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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).2011-05-07
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    @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.2011-05-07
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    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?2011-05-07
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    @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.2011-05-07
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    I asked this question once, and the answer I got was something to the effect of "use the compact-open topology".2011-05-07

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