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

The symplectic form $\omega$ is given by (with the restricted $a_{\parallel,I}$ on ${\partial \mathcal M}$ ), $$ \omega=\frac{K_{IJ}}{4\pi} \int_{ \mathcal M} (\delta a_{\parallel,I}) \wedge d (\delta a_{\parallel,J}). $$ for a varyation of the differential of this 1-form on an abelian Chern-Simons action $\delta S_{bluk}$.

For the well-known abelian gauge group such as the bulk gauge group $\text{U(1)}^N \cong \mathbb{T}_\Lambda$ as the torus, is the quotient space of $N$-dimensional vector space $\bf{V}$ by a subgroup $\Lambda \cong \mathbb Z^N$. Locally the gauge field $a$ is a 1-form, which has values in the Lie algebra of $\mathbb{T}_\Lambda$, we can denote this Lie algebra $\mathbf{t}_\Lambda$ as the vector space $\mathbf{t}_\Lambda =\Lambda \otimes \mathbb{R}$.

For topological gapped boundary, $a_{\parallel,I} \quad$ 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.

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

See a previous post for other background information: Lagrangian subgroups/submanifolds, 2d topological boundary and 3d "non-abelian" Chern–Simons theory

First, the symplectic form $\omega$ is given by (with the restricted $a_{\parallel,I}$ on ${\partial \mathcal M}$ ), $$ \omega=\frac{K_{IJ}}{4\pi} \int_{ \mathcal M} (\delta a_{\parallel,I}) \wedge d (\delta a_{\parallel,J}). $$ for a varyation of the differential of this 1-form on an abelian Chern-Simons action $\delta S_{bluk}$.

For the well-known abelian gauge group such as the bulk gauge group $\text{U(1)}^N \cong \mathbb{T}_\Lambda$ as the torus, is the quotient space of $N$-dimensional vector space $\bf{V}$ by a subgroup $\Lambda \cong \mathbb Z^N$. Locally the gauge field $a$ is a 1-form, which has values in the Lie algebra of $\mathbb{T}_\Lambda$, we can denote this Lie algebra $\mathbf{t}_\Lambda$ as the vector space $\mathbf{t}_\Lambda =\Lambda \otimes \mathbb{R}$.

For topological gapped boundary, $a_{\parallel,I} \quad$ 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 are very welcome.

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

The symplectic form $\omega$ is given by (with the restricted $a_{\parallel,I}$ on ${\partial \mathcal M}$ ), $$ \omega=\frac{K_{IJ}}{4\pi} \int_{ \mathcal M} (\delta a_{\parallel,I}) \wedge d (\delta a_{\parallel,J}). $$ for a varyation of the differential of this 1-form on a abelian Chern-Simons action $\delta S_{bluk}$.

For the well-known abelian gauge group such as the bulk gauge group $\text{U(1)}^N \cong \mathbb{T}_\Lambda$ as the torus, is the quotient space of $N$-dimensional vector space $\bf{V}$ by a subgroup $\Lambda \cong \mathbb Z^N$. Locally the gauge field $a$ is a 1-form, which has values in the Lie algebra of $\mathbb{T}_\Lambda$, we can denote this Lie algebra $\mathbf{t}_\Lambda$ as the vector space $\mathbf{t}_\Lambda =\Lambda \otimes \mathbb{R}$.

For topological gapped boundary, $a_{\parallel,I} \quad$ 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.

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

The symplectic form $\omega$ is given by (with the restricted $a_{\parallel,I}$ on ${\partial \mathcal M}$ ), $$ \omega=\frac{K_{IJ}}{4\pi} \int_{ \mathcal M} (\delta a_{\parallel,I}) \wedge d (\delta a_{\parallel,J}). $$ for a varyation of the differential of this 1-form on an abelian Chern-Simons action $\delta S_{bluk}$.

For the well-known abelian gauge group such as the bulk gauge group $\text{U(1)}^N \cong \mathbb{T}_\Lambda$ as the torus, is the quotient space of $N$-dimensional vector space $\bf{V}$ by a subgroup $\Lambda \cong \mathbb Z^N$. Locally the gauge field $a$ is a 1-form, which has values in the Lie algebra of $\mathbb{T}_\Lambda$, we can denote this Lie algebra $\mathbf{t}_\Lambda$ as the vector space $\mathbf{t}_\Lambda =\Lambda \otimes \mathbb{R}$.

For topological gapped boundary, $a_{\parallel,I} \quad$ 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.

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.

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

The symplectic form $\omega$ is given by (with the restricted $a_{\parallel,I}$ on ${\partial \mathcal M}$ ), $$ \omega=\frac{K_{IJ}}{4\pi} \int_{ \mathcal M} (\delta a_{\parallel,I}) \wedge d (\delta a_{\parallel,J}). $$ for a varyation of the differential of this 1-form on a abelian Chern-Simons action $\delta S_{bluk}$.

For the well-known abelian gauge group such as the bulk gauge group $\text{U(1)}^N \cong \mathbb{T}_\Lambda$ as the torus, is the quotient space of $N$-dimensional vector space $\bf{V}$ by a subgroup $\Lambda \cong \mathbb Z^N$. Locally the gauge field $a$ is a 1-form, which has values in the Lie algebra of $\mathbb{T}_\Lambda$, we can denote this Lie algebra $\mathbf{t}_\Lambda$ as the vector space $\mathbf{t}_\Lambda =\Lambda \otimes \mathbb{R}$.

For topological gapped boundary, $a_{\parallel,I} \quad$ 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.