2
$\begingroup$

Given a finite $p$-group $A$, with a non-degenerate quadratic form $q:A\rightarrow \mathbb Q/2\mathbb Z$ (that is a map satisfying $q(na)=n^2q(a)$ for all $n\in \mathbb Z,a\in A$), an important result of Nikulin (see his paper "Integral symmetric bilinear forms and some of their applications") is that there exists a unique $p$-adic lattice $K(q_p)$ of rank $l(A)$ (that is the minimal number of generators of $A$) whose discriminant-form is isomorphic to $q$ except in two exceptional cases when $p=2$. Here the discriminant form is the quadratic form obtained from extending $q_p$, the quadratic form on $K(q_p)$, to $K(q_p)^*$ and then restricting to $K(q_p)^*/K(q_p)$.

My question is, when $p=2$, how can you determine what $K(q_2)$ is? It would seem that we need more information than just the values taken by the discriminant form. For example, consider the orthogonal sum $q_1^{(2)}(2)\oplus q_1^{(2)}(2)$, where $q_1^{(2)}(2)$ is the quadratic form on the rank 1 $2$-adic lattice given by $q(v)=2v^2$. The values taken by the discriminant form are easily seen to be $0,1,1,1$ on $\mathbb Z/2\oplus \mathbb Z/2$. Similarly, consider the rank 2 $2$-adic lattice with quadratic form $q(x,y)=4xy$. The discriminant group is again $\mathbb Z/2\oplus \mathbb Z/2$, and the discriminant form takes the same values $0,1,1,1$. According to the Proposition 1.11.2 in that paper, these two lattices have different signs, so how can one differentiate between them?

That is, given just the discriminant group and quadratic form, how can we determine which $2$-adic lattice we're dealing with?

$\endgroup$
1
$\begingroup$

As the cited paper explains (and as you show yourself), there is no way to do that: in some sense, this is the only exception when two distinct lattices have isomorphic discriminant forms. You need extra data. For example, you can use the discriminant (strange name; this is the determinant of the matrix of the lattice). The point is that, in the presence of vectors of order $2$ and square $\frac12\bmod\mathbb{Z}$, the discriminant form determines the discriminant only modulo $4$, whereas in fact it is defined modulo $8$. That is how you get the magic number $5$ in Nikulin's paper.

Another nice treatment of these things is found in Miranda--Morrison's book. But be careful: they use slightly different conventions (so that all lattices are even).

$\endgroup$
  • 1
    $\begingroup$ But how could you determine the discriminant without knowing the lattices ahead of time? In the specific situation I'm in, I have quadratic forms on some finite groups, and I need to know which p-adic lattice corresponds to my specific p-group and its quadratic form. Is the answer really just that I can't determine this? $\endgroup$ – HNuer Feb 5 '14 at 22:05
  • $\begingroup$ @HNuer: Yes, the answer is "you cannot", of course unless you know something. The universal counterexample, easier than yours, is the pair $[2]$ and $[10]$ of $(1\times1)$-matrices: they are just different! More details on your particular problem may help: there may be extra data hidden somewhere. $\endgroup$ – Alex Degtyarev Feb 5 '14 at 22:44

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.