Let $Vect\_{\mathbb{R}}$ be the category of (say, finite dimensional) vector spaces over $\mathbb{R}$. The automorphism group of the object $\mathbb{R}^n\in Vect\_{\mathbb{R}}$, is $GL\_n(\mathbb{R})$. We usually like to think of it as a *topological* group. For example $BGL\_n(\mathbb{R})$ classifies real $n$-dimensional vector bundles. This suggests that perhaps we can make $Vect\_{\mathbb{R}}$ into a topological category in such a way that the topological group $BGL\_n(\mathbb{R})$ will be the *space* of automorphisms of $\mathbb{R}^n$. Namely, we will get an $\infty$-category such that the $BGL\_n(\mathbb{R})$-s are the connected components of its space of objects.


The obvious idea is just to take the hom-sets of linear maps with their natural topology, but note that this makes all the mapping spaces *contractible*. Of course one can make an artificial definition, like taking only isomorphisms or chopping the mapping space into connected components according to the rank, but this has other disadvantages. For example, if we let $Vect\_{\mathbb{C}}$ be the category of (finite dimensional) vector spaces over $\mathbb{C}$, we have the extension/restriction of scalars adjunction 
$$
Vect\_{\mathbb{R}}\leftrightarrows Vect\_{\mathbb{C}}
$$
and we would like to make it into an $\infty$-adjunction. The isomorphism of hom-sets that comes from the adjunction does not respect ranks. So my question is basically, 



> 
> can one enrich $Vect\_{\mathbb{R}}$ and $Vect\_{\mathbb{C}}$ over topological spaces in such a way that the restriction/extension of scalars will induce an $\infty$-adjunction and such that the automorphisms of $\mathbb{R}^n$ (resp. $\mathbb{C}^n$) will be equivalent to $GL\_n(\mathbb{R})$ (resp. $GL\_n(\mathbb{C})$)?
> 
> 
> 



real and complex vector spaces as topological categories
I think the answer is no. Suppose there exists an enrichment satisfying your requirements, and let $U: Vect\_{\mathbb{C}} \to Vect\_{\mathbb{R}}$ be the forgetful functor. Let $C \subseteq Map(\mathbb{R}^n,U(\mathbb{C}^n))$ be the subspace consisting of those maps which are adjoint to equivalences $\mathbb{R}^n \otimes \mathbb{C} \to \mathbb{C}^n$. Then $C$ is a connected component of the mapping space, and since the connected group $GL(U(\mathbb{C}^n))$ acts on this mapping space by post-composition it must preserve this component. On the other hand, if one restricts this action to $GL(\mathbb{C}^n) \hookrightarrow GL(U(\mathbb{C}^n))$ then $C$ becomes a principal homogeneous space. This implies that for every $n$ the subspace $GL\_n(\mathbb{C}) \cong GL(\mathbb{C}^n) \hookrightarrow GL(U(\mathbb{C}^n)) \cong GL\_{2n}(\mathbb{R})$ is a retract up to homotopy. At $n=1$ this happens to be ok, as $GL\_1(\mathbb{C}) \simeq S^1$ is indeed a retract of $GL\_2(\mathbb{R}) \simeq O(2) \simeq S^1 \coprod S^1$, but I would bet you would be able to find an $n$ where this inclusion is not a homotopy retract.


**-- Edit --** 


As pointed out in the comments below:


1) Already when $n=2$ the map $GL(\mathbb{C}^n) \to GL(U(\mathbb{C}^n))$ is not a homotopy retract.


2) $GL(U(\mathbb{C}^n))$ is not connected, but one can replace it with the connected component of the identify $GL^0(U(\mathbb{C}^n)) \subseteq GL(U(\mathbb{C}^n))$, replace $C$ with the corresponding $C^0 \subseteq C$, and continue the argument as before (since $GL(\mathbb{C}^n)$ is connected its image in $GL(U(\mathbb{C}^n))$ lies in $GL^0(U(\mathbb{C}^n))$, and the inclusion $GL(\mathbb{C}^n) \hookrightarrow GL^0(U(\mathbb{C}^n))$ is again not a retract already for $n=2$).


