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?