A couple of day ago, I was lamenting to a friend about the fact that I have no idea what vertex algebroids are. During our discussion, I came up with a guess of what a vertex algebroid might be. I'm wondering whether this guess has any merit.

Question: is the definition below equivalent to what people call "vertex algebroid"?

Let me first tell you what I know about vertex algebras:

Vertex algebras

Let $\mathbb D=Spec(\mathbb C[[t]])$ be the formal disk, and let $\mathcal M_n$ denote the moduli space of n-tuply punctured formal discs.

(While you probably have some intuition as to what it means to puncture $\mathbb D$ once, but you might be puzzled by the idea of puncturing it more than once. Indeed, the moduli space of twice punctured formal discs doesn't have any $\mathbb C$-points, so it's not possible to exhibit any twice punctured formal disc! However, given the once punctured formal disc $\mathbb D^*=Spec( \mathbb C[[t]][t ^ {-1}])$, there's $\mathbb D^*$ many places where one can perform the second puncture. So that produces a map $\mathbb D^*\to \mathcal M_2$, showing that $\mathcal M_2$ is at least non-trivial.)

The collection of all $\mathcal M_n$ looks a lot like the little discs operads:alt text
But if you think about it, you'll see that it's not quite an operad: it's only a partially defined operad. The operad multiplication $$ \mathcal M_n\times \mathcal M_{k_1}\times\ldots\times\mathcal M_{k_n}\to \mathcal M_{k_1+\ldots+k_n} $$ is only defined on a "subset" of $\mathcal M_n\times \mathcal M_{k_1}\times\ldots\times\mathcal M_{k_n}$. But that doesn't matter so much: once can still define the notion of an algebra over this partially defined operad, and it turns out that an algebra over $\mathcal M_\bullet$ is exactly the same thing as a vertex algebra.

Vertex algebroids?

Fix a smooth variety $X$ and consider the moduli space of n-tuply punctured formal discs equipped with a map to $X$. Those again form a partial operad (probably it's better to call it a colored partial operad), and one can consider algebras over it.

Could it be that those algebras are equivalent to vertex algebroids over $X$?

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    $\begingroup$ Is there a reference for the “partial operad” point of view? $\endgroup$ – Dmitri Pavlov Feb 9 '13 at 14:42
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    $\begingroup$ Huang in his book Two-Dimensional Conformal Geometry and Vertex Operator Algebras talks about partial operads. Those give what he calls "geometric vertex algebras". He then shows (if I remember correctly) that the category of geometric vertex algebras is equivalent to the category of vertex algebras. Note however that his punctured discs are not formal: they have non-zero spatial extent. If you let the punctured discs be formal, then you get something that is much closer to the actual definition of vertex algebras. $\endgroup$ – André Henriques Feb 9 '13 at 23:07
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    $\begingroup$ Dmitri, while not in the context of the question at hand, partial operads were used in E infty ring spaces and E infty ring spectra, VII\S1, SLN 577, 1977. I defined the little convex bodies'' partial operads there for purposes of multiplicative infinite loop space theory (one point being that the little discs operads do not suspend properly). In that context Steiner came up with a better solution using honest operads, in A canonical operad pair''. Further off, partial DGAs and their modules come up in mixed Tate motives (Kriz-May, Operads, algebras, modules, and motives''. $\endgroup$ – Peter May Feb 10 '13 at 1:54
  • $\begingroup$ It also could actually be the case that vertex algebras themselves already are (encodings of) "somethingoids" :D $\endgroup$ – მამუკა ჯიბლაძე May 10 '16 at 8:37

With David Ben Zvi's permission, I post the following answer that I got from him by email:

``this feels wrong to me, but not sure. A vertex algebroid is just a truncation of the notion of sheaf of vertex algebras, just as a Lie algebra is gotten out of the ≤1 filtered piece of the enveloping algebra.. so unless your definition is equivalent to a sheaf of vertex algebras it doesn't seem right. what you write is a lot like Segal's notion of elliptic object right? which is close to a sheaf of vertex algebras I guess...or more precisely a CFT with spacetime endowed to maps to a target.. anyway not sure the "oid" is so significant a definition. I'll think about it a bit more..''


