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Denote the category of $G$-graded vector spaces by SVect, where $G$ is an abelian group. Then morphisms $f:V\to W$ (where $V$ and $W$ are two $G$-graded vector spaces) should be $f(V_g) \subseteq W_g$ for any $g\in G$ in many references. Why don't we take morphisms as other homomorphisms between $V$ and $W$ such as $f(V_{g}) \subseteq W_{g+h}$ for any $g\in G$? In other words, why we use $Hom_{G}(V,W)=\{f|f(V_g) \subseteq W_g\}$ other than $HOM(V,W)=\{f|f(V_g) \subseteq W_{g+h}\forall h\in G\}$ where $HOM(V,W)$ is usually called internal hom in References. Does not SVect with $HOM(V,W)$ as morphisms form a category?

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It gives the wrong notion of isomorphism. Of course the internal hom is useful, but as an internal hom. –  Qiaochu Yuan Jul 11 '12 at 2:57
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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... –  Mariano Suárez-Alvarez Jul 11 '12 at 5:03
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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? –  Mark Grant Jul 11 '12 at 5:04
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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. –  Fernando Muro Jul 11 '12 at 9:54
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3 Answers 3

The mathematics becomes clearer if you add more structure. The category of graded vector spaces with morphisms of degree $0$ is fine. But it has a tensor product and also an internal hom that makes it into a closed symmetric monoidal category. To define the internal hom between graded vector spaces, $HOM(V,W)$, you must of course use morphisms that vary the grading, since it must be a graded vector space. The category of graded vector spaces is enriched over the ordinary category of ungraded vector spaces, and the underlying category of this enriched category is the category of graded vector spaces and morphisms of degree zero that you started with.

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Yes, it seems if we want an isomorphism between $Hom(U\otimes V, W)$ and $Hom(U, Hom(V,W))$ like in the ordinary case, so $Hom(V,W)$ should be an object in this category and we must replace with HOM. –  tzhang Jul 14 '12 at 7:39
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You might want to have a look at Section 3 of Chapter 2 of the Introduction of Saul Lubkin's Cohomology of Completions (North-Holland Mathematics Studies 42, 1980), where the differences between the two possibilities of defining a category of graded vector spaces suggested by you (with the obvious correction suggested by Mark) are discussed in detail and in a very general setting.

Also related, less abstract, but still very interesting are the following two papers (for whose authors names I am not capable of producing here the necessary diacritics):

J. L. Gomez-Pardo, C. Nastasescu, Topological aspects of graded rings, Comm. Algebra 21 (1993), 4481-4493;

J. L. Gomez-Pardo, G. Militaru, C. Nastasescu, When is HOM$_R(M,-)$ equal to Hom$_R(M,-)$ in the category $R-gr$? Comm. Algebra 22 (1994), 3171-3181.

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For the sake of having an answer that's not in the form of a comment...

What you describe are graded morphisms of degree $h$. Presumably you mean to allow all $h$ (otherwise it's not a category). In this category, two objects are isomorphic if they differ by a grading shift. In most applications it's not desirable to consider $V$ to be isomorphic to a shifted version of $V$, so in most applications morphisms are required to have degree zero. But in some applications the category you describe might be the right choice.

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Yes, all $h$ allowed. By the way you mean it is a category with HOM!? But Mariano Suárez-Alvarez and Yuan suggest it gives wrong notion of isomorphism. What's wrong with it? That is my questions. –  tzhang Jul 14 '12 at 7:14
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