Lie groups with same algebra

Question

I had a problem when considering symmetry breaking in an SO(4) gauge theory:

$\mathcal{L} = \left| D_\mu\phi \right|^2$

where $D_\mu$ is the SO(4) covariant derivative. Then assuming there is some potential that has a minimum such that we can choose the ground state to be:

$\langle \phi \rangle = \begin{pmatrix} 0 & 0 & 0 & v \end{pmatrix}^{T}$

After this I found the unbroken generators which have to generate a subgroup of SO(4) and that their generators fulfill the $\mathfrak{su}(2)$ algebra. Now I wanted to conclude that therefore the unbroken subgroup is SU(2). But there are multiple groups that have this same algebra, e.g. SO(3) does too. How do I know which one is the correct subgroup? Is there any way to see this from the explicit form of the generators? (e.g. the dimension of the representation)

Answer 1

The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1.

In general, you need to know the Lie group itself to find the correct subgroup (i.e. you can't just find the subgroup from the algebra alone). This is exactly because of cases like $SU(2)$ and $SO(3)$ that have isomorphic tangent spaces, but which have different global properties.

Answer 2

To a given Lie algebra $\mathfrak g$ there is a unique group $\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$.

The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde G\rightarrow G$. Once you have defined a particular representation you are able to compute this kernel. For example, you start with an $\mathfrak{su}(2)$ algebra. Then if you choose the adjoint representation you can show that $Ker(\rho)=\mathbb Z_2$ and the group will be $G=SU(2)/\mathbb Z_2=SO(3)$. On the other hand, if you choose the defining representation you get $Ker(\rho)=\mathbb 1$ and $G=SU(2)/\mathbb 1=SU(2)$.

There are some technical details needed to compute those kernel. In general, $$Ker(\rho)\subset\mathcal Z(\tilde G),$$ where $\mathcal Z(\tilde G)$ is the center of $\tilde G$, and this center is a finite group which can be obtained from the extended Dynkin diagram.

Same references: Cornwell, group theory in physics, 1984; Olive, Turok, Nucl Phys B215, 1983, p470;