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Let $K$ be a topological field. Let $V$ be a topological vector space over $K$ (if it makes things convenient, you may assume it is finite dimensional).

Naive Question: Is there a canonical way of defining a topology on $\text{GL}(V)$?

Attempted Focusing of Naive Question: Let $\mathcal{C}$ be the subcategory of $\mathsf{TVect}_K$ with the same objects (topological $K$-vector spaces) but where the morphisms are just the isomorphisms from $\mathsf{TVect}_K$. Can we define a functor $A:\mathcal{C}\rightarrow\mathsf{TGrp}$ such that

  1. $B(A(V))=\text{GL}(V)$ for all $V\in \mathcal{C}$, where $B:\mathsf{TGrp}\rightarrow\mathsf{Grp}$ is the forgetful functor

  2. Let $X=K^d$ with the standard (product) topology. Then $A(X)\cong\!\! \text{GL}_d(K)$, where $\text{GL}_d(K)$ is given the subspace topology from $K^{d^2}$

  3. Let $Y=K^d$ with the trivial (indiscrete) topology. Then $A(Y)\cong\!\!\text{GL}_d(K)$, where $\text{GL}_d(K)$ is given the trivial topology

In short, I want to avoid "silly" answers, like $A(V)=\text{GL}(V)$ with the trivial topology for all $V$. I'm not sure if my above conditions will sufficiently rule out that kind of thing, but if you see a "silly" answer, I encourage you to post it so that either I can hone my question better, or I can see why there is no well-defined question to ask here.

Motivation: In my differential topology class today, there was a lot of debate about what the topology on the Grassmanian $\text{Gr}(r,V)$ was (for $V$ an $\mathbb{R}$-vector space). The professor ultimately gave what was, in my opinion, an unaesthetic answer that depended on choosing a basis for $V$ (which I avoid when possible) and, now that $V$ and $\mathbb{R}^n$ are identified via the choice of basis, using the inner product structure on $\mathbb{R}^n$ (which I also avoid when possible) to define a metric, and hence topology, on the set of $r$-dimensional subspaces. In particular, I saw no reason there should fail to be a topology on $\text{Gr}(r,V)$ in the absence of an inner product structure on $V$, so my motivation here is to define a canonical topology on $\text{Gr}(r,V)$ for any topological vector space $V$. Looking at the Wikipedia page on Grassmanians, it seems the natural way of going about this would just be to have a topology on $\text{GL}(V)$, and then put the quotient topology on $\text{Gr}(r,V)=\text{GL}(V)/H$ where $H=\text{Stab}(W)$ for some $r$-dimensional subspace $W\subset V$. This raised the question of what exactly the topology was on $\text{GL}(V)$, which is what I'm asking here now.

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I'm not sure my comment is helping, but whatever norm you put on a finite dimensional vector space, I think the topology it induces turns out to be the same (so the same goes for $\textrm{End}(V)$, of which $\textrm{GL}(V)$ is an open subset). – Joel Cohen May 7 '11 at 1:20
@Joel: Well, my question is about topological vector spaces over some topological field $K$, where the topology is not necessarily induced by a norm (also note that $K$ is not necessarily $\mathbb{R}$ or $\mathbb{C}$). So I would hope to have a definition of a topology for $\text{GL}(V)$ even when $V$ has e.g. the trivial topology. But your comment may still be helpful, I am not sure where this question will lead. – Zev Chonoles May 7 '11 at 1:41
Forgive me if my question is naïve, but setting aside topological questions for a moment, how do you define a functor $\textrm{GL} : \textrm{Vect}_K \to \textrm{Grp}$ on morphisms? – Joel Cohen May 7 '11 at 3:06
@Joel: Ah, excellent point - I suppose my question only makes sense if $\mathsf{Vect}_K$ is restricted to the subcategory where the only morphisms are isomorphisms. I will make an edit. – Zev Chonoles May 7 '11 at 3:17
I asked this question once, and the answer I got was something to the effect of "use the compact-open topology". – Zhen Lin May 7 '11 at 8:00

My impression is that the Grassmannian, as a moduli space, ought to be given the unique topology (and, since you said this came up in a differential geometry class, smooth structure) that allows it to represent the appropriate moduli functor. Unfortunately I don't know what this moduli functor is in the smooth category; in the algebraic category it's described at this MO question.

