# Closed subgroups of $\mathrm{SO}(4)$

My question is quite simple : we know all closed subgroups of $\mathrm{SO}(3)$; is it also known what are the closed subgroups of $\mathrm{SO}(4)$?

• It so happens that SO(4) is almost SO(3) x SO(3) - close enough that one can use the SO(3) result to deal with SO(4). – Noam D. Elkies May 26 '16 at 12:23
• I've seen variations on this topic appear in textbooks. As Noam mentions, you lift the group to the double cover which is isomorphic to $S^3 \times S^3$. I believe the description of the closed subgroups of this was given by Hopf, back in the 1930's. – Ryan Budney May 26 '16 at 21:48

There is an epimorphism $\mathrm{SU}(2) \times \mathrm{SU}(2) \to \mathrm{SO}(4)$ with the kernel $\langle(−I, −I)\rangle$. Since $\mathrm{SU}(2)$ is isomorphic to the unit quaternions, the epimorphism is given by $(u,v)\mapsto R_{u,v}$ where $R_{u,v}$ is the rotation of $\mathbb{R}^4$ given by $R_{u,v}(q)=v^{-1}qu$ for any quaternion $q$.

And $\mathrm{SU}(2)$ maps onto $\mathrm{SO}(3)$ with kernel $\langle -I\rangle$; again use quaternions.

As you say, you know the closed subgroups of $\mathrm{SO}(3)$, and so this gives the closed subgroups of $\mathrm{SU}(2)$ and so those of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ (via Goursat's Lemma) and finally those of $\mathrm{SO}(4)$.

More generally, as it relates to semisimple subgroups, all simple subgroups of real Lie groups are known, as described here:

Karpelevič, F. I. The simple subalgebras of the real Lie algebras. Trudy Moskov. Mat. Obšč. 4 (1955), 3–112.

Karpelevič, F. I. Classification of the simple subalgebras of the real forms of classical algebras. Doklady Akad. Nauk SSSR (N.S.) 93, (1953). 613–616.

Karpelevič, F. I. Classification of the simple subgroups of the real forms of the group of complex unimodular matrices. Doklady Akad. Nauk SSSR (N.S.) 85, (1952). 1205–1208.

• "this gives the closed subgroups of $SU(2)$ and hence of $SU(2)\times SU(2)$": this is not correct. Understanding the closed subgroups of $G$ and $H$ is just a first step before a description of subgroups of $G\times H$. "More generally, all simple subgroups" is also weird: the question is not just about simple subgroups. – YCor May 26 '16 at 21:27
• I added some further detail and clarification. – Sean Lawton May 26 '16 at 21:49
• Yes, although Goursat's lemma reduces to a careful description of all quotients of all finite subgroups. Also it does not reduce to Goursat's lemma in all case, because in principle you could find closed subgroup with some non-closed projection (actually in the precise case of $SU(2)\times SU(2)$ are aren't, but there would be in $SU(3)\times SU(3)$.) – YCor May 26 '16 at 22:11
• I don't know if the OP intended to leave this door open, but discrete subgroups are also closed and those can't be classified just by looking at the Lie algebra (AFAIK). – Igor Khavkine May 26 '16 at 23:00
• References to the classification of finite subgroups can be found in mathoverflow.net/questions/37136/…. – Friedrich Knop May 27 '16 at 15:28

There is a complete list in the paper Mendes, Roberto de Maria Nunes. "Symmetries of spherical harmonics." Transactions of the American Mathematical Society 204 (1975): 161-178. As in the above answer this paper uses the 2 to 1 homomorphism $$\phi:S^3 \times S^3 \rightarrow SO(4)$$ . To summarize the construction in the paper for $$G$$ as subgroup of $$SO(4)$$ let L and R be the projections of $$\phi^{-1}G$$ on to the two components, these are closed subgroups of $$S^3$$ , and $$L_K$$ and $$R_K$$ are the normal subgroups $$(L,1)$$ and $$(1,R)$$ of $$L$$ and $$R$$. We can then define an isomorphism $$\Phi: L/L_K \rightarrow R/R_K$$ by $$\Phi(l L_K)= rR_k$$ where $$(l,r) \in \phi^{-1}G$$.

Given $$L,R, L_K,R_K,\Phi$$ we can define $$G$$ by its action on a quaternion $$q$$ as $$G = \{ g \in SO(4): g(q) = l q r^{-1} , l \in L, r \in R, \Phi(l L_K)= r R_K \}$$ All the closed subgroups of $$SO(4)$$ are enumerated as pairs $$(L/L_K; R/R_K)$$ a table in the paper.