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).

  • $\begingroup$ 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 Tamaroff Oct 14 '20 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 '20 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 '20 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 '20 at 12:45
  • 1
    $\begingroup$ @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 '20 at 15:25

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!


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.

  • $\begingroup$ 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 '20 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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