Let $(G,+)$ be a finite Abelian group. We say $q\colon G\to \mathbb{T}$ is a non-degenerated quadratic form, if $q(-a)=q(a)$ and the symmetric function $$ b(g,h) =q(g+h)q(g)^{-1}q(h)^{-1} $$ is a non-degenenerate bicharacter on $G$, i.e. $b(g+h,k)=b(g,k)b(h,k)$ and $b(k,g)=1$ for all $k\in G$ implies $g=0$.

Let $(\Gamma,\langle\,\cdot\,,\,\cdot\,\rangle)$ be a positive even lattice, i.e. a free abelian group with an symmetric bilinear form $\langle\,\cdot\,,\,\cdot\,\rangle\colon \Gamma \times \Gamma \to \mathbb{Z}$, such that $\langle a,a\rangle \in 2\mathbb N$ for all $a \in \Gamma$ and $\langle a,a\rangle=0$ if and only if $a=0$. Let $\Gamma^\ast =\mathrm{Hom}(\Gamma,\mathbb Z)$ be the dual lattice.

Then $G_\Gamma=\Gamma^\ast/\Gamma$ is a finite group and $q_\Gamma([a])=\exp(\pi i \langle a,a\rangle)$ is a non-degenerate quadratic form on $G_\Gamma$.

Question: Is the map from positive even lattices to finite Abelian groups with non-degenerate quadratic forms $$ (\Gamma,\langle\,\cdot\,,\,\cdot\,\rangle)\longmapsto (G_\Gamma,q_\Gamma) $$ surjective (up to the obvious equivalences)? It is for sure not injective, because all self-dual lattices map to the trivial group.

If one drops the assumption of positiveness the answer is, apparently, yes.

Motivation: A pointed unitary modular tensor category is completely characterized by the fusion rules which give a finite Abelian group $G$ and a non-degenerate quadratic form on $G$ given by the twists (basically type I Reidmeister move). Every even lattice $\Gamma$ gives a unitary modular tensor category using, for example, the lattice Vertex Operator Algebra or Conformal Net associated with $\Gamma$ and consider its representation category, which turns out to be a modular tensor category which is characterized by $(G_\Gamma,q_\Gamma)$.

So the question can be reformulated to: Do all unitary pointed modular tensor categories come from positive even lattice CFTs?


Edited: I have missed your "positive". The signature of the lattice modulo 8 depends on the form only (some people call this Brown invariant and van der Blij theorem; Nikulin below calls this just the signature of the form).

Otherwise (given the right signature), I would suggest that the map is surjective, and the easiest proof would be the known classification of forms and an explicit construction for each class. For more details (e.g., the classification) see Nikulin, V. V. Integer symmetric bilinear forms and some of their geometric applications. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 43 (1979), no. 1, 111–177, 238 [MR0525944]

Edit The existence of some positive form is given by Theorem 1.10.1 (which does not assume that the form is indefinite). For the construction, the two forms on $\mathbb{Z}/2$ are $[2]$ and $E_7$. Then, do induction on $p$ to construct the two forms on $\mathrm{Z}/p$. First, take $[2pq]$ with appropriate $q<p$ to get the correct $p$-primary part; then, use Nikulin's gluing and induction to kill the extra $2$- and $q$-primary parts. For the two remaining series of $(2\times2)$-forms, one can probably start from $D_4(2^k)$ or $E_8(2^k)$ and pass to appropriate finite index extensions. (Though, I did not work out the details.)

  • $\begingroup$ Thanks. I know about this paper and a paper by Wall (I had no time to add the references yet) and both show the surjectivity if one drops positivity. I would be really interested in a construction starting from a quadratic form. $\endgroup$ Mar 16 '16 at 21:55
  • $\begingroup$ In fact, one does not need to drop the positivity. Nikulin's Theorem 1.10.1 give the existence of a positive lattice. It suffices to assume that $\mathrm{rk}>\ell$; however, because of the Brown invariant, one may have to increase $\mathrm{rk}$ up to $\ell+8$, I guess. As to the explicit construction, one can probably go case by case: I will edit the answer. $\endgroup$ Mar 17 '16 at 7:00
  • $\begingroup$ Ah thanks. I guess the following is a translation: iopscience.iop.org/article/10.1070/IM1980v014n01ABEH001060/meta $\endgroup$ Mar 17 '16 at 14:28

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