# What's the meaning of defining a functor in a natural way?

I have a question: Given a group $$\mathbf{G}$$, there is homomorphism $$\rho$$ $$\colon$$ $$\mathbf{G}$$ $$\to$$ $$\mathbf{GL(V)}$$. BTW, $$\rho$$ is a representation of a group $$\mathbf{G}$$ on a vector space.

Now the task is to define a functor $$F$$ $$\colon$$ $$\mathbf{G}$$ $$\to$$ $$\mathbf{Vec}_K$$ in a natural way.

The second part is a converse way, that is, given a functor $$F$$ $$\colon$$ $$\mathbf{G}$$ $$\to$$ $$\mathbf{Vec}_K$$, define a representation of $$\mathbf{G}$$ on a vector space in a natural way.

Can anyone give me ideas?

• Try writing out exactly what the data of a functor $F:\mathbf{G}\to \mathbf{Vect}_K$ are.
– jgon
Sep 7, 2020 at 21:26
• What exactly are the objects and morphisms of $\mathbf{G}$? What does a functor do to objects and morphisms of the domain category?
– jgon
Sep 7, 2020 at 23:04

It seems that what you are after is the correspondence between representations of $$G$$ and functors from the groupoid G to the category of vector spaces over $$k$$. (By "the groupoid $$G$$" I mean the category with one object "$$\bullet$$", and $$\mathrm{Hom}(\bullet,\bullet) = G$$.)

To define a representation $$\rho:G\to \mathrm{GL}_k(V)$$, you need the following data:

1. a vector space $$V$$;
2. for each $$g\in G$$, a linear map $$\rho(g):V\to V$$ (subject to the relevant compatibility conditions making $$\rho$$ a group homomorphism).

To define a functor $$F:G\to\mathrm{Vec}(k)$$, you need:

1. a vector space $$F(\bullet)$$;
2. for each $$g\in\mathrm{Hom}(\bullet,\bullet) = G$$, a linear map $$F(g): F(\bullet)\to F(\bullet)$$ (subject to the relevant compatibility conditions making $$F$$ a functor).

What would be the natural thing to do?

• I have a question in mind now: 1) To define a group $\mathbf{G}$ representation on $\mathbf{GL(V)}$. You define a linear map $\rho(g)$ $\colon$ $\mathbf{V}$ $\to$ $\mathbf{V}$. Does this linear transformation preserves basis? I would prefer to rewrite a few days later based on what I understand. Thank you. Sep 9, 2020 at 0:53
• For any $g\in G$, the map $\rho(g)$ is invertible (since it is in $\mathrm{GL}(V)$). Hence it sends any basis of $V$ to a basis of $V$ (but not the same basis, unless $\rho(g)=\mathrm{id}$). Sep 9, 2020 at 1:34
• Is it possible for $\rho(g)$ is not invertible? In other words, whether there is a chance it is not $GL(V)$ group, but others? Sep 13, 2020 at 21:45
• By the common definition, the codomain of $\rho$ is $\mathrm{GL}(V)$, so the linear map $\rho(g):V\to V$ is invertible for all $g\in G$. This is, however, not strictly necessary: You may instead say that $\rho: G \to \mathrm{End}_k(V)$ is a morphism of monoids, where $\mathrm{End}_k(V)$ is the monoid of all linear maps $V\to V$ (with composition as multiplication). Explicitly, you require $\rho(gh) = \rho(g)\circ\rho(h)$ for all $g,h\in G$, and $\rho(1_G) = \mathrm{id}_V$. Then $\rho(g)\circ\rho(g^{-1}) = \rho(gg^{-1}) = \mathrm{id}_V$, so $\rho(g)$ is still invertible. Sep 22, 2020 at 23:14
• Sounds interesting:-) I just get in touch with group action with connection of endomorphism. Oct 21, 2020 at 14:46