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A metric group is a finite abelian group $G$ with a quadratic function $$q:G\rightarrow \mathbb R/\mathbb Z\;,$$ that is, $$M(a,b):= q(a+b)-q(a)-q(b)$$ is bilinear in $a$ and $b$ [edit: and non-degenerate]. A Lagrangian subgroup $H\subset G$ is one with $q(h)=0$ $\forall h\in H$, and $\lvert H\rvert^2=\lvert G\rvert$.

Question: Is it true that $H$ is always isomorphic to $G/H$, and how can one see this? Can one somehow relate $G/H$ to the $\mathbb R/\mathbb Z$-valued representations of $H$ via $M$ (or vice versa)?

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    $\begingroup$ Do you assume the bilinear form is nondegenerate? If not, take $q = 0$ and you get easy counterexamples. $\endgroup$
    – Sean Eberhard
    Jun 28 at 13:08
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    $\begingroup$ Each element of $G$ induces a map $H \to \mathbb R / \mathbb Z$, and this map $G \to H^*$ factors through $G/H$ since the bilinear form is zero on $H$. Assuming the bilinear form is nondegenerate, the image of $G/H \to H^*$ separates the points of $H$, so it is a surjective homomorphism and it follows that $G/H$ is canonically isomorphic to $H^*$ since $|G/H| = |H| = |H^*|$. $\endgroup$
    – Sean Eberhard
    Jun 28 at 13:11
  • $\begingroup$ @SeanEberhard Thanks, I had also just noticed that $\rho_a=M(\bullet, a)$ is exactly the $\mathbb{R}/\mathbb{Z}$ representation I was looking for. And yes, I forgot to say $M$ is non-degenerate. $\endgroup$
    – Andi Bauer
    Jun 28 at 13:59

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