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Other than subgroups of SU(3), what are the Lie subgroups of SU(4)? Assume that the subgroup is closed but not necessarily connected.

Additionally, which of these subgroups admit four dimensional irreducible representations?

Any references would be much appreciated.

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you maybe mean $S(U(3)\times U(1))$ (which is one dimension bigger than $SU(3)$). Probably you want to ask about irreducible subgroups (= irreducible on $\mathbf{C}^4$). Maybe you want connected, or are you asking about finite subgroups as well. – YCor Jan 10 '13 at 0:06
you could take $H=S(U(2)\times U(2))$, where each $U(2)$ acts on $(\mathbb C)^2$ the standard representation, and the representation of $H$ is the tensor product of these standard representations. This is also $SO(4)\subset SU(4)$. This is not contained in $SU(3)$. You could take finite subgroups of $H$ as well esentially of the form $A\times A$ where $A$ acts irreducibly on ${\mathbb C}^2$. There is also the maximal compact of $Sp_4({\mathbb C})\subset SL_4({\mathbb C})$ which acts irreducibly on ${\mathbb C}^4$. – Venkataramana Jan 10 '13 at 0:18
This question is worded very much like a homework problem. It would be improved if you included some of your background and motivation, and elaborated a bit on what applications you have in mind, so that references can be more directed. Please read for further recommendations on what makes a good MathOverflow question. – Theo Johnson-Freyd Jan 10 '13 at 0:21
Maybe this could help: B. Gruber and M. Ramek, “Boson and Fermion Operator Realizations of su(4) and its Semisimple Subalgebras”, in “Symmetries in Science VII”, ed. B. Gruber and T. Otsuka, Plenum Press, New York, 1994. I think they also wrote a computer program that tells you su(4) reps decompose as reps of its subalgebras. – Uwe Franz Jan 10 '13 at 16:21
Thank you for the comments. To be more specific, this arose in the context of a universality proof for a quantum gate set. I have a finite set $S$ of four-by-four special unitary matrices, and I want to know what group of matrices $G$ is densely generated by taking finite products from $S$. I can already show that $G$ is continuous, and the representation of $G$ as special unitary matrices over $\mathbb{C}^4$ is irreducible. – Adam Bouland Jan 10 '13 at 18:03

Dynkin published a paper in the fifties, where he determines the maximal connected subgroups of SU(n). Dynkin, E. B. Maximal subgroups of the classical groups. (Russian) Trudy Moskov. Mat. Obšč. 1, (1952). 39–166.

Maximal non connected subgroups has been computed by Seitz in the nineties. Liebeck, Martin W.; Seitz, Gary M. On the subgroup structure of classical groups. Invent. Math. 134 (1998), no. 2, 427–453. (Reviewe

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Thank you for the references. These classify the maximal subgroups of SU(n). However in my application I have no reason to believe the subgroup generated by the quantum gate set is maximal. Do you know of any references listing all subgroups of SU(4), rather than only the maximal subgroups? – Adam Bouland Feb 19 '13 at 19:08

Complementing the answer of Jorge Vargas, there is a more recent paper on the classification of maximal (non-necessarily connected) subgroups of compact Lie groups by F. Antoneli, M. Forger & P.A. Gaviria: Maximal Subgroups of Compact Lie Groups; to appear in J. Lie Theory. I think there are explicit lists for the classical groups.

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Thank you for the reference - this does give an explicit list of maximal subgroups of SU(n). As mentioned in my comment on Jorge's answer, in my application I have no reason to believe the subgroup is maximal. Do you know of any references listing all subgroups of SU(4), rather than only the maximal subgroups? – Adam Bouland Feb 19 '13 at 19:10
Adam: see, you do this just by iteration. First list the maximal subgroups of $SU_n$. Then, for each one in this list, you repeat the procedure and list all of the maximal subgroups of it and so on. Eventually you will have all the subgroups of $SU_n$ listed. – Claudio Gorodski Feb 19 '13 at 19:38

Hola Adam in the papers of Seitz it should be a complete list of the subgroups of SU_4, best regards jorge

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