2
$\begingroup$

Let $U=\cup_{n=1}^{\infty} U(n)$ be endowed with the weak topology, let $BU$ be its classifying space. Then for any compact CW complex $X$, $[X,BU]$ classifies all the vector bundles on $X$ up to stable isomorphisms, which is in one-one correspondence with $\tilde{K}(X)$.

On the other hand, let $\mathrm{Fred}_0(H)$ be the set of zero-index Fredholm operators of a given Hilbert space $H$ (with the strong topology), then $\tilde{K}(X)\cong [X,\textrm{Fred}_0(H)]$.

Since all these correspondences are natural with respect to $X$, it is reasonable to expect a connection between these two spaces. What is the relationship between them? Thanks a lot.

$\endgroup$
2
  • $\begingroup$ At least formally (up to restrictions on $X$) one expects from the Yoneda lemma that the two spaces are homotopy equivalent. $\endgroup$ Nov 11, 2014 at 5:08
  • 2
    $\begingroup$ Even better would be: what is a map that makes them homotopy equivalent? $\endgroup$
    – David Roberts
    Nov 11, 2014 at 6:16

1 Answer 1

8
$\begingroup$

See a paper by G. Segal, "K-homology theory and algebraic K-theory" (I'm sure it's not the original source, though). There is a homotopy equivalence between $BU\times \Bbb Z$ and the space $Fred$ of Fredholm operators. We can consider $BU \times \Bbb Z$ as a classifying space of the category of virtual vector bundles. Its objects a pairs $(V_1, V_2)$ of finite-dimensional vector spaces (morally a formal difference $V_1 - V_2$). Its morphisms are generated by formal equivalences $(V_1 , V_2) \simeq (V_1 \oplus W , V_2 \oplus W)$ and morphisms of pairs of vector spaces $f_{1,2} : V_{1,2} \to V^{\prime}_{1,2}$. It is a particular case of Quillen's $S^{-1}S$-construction. The classifying space of this category is $BU \times \Bbb Z$, where the index $\Bbb Z$ corresponds to the difference of dimensions $\dim V_1 - \dim V_2$.

More concretely, we can consider these vector spaces as embedded into some fixed countable-dimensional vector space, the classifying space of such category will be the same.

For any Fredholm operator $D : Fred$ its kernel and cokernel are finite-dimensional vector spaces, thus we get a pair $(\ker D, \mathrm{coker}\ D)$ of vector subspaces in $\Bbb R^\infty$. Moreover, a Fredholm operator is essentially uniquely determined by this pair, since the space of Fredholm operators with prescribed kernel and cokernel is nonempty and contractible. Any deformation of $D$ will induce an isometry of these vector subspaces. In general we can also get "points of discontinuity" where the dimensions of kernel and cokernel jump, i.e. there could exist some vector space $W$ such that $(\ker D^\prime, \mathrm{coker}\ D^\prime) = (\ker D \oplus W, \mathrm{coker}\ D \oplus W)$. Thus we see that the homotopies of Fredholm operators correspond to morphisms in the category of virtual vector bundles, and the topological categories $Fred$ and $BU \times \Bbb Z$ are equivalent. Their classifying spaces are thus also homotopy equivalent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.