When looking at the standard $3\times3$ representation of $SU(3)$, one immediately recognizes some subgroups isomorphic to $SU(2)$:

  • There is the subgroup acting on the first two elements of the vector, keeping the third one unchanged (and of course, the same for leaving any of the other elements intact). Indeed, it is not hard to see that for any unit vector, there's an $SU(2)$ subgroup acting only on the two vectors othogonal on it. It is also not hard to see that all those subgroups are conjugate to each other.
  • Since the $3\times3$ representation of $SU(2)$ also consists of special unitary matrices, it of course also corresponds a subgroup of $SU(3)$. The elements of this representation can e.g. be written as $\exp(\mathrm i\alpha L_z)\exp(\mathrm i\beta L_x)\exp(\mathrm i\gamma L_z)$ where $L_x=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}$ and $L_z=\operatorname{diag}(1,0,-1)$.

Now a natural question is whether this last subgroup is conjugate to the previous ones. I've come to the conclusion that it is not, with the following argumentation:

It of course suffices to check conjugacy with the subgroup acting on the first two elements (because conjugacy is an equivalence relation). Now the elements of that subgroup all commute with the subgroup of transformations of the form $\operatorname{diag}(\mathrm e^{\mathrm i\phi}, \mathrm e^{\mathrm i\phi}, \mathrm e^{-2\mathrm i\phi})$. However, the members of the $3\times3$ representation of $SU(2)$ only commute with multiples of the unit matrix (three of which are members of $SU(3)$). Therefore the two subgroups cannot be conjugate to each other.

Now my question is: Is my reasoning correct?

  • $\begingroup$ What is the $3\times3$ rep of $SU(2)$ in the second bullet point? $\endgroup$
    – anon
    Aug 2, 2012 at 19:16
  • $\begingroup$ @anon: I hope my edit answers your question. $\endgroup$
    – celtschk
    Aug 2, 2012 at 19:31

2 Answers 2


Your reasoning is correct: If the two subgroups were conjugate, then the centralizers were conjugate, too. This follows from the identity $C_G(H^g) = C_G(H)^g$, where $C_G(H)$ is the centralizer of $H$ in the group $G$. Since the centralizers have different cardinalities, they can not be conjugate.


Here's another way to see they're not conjugate. First, convince your self that conjugate subgroups are diffeomorphic.

In the $3\times 3$ representation of $SU(2)$, $-I$ acts trivially, so the induced map $SU(2)\rightarrow SU(3)$ actually factors through $SU(2)/-I = SO(3)$. This implies that the image of $SU(2)$ in $SU(3)$ is actually $SO(3)$, which is not even homotopy equivalent to $SU(2)$ (look at fundamental groups).


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