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 an easy corollary is that all infinite dimensional (seperable Hilbert) vector bundles are trivial. But, the space $\Phi(\mathbb{H})$, which I have introduces as the analogon of $\Phi(V)$ in infinite dimensions, is not contractible. Most invariants that I know in infinite dimensions actually use the fact that $\Phi(\mathbb H)$ is non-trivial.

By the Atiyah-Jänich theorem $\Phi(\mathbb{H})$ classifies topological $KO$-theory and the homotopy type of $\Phi(\mathbb{H})$ is closely related to the homotopy type of $BGL(V)$, 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 $BGL(V_\infty)\times \mathbb{Z}$.

Hence the topology $GL(V)$ resembles that of $\Phi(\mathbb{H})$ and the topology of $\Phi(V)$ resembles that of $BGL(V)$.

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.

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.

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 an easy corollary is that all infinite dimensional (seperable Hilbert) vector bundles are trivial. But, the space $\Phi(\mathbb{H})$, which I have introduces as the analogon of $\Phi(V)$ in infinite dimensions, is not contractible. Most invariants that I know in infinite dimensions actually use the fact that $\Phi(\mathbb H)$ is non-trivial.

By the Atiyah-Jänich theorem $\Phi(\mathbb{H})$ classifies topological $KO$-theory and the homotopy type of $\Phi(\mathbb{H})$ is closely related to the homotopy type of $GL(V)$, 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 space $GL(V_\infty):=\lim_{i\rightarrow \infty} GL(V_i)$. The fact that I find miraculous is that the homotopy type of $\Phi(\mathbb{H})$ is then that of $GL(V_\infty)\times \mathbb{Z}$.

Hence the topology $GL(V)$ resembles that of $\Phi(\mathbb{H})$ and the topology of $\Phi(V)$ resembles that of $GL(V)$.

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.

Ps: I am aware of the space $GL_c(\mathbb{H})$ of invertible bounded linear transformations that differ from the identity by a compact operator, which has the homotopy type of $GL(V_\infty)$.

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 an easy corollary is that all infinite dimensional (seperable Hilbert) vector bundles are trivial. But, the space $\Phi(\mathbb{H})$, which I have introduces as the analogon of $\Phi(V)$ in infinite dimensions, is not contractible. Most invariants that I know in infinite dimensions actually use the fact that $\Phi(\mathbb H)$ is non-trivial.

By the Atiyah-Jänich theorem $\Phi(\mathbb{H})$ classifies topological $KO$-theory and the homotopy type of $\Phi(\mathbb{H})$ is closely related to the homotopy type of $BGL(V)$, 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 $BGL(V_\infty)\times \mathbb{Z}$.

Hence the topology $GL(V)$ resembles that of $\Phi(\mathbb{H})$ and the topology of $\Phi(V)$ resembles that of $BGL(V)$.

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.