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YCor
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Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$$\bar{q}:G/\mathrm{Rad}(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/\mathrm{Rad}(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

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Bipolar Minds
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Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=[\frac{n}{2}]$$q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=[\frac{n}{2}]$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

added 7 characters in body
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Bipolar Minds
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Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=(-1)^n$$q(n)=[\frac{n}{2}]$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=(-1)^n$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/Rad(b) \to \mathbb{Q/Z}$, s.t. $\pi^*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=[\frac{n}{2}]$). Hence my question:

Is there some definition/construction of the kernel of quadratic forms on abelian groups?

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Bipolar Minds
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