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In p.32 of Milnor's Characteristic Classes, Milnor defines a "continuous" functor from the category $\mathfrak{V}$ consisting of all finite dimensional vector spaces and all isomorphisms between such vector spaces, to $\mathfrak{V}$ itself. He says that the set of all isomorphisms from one finite dimensional vector space to another has a natural topology. How does this topology defined? I can understand how the set of all isomorphisms from one finite dimensional vector space to another has a natural topology. Thanks in advance.

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    $\begingroup$ consider the operator norm $\endgroup$ May 17, 2020 at 3:32
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    $\begingroup$ What is the field of scalars? It must at least be a topological field. $\endgroup$ May 17, 2020 at 4:41
  • $\begingroup$ "I can understand how the set of of all isomorphisms from one finite dimensional vector space to another has a natural topology." So what is your question? $\endgroup$
    – Paul Frost
    May 17, 2020 at 8:11
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    $\begingroup$ @PaulFrost I think a " 't " is missing somewhere. $\endgroup$ May 17, 2020 at 10:12

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Given the context, I'll assume you mean "finite dimensional vector spaces over $\mathbb R$", or over $\mathbb C$ (luckily, the answer is the same for both).

What I'm going to say would work for pretty much any complete metric field, but for the sake of concreteness (and not having to check hypotheses because I'm not sure about them), I'm going to work over $\mathbb{K=R}$ or $\mathbb C$.

Then for any finite dimensional vector space $V$ there is a canonical topology, which is essentially the unique Hausdorff topology which makes it into a topological vector space. This comes from any norm on $V$ (they're all equivalent anyways).

In particular, you can apply this to $V= \hom(E,F)$ for any finite dimensional vector spaces $E,F$. This topology agrees with the compact-open topology (restricted to $V$, because it's actually defined on all continuous maps $E\to F$), and also with the product topology, and the simple convergence topology etc. basically with any reasonable topology you could put on $V$.

This is essentially because a linear map is entirely determined by what it does on a basis of $E$, so for any basis $B$, the restriction $\hom(E,F)\to map(B,F)$ is a bijection (and in fact an isomorphism of vector spaces).

You can try to prove that they're all equivalent, but it's not so important; the point here is that there is a canonical topology on $\hom(E,F)$. You can then simply take the subspace topology on $\mathrm{Isom}(E,F)$, the subset of isomorphisms, which therefore has a natural topology.

When $E=F= \mathbb K^n$, this is the usual topology on $GL_n(\mathbb K)$, when $E\not\cong F$, this is the only topology on the empty-set, and when $E\cong F$ this is a topology which is homeomorphic to $GL_n(\mathbb K)$

It's then easy to check that this topology is compatible with composition and inversion.

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It's a finite dimensional vector space over $\Bbb R$. This has essentially one vector space topology namely the one that makes it homeomorphic to $\Bbb R^N$ where $N$ is the dimension of the space of isomorphisms. Any linear bijection between $\Bbb R^N$ and that space will be a homeomorphism; a classic fact of linear spaces. I suppose Milnor might be referring to that.

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  • $\begingroup$ I think the OP is interested in the space of isomorphisms $V\to W$, which is not a vector space. ($A + (-1)\cdot A$ is not invertible). $\endgroup$ May 17, 2020 at 9:52
  • $\begingroup$ @PawełCzyż The compact-open topology seems natural in that case. $\endgroup$ May 17, 2020 at 10:13
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    $\begingroup$ @PawełCzyż I think it coincides with the subspace topology wrt $L(V,W)$. $\endgroup$ May 17, 2020 at 10:20
  • $\begingroup$ Now I see what you meant – thank you! $\endgroup$ May 17, 2020 at 12:44

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