lagrangian subspace and Heisenberg group

Question

Let $(V,\omega)$ be a symplectic vector space. Also we assume $L\subset V$ be a Lagrangian subspace., and $H(V)$ be Heisenberg group, then why $L\bigoplus U(1)\subset H(V)$ is maximal abelian subgroup.?

Answer 1

First let us agree on the group law: for $v_1,v_2\in V$ and $z_1,z_2\in S^1=U(1)$, we have $(v_1,z_1)(v_2,z_2)=(v_1+v_2,z_1z_2e^{2\pi i \omega(v_1,v_2)})$.

Now let us contemplate $A:=L\times U(1)$. This is certainly abelian: if $v_1,v_2\in L$ then $\omega(v_1,v_2)=0$ so that $$ (v_1,z_1)(v_2,z_2)=(v_1+v_2,z_1z_2)=(v_2,z_2)(v_1,z_1). $$ So we see that the group law on $A$ amounts to that of the direct product of $L$ (under addition) and $U(1)$. In particular, $A$ is a subgroup.

Finally, we show that $A$ is maximal: suppose that $A\subset B$ with $B$ a strictly larger abelian subgroup and let $(w,z)\in B\setminus A$. Now $L\oplus \langle w\rangle$ is not isotropic for $\omega$ because $L$ is maximal for that property so there is some $v\in L$ such that $\omega(v,w)\neq 0$. Scaling $v$ if necessary, we fix things so that $\omega(v,w)\not\in\tfrac12\mathbb{Z}$. We can now get a contradiction: $(v,1)\in A$ and, by hypothesis, commutes with $(w,z)$. That is $$ (v+w,ze^{2\pi i\omega(v,w)})=(w+v,ze^{2\pi i\omega(w,v)})$$ so that $e^{4\pi i\omega(v,w)}=1$ or $\omega(v,w)\in\tfrac12\mathbb{Z}$: a contradiction.