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Let $V$ be a finite dimensional real vector space. Let $GL(V)$ be the set of invertible linear transformations, and $\Phi(V)$ be the set of all linear transformations. We can also characterize $\Phi(V)$ as those linear transformations with a finite dimensional kernel and a finite dimensional cokernel, i.e. these are exactly the Fredholm operators in finite dimensions.

If $\mathbb{H}$ denotes an infinite dimensional separable Hilbert space, we can also consider the space of all bounded linear transformations $GL(\mathbb{H})$, and the set of all bounded linear transformations which have finite dimensional kernel and cokernel. The latter is the space of (bounded) Fredholm operators $\Phi(\mathbb{H})$.

Many invariants in finite dimensional geometric topology use, either implicitly or explicitly, the fact that $GL(V)$ has non-trivial topology. For example, it is the reason that vector bundles can be non-trivial. In contrast, $\Phi(V)$ is contractible.

In infinite dimensional topology all these geometric invariants vanish. Kuiper's theorem states that $GL(\mathbb H)$ is contractible, and a corollary is that all infinite dimensional (seperable Hilbert) vector bundles are trivial. But, the space of Fredholm operators $\Phi(\mathbb{H})$, which I have introduced as the analogon of $\Phi(V)$ in infinite dimensions, is not contractible. Most invariants that I know in infinite dimensional topology actually use the fact that the homotopy type of $\Phi(\mathbb H)$ is non-trivial.

By the Atiyah-Jänich theorem $\Phi(\mathbb{H})$ classifies topological $KO$-theory and therefore the homotopy type of $\Phi(\mathbb{H})$ is closely related to the homotopy type of the classifying space $BGL(V)$, when the dimension of $V$ is large. Namely, take a sequence of finite dimensional vector spaces $V_i$ of increasing dimension, along with compatible inclusions. Then we can define the limiting group $GL(V_\infty):=\lim_{i\rightarrow \infty} GL(V_i)$. Then the homotopy type of $\Phi(\mathbb{H})$ is then that of $BGL(V_\infty)\times \mathbb{Z}$.

Now the classifying space of $GL(\mathbb{H})$ has the homotopy type of a point. Any group acts freely on itself, and in this case the group is contractible. The quotient is just a point, but also a model of $BGL(\mathbb{H})$.

Hence the topology $BGL(V)$ resembles that of $\Phi(\mathbb{H})$ for $V$ sufficiently large, and the topology of $\Phi(V)$ resembles that of $BGL(\mathbb H)$.

Now I think I understand the proofs of these facts, but I still find it miraculous that there is this "switch" when passing to infinite dimensions. My question is if there is some big picture reason why one should expect this.

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    $\begingroup$ I guess I would say: $\Phi(V)$ is completely unconstrained when $V$ is finite-dimensional, while it is constrained by finite-dimensionality once you reach infinite dimensions, and the interesting topology accounts for how kernel and cokernel move around inside an infinite-dimensional space. On the other hand, when $V$ is infinite-dimensional the space $GL(V)$ is unconstrained because of the existence of a swindle. $\endgroup$
    – mme
    Commented Aug 17, 2019 at 20:16
  • $\begingroup$ My vague memory is that this information is better organized in terms of Grassmannians than groups, but I don't have anything specific to say. $\endgroup$ Commented Sep 17, 2019 at 16:44
  • $\begingroup$ The contractibility of GL(H), looking closely to the construction, is due to the fact that in an infinite dimensional Hilbert space there is "enough room" to perform the maneuver, just like in the Hilbert Hotel there are enough rooms to host a new guest $\endgroup$ Commented Dec 26, 2023 at 21:23

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