Timeline for []-infinity algebra and Projective representation
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21 events
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| Nov 20, 2011 at 6:03 | comment | added | Theo Johnson-Freyd | Oh, I'm sorry, yes of course there is a very standard notion of "projective representation" of a group, as an action $G \to \operatorname{PGL}(V)$, where $V$ is a vector space and $\operatorname{PGL}(V) = \operatorname{GL}(V)/\text{center})$; these are generally classified in terms of 2-cocycles. What confused me is that this notion is usually fairly specific to things that are not already linear (groups, bordism categories categories), whereas your first paragraph had put me in mind of modules for (linear) algebraic objects. | |
| Nov 19, 2011 at 14:56 | comment | added | Qfwfq | Is the above definition of projective representation somehow related to a homomorphism $\rho : G \to \mathrm{PGL}(n,\Bbbk)$ or is "projective" meant in another sense? | |
| Nov 19, 2011 at 10:30 | comment | added | Ma Ming | @S. Carnahn The traditional projective representation of G a group, a linear representation up to scalars in a coherent way, is just a group representation in certain 2-category, a representation upto 2-morphisms in a coherent way, which is just a non-strict functor from one object category BG to some 2-category. That why I say the target is replaced by higher stuff. | |
| Nov 19, 2011 at 8:03 | comment | added | Vladimir Dotsenko | @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. | |
| Nov 19, 2011 at 5:17 | comment | added | S. Carnahan♦ | 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. | |
| Nov 18, 2011 at 22:19 | history | edited | Ma Ming | CC BY-SA 3.0 |
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| Nov 18, 2011 at 21:38 | history | edited | Ma Ming | CC BY-SA 3.0 |
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| Nov 18, 2011 at 21:24 | comment | added | Ma Ming | @Theo By projective representation(related term, twisted action) of G a group, I think it as a (non-strict) 2-functor from the one object category associated to G to a 2-category. | |
| Nov 18, 2011 at 21:09 | comment | added | Theo Johnson-Freyd | @Qfwfq: It's important to keep in mind that an operad is nothing more nor less than an associative algebra object in a certain (non-symmetric!) monoidal category $\mathcal S$. This category has the property that every symmetric monoidal category $(\mathcal C,\otimes)$ is naturally enriched over $\mathcal S$, and so it makes sense to talk about "modules" for the operad in $\mathcal C$. But these are usually called "algebras" for the operad, which is really a shame, as the language obscures the underlying structure. | |
| Nov 18, 2011 at 21:06 | comment | added | Theo Johnson-Freyd | That said, I don't understand your question. Usually, one thinks of "projective" as a property that some (non-$\infty$ized) representations have: it has something to do with certain maps factoring. Projective modules are important because they satisfy a version of the "axiom of choice", which gives category of chain complexes of projective modules certain properties making it better than the category of (chain complexes of) all modules. | |
| Nov 18, 2011 at 21:03 | comment | added | Theo Johnson-Freyd | (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 mathoverflow.net/questions/80803 . | |
| Nov 18, 2011 at 21:00 | comment | added | Theo Johnson-Freyd | @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) | |
| Nov 18, 2011 at 20:07 | comment | added | Ma Ming | @Qfwfq Yes, some use module, some use algebra (seems more common), to compare I chose module. | |
| Nov 18, 2011 at 19:48 | comment | added | Qfwfq | I'm by no means an expert, but I think the right terminology in the context you mention would be « algebra for an operad $T$ », not « module for an operad $T$ » | |
| Nov 18, 2011 at 16:43 | comment | added | Ma Ming | @Qianchu The grading is encoded by the action of $R^{1|0}$, the differential is encoded by the action of $R^{0|1}$, we shall mean the same thing. | |
| Nov 18, 2011 at 15:52 | comment | added | Qiaochu Yuan | 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 mathoverflow.net/questions/59357/… . | |
| Nov 18, 2011 at 15:39 | history | edited | Ma Ming | CC BY-SA 3.0 |
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| Nov 18, 2011 at 15:06 | history | edited | Ma Ming |
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| Nov 18, 2011 at 14:50 | history | edited | Ma Ming | CC BY-SA 3.0 |
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| Nov 18, 2011 at 14:44 | history | edited | Ma Ming | CC BY-SA 3.0 |
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| Nov 18, 2011 at 14:39 | history | asked | Ma Ming | CC BY-SA 3.0 |