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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. $GL_n(\mathbb{C})$)?

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    $\begingroup$ I don't understand what the rank condition has to do with your actual question (the "real" rank seems to depend on the real structure one chooses on the complex vector space that is the source of a morphism. It gives additional structure). I also have the impression that the "classical" enrichments do the job even in the strict categorical sense. Finally, although the mapping spaces are contractible, they carry more structure. For example, the subspaces of mono- or epimorphisms are typically not contractible. Maybe some additional motivation would help me to see your point. $\endgroup$
    – Sebastian Goette
    Commented Feb 23, 2016 at 13:20
  • $\begingroup$ @SebastianGoette In $\infty$-categories there's no notion of monomorphism or epimorphism. The question is asking whether there's a topological enrichment of $Vect$ such that the interior of the resulting $\infty$-category is $\coprod_n BO_n$ and such that the natural adjunction between real and complex vector spaces can be upgraded to an adjunction of $\infty$-categories. It seems a very reasonable question to me. $\endgroup$
    – Denis Nardin
    Commented Feb 23, 2016 at 15:09
  • $\begingroup$ The usual approach is to utilize the topological category given by the isometric embeddings. This works well enough but it does not satisfy the condition on the adjunction. $\endgroup$
    – Denis Nardin
    Commented Feb 23, 2016 at 15:11
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    $\begingroup$ If you take the $\infty$-category associated to the usual enrichment then the adjunction works but all the groups of self-equivalences are trivial (in fact, the entire $\infty$-category is equivalent to a point). If you first take the maximal sub-groupoids of $Vect_{\mathbb{C}}$ and $Vect_{\mathbb{R}}$ (with their usual enrichment), and then take the associated $\infty$-categories then the groups of self-equivalences are what you want, but you don't have an adjunction any more. The question is if there exists an $\infty$-category which gives both. $\endgroup$
    – Yonatan Harpaz
    Commented Feb 23, 2016 at 15:49
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    $\begingroup$ I thought that people usually pass to $\infty$-categories if strict categories do not work for a certain problem. Here, I have the impression that all the OP asks for holds in strict categories, enriched over $Top$. Hence, I asked for additional motivation in my first comment. $\endgroup$
    – Sebastian Goette
    Commented Feb 23, 2016 at 16:34

1 Answer 1

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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 $n=2$).

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  • $\begingroup$ At the limit the map $U\to O$ is certainly not a retract because it is not injective on homotopy groups (in particular on $\pi_1$, where $\pi_1U=\mathbb{Z}$ and $\pi_1O=\mathbb{Z}/2$). So for $n\gg0$ the map $U_n\to O_{2n}$ cannot be a retraction. In fact I think this proves it even for $n=2$ $\endgroup$
    – Denis Nardin
    Commented Feb 23, 2016 at 16:07
  • $\begingroup$ Yes, I believe you're right. Indeed, $\pi_1(O(4)) = \pi_1(O(3))$ is already $\mathbb{Z}/2$. $\endgroup$
    – Yonatan Harpaz
    Commented Feb 23, 2016 at 19:31
  • $\begingroup$ NIce! thanks for the answer Yonatan, I kind of suspected the answer is no, and it is good to see a proof (a small nitpick, the group $GL(U(\mathbb{C}^n))$ is not connected, but you can restrict to $SL$ and continue the argument without change). $\endgroup$
    – KotelKanim
    Commented Feb 24, 2016 at 8:27
  • $\begingroup$ Right! Silly of me. I added an edit to correct it. By the way, the argument above seems to work quite generally. If $f: C \rightleftarrows D: g$ is an adjunction of $\infty$-categories then the induced map $g_*:Aut^0(X) \to Aut^0(Y)$ is a retract inclusion if $X$ is in the essential image of $f$, and vice-versa. $\endgroup$
    – Yonatan Harpaz
    Commented Feb 24, 2016 at 8:56

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