Non-degenerate alternating bilinear form on a finite abelian group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:30:33Z http://mathoverflow.net/feeds/question/58825 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58825/non-degenerate-alternating-bilinear-form-on-a-finite-abelian-group Non-degenerate alternating bilinear form on a finite abelian group Giuseppe 2011-03-18T11:10:18Z 2011-07-29T19:15:53Z <p>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...</p> <p>Let $A$ be a finite abelian group, and let </p> <p>$\psi : A \times A \to \mathbb{Q}/\mathbb{Z}$</p> <p>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 $0$.</p> <blockquote> <p>Why must $A$ have square cardinality?</p> </blockquote> <p>I believe it will follow from the following theorem in Linear algebra:</p> <p>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. </p> <p>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. </p> <p>Any pointers anyone? </p> http://mathoverflow.net/questions/58825/non-degenerate-alternating-bilinear-form-on-a-finite-abelian-group/58828#58828 Answer by Francesco Polizzi for Non-degenerate alternating bilinear form on a finite abelian group Francesco Polizzi 2011-03-18T11:47:01Z 2011-03-18T15:38:57Z <p>Actually, one can show the following stronger result:</p> <p><strong>Proposition</strong></p> <p>Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a <em>lagrangian decomposition</em>, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that </p> <p>$A \cong G \times \widehat{G}$,</p> <p>where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$. </p> <p>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:</p> <p>$\psi((x, \chi), (y, \eta))=\chi(y)\eta(x)^{-1}$.</p> <p>An easy proof, by induction on the order of the group, can be found in Lemma 5.2 of <a href="http://arxiv.org/abs/0708.2758" rel="nofollow">A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758 </a></p> <p><strong>Remark.</strong> 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:</p> <p>$\omega(x \oplus \chi, y \oplus \eta) = \chi(y) - \eta(x)$. </p> <p>In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.</p> http://mathoverflow.net/questions/58825/non-degenerate-alternating-bilinear-form-on-a-finite-abelian-group/71513#71513 Answer by mbeck for Non-degenerate alternating bilinear form on a finite abelian group mbeck 2011-07-28T19:00:53Z 2011-07-29T19:15:53Z <p>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?</p> <p>Thanks a lot!</p>