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Do Fredholm operators show up in K-theory? Why or why not? The idea of infinite Grassmannians classifying vector bundles is pretty straightforward, but why would adding in additive inverses and what not give you this? Is it a generalization of some nice finite-dimensional concept? Does it have a deep connection to (finite-dimensional) vector bundles?

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    $\begingroup$ A Fredholm operator describes a (virtual) vector space, namely its kernel minus its cokernel. So you might expect that a bundle of Fredholm operators describes a bundle of virtual vector spaces, hence a class in K-theory. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 5, 2015 at 7:20
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    $\begingroup$ Also Fredholm operators show up very naturally in K-homology, the dual generalized homology theory, and of course the space of Fredholm operator on a separable Hilbert space is the classifing space for $K_0$. $\endgroup$
    – Simon Henry
    Commented Aug 5, 2015 at 7:42
  • $\begingroup$ @user241357 I think fredholm index appear in operator K theory of C* algebra and six term exact sequence. It can be fount at page 55 of toknotes.mimuw.edu.pl/sem1/files/K_theory.pdf $\endgroup$
    – Ali Taghavi
    Commented Aug 5, 2015 at 13:06

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A classical connection is the Atiyah-Jänich Theorem, see

Klaus Jänich: Vektorraumbündel und der Raum der Fredholm-Operatoren. Math. Ann. 161 (1965) 129–142.

Let ${\mathcal{F}}$ be the space of Fredholm operators with the operator norm, and $X$ any compact space. To a map $F\colon X\to {\mathcal{F}}$ one can associate the virtual vector bundle $(ker(F(x)))_x-(coker(F(x)))_x$ from Qiaochu Yuan's comment. (In general, there might be points where dimension of kernel and cokernel jump, but upon homotopy of $F$ one can assume that kernel and cokernel are indeed vector bundles.)

Jänich proves that this yields a bijection $$\left[X,{\mathcal{F}}\right]\cong K(X),$$ where $\left[.,.\right]$ means homotopy classes.

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  • $\begingroup$ Just to be pedantic: This is true for the space of fredholm operators on a Hilbert space, or a bit more generally for Banach spaces $E$ that are Kuiper (GL(E) is contractible) and stable ($E\times \mathbb{R}\cong E$). $\endgroup$
    – Thomas Rot
    Commented Jul 6, 2017 at 14:49
  • $\begingroup$ I didn't know that it was true in that generality, I was actually thinking of "the" Hilbert space $L^2(S^1)$. $\endgroup$
    – ThiKu
    Commented Jul 7, 2017 at 7:46
  • $\begingroup$ That is the only Banach space I can imagine;). The reference for this more general statement is in Korschorke "Infinite dimensional K-theory". $\endgroup$
    – Thomas Rot
    Commented Jul 7, 2017 at 9:30

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