# Maximal Closed Subgroups of $SU_4$

What are the maximal closed subgroups of $$SU_4$$?

• In my comment that you linked to, I was specifically focusing on simple subgroups of $SU(4)$ containing $SO(4)$. In particular, I think your list should also have a finite extension of $Sp(2)$, included via the identification of $\mathbb{H}^2$ with $\mathbb{C}^4$. The extension at least contains $iI$, but I don't know if it has more than 2 components. Sep 20, 2022 at 22:38
• @JasonDeVito this is exactly the sort of feedback I was looking for I'll incorporate all those corrections into my question Sep 21, 2022 at 14:00
• A minor thing to add to my previous comment: I can now show the maximal proper extension of $Sp(2)$ does indeed just have two components. Would you like me to write it up as an answer? Also, $SU(3)$ doesn't have an irreducible rep of dim 4, but it does have a 4 dim rep: the standard rep + trivial. The corresponding subgroup of $SU(4)$ is the block $SU(3)$, so is not maximal, being contained in $U(3)$. Sep 22, 2022 at 23:53
• @JasonDeVito Yes, please! And ya you are right that $SU(3)$ has a 4d rep but all reducible representations are non maximal since they are contained in $S(U_3 \times U_1)$ or $S(U_2 \times U_2 )$ as you say ( in my question I use irrep as short hand for irreducible representation, I think the only thing I claim is that $SU_3$ has no 4d irreps?). Sep 23, 2022 at 0:04
• Sorry to be confusing. There is nothing wrong in what you wrote above, I was just clarifying. I hadn't though about it previously, but it seems the following is true: For $G\in \{SU(n), SO(n), Sp(n)\}$, a closed subgroup $H\subseteq G$ is not maximal if the corresponding $n$-dim representation is reducible. In particular, in terms of classifying maximal subgroups, one only needs to consider irreducible reps. You probably already knew this ;-). I'll write the answer about $Sp(2)$ now. Sep 23, 2022 at 1:28

I claim that $$N:=N_{SU(4)}(Sp(2))$$ is a maximal subgroup of $$Sp(2)$$, and that $$N = Sp(2) \cup iI Sp(2)$$.

To see this, consider the double covering $$\pi:SU(4)\rightarrow SU(4)/\{\pm I\}\cong SO(6)$$. Note that $$-I\in Sp(2)\subseteq SU(4)$$, so $$\pi|_{Sp(2)}$$ is the double covering $$Sp(2)\rightarrow SO(5)$$. Up to conjugacy, there is an essentially unique $$SO(5)\subseteq SO(6)$$, the usual block form.

So, instead of studying $$Sp(2)\subseteq SU(4)$$, we'll study $$SO(5)\subseteq SO(6)$$ and pull the information back via $$\pi$$.

Proposition: The only proper subgroup of $$SO(6)$$ which properly contains $$SO(5)$$ is $$O(5) = \{ \operatorname{diag}(A,\det(A)):A\in O(5)\}$$.

Proof: The isotropy action of $$SO(5)$$ on $$S^5 = SO(6)/SO(5)$$ is transitive on the unit sphere in $$T_{I SO(5)} S^5$$, so is, in particular, irreducible. This, then, implies that $$SO(5)$$ is maximal among connected groups: if $$SO(5)\subseteq K\subseteq SO(6)$$, then on the Lie algebra level, the adjoint action of $$\mathfrak{so}(5)$$ would preserve both $$\mathfrak{k}$$ and $$\mathfrak{k}^\bot$$, contradicting irreducibility. (Here, I'm using the fact that we can naturally identify the isotropy action of $$H$$ on $$T_{I SO(5)} SO(6)/SO(5)$$ with $$\mathfrak{so}(5)^\bot\subseteq \mathfrak{so}(6)$$.)

Since we now know that $$SO(5)$$ is maximal among connected subgroups of $$SO(6)$$, and the identity component of a Lie group is always a normal subgroup, it now follows that $$N_{SO(6)}(SO(5))$$ is a maximal subgroup of $$SO(6)$$.

Of course, $$O(5)\subseteq N_{SO(6)}(SO(5))$$, but why is the reverse inclusion true? Well, every matrix $$B\in SO(5)$$ fixes the basis vector $$e_6$$ of $$\mathbb{R}^6$$, and $$\operatorname{span}\{e_6\}$$ is the unique subspace of $$\mathbb{R}^6$$ fixed by all of $$SO(5)$$. A simple computation reveals that for any $$C\in N_{SO(6)}(SO(5))$$, that $$CSO(5)C^{-1} = SO(5)$$ fixes $$Ce_6$$. It follows that $$Ce_6 \in \operatorname{span}\{e_6\}$$. Moreover, since $$C\in SO(6)$$, we must in fact that $$Ce_6 = \pm e_6$$. In either case, the fact that $$CC^t = I$$ now implies that $$C\in O(5)$$. $$\square$$

Now, let's pull that information back to to better understand $$N = N_{SU(4)}(Sp(2))$$. It's not too hard to see that $$\pi|_N:N\rightarrow N_{SO(6)}(SO(5)$$ is a double covering, with $$\pi$$ mapping $$Sp(2)$$ and $$iI Sp(2)$$ to the two different components of $$N_{SO(6)}(SO(5))$$.

