MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a very vague question.

We know that some algebra structures can be viewed as modules of some fantastic stuff, call T. Such examples include: Abelian groups are $\mathbb{Z}$-modules, chain complexes are modules of $\mathbb{R}^{1|1}$(it is a super Lie group, I shall not call it as dual numbers, since the multiplication here differs from the usual one), more generally, modules of a operad T (Lie algebra and associative algebra are of this kind). To get the infinitified version(L-infinity algebra, A-infinity algebra, etc.) one uses an infinity operad. Note that module is basically a morphism from T to some Auto( ). And the infinitified version basically replace the source thing by a higher thing.

Projective representation goes differently. A representation is the same as a module, a morphism from T to some Auto( ). Projective representation of G, which can be think of as a non-strict functor from BG, the one object category associated with G, to some 2-category. A lot of constructions come from similar ways, they basically replaces the target by a higher thing. Other examples: 2-bundle as a functor Cech groupoid to a 2-category associated to a 2-group, and representation (of a Lie algebra) up to homotopy should also be this kind.

A nature question: whether these two constructions are the same? It seems obvious not. On the other hand, I feel they should have some relations.

To make my question more clear and concrete, consider Lie algebra and L-infinity algebra. Lie algebras are modules of the operad $\mathcal{Lie}$, view as certain map from $\mathcal{Lie}$ to some Auto(). L-infinity algebras are modules of operad $\mathcal{Lie}^\infty$, view as certain map from $\mathcal{Lie}^\infty$ to some Auto(). My question is whether it is possible to view L-infinity algebras as certain map from $\mathcal{Lie}$ to infinitified version of Auto().

share|cite|improve this question
Your description of chain complexes doesn't describe the grading. One way to describe them as modules over something is as modules over a certain (graded, super) Lie algebra: see Theo Johnson-Freyd's answer at… . – Qiaochu Yuan Nov 18 '11 at 15:52
@Ma Ming: You are correct that the structure of a chain complex can be encoded by an action of a certain super Lie group, whose underlying manifold is $\mathbb R^{1|1}$. I don't like writing the group this way however, just like I don't like to write the group of strictly upper triangular $3\times 3$ matrices as "$\mathbb R^3$", even though this is the underlying manifold. Better is to write the group as $\mathbb R^{1|0} \ltimes \mathbb R^{0|1}$, where the action is $t\cdot \theta = e^t\theta$. But there are some subtleties when trying to define chain complexes as (continued) – Theo Johnson-Freyd Nov 18 '11 at 21:00
(continuation) representations of $\mathbb R^{1|0} \ltimes \mathbb R^{0|1}$. First, $\mathbb R^{1|0}$ has too many representations: some without integer eigenvalues; some that are not semisimple. So better is to use the affine algebraic supergroup $G = \mathbb G_m \ltimes \mathbb G_a^{0|1}$ — it's real points are the Lie group $\mathbb R^\times \ltimes \mathbb R^{0|1}$. Then $G$-mod is equivalent to the category of chain complexes of supervector spaces. Getting the category of chain complexes of regular vector spaces is more subtly; see . – Theo Johnson-Freyd Nov 18 '11 at 21:03
Your use of the term "projective representation" is not compatible with traditional uses. You seem to be looking at representations of 2-groups, which are described by 3-cocycles. Projective representations are described by 2-cocycles, and do not need 2-categories. You can get projective representations from actions of centralizers in 2-groups, but there is a delooping that you seem to be ignoring. – S. Carnahan Nov 19 '11 at 5:17
@Theo, Qfwfq: As Theo says, operads are associative algebras in a certain monoidal category. It sort of makes sense to define modules over associative algebras in such a way that an algebra is naturally a bimodule over itself. This leads to a definition of a module over a given operad which contains what is called algebras over that operad, but much much more, and I think the names 'algebras' and 'modules' are very much justified and do not obscure anything. Alas I have nothing to say about the OP's question right now. – Vladimir Dotsenko Nov 19 '11 at 8:03

Your Answer


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

Browse other questions tagged or ask your own question.