# Modular tensor category associated to an even integral lattice and the lattice automorphism

Let $(L,\langle -,-\rangle)$ be an even integral lattice, and let $(A,q)$ be the associated discriminant form: $$A=L^*/L, \quad q(a)=e^{\pi i \langle a,a\rangle}.$$ We let $\hat L$ to be the extension of $L$ by $\{\pm1\}$ such that $\hat a\hat b=(-1)^{\langle{a,b}\rangle} \hat b\hat a$.

Then we have three natural "isometry groups", $O(L)$, $O(A)$ and $O(\hat L)$; the first two are subgroups of automorphism groups preserving $\langle,\rangle$ or $q$; the last is a natural extension of $O(L)$ by $\mathrm{Hom}(L,\{\pm1\})$.

Let us now consider the modular tensor category $\mathcal{C}(A,q)$ associated to $(A,q)$. This is also the category of modules of the lattice VOA $V_L$ constructed from $L$.

Let $\mathrm{Aut}(\mathcal{C})$ be the group of braided auto-equivalences of a modular category $\mathcal{C}$ up to natural transformations. $\mathrm{Aut}(\mathcal{C}(A,q))$ is known to be equal to $O(A)$.

So we have a natural sequence of group homomorphisms $$O(\hat L)\to O(L)\to O(A)=\mathrm{Aut}(\mathcal{C}(A,q)) .$$

In [ENO, arXiv:0909.3140], extendsion of a modular tensor category $\mathcal{C}$ by any group $G$ with a homomorphism to $\mathrm{Aut}(\mathcal{C})$ was studied, and two obstructions are identified. The first obstruction takes values in $H^3(G,\mathrm{Inv}(\mathcal{C}))$ where $\mathrm{Inv}(\mathcal{C})$ is the group of invertible objects of $\mathcal{C}$.

So it seems natural to study the obstructions to extending $\mathcal{C}(A,q)$ by any subgroup of $O(A)$, $O(L)$ or $O(\hat L)$. The first obstruction takes values in $H^3(G,A)$. (I believe the first obstruction vanishes for (subgroups of) $O(\hat L)$.)

Does anybody know of any references concerning this point? I would be happy with any partial results, or any starting point for me to explore the literature. (Also, does the extension theory of ENO appear somewhere in the theory of lattice VOA?)

• I'm not so sure that we have a group homomorphism $O(L)\to Aut(C(A,q))$. We certainly don't have a group homomorphism $O(L)\to Aut(V_L)$. In other words, $V_L$ does not depend functorially on $L$. In order to construct $V_L$ you need not just $L$, but also a central extension of $L$ by $\mathbb Z/2$ with the property that $[\lambda_1,\lambda_2]=\langle \lambda_1,\lambda_2\rangle$ (mod 2). So it's not $O(L)$ but an extension of $O(L)$ by the group $Hom(L,\mathbb Z/2)$ which acts on $V_L$, and hence on $C(A,q)$. – André Henriques Apr 19 '18 at 8:41
• Thanks for the question. I wasn't clear about what I mean by a group "acting on" a modular category. I tried to clean up the question. – Yuji Tachikawa Apr 20 '18 at 2:29
• A general comment - since self-equivalences of a category form a group-up-to-isomorphisms rather than a group, all kinds of obstructions, unnatural choices, etc. might stem from forgetting this structure. – მამუკა ჯიბლაძე Apr 20 '18 at 3:27
• I think you can use conformal nets to show that for any positive $L$ and any finite subgroup $G$ of $Aut(V_L)$ there is a unitary $G$-crossed braided extension (that should be fine) which comes from the action you describe (needs to be checked)! Thus the obstructions should vanish... – Marcel Bischoff Oct 3 '19 at 3:27
• @MarcelBischoff Your $Aut(V_L)$ is $O(\hat L)$, right? This question originated from a physics paper arxiv.org/abs/1803.09336 which claims that the first obstruction always vanishes even for arbitrary subgroups of $O(A)$. Even to my physics standard of rigour their argument has gaps, but I thought the question interesting. – Yuji Tachikawa Oct 4 '19 at 4:15

Edit: I've thought about this question again, and I think the answer is more positive than what I said in an earlier version.

I will assume $L$ is positive-definite, since we need that to make $V_L$ into an honest VOA. I think what you say is still true using vertex algebras of indefinite lattices, but one has to be very careful with definitions to make fusion work well, and no one has done so in that setting yet.

Here is what we expect to be true: For subgroups of $\operatorname{Aut}(\mathcal{C}(A,q))$ in the image of the map from $O(\hat{L})$, we expect an extension to exist and to be given by the category of twisted $V_L$-modules (where the twisting ranges over automorphisms of $V_L$). The two obstructions $O_3$ and $O_4$ in the ENO paper concern associativity of tensor product, and are expected to vanish in all cases of interest to us. I have no idea what happens for elements not in the image of $O(\hat{L})$.

Here is what we know: If the twisting ranges over a solvable group $G$, then the existence of an extension essentially follows from the regularity (i.e., complete reducibility) properties of the fixed-point subVOA $(V_L)^G$ established in C-Miyamoto, together with Huang's work on modular tensor structure. There is some technical work involving compatibility of twisted intertwining operators, but I don't think anything important is missing. In other words, if the determinant of $L$ is small, you get an extension unconditionally. If the determinant is large, then the existence of an extension given by twisted $V_L$-modules is conditional on the conjecture that taking fixed points under any finite group action preserves complete reducibility.

The technical work with intertwining operators comes from the fact that a full definition was only given in Huang's preprint from last year. On the other hand, you can simplify the definition significantly under the assumption that your twisted modules are just finite direct sums of irreducible $(V_L)^G$-modules, and various conjectured properties do not need new proofs.

• Are the conjectures that you mention subsumed by what is stated in the preprint you link to? – მამუკა ჯიბლაძე Apr 20 '18 at 8:08
• @მამუკაჯიბლაძე The preprint mentions a conjecture in the introduction, and that conjecture implies the category of twisted modules is a $O(\hat{L})$-crossed braided tensor category. However, there is no mention of obstructions, and no substantial progress on the conjecture itself. – S. Carnahan Apr 20 '18 at 21:45
• Dear @S.Carnahan, thanks for a detailed answer, and sorry for my slow response. – Yuji Tachikawa Apr 25 '18 at 6:38