My post here concerns the concept of Lagrangian subgroup for a non-abelian Lie group, such as a semi-simple non-abelian Lie for gauge theory.

For topological gapped boundary, $a_{\parallel,I}$ lies in a Lagrangian subspace of $\mathbf{t}_\Lambda$ implies that the boundary gauge group ($ \equiv \mathbb{T}_{\Lambda_{0}}$) is a Lagrangian subgroup.

We can rephrase it in terms of the exact sequence for the vector space of Abelian group $$\Lambda \cong \mathbb Z^N$$ and its subgroup $\Lambda_0$:

These form an exact sequence: $$ 0 \to \Lambda_0 \overset{\mathbf{h}}{\to} \Lambda \to \Lambda/\Lambda_0 \to 0. $$ Here $0$ means the trivial Abelian group with only the identity, or the zero-dimensional vector space. Here $0$ means the trivial zero-dimensional vector space and $\mathbf{h}$ is an injective map from $\Lambda_0$ to $\Lambda$.

We can also rephrase it in terms of the exact sequence for the vector space of Lie algebra by

$$0 \to \mathbf{t}_{(\Lambda/\Lambda_0)}^* \to \mathbf{t}_{\Lambda}^* \to \mathbf{t}_{\Lambda_0}^* \to 0.$$

My Question: For topological gapped boundary, do we have a direct generalization such that the $a_{\parallel,I}$ lies in a Lagrangian subspace of $\mathbf{t}_\Lambda$ implies that the boundary gauge group is still a Lagrangian subgroup? How do we define the Lagrangian subgroup of a nonabelian Lie group?

• References along this direction is very welcome.