Topology on the general linear group of a topological vector space 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 


*

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

*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 $K^{d^2}$

*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. 
 A: 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. 
A: 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.  
A: 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 https://mathoverflow.net/questions/214/how-to-topologize-xr-when-r-is-a-topological-ring/219#219.
