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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!!

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Because it's obviously satisfied in a category $Rep_k(G)$. –  YBL Oct 9 '10 at 8:29
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2 Answers 2

up vote 5 down vote accepted

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".

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Nice answer. After all I think $End(1)=k$ is a more natural condition than $C \neq 0$. The first property is (a priori) preserved under more operations as the second one. –  Martin Brandenburg Jan 5 '11 at 17:58
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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.

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Actually I was trying to find that where the condition $\mathrm{End}(1) = k$ is used in the proof of equivalence. –  Yashica Oct 9 '10 at 12:41
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