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Jul 14, 2012 at 6:40 history edited tzhang CC BY-SA 3.0
added 14 characters in body
Jul 11, 2012 at 21:56 comment added Matthias Ludewig If one defines the morphisms to include the "any h case", then one obtains a grading on HOM(V, W) that turns it into a graded algebra - so this should be quite natural!
Jul 11, 2012 at 18:10 answer added Peter May timeline score: 9
Jul 11, 2012 at 13:02 answer added Fred Rohrer timeline score: 3
Jul 11, 2012 at 12:04 answer added Kevin Walker timeline score: 1
Jul 11, 2012 at 9:54 comment added Fernando Muro You can actually define morphisms as homomorphisms on the underlying vector space. The definition of morphisms should be motivated by the application you have in mind. There are different possible choices.
Jul 11, 2012 at 7:45 comment added Martin Brandenburg tea.mathoverflow.net/discussion/1392/…
Jul 11, 2012 at 5:04 comment added Mark Grant Your HOM consists of morphisms of degree $h$. If you compose two such I think you get something of degree $2h$, so this will not be a category. Or do you mean to let $h$ be any element of $H$ in the definition of HOM?
Jul 11, 2012 at 5:03 comment added Mariano Suárez-Álvarez Well, this gives the wrong notion of isomorphism if it gives the wrong notion of isomorphism. It could well be the correct one—one needs a context to know! It is not like one does not see in nature categories with the same objects but different morphisms, after all...
Jul 11, 2012 at 4:46 comment added S. Carnahan I agree with Qiaochu. Your category is the Tannakian category $Rep(D(G))$. Here $D(G)$ is the diagonalizable group scheme $\underline{Hom}(G, \mathbb{G}_m)$ (see SGA3).
Jul 11, 2012 at 4:00 history edited MTS CC BY-SA 3.0
Fixed some latex
Jul 11, 2012 at 2:57 comment added Qiaochu Yuan It gives the wrong notion of isomorphism. Of course the internal hom is useful, but as an internal hom.
Jul 11, 2012 at 2:55 history asked tzhang CC BY-SA 3.0