Now, let $$g\in N$$ be arbitrary. Since $$\pi|_{Sp(2)\cup iI Sp(2)}$$ is surjective onto $$N_{SO(6)}(SO(5))$$, there is an $$h\in Sp(2)\cup iI Sp(2)$$ with $$\pi(g) = \pi(h)$$. Then $$gh^{-1}\in \ker \pi = \pm I$$, so $$g = \pm I h$$. Since both $$h, \pm I\in Sp(2)\cup iI Sp(2)$$, it follows that $$g\in Sp(2)\cup iI Sp(2)$$ as well.

• very cool argument using the "accidental" isomorphisms $SU(4) \cong Spin(6)$ and $Sp(2) \cong Spin(5)$, love it! Sep 23, 2022 at 2:16
• Ok I finally posted a full classification that I'm pretty confident in. One last sanity check I wanted to make, $SO_4$ is a subgroup of $Sp_2$ right? Feb 26 at 17:03
• @IanGershonTeixeira: No, I don't think $SO(4)$ is a subgroup of $Sp(2)$. The complex $4$-dim representation of $SU(2)\times SU(2)$ given by $SU(2)\rightarrow SU(2)\rightarrow SO(4)\subseteq SU(4)$ is irreducible, so it is precisely one of real, quaternionic, or complex. As it's obviously real, it cannot be quaternionic. Feb 27 at 3:59

The full list of maximal subgroups is:

Type I (normalizer of maximal connected subgroup) \begin{align*} & U_3 \cong S(U_3 \times U_1) \\ & S(U_2 \times U_2):2 \\ & 4 \circ_2 Sp_2 \end{align*} Type II (finite maximal closed subgroup) \begin{align*} &4\circ_2 2.A_7 \\ &4\circ_2 Sp(4,3) \\ &N(2^{2(2)+1}) \end{align*} Type III (normalizer of a subgroup which is connected but not maximal connected) \begin{align*} & N(T^3)=S(U_1 \times U_1 \times U_1 \times U_1) : S_4\\ & SO_4(\mathbb{R})\cdot 4 \\ \end{align*}

Note that $$S(U_1 \times U_1 \times U_1 \times U_1)$$ is contained in $$S(U_3 \times U_1)$$ above. And $$SO_4(\mathbb{R})$$ is contained in $$Sp_2$$. However $$SO_4(\mathbb{R}) \cdot 4$$ is not contained in $$4 \circ_2 Sp_2$$.

Note on notation. $$:$$ means split extension (semidirect product). $$\cdot$$ means nonsplit extension. $$\circ$$ denotes central product, in all cases here we have $$4 \circ_2 H$$ is just the group generated by $$H$$ and $$iI$$ but that group is not a direct product since already $$-I \in H$$, we get a central product essentially with two $$H$$ components.

Here all the $$N$$ denote normalizer. Recall that a positive dimensional (type I and type III above) maximal subgroup of a simple Lie group equals the full normalizer of its identity component.

https://arxiv.org/pdf/math/0605784.pdf classifies all maximal closed subgroups of $$SU_n$$ whose identity component is not simple (here trivial counts as simple). According to table 5 the maximal closed subgroups of $$SU_4$$ of this type are:

The normalizer of the maximal torus (row 4 table 5, $$\ell=4, p=1$$) $$N(T)=S(U_1 \times U_1 \times U_1 \times U_1) : S_4$$ As well as (row 1 table 5, $$p=3,q=1$$ ) $$S(U_3 \times U_1 )\cong U_3$$ and the normalizer of $$S(U_2 \times U_2)= \{\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}:A,B\in U_2,det(A)det(B)=1 \}$$ which is a split extension (row 1 table 5 $$p=q=2$$) $$< S(U_2 \times U_2),\begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}> \cong S(U_2 \times U_2):2$$ where the normalizing matrix swaps the two blocks in the direct sum. This is the lift of $$S(O_4 \times O_2 )$$ through the double cover $$SU_4 \to SO_6$$.

Next there is (row 3 table 5, $$p=2$$) $$$$ where the normalizing matrix swaps the two tensor factors and applies a global phase. Here the identity component $$SU_2 \otimes SU_2$$ contains $$-I=(\zeta_8 SWAP2)^4$$, but does not contain $$iI=(\zeta_8 SWAP2)^2$$, so the full normalizer is the nonsplit extension $$SU_2 \otimes SU_2 \cdot 4$$. This subgroup is conjugate to the normalizer of the standard $$SO_4$$ subgroup $$SO_4(\mathbb{R}) \cdot 4$$ and we prefer to write it that way. This is the lift of $$S(O_3 \times O_3)\cdot 2$$ through the double cover $$SU_4 \to SO_6$$. For more details see

$SO_4(\mathbb{R})$ and $SU_2 \otimes SU_2$ subgroups of $SU_4$

Next, we consider maximal closed subgroups with nontrivial simple connected component. [Credit to Jason for pretty much all the group with simple identity component stuff]

By dimension, such a subgroup would be isogeneous to $$SU_2,SU_3,Sp_2$$ or $$G_2$$.

