The condition End(1) = k in Tannakian Categories A neutral Tannakian category over a field $k$ is a rigid $k$-linear abelian tensor category
$\mathcal{C}$ whose unit $1$ satisfies $\mathrm{End}(1) \simeq k$, and is
moreover equipped with an exact faithful tensor functor $\omega : \mathcal{C} \rightarrow 
\mathrm{Vect}_k$ into the category of finite dimensional $k$-vector spaces.
Question:- Why the condition $\mathrm{End}(1) \simeq k$ is necessary to get equivalence with the category of finite dimensional representations of some affine group scheme?
Thanks in advance!!
 A: If you say the other axioms correctly, then the condition on $\operatorname{End}(1)$ is redundant.  Indeed, the word "tensor functor" implies that $\omega: 1 \mapsto k$, and the word "faithful" implies that $\operatorname{End}(1) \hookrightarrow \operatorname{End}(\omega(1))$.  What you should include that you don't on your list is that $\omega$ be $k$-linear.  You should also demand that $\mathcal C$ be a nontrivial category; then you cannot have $\operatorname{End}(1) = 0$, as $\operatorname{End}(1)$ acts on all other homsets via the $1$ action and in particular $\operatorname{id}\in \operatorname{End}(1)$ acts as the identity on all other homsets.  With all of this, it follows that $\operatorname{End}(1) = k$.
Conversely, you can see the condition that $\operatorname{End}(1) = k$ as being a "nontriviality" condition.  It is necessary only to assure that $\mathcal C \neq 0$.  In particular, no group has zero representation theory, as every group has a trivial representation on $k$.
If you believed in "the empty group", then you would not need this restriction: the zero category is the category of representations of the zero ring, which is "the group ring of the empty group".
A: Assuming you have such an equivalence of categories, the object $1$ is sent to the trivial representation of your affine $k$-group $G$.  This is the representation that factors through the canonical homomorphism from $G$ to the trivial group, and its endomorphisms are the endomorphisms of the one dimensional $k$ vector space, i.e., the ring $k$.  Since the condition $\operatorname{End}(1) = k$ is implied by the existence of such an equivalence, it is a necessary condition.
