# Factorization and vertex algebra cohomology

A chiral algebra on a smooth curve $$X$$, in the sense of Beilinson-Drinfeld, is a right $$D_{X}$$-module with a chiral bracket, which is a map $$\mathcal{V}^{\boxtimes 2}(\infty\Delta)\rightarrow \Delta_{*}\mathcal{V}$$ satisfying certain conditions making it look like a Lie bracket. In fact $$\mathcal{V}$$ is a Lie algebra with respect to a certain pseudo-tensor structure on $$D_{X}$$-modules. A pseudo-tensor structure is what one obtains when one forgets the tensor product but remembers the associated multilinear maps. Multi-linear maps from a collection $$\{\mathcal{V}_{i\in I}\}$$ to $$\mathcal{W}$$ are defined to be $$\mathrm{Hom}_{D_{X^{I}}}(\boxtimes_{i\in I}\mathcal{V}_{i}(\infty\Delta_{\mathrm{big}}),\Delta_{*}\mathcal{W})$$. One thus obtains a complex of vector spaces $$C^{*}_{CE}(\mathcal{V})$$ via the appropriate construction of Chevalley-Eilenberg cochains, which makes sense in a pseudo-tensor category. Notably this formalism does not produce CE chains.

If $$X$$ is taken to a be a formal disc $$D$$, then we obtain a vertex algebra. I believe in this case the above construction produces the vertex algebra cohomology (with coefficients in the adjoint module) studied in the work of Bakalov, de Sole, Heluani and Kac (arXiv link). Is this true? If not what is the precise relation?

Now if $$\mathcal{V}$$ is a chiral algebra, then BD construct its factorization homology. This is constructed as de Rham homology of the associated $$D$$-module on the Ran space of $$X$$. In chiral terms I believe it is gotten by taking (the colimit over finite sets $$I$$) of the Chevalley Cousins complexes on $$X^{I}$$, cf. these notes by Rozenblyum (on Gaitsgory's webpage).

This can also be performed on a formal disc $$D$$*, in which case one obtains what one might call Vertex Homology. Has this been studied anywhere? Is there a relation with the work of BdSHK cited above, in particular does the BdSHK complex act on it?

For example, if I'm not mistaken we compute $$H^{\mathrm{vtx}}_{0}(V)=V/\operatorname{Res}(V)$$, where $$\operatorname{Res}(V)$$ is the image of all the residues $$\operatorname{res}_{z=0}(v(z))$$ of the fields, which has a very similar flavour to the complex of BdSH.

*Edit, in accordance with David Ben-Zvi's comments it seems that vertex algebra homology as defined in this question would be better defined over a punctured disc $$D^{*}$$.

• Factorization homology of a vertex algebra on the disc recovers the vertex algebra itself. Its factorization algebra on the punctured disc recovers the universal enveloping algebra of the vertex algebra (the Lie algebra of Fourier coefficients), and the former is a module for the latter (in fact modules for the latter are the same as vertex/chiral modules for the vertex algebra). So it sounds like the vertex algebra cohomology you're asking about is a form of coinvariants for the enveloping algebra? – David Ben-Zvi Feb 7 at 15:48
• @ David, perhaps I'm using non standard terminology but I would have thought that factorization homology is already a sort of coinvariants construction as it is the derived functor of conformal blocks, which is fairly clearly a sort of coinvariants. In particular I don't think I understand your first sentence, do you have a reference handy? – EBz Feb 7 at 17:25
• Eg if one computes H^{fact}_{0}(X, V) for a curve X one can compute it by first choosing a point x and then computing coinv for the action of DR(X\x, V) on the fibre V|_{x}, this isn't immediate but it's not too hard. – EBz Feb 7 at 17:28
• I don't remember exact references but this is in Beilinson-Drinfeld Chiral Algebras (and in my book with Frenkel in the underived case). I find it much easier to understand in the topological context (cf Ayala-Francis, Factorization Homology), where fact.homology is DEFINED by the property that it attaches the (here $E_2$ rather than vertex) algebra to the disc and you extend by colimits over disc embeddings. – David Ben-Zvi Feb 7 at 20:55
• The key property of factorization homology is the Mayer-Vietoris formula, explaining how to glue over a cylinder. In the conformal case that's exactly the formula you mention: blocks on surface is given by relative tensor product of the punctured curve and the disc over the punctured disc, which becomes coinvariants for the "out" fields on the disc. This is consistent, in that if you apply your formula to the disc, you get relative tensor product of punctured disc (ie enveloping algebra) with the disc over the punctured disc, ie the vertex algebra again (hooray for circular reasoning :-) ) – David Ben-Zvi Feb 7 at 20:58

