What are the maximal closed subgroups of $ SU_4 $?
The full list of maximal subgroups I'm proposing 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.
Next there is (row 3 table 5, $ p=2 $)
$$
<SU_2 \otimes SU_2, \zeta_8\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}>
$$
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. For more details see
Is $ SU_2 \otimes SU_2 $ conjugate to $ SO_4(\mathbb{R}) $ in $ SU_4 $?
Next, we consider maximal closed subgroups with nontrivial simple connected component.
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)
[Credit to Jason]
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
$$
4 \circ_2 Sp_2=<iI,Sp_2>
$$
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
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 two designs are 3 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 $.
Best Answer
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.