There is no 4d irreps of $$SU_3$$ since the dimension of $$SU_3$$ irreps are given by the formula $$\frac{(m_1+1)(m_2+1)(m_1+m_2+2)}{2}$$

Similarly there are no 4d irreps of $$G_2$$ since the dimensions are given https://en.wikipedia.org/wiki/G2_(mathematics)

The symplectic subgroup $$O_5(\mathbb{R})= 2 \times SO_5(\mathbb{R})$$ is a maximal subgroup of $$SO_6(\mathbb{R})$$. Lifting through the double cover $$SU_4 \to SO_6(\mathbb{R})$$ we have that $$4 \circ_2 Sp_2=$$ is maximal subgroup of $$SU_4$$.

Every irreducible $$SU_2$$ subgroups of $$SU_4$$ is contained in a conjugate of $$Sp_2$$. See

Understanding the 4 dimensional irrep of $SU_2$

Indeed the containment $$SU(2)_{irr} \subset Sp(2) \subset SU(4)$$ is the lift of $$SO(3)_{irr} \subset SO(5) \subset SO(6)$$ through the double cover $$SU(4) \to SO(6)$$. Here $$SU(2)_{irr}$$ is the image of the 4d irrep of $$SU(2)$$ and $$SO(3)_{irr}$$ is the image of the 5d irrep of $$SO(3)$$. Similarly we have that $$N(SU(2)_{irr})=4 \circ_2 SU(2)_{irr} \subset N(Sp(2))=4 \circ_2 Sp_2 \subset SU(4)$$ is the lift through the double cover of $$N(SO(3)_{irr})=O(3)_{irr} \subset N(SO_5)=O(5)=S(O_5 \times O_1) \subset SO(6)$$. So in particular there is no maximal subgroup of $$SU_4$$ with simple connected component isogeneous to $$SU_2$$. All such groups are $$4 \circ_2 SU(2)_{irr} \subset 4 \circ_2 Sp_2$$ and thus not maximal.

Finally we consider subgroups with trivial connected component. These are finite since $$SU_4$$ is compact. To be maximal they must at least be primitive. Primitive finite subgroups of $$SU_4$$ are classified by work of Blichfeldt 1911 which was rewritten in modern notation here https://arxiv.org/abs/hep-th/9905212 From this we conclude there are $$4$$ finite groups maximal among the finite subgroups of $$SU_4$$. The central product $$4 \circ_2 2.A_7$$ of order $$4(2,520)=10,080$$ (maximal closed since it is maximal finite and a 3-design) the central product $$4 \circ_2 Sp(4,3)$$ of order $$4(25,920)=103,680$$ (maximal closed since it is maximal finite and a 3-design). $$N(2^{2(2)+1})$$ is the normalizer of an extraspecial 2 group of order $$32$$. This group has order $$4(11,520)=46,080$$ (maximal closed since it is maximal finite and a 3-design). This group is know as the 2 qubit Clifford group in quantum computing. For details see

https://quantumcomputing.stackexchange.com/questions/25591/is-the-clifford-group-a-semidirect-product?noredirect=1&lq=1

Note that this group has order $$6!2^6=46,080$$ and is the lift through the double cover $$SU_4 \twoheadrightarrow SO_6$$ of $$W(D_6)$$. Here $$W(D_6)$$ is the subgroup of $$SO_6$$ of signed permutation matrices, the Weyl group of $$D_6$$, which has order $$6!2^6/2$$. Finally, $$4\circ_2 2.S_6$$ is maximal among the finite subgroups but is actually contained in the group $$N(Sp_2)$$ described above. To see this observe that there is a faithful 4d irrep of $$2.S_6$$ which is quaternionic (Schur indicator -1) so $$2.S_6$$ is a subgroup of $$Sp_2$$. Thus by adding in $$iI$$ we have that $$4\circ_2 2.S_6$$ is a subgroup of $$N(Sp_2)$$.

For references on designs and maximality see Finite maximal closed subgroups of Lie groups

So the maximal closed subgroups with trivial identity component are the $$3$$ finite groups: $$4\circ_22.A_7, 4\circ_2 Sp(4,3), N(2^{2(2)+1})$$. This is consistent with the fact that a maximal $$2$$-design group is maximal closed ( all $$3$$ designs are $$2$$ designs).

Note: $$2.A_7$$ denotes PerfectGroup(5040,1), the unique perfect group of that order.

Note: \begin{align*} & 4\circ_2 2.A_7 \\ & 4\circ_2 Sp(4,3) \\ & N(2^{2(2)+1})\\ \end{align*} are all 2-designs (at least). The other 2- designs are three other subgroups of $$N(2^{2(2)+1})$$. These six groups are all Lie primitive (not contained in any proper positive dimensional closed subgroup). There is in addition one more Lie primitive group: it corresponds to the $$GL(3,2)$$ subgroup of $$A_7$$/ corresponds to the $$SL(2,7)$$ subgroup of $$2.A_7$$.

• Literally brilliant. Feb 27 at 0:33