Non-degenerate alternating bilinear form on a finite abelian group - MathOverflow

Non-degenerate alternating bilinear form on a finite abelian group

2011-03-18T11:10:18Z

I asked this question on math.stackexchange yesterday, but nobody has helped so far, and only 44 people have seen it! So I hope people do not mind me asking it here...

Let $A$ be a finite abelian group, and let

$ \psi : A \times A \to \mathbb{Q}/\mathbb{Z} $

be an alternating, non-degenerate bilinear form on $A$. Maybe I should say what I mean by these words; bilinear means it is linear in each argument separately; alternating means that $\psi(a,a) = 0$ for all $a$; non-degenerate means that, if $\psi(a,b) = 0$ for all $b$, then $a$ must be $0$.

Why must $A$ have square cardinality?

I believe it will follow from the following theorem in Linear algebra:

Theorem. Let $V$ be a finite dimensional vector space over a field $K$ that has an alternating, non-degenerate bilinear form on it (from $V \times V \to K$). Then dim $V$ is even.

My idea was to proceed as follows: If the size of $A$ is not square, then for some prime $p$, $A(p)$ is not square, where $A(p)$ means the $p$-primary part of $A$. The original $\psi$ induces a map on $A(p)$ that is non-degenerate, alternating and bilinear. I then wanted to say that $A(p)$ is an $\mathbb{F}_p$-vector space, and then applying the theorem I am done, but this is not true, e.g, $\mathbb{Z}/25\mathbb{Z}$ is not an $\mathbb{F}_5$-vector space.

Any pointers anyone?

Answer by Francesco Polizzi for Non-degenerate alternating bilinear form on a finite abelian group

2011-03-18T11:47:01Z

Actually, one can show the following stronger result:

Proposition

Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that

$A \cong G \times \widehat{G}$,

where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$.

Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following shape:

$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.

An easy proof, by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758

Remark. It is interesting to notice the analogy with symplectic vector spaces. In fact, any symplectic vector space $(V, \omega)$ can be written as $V = W \oplus W^{*}$, where $W$ is a lagrangian (=isotropic of maximal dimension) subspace for $\omega$. In particular, $\dim V = 2 \dim W$. Moreover, with respect to this decomposition, $\omega$ has the following shape:

$\omega(x \oplus \chi, y \oplus \eta) = \chi(y) - \eta(x)$.

In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.

Answer by mbeck for Non-degenerate alternating bilinear form on a finite abelian group

2011-07-28T19:00:53Z

I'm very glad about this topic, because I have a similar problem and Lemma 5.2 by Davydov would be the solution. But I have a problem with the proof. Davydov says that the inclusion $\langle a \rangle \to A$ and the surjection $A \to \widehat{\langle a \rangle}$ split. Does that mean that the short exact sequence $$ 0 \to \langle a \rangle \to A \to \widehat{\langle a \rangle} \to 0 $$ splits? But then the Splitting Lemma provides an isomorphism $A \cong \langle a \rangle \oplus \widehat{\langle a \rangle}$, but Davydov states that $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?

Thanks a lot!