# (Topological) K-theory for commutative $C^*$-algebras: operator and standard approaches

Let $A$ be a commutative unital $C^*$ algebra. Then $A=C(X)$ for some compact Hausdorff space $X$. Topological $K$-theory group (namely $K_0$) is defined in terms of vector bundles as a Grothendieck group of the semigroup of isomorphism classes of vector bundles (with respect to the Whitney sum of bundles). On the other side operator $K$-theory ($K_0$ group) is defined in terms of projections (self adjoint idempotents) namely one takes classes of projections in $M_{\infty}(A)$ with respect to relation $\sim$ where $x \sim y$ if there is a homotopy of projections in $M_{\infty}(A)$ connecting $x$ and $y$ (this is equivalent to the unitary equivalence and also to the Murray-von Neumann equivalence when we admit arbitrary large matrices). There is also the theorem of Serre-Swan which states that each projective finitely generated $C(X)$ module is of the form $\Gamma(E)$ for some vector bundle $E \to X$. In fact this theorems gives the equivalence of categories of finitely generated projective $C(X)$ modules and vector bundles over $X$. Usually authors say that thanks to this theorem the two aproaches: (classiccal) topological and operator theoretic are the same. However there are some details which (at least for me) are not quite clear:

First problem: each vector bundle gives rise to a finitely generated projective $C(X)$ module which further gives rise to some idempotent in $M_n(C(X))$. This idempotent needs not to be self adjoint. So how to construct projection giving an operator theoretic $K$-theory class, out of the vector bundle?

Suppose that we somehow overcame this diffulty and constructed the isomorphism of $K^0(X) \cong K_0(C(X))$. Still it is not obvious that this isomorphism behaves nicely at the level of maps:

Second problem Is this isomorphism natural? In other words: given continuous $f:X \to Y$ one has a morphism $T_f:C(Y) \to C(X)$ $T_f(g)=g \circ f$ (and each (unital, star) morphism between $C(Y)$ and $C(X)$ is of this form) and further one has $f^*:K^0(Y) \to K^0(X)$ given by the pullback bundle construction and $(T_f)_*:K_0(C(Y)) \to K_0(C(X))$ acting on the each entry of the (class of) projection. Then the question is whether the obvious diagram commutes?

Maybe thinks would become more clear if I would see the explicit formula for the isomorphism.

Apologies I'm pretty convinced that this is not a research level question-however it seems to me that this is a kind of folklore which is hard to find carefully and properly explained in the literature. So my hope is that somebody will find useful to have a careful explanation of this issue.

• For problem 1 just put a metric on your vector bundle, embed the vector bundle isometrically in some trivial bundle and take the corresponding orthogonal projection. For problem 2, pullback the whole thing (metric, trivial bundle, isometrically embedding etc.). – Denis Nardin May 5 '16 at 23:53
• In fact, even better: show that isomorphism classes of vector bundles of rank $k$ are the same things as equivalence classes of vector bundles of rank $k$ together with an embedding in $X\times \mathbb{R}^\infty$. Then you can send $(V,\eta:V\to X\times \mathbb{R}^\infty)$ to the function $X\to M_\infty(\mathbb{R})$ sending every point $x\in X$ to the orthogonal projection onto $V_x$. – Denis Nardin May 6 '16 at 0:01
• Blackadar's "K-theory for Operator Algebras" provides textbook answers to both of these questions. – Paul Siegel May 6 '16 at 2:36
• @Denis Nardin thank you for your comment, now I understand how to proceed to solve the fisrt problem. However could you provide some more details for the second one? – truebaran May 6 '16 at 23:16
• @Paul Siegel, I haven't found the answer in Blackadar's book (only exercise 9.4.3. at the end of chapter IV which does not answet my question: it only states that the Gelfand transform induces the isomorphism) – truebaran May 6 '16 at 23:16