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Definition A bilinear form, or bicharacter on an abelian group $A$ with values in another abelian group $E$ is a function $b\colon A \times A \to E$ satisfying: $$b(x,y+z) = b(x,y) + b(x,z)$$ $$b(x+y,z) = b(x,z) + b(y,z)$$ It is said to be symmetric if $b(x,y) = b(y,z)$.

Definition A quadratic form on an abelian group $A$ with values in an abelian group $E$ is a function $q\colon A \to E$, such that $q(x) = q(-x)$, and the form assigned to it $b(x,y) := q(x+y) - q(x) - q(y)$ is bilinear.

It is obvious that the bilinear form assigned above is also symmetric, so it is a natural question to ask which bilinear forms come from a quadratic form.

In this article, it is stated that "if the order of the group $A$ is odd, the assignment [...] defines a bijection between symmetric bilinear forms and quadratic forms, but in general, it is not a bijection."

How do I see this? How can I recover the quadratic form from the bilinear form in the case of odd order? And how does the assignment fail to be bijective? Is it still surjective or injective? How can I see which bilinear forms will not come from a quadratic form?

Turion
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2 Answers2

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As has been explained in this question, the function $f\colon E \to E$ given by $f(x) = x + x$ is injective iff $E$ has odd order.

This means that exactly if $E$ has odd order, we can form the inverse of the assignment, which is $q(x) = \frac{1}{2}b(x,x)$.

If $E$ doesn't have odd order, there will be elements $x \in E$ such that $f^{-1}(x)$ consists of multiple elements, and if $b(y,y) = x$, there are potentially several quadratic forms. If $f^{-1}(x)$ has no elements at all, there could even be no quadratic form corresponding to $b$.

Community
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Turion
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Although it is not a bijection, there is a surjection from quadratic forms to symmetric bilinear forms on finite abelian groups. This can be proven using decomposition into orthogonal irreduibles, as is done in the appendix here: https://arxiv.org/pdf/1405.7950.pdf

Ryan T Johnson
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    That's interesting! Is my last sentence "If $f^{−1}(x)$ has no elements at all, there could even be no quadratic form corresponding to b." in my answer wrong then by any chance? – Turion Feb 04 '22 at 09:02
  • This problem can be avoided by letting the quadratic form go to a slightly bigger codomain than the bilinear form. For example, if $G = \mathbb{F}_2 = \mathbb{Z} / 2\mathbb{Z}$ and $\hat{\cdot}$ is the natural lifting from $\mathbb{F}_2$ to $\mathbb{Z}$ then define a bilinear form $b : \mathbb{F}_2 \times \mathbb{F}_2 \to {1,-1}$ by $b(x,y) = (-1)^{\hat{x} \hat y}$. Then define $q : \mathbb{F}_2 \to {1,-1,i,-i}$ by $q(x) = i^{\hat{x}^2}$. You can check that $\frac{q(x+y)}{q(x)q(y)} = b(x,y)$. – Ryan T Johnson Mar 03 '22 at 16:55