## 1 Answer

If $$X$$ is taken to a be a formal disc $$D$$, then we obtain a vertex algebra. I believe in this case the above construction produces the vertex algebra cohomology (with coefficients in the adjoint module) studied in the work of Bakalov, de Sole, Heluani and Kac. Is this true? If not what is the precise relation?

This is not exactly so. If you take $$X=\mathbb{A}^1$$ and your D-modules are translation equivariant and you consider the translation equivariant CE complex you will get the same complex that we studied.

Chiral algebras on the disc are not the same thing as vertex algebras. If you consider chiral algebras equivariant with respect to the action of the group $$Aut_\mathcal{O}$$ of changes of coordinates then they are the same as quasi-conformal vertex algebras (notation of Frenkel Ben-Zvi). The $$Aut_\mathcal{O}$$-equivariant complex as you describe on the punctured disk will be equivalent to ours in the quasi-conformal setting.

This can also be performed on a formal disc $$D^*$$, in which case one obtains what one might call Vertex Homology. Has this been studied anywhere? Is there a relation with the work of BdSHK cited above, in particular does the BdSHK complex act on it?

Again if you want something that should be called "vertex homology" then you should consider either $$Aut_{\mathcal O}$$ equivariant objects on the disk or translation equivariant objects on the line.

There are a few statements in this answer that are plain wrong cause I had in mind translation invariants sections on the disk. I'll strike them through.

Factorization homology of a unital chiral algebra vanishes on an open proper subset of a curve. Factorization homology of the restriction of a chiral algebra to an affine proper subset of a curve vanishes. This is the content of Lemma 4.3.4 in Beilinson and Drinfeld's book.

I also believe that David's comments about Factorization homology on the disk are wrong. There's a big difference between the topological setting and the algebraic setting and between $$E_2$$ algebras and the chiral/factorization algebras of Beilinson and Drinfeld. In particular, the chiral homology of the disk with coefficient on any unital chiral algebra vanishes. This is because chiral homology is not just the CE complex of a Lie algebra on the Ran space, but also it's de Rham complex. In particular as you point out the degree zero homology of this complex is $$V/V_{(-1)}V = 0$$ and this is because we have the vacuum vector to play around. Therefore the image of $$\mathbb{1}$$ in chiral homology vanishes and then by the last comment before the mentioned Lemma 4.3.4 the whole complex vanishes.

Chiral homology of a curve is really a global object and you cannot detect any of it by looking at formal disks (see however 2.4.12 in BD in agreement with David's comments). The complex studied in BdSHK is purely a local object, or rather a translation equivariant complex on the line. I am not aware of any direct connection between them.

There are however a couple of instances where you can compute the global chiral homology by local considerations: the case of $$X=\mathbb{P}^1$$ and the case of an elliptic curve. Both cases because you have global coordinates. In degree $$0$$ this is the work of Zhu way before chiral homology was defined. In the elliptic curve case we have a preprint with J. van Ekeren describing the nodal curve limit of the first chiral homology group and we are supposed to post a continuation describing the general elliptic curve soon.

Edit: To reflect David's comments. Indeed by $$U\subset X$$ open they really mean an open affine, so that the example of a disconnected curve is not allowed. The point needed is the vanishing of the de Rham cohomology. They use this lemma heavily in proving the quasi-isomorphism of the complex with supports in 4.4.2--4.4.3.

