Applying Lemma 2.12 of Deligne's/ Milne's "Tannakian Categories" on an irreducible representation

Question

since this is my first question here, I'm not very certain whether I state it properly. I'm thankful for any helpful remarks.

Currently I'm trying to understand the above-mentioned article which can be found under http://www.jmilne.org/math/xnotes/tc.pdf and right now I'm a bit stuck understanding the proof of the main theorem, 2.11, in particular I'm trying to get familiar with the basic constructions and lemma 2.12 (on page 21).

To get more comfortable, I tried to apply this lemma to the most relevant category, Rep(G) itself, and in particular on some representation $X:= (V, \rho)$, where $V$ is some finite-dimensional vector space over a field $k$.

My first question would be whether $\underline{\text{Hom}}(\omega(X),X) = V^\vee \otimes V$, and if so, how I would be able to verify this. Strictly using the definitions given in the article and if I'm not mistaken, $\underline{\text{Hom}}(\omega(X),X) = V^\vee \otimes X = \oplus_{n=1}^{\text{dim}_k(V)} (V, \rho)$ and considering that this should again be a representation, my guess would have been that not $\underline{\text{Hom}}(\omega(X),X)$, but $\omega(\underline{\text{Hom}}(\omega(X),X)) = V^\vee \otimes V$.

Another question is what happens in this lemma if $X$ even is irreducible. I was pointed towards the question whether the subobject $P' \subset \underline{\text{Hom}}(\omega(X),X) \stackrel{?}{=} V^\vee \otimes V \cong \text{Aut}_k(V)$ with $g.(v_i^\vee \otimes v_j):= v_i^\vee \otimes g.v_j$ containing $\text{id}_V \sim \sum_{i \in I} v_i^\vee \otimes v_i$ wouldn't automatically be $\underline{\text{Hom}}(\omega(X),X)$ itself. ($(v_i)_{i \in I}$ is meant to be a basis of $V$ here.)

My first intuition was to look at the problem component-wise: if $v_i^\vee \otimes v_j$ is contained in the subrepresentation, then $v_i^\vee \otimes v_{l}$ can also be found for every other $l$ since $V$ is irreducible. Matrixwise this should mean that if a matrix with just one component is contained in the subrepresentation, we can push this component from left to right and hence create every entry in this column as we wish. But that didn't get me much further since we don't have those matrices with just one component, but just the identity matrix.

Does anyone have any ideas how I could tackle this problem? Thanks in advance!

Answer 1

To answer your first question, $\underline{\mathrm{Hom}}(\omega(X),X)$ is the representation of $G$ with underlying vector space $V^{\vee}\otimes V$, but where $G$ acts only on the second factor. So you are correct that $\omega(\underline{\mathrm{Hom}}(\omega(X),X))=V^{\vee}\otimes V$ (but generally $\underline{\mathrm{Hom}}(\omega(X),X)\neq X^{\vee}\otimes X$).

For the second question: in general $P'\subsetneq\underline{\mathrm{Hom}}(\omega(X),X)$, even when $X$ is irreducible. By construction $\omega(P') \subset V^{\vee}\otimes V=\mathrm{End}(V)$ contains $\mathrm{Id}_V$ and is closed under postcomposition with the $G$ action. If $G$ is a finite group, this means $\omega(P')$ contains the image of the group algebra $k[G]$ in $\mathrm{End}(V)$, and since $P'$ is defined to be the smallest subobject of $\underline{\mathrm{Hom}}(\omega(X),X)$ satisfying these conditions, $\omega(P')$ is exactly the image of $k[G]$ in $\mathrm{End}(V)$ (which carries a natural $G$-action). In general one can view $P'$ as playing the role of "the image of $k[G]$ in $\mathrm{End}(V)$," even though the group algebra $k[G]$ doesn't really make sense if $G$ is an affine group scheme.