$\DeclareMathOperator\Hom{Hom}$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?](https://mathoverflow.net/questions/450594/is-there-a-clear-pattern-for-the-degree-2n-cohomology-group-of-the-nth-eile)), 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 $Q(A)$.



Quadratic refinements of a bilinear form on finite abelian groups
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 $\mathrm{Hom}(A, \mathbb Z / 2 \mathbb Z)$.


