The condition End(1) = k in Tannakian Categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:36:42Z http://mathoverflow.net/feeds/question/41569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41569/the-condition-end1-k-in-tannakian-categories The condition End(1) = k in Tannakian Categories Yashica 2010-10-09T07:05:04Z 2010-10-09T18:49:11Z <p>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.</p> <p>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?</p> <p>Thanks in advance!!</p> http://mathoverflow.net/questions/41569/the-condition-end1-k-in-tannakian-categories/41571#41571 Answer by S. Carnahan for The condition End(1) = k in Tannakian Categories S. Carnahan 2010-10-09T08:29:45Z 2010-10-09T08:29:45Z <p>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.</p> http://mathoverflow.net/questions/41569/the-condition-end1-k-in-tannakian-categories/41623#41623 Answer by Theo Johnson-Freyd for The condition End(1) = k in Tannakian Categories Theo Johnson-Freyd 2010-10-09T18:49:11Z 2010-10-09T18:49:11Z <p>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 $\operatorname{End}(1) = k$.</p> <p>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 $k$.</p> <p>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".</p>