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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 $\bar{q}:G/\mathrm{Rad}(b) \to \mathbb{Q/Z}$, 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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    $\begingroup$ What property should the kernel $K$ have? The end of your first paragraph suggests you might want something like $K$ being minimal with respect to the requirement that there is a quadratic form $\bar q$ on $G/K$ that pulls back to $q$. Is that, together presumably with some sort of uniqueness, correct? (Also, a notational suggestion: old-fashioned texts often used $[\cdot]$ for the greatest-integer function, so $q(n) = n/2 + \mathbb Z$ might be clearer than $q(n) = [n/2]$.) $\endgroup$
    – LSpice
    Feb 18 at 23:32
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    $\begingroup$ @LSpice Correct! $\endgroup$
    – Bipolar Minds
    Feb 18 at 23:35
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    $\begingroup$ A "kernel" can be defined for an arbitrary map $f$ from an abelian group $G$ to a set $X$, namely $K=\{g:\forall g':f(g'+g)=f(g')\}$. Thus $G/K_G$ is the largest quotient group of $G$ through which $f$ factors. This extends the usual kernel of quadratic maps on vector spaces. $\endgroup$
    – YCor
    Feb 19 at 10:59
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    $\begingroup$ In the theory of quadratic forms over a field of characteristic 2, the radical of a quadratic form $q$ is sometimes defined as $R(q)=\{x\in \mathrm{rad}(b):q(x)=0\}$, where $b$ is the polar form of $q$. This should work in your situation as well: $R(q)$ is a subgroup of $G$ and $q$ factors via a $\mathbb{Q}/\mathbb{Z}$-quadratic map on $G/R(q)$. (This should coincide with @YCor's suggestion, but is perhaps conceptually clearer.) $\endgroup$
    – Uriya First
    Feb 19 at 13:09
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    $\begingroup$ Ah, so its the (usual) kernel of $q|_{\operatorname{rad(b)}}$ as $q$ is a homomorphism on $\operatorname{rad(b)}$ $\endgroup$
    – Bipolar Minds
    Feb 19 at 17:51

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As requested, I post my comment (with mild changes) as answer:

In the theory of quadratic forms over a field of characteristic $2$, the radical of a quadratic form $q$ is sometimes defined as $R(q)=\{x\in \mathrm{rad}(b):q(x)=0\}$, where $b$ is the polar form of $q$. This should work in your situation as well: $R(q)$ is a subgroup of $G$ and $q$ factors via a $\mathbb{Q}/\mathbb{Z}$-valued quadratic map on $G/R(q)$.

The subgroup $R(q)$ should coincide with the kernel suggested in @YCor's comment.

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