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Does anyone know what the Lie subgroups of $\text{U}(3)$ are?

Thomas
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    This is no homework :) I am studying symmetry groups that occur in particle physics and I have unfortunately never had the chance to do a course on Lie groups. That is why I got stuck. I am pretty sure that $\text{U}(1)$ is a subgroup and other subgroups probably include low-dimensional unitary and orthogonal groups, right? And there might be other subgroups that I am not familiar with... – Thomas Nov 18 '13 at 21:10
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    This is almost the same as http://math.stackexchange.com/questions/497853 – Moishe Kohan Nov 18 '13 at 21:19
  • Thanks for pointing that out Studiosus. I already saw that post actually, but it is about $\text{SU}(3)$ rather than $\text{U}(3)$ and I do not know how to use those results for this problem. Btw, I am especially interested to know if there could possibly be a Lie subgroup of $\text{U}(3)$ that has dimension 4. – Thomas Nov 18 '13 at 21:27
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    Since $SU(3)$ is a subgroup of $U(3)$, its subgroups are subgroups of $U(3)$ too. One of them, according to the linked question, is $SU(2)\times U(1)$, which has dimension 4. – hmakholm left over Monica Nov 18 '13 at 21:55
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    There are many nonconjugate subgroups of $U(3)$ of dimension 4. First, there is $SO(3)\times S^1$ where the $S^1$ is the diagonal matrices in $U(3)$. There are also infinitely many nonconjugate subgroups of the form $SU(2)\times S^1$ or $U(2)$. They are given as the image of the map $f:SU(2)\times S^1\rightarrow U(3)$ given by $f(A,z) = diag(z^aA, z^b, z^c)$ where $\gcd(a,b,c) = 1$. I haven't worked out the details, but I suspect there are infinitely many conjugacy classes of subgroups among these examples. – Jason DeVito - on hiatus Nov 18 '13 at 21:56

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