In any case, for the case of $\mathbb{R}$-vector spaces what's wrong with the subspace topology induced from $\text{Hom}_{\mathbb{R}}(V, V)$ when $V$ is finite-dimensional? Isn't there a unique Hausdorff topology on a finite-dimensional $\mathbb{R}$-vector space making addition continuous and compatible with scalar multiplication?

Edit: Here's a sketch.

Proposition: Let $V$ be a finite-dimensional real vector space of dimension $n$ equipped with a Hausdorff topology such that addition and scalar multiplication $\mathbb{R} \times V \to V$ are continuous. Then $V \cong \mathbb{R}^n$ with the product topology.

Proof. Let $e_1, ... e_n$ be a basis for $V$. By assumption the function

$$f : \mathbb{R}^n \ni (x_1, ... x_n) \mapsto x_1 e_1 + ... + x_n e_n \in V$$

is a continuous bijection. For every real $r$, the restriction of $f$ to the hypercube $[-r, r]^n$ is a continuous bijection from a compact space onto its Hausdorff image, hence is a homeomorphism.

I don't really see any way to avoid bases here; there's no way to construct $f$ canonically anyway.

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We were having enough trouble coming up with a topology on it, much less a smooth structure, so I wanted to restrict my question to the topology only. Also, I like the idea of defining the topology via a moduli functor (side note: it certainly would be interesting to see how far that approach can be taken for manifolds), but still this would only handle topological vector spaces of the form $\mathbb{R}^n$ and $\mathbb{C}^n$. I was hoping there would be a general construction for not-necessarily-Hausdorff topological vector spaces, and for arbitrary topological fields. – Zev Chonoles May 7 '11 at 6:26
Although, given any topological vector space $V$, I suppose we could just consider a (significantly altered) concept of manifold as just a topological space that is locally like $V$ (not necessarily Hausdorff, since I would prefer not to require that of $V$), and perhaps the moduli space approach on these structures could yield a topology. – Zev Chonoles May 7 '11 at 6:30
Also, it is quite plausible to me that there is a unique Hausdorff topology on a finite dimensional $\mathbb{R}$-vector space making it a topological $\mathbb{R}$-vector space, though I don't believe I've heard explicitly of such a result. Again, ideally I am looking for arbitrary topological vector spaces (or, I suppose, an explanation of whatever the minimum amount of structure on $K$ and $V$ is necessary to get an induced topology on $\text{GL}(V)$ in a canonical way). – Zev Chonoles May 7 '11 at 6:39
@Zev: ah, I thought that result was well-known. I'll include a proof. (I don't really know if it makes sense to expect a result for arbitrary $K$. I don't think the analogue of the above result holds in general.) – Qiaochu Yuan May 7 '11 at 6:48
@ Qiaochu : Your proof works for any locally compact field $K$ (which includes finite fields, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_p$ ...). You just have to replace $[-r,r]$ with compact neighborhoods covering $K$. – Joel Cohen May 7 '11 at 15:13

If $V$ is a topological vector space you can give $GL(V)$ the compact-open topology. This is the usual topology for these spaces. Then the Grassmannians are topologized as quotient objects.

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I agree that this works well if $V$ is finite dimensional and $K$ is locally compact. Are you saying this is the good topology in general? Why is $\operatorname{GL}(V)$ a topological group? Don't you need to "Kellify" or something (this is a very naive comment, I haven't thought about the details recently)? – t.b. May 7 '11 at 9:37

Suppose V is finite dimensional. Then K only needs to have the structure of a topological ring in order for GL(V) to have a canonical topology. This comes from the structure of GL_n as an algebraic variety. Now this question becomes a special case of a MO question

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Thanks for your answer, bartogian, this is an interesting idea. However, correct me if I'm wrong, but doesn't this give the topology to $GL(V)$ only when $V\cong K^{\dim(V)}$ as topological vector spaces (in order to identify $GL_n(K)$ with $GL(V)$)? I had been hoping it would be possible to give $GL(V)$ a topology even when $V$ didn't necessarily have the standard topology. – Zev Chonoles Jun 21 '11 at 5:04
Good catch. I missed that this approach only works when V has the standard topology. – bartogian Jun 21 '11 at 5:39

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