Commutative vertex algebras are not a counter-example to this statement as the functor of coinvariants on the disk is canonically zero as long as your algebras are unital. Here the very naive notion of coinvariants is clear, the Cousin complex on $$D^2$$ is already enough to compute the homology in degree $$0$$ and this group is $$V/V_{(-1)}V$$ (think of homologies with support at the origin).

As for the $$E_2$$ algebras business: here is where things get sloppy. As far as I know the folklore is that $$E_2$$ algebras are essentially the same as topological vertex algebras. With this I have no objection, however notice that the latter is a vertex algebra together with a BRST differential, and that this equivalence is inherently derived. Since the BRST cohomology of a topological vertex algebra is typically stupid (it lies in conformal weight zero and is a commutative algebra) then it's no wonder that the chiral homology viewed as a complex with this extra BRST differential is silly. However the complex without taking this differential into account is still very interesting, you capture irreducible modules and Jacobi elliptic functions on genus $$1$$ (at least in rational cases) and so forth.

The second point is more subtle: I have seen in many places the analogy drawn between Factorization homology in the topological world and chiral homology in the algebraic setting. I may be wrong and very much outdated in the literature that I can cope with, but I have not seen anything written to prove an actual dictionary. There is no formal way of constructing a factorization algebra (in the topological meaning) coming from a factorization algebra in the algebraic world. Folklore expected that if one takes the complex that computes chiral homology on the Ran space of $$X$$ á la BD and one takes an analitification of this and then sections with compact support, this is something that looks like a Factorization algebra in the topological setting. As far as I know the problem is that the process of analytification (or tensoring with the analytic de Rham complex) does not commute with tensor products and this breaks down being a Lie/Commutative algebra.

I don't claim that having a topological intuition is wrong, just that at least when trying to translate that intuition into an actual theorem in BD's setting the situation is not so transparent even in the simplest of cases (for example the zeoro-th Hochshild Homology of higher Zhu algebras equals zero-th chiral homology of vertex algebras, this is a theorem that requires a lot of theory and a bunch of further assumptions that have no topological counterparts).

• This is fantastic, and really clarifies things for me, many thanks! – EBz Feb 10 at 14:00
• Thanks Reimundo! I'm clearly profoundly rusty on this, but I think there's some disconnect here (maybe even just a shift). In the special case of a commutative chiral algebra (not so far from the special case of $E_2$ algebras), ie functions on the D-scheme of jets of maps into a target, BD explain that chiral homology is just functions on the mapping space - so we're not getting zero on the disc (except for the issue of left vs right D-modules, i.e. I may be off by a shift). Certainly in this case chiral homology makes sense locally as well as globally and agrees with the $E_2$ notion – David Ben-Zvi Feb 10 at 16:11
• I'm also not sure I agree with the interpretation of their Lemma 4.3.4, which is about chiral homology of a proper curve with coefficients in $j_*j^*$. To take an extreme example, I could ask for factorization homology (say in topology) of a disjoint union of surfaces where I take zero on one and whatever you want on the other, and will always get zero.. – David Ben-Zvi Feb 10 at 16:13
• The point I was making about a disjoint union is that factorization homology is a version of a "restricted tensor product" over points of the curve, so it's not surprising that if you put 0 at one point you get 0. That's different from taking restricted tensor product over a subset of the curve, which is what I think of fact.homology on an open curve (or disc) as being. – David Ben-Zvi Feb 11 at 2:35
• I don't understand your claim in the commutative case. Do you disagree with the statement that chiral homology/conformal blocks of an induced commutative fact.algebra - ie functions on jets of maps into some target - is functions on the mapping space? (that's a theorem in BD - or am I misinterpreting something?). So you get something nonzero on the disc -it's just shifted into a different cohomological degree because we're thinking of right not left D-modules. – David Ben-Zvi Feb 11 at 2:38