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

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).

• 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? Oct 14, 2020 at 9:40
• @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. Oct 14, 2020 at 10:39
• 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). Oct 14, 2020 at 11:41
• @JackL. A bicharacter $B$ is non-degenerate if for all non-identity $g ∈ G$ there exists some elements $h,h’ ∈ G$ such that $B(g, h), B(h’,g) \neq 1$. Oct 14, 2020 at 12:45
• @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? Oct 14, 2020 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.
• 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. Oct 16, 2020 at 19:48