Lie groups with same algebra

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)

• That's an exercise in calculating the isotropy group. – Qmechanic Apr 11 '16 at 16:17
• Sure. Find the subgroup generated by unbroken generators. Which subgroup seems more natural to you? (And which probably has a topological obstacle?) – TotallyRhombus Apr 11 '16 at 16:18
• @fs137 what is a topological obstacle? – Wolpertinger Apr 11 '16 at 16:51
• I guess I should have written "obstacle associated with global rather than local properties". If you exponentiate within $SO(4)$ the particular generators that leave $\langle \phi\rangle$ invariant, then you will find that the closure gives $SO(3)$ and not $SU(2)$, but the reason for this has more to do with differential geometry than topology. – TotallyRhombus Apr 11 '16 at 17:15

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.

• that essentially answers my question thank you! may I ask one more thing: if I have an explicit representation of the Lie algebra (i.e. in the example I gave I had the unbroken generators corresponding to the fundamental representation of the group), can I then know what group it is just by exponentiating? as in the matrices that I get from that are they a representation of SO(3) only or could they be a representation of SU(2) as well? (my guess is they would only represent SO(3)) – Wolpertinger Apr 11 '16 at 16:39
• It turns out that $SU(2)$ is a double cover of $SO(3)$, and there is a surjective homomorphism from $SU(2)$ to $SO(3)$. So you can view a representation of $SO(3)$ as a representation of $SU(2)$. Note $SO(4)$ contains a copy of $SU(2)$, but the subgroup of $SO(4)$ that preserves a given vector is not equivalent to this group. – TotallyRhombus Apr 11 '16 at 17:07

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;

• Your answer is nicely written, but does not address the question posed. – TotallyRhombus Apr 11 '16 at 17:09
• I think it addresses. The question basically is: Given a set of unbroken generators, forming a Lie algebra, and its representation, which is the associated Lie group? I just gave the answer without mention the fact $G$ is a group or a subgroup of something else. The math behind it is the same. – Diracology Apr 11 '16 at 17:13
• Your answer describes how one could obtain the set of all groups that are consistent with a particular Lie algebra, but does not describe how to choose the correct group. – TotallyRhombus Apr 11 '16 at 17:21
• I see your point. But my answer not only describes how to get all groups associated to a given Lie algebra. It also describes how to find the particular and "physical" group in a particular spontaneous symmetry breaking (SSB). Once you define a representation of the gauge algebra and a particular SSB you know the representation of the subalgebra. Then you obtain the unbroken group in the way I described above. – Diracology Apr 11 '16 at 17:24
• I suppose that since the symmetry is local, it shouldn't matter in perturbation theory whether we regard the unbroken symmetry group as $SU(2)$ or $SO(3)$. However, it seems that the relationship between $SO(3)$ and $SU(2)$ lattice gauge theories is more subtle than the "naive continuum limit" suggests: arxiv.org/pdf/hep-lat/0211004.pdf. If there is a significant nonperturbative difference between $SU(2)$ and $SO(3)$, then the embedding of the unbroken group in $SO(4)$ matters. – TotallyRhombus Apr 13 '16 at 16:45