same in matrix algebras.

See Wegge-Olsen 5.B: Every idempotent is homotopic to a projection.

2) "Why it is enough to prove this to conclude that the inclusion induces iso on $K_0$ groups?"

By item 2) we may equivalently switch from invertible equivalence of idempotents to homotopy equivalence of idempotents in $Q$.

If $i$ is a homotopy equivalence on the level $i:Q({\mathcal A}) \rightarrow Q(A)$ then there exists a continuous inverse $k$ such that $i\circ k \sim id$ and $k \circ i \sim id$.

$i$ on the level $K_0$ is injective:

Suppose $$i(e) - i(f)=0$$ for $e,f \in Q({\mathcal A})$. Then $i(e) \oplus m \sim i(f) \oplus m$ (homotopy is $\sim$; Grothendieck group) for some $m \in Q(A)$. Note $m \sim i \circ k (m)$. Hence $$i(e) \oplus i(k(m)) \sim i(f) \oplus i(k(m)),$$ or $$i(e \oplus k(m)) \sim i(f \oplus k(m)).$$ Since $k$ is continuous, it respects homotopy (pathes)!. Thus $$e \oplus k(m) \sim k(i(e \oplus k(m))) \sim k(i(f \oplus k(m))) \sim f \oplus k(m).$$ Hence $$e - f = 0.$$

Surjectivity of $i$ is similar.

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