# Functors that aren't continuous/smooth

Let $$\mathbf{FinVec}$$ denote the category of finite-dimensional real vector spaces and linear maps, let $$F\colon\mathbf{FinVec}\rightarrow\mathbf{FinVec}$$ be a functor. For each pair of objects $$V,W$$, the Hom-set $$\operatorname{Hom}(V,W)$$ is itself a finite-dimensional vector space and we equip it with the unique structure of a topological space/of a smooth manifold that is compatible with its vector space structure. The composition maps are then continuous/smooth. In this way, we view $$\mathbf{FinVec}$$ as enriched over the category $$\mathbf{Top}$$/$$\mathbf{Diff}$$ respectively. It then makes sense to ask whether the functor $$F$$ is enriched over $$\mathbf{Top}$$/$$\mathbf{Diff}$$. That is just to ask whether the induced map $$F\colon\operatorname{Hom}(V,W)\rightarrow\operatorname{Hom}(FV,FW)$$ is continuous/smooth for all $$V,W$$.

(Irrelevant to the rest of the question, but the reason this is a useful notion is that it is the natural hypothesis assuring that the functor induces another functor on the category of topological/smooth vector bundles, applying the original functor fiberwise.)

In the case $$V=W$$, by functoriality, we can restrict $$F$$ to a group homomorphism $$\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$$. If the functor is continuous/smooth, so is this map. Note that $$\operatorname{GL}(V)$$ is a Lie group. It is a general result that a measurable group homomorphism between Lie groups is automatically smooth. This begs the following questions:

1. What's an argument for the existence of functors that aren't continuous? Can we construct one? Given the above, it either happens that the restrictions to $$\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$$ are continuous and continuity fails elsewhere, or the induced maps will be non-measurable, so that the construction will necessarily have to involve choice to some extent.
2. Are there continuous functors that aren't smooth? Given the above, a continuous functor will have smooth restrictions to $$\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$$, but smoothness may fail elsewhere.
• Many of the functors you encounter will be $\mathbb{R}$-linear – which is better than continuous or even smooth! Oct 30 '20 at 3:36
• I mean yeah there ought to be some terrible choicy thing you can do I would expect, like with the $\Bbb{Q}$-vector space automorphisms of $\Bbb{R}$, but I do sort of wonder why we would care about nonlinear functors. It's sort of like taking all group homomorphisms between Lie groups. Like you could do that, but why?
– jgon
Oct 30 '20 at 4:54
• Over an arbitrary field $k$ there are endofunctors coming from twisting by endomorphisms $k \to k$ (and for $k = \mathbb{C}$ most of these are non-measurable) but there aren't any such endomorphisms for $k = \mathbb{R}$. Over $\mathbb{R}$ every endofunctor I know how to construct is a Schur functor and so is even polynomial (and in particular smooth and in particular continuous). Oct 30 '20 at 8:52
• Ah, this question came up previously and I said the same thing then: math.stackexchange.com/questions/3016674/… Oct 30 '20 at 9:06

Reinstate Monica gives a construction in the comments to the question I linked in the comments (I can't close as a duplicate because that question has no upvoted or accepted answers): first, the category $$\text{FinVec}(\mathbb{C})$$ of finite-dimensional complex vector spaces has discontinuous endofunctors given by pulling back along discontinuous automorphisms $$\mathbb{C} \to \mathbb{C}$$. Second, these give rise to discontinuous endofunctors of $$\text{FinVect}(\mathbb{R})$$ given by complexifying, applying one of these automorphisms, then restricting back down to a real vector space of twice the dimension.
I suspect every continuous endofunctor is smooth, and in fact I suspect every continuous endofunctor is a (possibly infinite) sum of Schur functors, and hence even polynomial (by which I mean the induced maps $$\text{Hom}(V, W) \to \text{Hom}(F(V), F(W))$$ are polynomial; there are a few other possible definitions and I don't think they agree in general). The "possibly infinite" is to handle examples like the exterior algebra $$V \mapsto \wedge^{\bullet}(V)$$.