The following is a somewhat well-known fact: Given an even lattice $L$ with the pairing $\langle,\rangle: L\times L\to \mathbb{Z}$, we extend the pairing to $L\otimes \mathbb{Q}$ by tensoring with $\mathbb{Q}$. Then the discriminant group $A=L^*/L$ comes with a non-degenerate quadratic form $q: A\to \mathbb{Q}/\mathbb{Z}$ given by $q(x+L)=\langle x,x\rangle/2$. Conversely, any such pair $(A,q)$ is known to come from some even lattice $L$, see e.g. Wall, "Quadratic forms on finite groups".

My question is the $G$-symmetric case. Namely, suppose there is a finite abelian group $A$ with a non-degenerate quadratic form $q$ which is invariant under a subgroup $G$ on $Aut(A)$. Is there always a lattice $L$ with the pairing $\langle,\rangle$ which is invariant under $G\subset Aut(L)$ which gives rise to $(A,q)$?

(My motivation is to study the obstructions assocaited to $G$ actions on the modular tensor category associated to the metric group $(A,q)$, but I guess there can be various other applications, if the question I asked has a positive answer.)

  • $\begingroup$ You need some more information here. Every group $G$ can act on any lattice and any group $A$, by just acting trivially. Do you want faithful actions? Do you want $G$ to embed into the automorhpism group of $L$ or $A$ (i.e. some sort of linear action)? $\endgroup$ – Dirk Apr 16 '18 at 7:45
  • $\begingroup$ Sorry and thanks. I want G to be a subgroup of Aut(A) or Aut(L). $\endgroup$ – Yuji Tachikawa Apr 16 '18 at 9:33

The group of automorphisms of $A$ which preserve the quadratic form $q$ is known as the orthogonal group $O(A,q)$. Likewise, if $L$ is a free $\mathbb{Z}$-module of finite rank with an even $\mathbb{Z}$-valued quadratic form $Q$, there is an orthogonal group $O(L,Q)$. When $(A,q)$ is the discriminant form of $(L,Q)$, there is a natural homomorphism $O(L,Q) \to O(A,q)$.

It is a consequence of the Strong Approximation Theorem for the orthogonal group that this homomorphism is surjective when $L$ is indefinite and of sufficiently large rank. (I believe that this is proven in several of the standard books about integral quadratic forms.) Note that adding a copy of the hyperbolic plane (resp. the $E_8$ lattice) to $L$ does not change the disriminant form, but makes $L$ indefinite (resp. increases its rank). Thus, if we are given $(A,q)$, an $L$ can be found with this property.

This doesn't quite answer your question, because you want to know if the surjection $O(L,Q) \to O(A,q)$ can be split. I don't know, but I am inclined to doubt it. Consider the analogous problem for the special linear group instead of the orthogonal group. We have surjections $SL(n,\mathbb{Z}) \to SL(n,\mathbb{F}_p)$ which cannot be split for $p$ sufficiently large.


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