Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The category of $\mathbb{Z}$-graded abelian groups is equivalent to the category of comodules over the commutative Hopf algebra (over $\mathbb{Z}$) $A=\mathbb{Z}[t,t^{-1}]$, with comultiplication $t\mapsto t\otimes t$ and counit $t\mapsto 1$. Explicitly, the correspondence is given as follows: given an $A$-comodule $M$ with structure map $\psi:M\to A\otimes M$, let $M_n=\{x\in M:\psi(x)=t^n\otimes x\}$. Then the coassociativity and counitality of $\psi$ can be easily checked to imply that $M$ is actually the direct sum of these $M_n$, and it can similarly be checked that a map $M\to M'$ of two such comodules commutes with the coaction of $A$ iff it maps $M_n$ to $M_n'$ for all $n$. The Hopf algebra $A$ is the ring of functions on the algebraic group $\mathbb{G}_m$, and this equivalence of categories is closely related to the correspondence between graded algebras and projective varieties (through the fact that projective varieties are quotients of certain quasiaffine varieties under an action of $\mathbb{G}_m$).

The commutativity of $A$ gives a natural symmetric monoidal structure on the category of $A$-comodules, given by $M\otimes_{\mathbb{Z}}M'$ with coaction $M\otimes M'\to(A\otimes M)\otimes(A\otimes M')\cong(A\otimes A)\otimes(M\otimes M')\to A\otimes M\otimes M'$, where the first map is given by the coactions on $M$ and $M'$ and the last map is given by the multiplication on $A$. This tensor product corresponds to the standard tensor product of graded modules, in which $(M\otimes M')_n=\bigoplus_{i+j=n}M_i\otimes M_j'$.

In this symmetric monoidal structure (whose existence only depends on the fact that $A$ is a commutative Hopf algebra), the swap isomorphism $M\otimes M'\to M'\otimes M$ is the standard swap map coming from the underlying symmetric monoidal tensor product of $\mathbb{Z}$-modules. This corresponds to the "even" symmetric monoidal structure on graded objects, in which the swap map comes from the standard swap map $M_i\otimes M_j'\to M_j'\otimes M_i$. Commutative monoid objects under this structure are what an algebraic geometer might call graded commutative rings. However, there is a second symmetric monoidal structure on graded abelian groups with the same tensor product but for which the swap map $M_i\otimes M_j'\to M_j'\otimes M_i$ is multiplied by a sign $(-1)^{ij}$. In this symmetric monoidal structure, a commutative monoid object is a "skew-commutative graded ring", or what a topologist would just call a graded commutative ring: even degree elements commute with everything, and odd degree elements anticommute with each other.

My question is: is there a natural interpretation of this second symmetric monoidal structure when we view graded abelian groups as $A$-comodules? Morally, I would say it can't come naturally just from the Hopf algebra structure of $A$, because we could reinterpret $A$ as the ring $\mathbb{Z}[s^2,s^{-2}]$ so that everything ought to be treated as having even grading, and then we get the original symmetric monoidal structure (this is like reinterpreting a $\mathbb{Z}$-graded ojbect as actually being $2\mathbb{Z}$-graded). Thus, I suppose I'm really asking: is there some additional structure you can put on $A$ that makes the "anticommutative" symmetric monoidal structure pop out naturally (and have a purely categorical description)?

share|improve this question
1  
Yes, it's called a triangular structure. See mathoverflow.net/questions/4640/… . –  Qiaochu Yuan Mar 13 '11 at 11:25
    
Great! If you post that as an answer I'll accept, though I would also like to have a conceptual description of the triangular structure (Would it correspond to anything if we had sets and groups instead of modules and Hopf algebras? Does it have an algebro-geometric interpretation on $\mathbb{G}_m$? Does it have a geometric interpretation in terms of TQFT?) –  Eric Wofsey Mar 13 '11 at 14:22

1 Answer 1

up vote 3 down vote accepted

As Qiaochu points out, you can include a little extra data in your Hopf algebra $\mathbb G_m$, and use it to realize the braiding. Here's the best way to say this, since you've decided to work with comodules over $\mathbb Z[t,t^{-1}]$. The point is that the comultiplication on $\mathbb Z[t,t^{-1}]$ makes the space of linear maps $\hom(\mathbb Z[t,t^{-1}],\mathbb Z)$ into an algebra under "convolution", and also the space of linear maps $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$; also, $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},A)$ is a bimodule for this algebra for any abelian group $A$. A triangular structure is an element $\beta \in \hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$ which squares to identity under convolution, and conjugates the multiplication to its opposite (which is, in this case, the same coproduct), and also should satisfy some other conditions. In your case, the map $\beta$ on a basis is $\beta(t^n,t^m) = (-1)^{nm}$, which should look familiar. (A quasitriangular structure is the same except that $\beta$ is required to be invertible but not square-$1$.)

The funny thing, of course, is why you would have to put this in by hand. Since $\mathbb Z[t,t^{-1}]$ is commutative, there is a canonical choice of $\beta$, the identity element in $\hom(\mathbb Z[t,t^{-1}]^{\otimes 2},\mathbb Z)$ (the tensor square of the counit). But you could have picked some other category (say, super-modules), and then built a commutative Hopf algebra where the canonical triangular structure was this one.

Note that $\beta$ is not an algebra homomorphism, so it's hard to see it from the geometric side. This is not surprising. The corresponding notion for groups and sets would be: let $G$ be a group; then a triangular structure for $G$ is a map of sets $1 \to G^{\times 2}$ that commutes with the (unique!) comultiplication $G \to G^{\times 2}$; such a map is just an ordered pair $(a,b)$ of central elements of $G$; but then the other conditions come in, which say that $(a,b,b) = (a^2,b,b)$ and something similar on the other side, and also that $a,b$ are invertible (self-inverse if you want triangular rather than just _quasitriangular), and clearly this means that $(a,b) = (1,1)$. So a group admits only one quasitriangular structure --- its canonical one. From the category-theory side, all I said was that the cartesian product (in $G$-sets), thought of as a monoidal structure, is braided in a unique way.

In the physics world, one does imagine that $\mathbb G_m$ acts on one's graded vector spaces; for a physicists, an action is the same as a conserved quantitiy, and there are many formulas that include explicit mention of this one, called "ghost number" in most situations. The action is a funny one — physicists work over $\mathbb C$ with its unitary structure, and this action is antiunitary rather than unitary. I don't know of a similar discussion in TQFT (of course, supermanifolds are very geometric).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.