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Let $G$ be a finite abelian group. A quadratic form on $G$ is a map $q: G \to \mathbb{C}^*$ such that $q(g) = q(g^{-1})$ and the symmetric function $b(g,h):= \frac{q(gh)}{q(g)q(h)}$ is a bicharacter, i.e. $b(g_1g_2, h) = b(g_1, h)b(g_2, h)$ for all $g, g_1, g_2, h \in G.$

The quadratic form $q$ is called non-degenerate if the corresponding bicharacter $b$ is non-degenerate.

Question: Is there a non-degenerate quadratic form on every finite abelian group?

Motivation: it is used to make pointed braided/modular tensor categories, see Chapter 8 of this book (in particular Sections 8.4, 8.13 and 8.14).

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  • $\begingroup$ You can realize any cyclic group as a subgroup of $S^1\subseteq \mathbb C^\times$. Isn't there any way to get going from there, using the bona-fide quadratic form $z\mapsto z^2$ in the circle? Do you have any ideas? $\endgroup$
    – Pedro
    Oct 14, 2020 at 9:40
  • $\begingroup$ @PedroTamaroff: Yes, I guess you mean the quadratic form $q: e^{i\theta} \mapsto e^{i\theta^2}$. Then $b(e^{i \theta_1},e^{i \theta_2}) = e^{2i \theta_1\theta_2}$ is a non-degenerate bicharacter. So it is ok for the cyclic groups. $\endgroup$
    – Sebastien Palcoux
    Oct 14, 2020 at 10:39
  • $\begingroup$ What do you mean when you say the bi-character is non-degenerate (do you mean non-trivial, as in it becomes a constant function?). Sorry, couldn’t figure it out from the statement (unless it’s obvious in a way I can’t seem to see). $\endgroup$
    – Jack L.
    Oct 14, 2020 at 11:41
  • $\begingroup$ @JackL. A bicharacter $B$ is non-degenerate if for all non-identity $g ∈ G$ there exists some elements $h,h’ ∈ G$ such that $B(g, h), B(h’,g) \neq 1$. $\endgroup$
    – Sebastien Palcoux
    Oct 14, 2020 at 12:45
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    $\begingroup$ @JackL. For the cyclic group $C_n$, it should be $q: e^{\frac{2\pi i k}{n}} \mapsto e^{\frac{2\pi i k^2}{n}}$, with $k$ integer. Let $A$ be an abelian group, it is of the form $\prod_i C_{n_i}$. Let $N$ be $lcm_i(n_i)$, then $C_{n_i}$ is a subgroup of $C_N$. By combining that with the idea in your answer, something natural can be written; but is it a non-degenerate quadratic form? If $N$ is odd, it should be ok, but if $N$ is even, it is degenerate because of the multiplicative constant $2$. Do you see a way to fix the even case? $\endgroup$
    – Sebastien Palcoux
    Oct 14, 2020 at 15:25

2 Answers 2

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Thanks to the Fundamental Theorem of Abelian Groups, let $$G:=\prod_{k=1}^{n}\{z:z^{m_k}=1\,,z\in\mathbb{S}\}\,,$$ and let $\chi(m)=2$ if $m$ is odd and $\chi(m)=1$ if $m$ is even. Then define $$q\colon G\to\mathbb{C}^*\,,~\,~\,~\,(e^{2\pi i\frac{a_k}{m_k}})_k\mapsto \exp(\pi i\sum_k \chi(m_k)\frac{a_k^2}{m_k})\,.$$ We observe that $q$ is a quadratic form because $$e^{\chi(m)\pi i\frac{(a+m)^2}{m}}=e^{\chi(m)\pi i\frac{a^2}{m}}e^{2\pi i(a\chi(m)+\frac{m\chi(m)}{2})}= e^{\chi(m)\pi i\frac{a^2}{m}}\,.$$ The associated bi-character then becomes $$B((e^{2\pi i\frac{a_k}{m_k}})_k,(e^{2\pi i\frac{b_k}{m_k}})_k)= \exp(2\pi i\sum_k \chi(m_k)\frac{a_kb_k}{m_k})\,.$$ Since the degeneracy of the bi-character $B$ is equivalent to $\sum_k\chi(m_k)\frac{a_kb_k}{m_k}\in\mathbb{Z}$ say for all admissible $(b_k)_k$, this forces $a_k\equiv 0\mod m_k$; thus $B$ is non-degenerate!

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Yes. It's necessary and sufficient to show that every finite abelian group admits a nondegenerate quadratic form valued in a finite cyclic group. The following slightly stronger statement is true: every finite abelian $p$-group admits a nondegenerate quadratic form valued in $C_{p^k}$ for some $k$ (this suffices by the Chinese remainder theorem). So let's prove this.

If $A$ is a finite abelian $p$-group for $p$ an odd prime, we can pick any isomorphism $A \cong A^{\ast}$ from $A$ to its Pontryagin dual $A^{\ast} = \text{Hom}(A, \mathbb{Q}/\mathbb{Z}) \cong \text{Hom}(A, \mathbb{Q}_p/\mathbb{Z}_p)$ and we'll get a nondegenerate bilinear form $B : A \times A \to C_{p^k}$ whose associated quadratic form $Q : A \to C_{p^k}$ is nondegenerate. As you say in the comments, this almost but doesn't quite work when $p = 2$.

When $p = 2$ the following slight modification works. Again pick an isomorphism $A \cong A^{\ast}$ to the Pontryagin dual and get a nondegenerate bilinear form $B : A \times A \to C_{2^k}$. Now we do something a bit funny. Consider the inclusion (not a group homomorphism!) $C_{2^k} \to C_{2^{k+1}}$ given by $k \mapsto k$, thinking of elements of $C_n$ as elements of $\mathbb{Z}/n$. Composing this inclusion with $B$ gives a map (not a bilinear map!) $B' : A \times A \to C_{2^{k+1}}$. Now I claim that the diagonal $Q(a) = B'(a, a)$ of this map is a nondegenerate quadratic form. We clearly have $Q(-a) = B'(-a, -a) = B'(a, a)$, and the associated bilinear form $Q(a + b) - Q(a) - Q(b)$ recovers $B$, now taking values in $2 C_{2^{k+1}} \cong C_{2^k}$.

In particular, when $A = C_2$ we get the quadratic form $Q : C_2 \to C_4$ given by $Q(0) = 0, Q(1) = 1$. This quadratic form can be interpreted as a cohomology class in $H^4(B^2 C_2, C_4)$ and so in turn a cohomology operation $H^2(-, C_2) \to H^4(-, C_4)$ which I believe is exactly the Pontryagin square.

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  • $\begingroup$ The generalization I guess is to observe that when you pass from a bilinear form $B$ to a quadratic form $Q$ to a bilinear form again you end up with $2B$, so for $B$ to induce a quadratic form it suffices for $2B$ to be a bilinear form, not $B$; so you can take pointwise square roots of a bilinear form (my apologies for switching back to thinking multiplicatively here) and the result will still induce a genuine quadratic form. Of course these will only be non-unique if there's $2$-torsion. $\endgroup$
    – Qiaochu Yuan
    Oct 16, 2020 at 19:48

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