0
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ Try writing out exactly what the data of a functor $F:\mathbf{G}\to \mathbf{Vect}_K$ are. $\endgroup$
    – jgon
    Sep 7, 2020 at 21:26
  • 1
    $\begingroup$ What exactly are the objects and morphisms of $\mathbf{G}$? What does a functor do to objects and morphisms of the domain category? $\endgroup$
    – jgon
    Sep 7, 2020 at 23:04

1 Answer 1

5
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Sep 9, 2020 at 0:53
  • $\begingroup$ 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 $\rho(g)=\mathrm{id}$). $\endgroup$
    – Erik D
    Sep 9, 2020 at 1:34
  • $\begingroup$ 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? $\endgroup$ Sep 13, 2020 at 21:45
  • $\begingroup$ 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. $\endgroup$
    – Erik D
    Sep 22, 2020 at 23:14
  • $\begingroup$ Sounds interesting:-) I just get in touch with group action with connection of endomorphism. $\endgroup$ Oct 21, 2020 at 14:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .