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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?

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  • $\begingroup$ 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 (cont.) $\endgroup$
    – user16796
    Commented Jul 28, 2011 at 19:00
  • $\begingroup$ No. This means that both $0→⟨a⟩→A→A/⟨a⟩→0$ and $0→⟨a⟩^{\perp}→A→\widehat{⟨a⟩}$ split. Now it is easy to see (for instance by using the Snake Lemma) that $A/⟨a⟩=⟨a⟩^{\perp}/⟨a⟩⊕ \widehat{⟨a⟩}$, so the claim follows. $\endgroup$
    – Francesco Polizzi
    Commented Jul 29, 2011 at 21:11
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    $\begingroup$ Thank you so much! You saved my day! That was the last gap in my diploma thesis :) If I have an acknowlegdement in my thesis, you will be mentionned! $\endgroup$
    – user16796
    Commented Jul 29, 2011 at 21:14
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    $\begingroup$ The continuation of user16796's comment (converted to a comment from an answer a long time ago, and slightly messed up in the process) is this: $${}$$ "...Davydov states $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?" $\endgroup$
    – David Roberts ♦
    Commented Jul 24 at 0:56

1 Answer 1

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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 form: $$\psi((x, \, \chi), \, (y, \, \eta))=\chi(y)\eta(x)^{-1}.$$ 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 form: $$\omega(x \oplus \chi, \, y \oplus \eta) = \chi(y) - \eta(x).$$ In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.

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  • $\begingroup$ That proof is quite swish, thank you. I still feel though that it should follow from the analogous result from linear algebra for vector spaces. $\endgroup$
    – Giuseppe
    Commented Mar 18, 2011 at 14:01
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    $\begingroup$ This proof is very natural because it really rephrases, in the contest of finite groups, the usual proof for symplectic vector spaces. In fact, any finite-dimensional symplectic vector space $V$ can be written as $W \oplus W^*$, where $W$ is a lagrangian (=isotropic of maximal dimension) subspace for the given symplectic form. This explains why $\dim V$ must be even. Look at en.wikipedia.org/wiki/Symplectic_vector_space, in particular the paragraph "Standard symplectic space" $\endgroup$
    – Francesco Polizzi
    Commented Mar 18, 2011 at 15:24
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    $\begingroup$ To me this result has a folkloric feel: I am certainly aware of it and have used it in my own work (I even use the phrase "Lagrangian decomposition" in this context, which I think I picked up from some notes of van der Geer and Moonen), at least since 2003. But I don't know where this result comes from originally: does anyone have a primary source? $\endgroup$
    – Pete L. Clark
    Commented Mar 19, 2011 at 19:33
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    $\begingroup$ This result comes up in the context of classifying Heisenberg central extensions of finite abelian groups. We discuss this aspect in the intro to our paper Locally Compact Abelian Groups with Symplectic Seld-duality, by Prasad, Shapiro and Vemuri (Adv. Math. 225 (2010) 2429-2454), arxiv.org/abs/0906.4397. It also seems to come up quite naturally in the theory of abelian varieties. It continues to hold for most reasonable classes of locally compact abelian groups, but fails in general. $\endgroup$
    – Amritanshu Prasad
    Commented Mar 20, 2011 at 0:12
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    $\begingroup$ Also, it might be noted (since this is not obvious from the notation) that in fact the group of characters of $G$ is isomorphic to $G$ (since $G$ is abelian), so even though it at first glance just looks like we get that the order of $A$ is square, we actually get that in some sense $A$ itself is square. $\endgroup$
    – Tobias Kildetoft
    Commented Jul 29, 2011 at 8:17

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