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If $X$ is separable complex Hilbert space and $\mathcal{F}$ the topological space of Fredholm operators on $X$, then it is well-known, that $$ \pi_0(\mathcal{F}) = \mathbb{Z}\, , $$ i.e. the connected components are classified by the index of the Fredholm operator.

But what is about higher homotopy groups? What is known about $\pi_n(\mathcal{F})$ for $n \in \mathbb{N}$?

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2 Answers 2

It is a theorem (due I think to Atiyah) that $\mathcal{F}$ is the classifying space for the topological K-theory functor: $$[X,\mathcal{F}] \cong K(X)$$ for any space $X$. The isomorphism is given as follows: given a map $T \colon X \to \mathcal{F}$, deform $T$ so that $\dim \ker(T(x))$ is constant and assign $T$ to the K-theory class $[\ker T] - [\text{coker}\, T]$. It follows that $\pi_n(\mathcal{F}) \cong K(S^n)$ which can be calculated using Bott periodicity.

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Thanks for the answer. So I can use the $K_0$-groups of $S^n$? –  Chandler Aug 19 at 17:20
    
That's correct. –  Paul Siegel Aug 19 at 17:38
    
Hi Paul, Maybe I'm missing something, but I think that $\pi_n (\mathcal{F})$ is isomorphic to the reduced groups $\widetilde{K} (S^n)$, not $K(S^n)$. Unreduced K-theory classifies (unbased) homotopy classes of (unbased) maps out of X, but for homotopy groups we're interested in based maps and based homotopy. Letting $\mathcal{F}_0$ denote the index 0 Fredholm operators (which form one connected component of $\mathcal{F}$), the bijection above restricts to $[X, \mathcal{F}_0] \cong \widetilde{K} (X)$. –  Dan Ramras Aug 20 at 4:59
    
(cont'd) So $\pi_1 (\mathcal{F}_0) = 1$ (for this, it's enough to check that each loop is (unbased) nullhomotopic, which follows from $[S^1, \mathcal{F}_0] = \widetilde{K} (S^1) = 0$) and now $\pi_n (\mathcal{F}) = \pi_n (\mathcal{F}_0) = \langle S^n, \mathcal{F}_0 \rangle = [S^n, \mathcal{F}_0] = \widetilde{K}(S^n)$. Here $\langle , \rangle$ means based homotopy classes of maps, which is the same as unbased homotopy classes when the range is simply connected. –  Dan Ramras Aug 20 at 5:02
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Oh, and a reference for the bijection in the above answer is the Appendix to Atiyah's book K-theory. –  Dan Ramras Aug 20 at 5:06

This answer is an elaboration of my comments on Paul Siegel's answer.

Atiyah showed that there is a natural bijection $$[X,\mathcal{F}] \cong K(X)$$ whenever $X$ is a compact space. Here the left-hand side is the set of unbased homotopy classes of maps into $\mathcal{F}$. This result is proven in the Appendix to Atiyah's book K-theory.

As I'll explain, it follows from this theorem that there is an isomorphism $$\pi_n (\mathcal{F}) \cong \widetilde{K} (S^n),$$ where $\widetilde{K}$ denotes reduced $K$-theory. So these groups are trivial for $n$ odd and infinite cyclic for $n$ even. Note that all path components of $\mathcal{F}$ are homotopy equivalent, because (as Atiyah explains) there is an associative product $\mathcal{F} \times \mathcal{F} \to \mathcal{F}$ that makes the set of path components of $\mathcal{F}$ into a group (namely the group $[\{*\}, \mathcal{F}] \cong K(\{*\}) \cong \mathbb{Z}$). So it doesn't matter what basepoint we choose for computing homotopy.

Now, to deduce the claimed calculation of homotopy groups, let $\mathcal{F}_0$ denote the index 0 Fredholm operators (which form one connected component of $\mathcal{F}$), Atiyah's bijection restricts to a bijection $[X,\mathcal{F}_0]\cong \widetilde{K}(X)$.

Next, I claim that $\mathcal{F}_0$ is simply connected. First, $\mathcal{F}_0$ is path connected by definition, so we just need to check that $\pi_1 (\mathcal{F}_0)=1$. For this, it's enough to check that each loop is (unbased) nullhomotopic, which follows from $[S^1,\mathcal{F}_0]\cong \widetilde{K}(S^1)=0$.

Now $\pi_n(\mathcal{F})=\pi_n(\mathcal{F}_0)=\langle S^n,\mathcal{F}_0\rangle \cong [S^n,\mathcal{F}_0] = \widetilde{K} (S^n)$. Here $\langle\, ,\rangle$ means based homotopy classes of (based) maps, which is the same as unbased homotopy classes when the range is simply connected (this is proven in Section 4.A of Hatcher's book Algebraic Topology, for instance).

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