This was meant to be a comment to David's answer but it grew too much. At first I thought with David's intuition that your definition seemed as a sheaf of vertex algebras on X. But then I realized that I couldn't see the restriction maps. On the other hand, over each point $x \in X$ --restricting to constant maps to X -- you obtain an algebra over that partial operad of punctured formal discs. In fact, your definition seems to agree with the definition hinted in 3.4.9 (ii) [Beilinson and Drinfeld, Chiral Algebras] of $X$-family of factorization algebras, which in turn are defined in 3.4.6 of loc. cit. I chose that definition instead of the standard one of 3.4.4-5 because there you see that replacing the category $C(X)$ ($X$ as in 3.4.6. of loc. cit. and not your target space that I will denote now $Y$) that fibers over affine schemes by a suitable category fibering over $Sch/Y$ (essentially replace $Z$ in that definition by maps $Z \rightarrow Y$) you obtain the notion of $Y$-family of factorization algebras. So your definition is a factorization version of $Y$-families of vertex algebras.

Now just to give references, as David pointed out, vertex algebroids are truncated sheaves of vertex algebras. This idea came from works of Gorbounov, Malikov, Schechtman and Vaintrob [Gerbes of chiral differential operators II]. You start with a sheaf $\mathcal{V}$ of $\mathbb{Z}_{\geq}$-graded vertex algebras on $X$ and look at the structure that you get from the filtered components $\mathcal{V}_{\leq 0}$ and $\mathcal{V}_{\leq 1}$. This definition is nicely distilled in [Bressler; Vertex algebroids I].

However, I thought this might give you a little intuition given that you think of vertex algebras as algebras over certain operad. It turns out [Beilinson Drinfeld - 3.3.3] that vertex algebras are Lie algebras in certain (compound) tensor category of $D$-modules on the disk $D$ (again my notation collides with yours). You can think of a Lie algebroid on $X$ as a sheaf of Lie algebras (in the category vector spaces) with an action on a given sheaf of commutative algebras $\mathcal{O}_X$ and a compatible $\mathcal{O}_X$-module structure. A $D$-module version of that statement gives rise to the notion of a $Lie^*$-algebroids and Chiral Lie algebroids (where we replace the tensor category of $\mathcal{O}_X$-modules with some category of $D_X$-modules where the tensor product is not representable). Granted this is very sketchy, but the relevant definitions are all in Beilinson and Drinfeld's book: 1.4.11 gives the general definition of a $Lie^*$ algebroid, 2.5.16 treats them in the $D$-module setting, these ought to be thought more as "conformal algebras" or even commutative vertex algebras rather than "vertex algebras". Finally in 3.9.6 comes the definition of a Chiral Lie algebroid which corresponds to vertex algebroids.

One last observation that perhaps someone here can comment, in the introduction to section 2.5, Beilinson and Drinfeld make a very interesting comment how how $Lie^*$ algebroids give rise to Lie algebroids on ind-schemes of sections over formal punctured discs. They refer to sections 4.6.10 and 4.6.11 for a global theory. Unfortunately those sections do not exist.

  • $\begingroup$ @Reimundo: The text appears a bit garbled to me. Perhaps you require some backticks around the math expressions? $\endgroup$ – Ricardo Andrade Mar 30 '13 at 23:49
  • $\begingroup$ @Ricardo: Thanks! I sporadically come back to MO and forget those quirks. $\endgroup$ – Reimundo Heluani Mar 31 '13 at 0:12
  • $\begingroup$ @Reimundo: No problem. The backtick issue is indeed an annoying peculiarity of mathoverflow. $\endgroup$ – Ricardo Andrade Mar 31 '13 at 1:49

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