I am currently facing this question: I have a locally compact abelian group $B$, which is $2$-regular ( $x\mapsto 2x$ is invertible), endowed with a bicharacter $h(x,y)$ so that $x\mapsto h(x,\cdot)$ is an isomorphism between $B$ and its dual group. For a subgroup $I$, its orthogonal $I^+$ is the set of elements $x$ in $B$ such that $h(x,y)=1$ for any $y$ in $I$. If $I \subset I^+$, $I$ is said isotropic.
I consider such an isotropic subgroup such that it is maximal for this property. I am wondering if by chance $I^+/I$ is compact. In the examples I have in mind this group is finite. If it's not true, does it become true if in addition $B$ has the structure of a topological unitary ring such that $h(xy,z)=h(x,yz)$ or if $h(x,y) =\chi(xy)$ where $\chi$ is some no trivial character on $B$?
An example where it is finite but not trivial: $B$ is a $p$-adic field, $\chi(x)=e^{2i\pi\{px\}_p}$, $I=Z_p$, $I^+=\frac{1}{p}Z_p$ where $Z_p$ are the $p$-adic integers.
Any idea, or simply any hint on the structure of $I^+/I$?
Thanks by advance!
I think I have a solution when $B$ is Hausdorff, with the conclusion that $I^+/I$ is finite, which is perfect for me.
First, one can show rather easily that $x\mapsto {\tilde h}(x,\cdot) = h(x,\cdot)$ is a well defined isomorphism between $I^{+}/I$ and its dual group. Also if $x$ is a element in a class of $I^{+}/I$, $h(x,x)=1$ implies that $x=0$. This comes from the fact that for any $y$ in $I$ and $n \in \mathbb{Z}$, $h(nx+y,mx+y)=h(x,x)^{nm}h(x,y)^{n+m}h(y,y)=1$ and the subgroup generated by $x$ and $I$ in $B$ would be isotropic, contradicting the fact that $I$ is maximal isotropic, except when it is the $0$ class.
Then if $K$ is any closed subgroup of $I^{+}/I$, then we have the decomposition $I^{+}/I = K \oplus K^{+}$. To prove that, consider $x$ in $K\cap K^{+}$. We have $h(x,x)=1$ and therefore $x=0$ from the above, and the sum $K + K^{+}$ is therefore direct. And since $K$ is closed $(K\cap K^{+})^{+} = K \oplus K^{+} = 0^{+} = I^{+}/I$.
Being an LCA group, $I^+/I$ is isomorphic to $\mathbb{R}^n \times H$ where H contains a compact open subgroup $O$ (see for instance the following lecture of Sidney Morris, corrolary 3, https://sidneymorris.net/Morris53.pdf ). But $n=0$ because $Q(x)=h(x,x)$ with $h$ being a continuous bicharacter, would necessarily be trivial on a non trivial element of the subgroup $\mathbb{R}^n \times \{0\}$. Therefore $I^+/I$ contains a compact open subgroup $O$. Since $O$ is compact, and therefore closed, with $I^+/I$ being self dual we may show that $O$ is also self dual. This comes from the fact that $I^+/I = O \oplus O^{+}$ from the above discussion, and ${\hat O}$ is canonically isomorphic to ${\widehat{I^+/I}}/O^\perp$, itself isomorphic to $I^+/I/O^{+}$, which is indeed isomorphic to $O$ thanks to $I^+/I = O \oplus O^{+}$. Consequently, $O$ being compact and self dual, must be finite. Then any coset of the form $x + {\cal O}$ for $x$ any element of $I^{+}/I$, is an open set containing $x$ thanks to the continuity of the group law. For any $y_k$ in ${\cal O}$ different from $0$, there exist $V_k$ and $W_k$, respectively neighborhoods of $x$ and $y_k$ that are disjoint because $B$ is Hausdorff, and therefore $I^{+}/I$ is also Hausdorff. Then $(\cap_k V_k) \cap {\cal O}$ is a neighborhood of $x$, since the number of $V_k$ is finite, which contains only $x$. Therefore
$I^{+}/I$ is discrete, and being self dual it is necessarily also compact, thus finite.
I hope this is correct...