Quadratic refinements of a bilinear form on finite abelian groups

Question

$\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$.

A quadratic function on $A$ is $q:A\rightarrow \mathbb{R}/\mathbb{Z}$ such that $q(-a)=q(a)$ and $$ \chi_q(a,b)=q(a+b)-q(a)-q(b) $$ is bilinear. Hence if $q$ is quadratic $\chi_q\in \text{Sym}(A)$. The map $q\rightarrow \chi_q$ is clearly linear and thus defines a homomorphism from the group $Q(A)$ of quadratic functions to $\text{Sym}(A)$.

1. Is this map surjective? In other words, is there always a quadratic refinement of a symmetric bilinear form valued in $\mathbb{R}/\mathbb{Z}$? I think the answer is yes but I am not sure about a concrete proof.

2. Assuming the answer to the first question is yes, what is the kernel of the map? I think it should be isomorphic to $\Hom(A,\mathbb{Z}_2)$ by looking at examples (and from other abstract arguments, see Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?), but It is not clear to me how to prove this, and even how to see $\Hom(A,\mathbb{Z}_2)$ as a subgroup of $Q(A)$.

Answer 1

1. By the classification of finite abelian groups we have $A \cong C_1 \oplus \cdots \oplus C_k$ for some cyclic groups $C_i = \langle g_i \rangle$ of prime power order $r_i$. Let $\chi: A \times A \to \mathbb R / \mathbb Z$ be a symmetric bilinear map. Observe that $\chi(g_i, g_j)$ has order dividing $\gcd(r_i, r_j)$ for all $i,j$. For each $i = 1, \dots, k$, let $q_i$ be a solution to $2q_i = \chi(g_i, g_i)$ such that $q_i$ has odd order if $r_i$ is odd (there is a unique choice if $r_i$ is odd and two choices if $r_i$ is even). Now define $q : A \to \mathbb R / \mathbb Z$ by $$q\left(\sum_{i=1}^k n_i g_i\right) = \sum_{i=1}^k n_i^2 q_i + \sum_{1 \le i < j \le k} \chi(g_i, g_j) n_i n_j.$$ This is well-defined by our careful choice of $q_i$ and it is easy to check that $\chi_q = \chi$.

2. As mentioned in the comments, the kernel consists of the homomorphisms $q : A \to \mathbb R / \mathbb Z$ taking half-integer values, which can obviously be identified with $\mathrm{Hom}(A, \mathbb Z / 2 \mathbb Z)$.