How to deal with unusual factor-group?

[tuu-tikki]

Homework Statement

Let symmetry group SU(3) be broken by two Higgs fields in adjoint representation, so as SU(3) is transitive on vacuum manifold M, and the fields in one of vacuums are diagonal:
$$v_1 \cdot$$ diag(1,1,-2); $$v_1 \cdot$$ diag(1,-1,0). Find the homotopy groups of vacuum manifold $$\pi_2$$ (M); $$\pi_1$$ (M).

Homework Equations

Theorem. M=G/H, if action G is transitive on M, H - stationary subgroup. Stationary subgroup consists of all the elements h$$\in$$ G which don't change the element of M: F(h) m = m, m$$\in$$ M. There also is a theorem that the resulting subgroup H doesn't depend on the chosen element m.

The Attempt at a Solution

So, to use the theorem, we have to find stationary subgroup H, and the manifold M would be isomorphic to SU(3)/H. One element of M (we'll call it m) is given right in the problem statement. So, we should find all h that F(h) m = m. The fields do transform on the adjoint representation of SU(3), so it is $$h m h^{-1} = m$$ which leads us to equation [h,m]=0, [,] is commutator. For the given in the problem statement vacuum m (two expressions for fields) one can find two constraints on all the elements h. In addition to this, elements $$h \in H$$ are subgroups of SU(3), so, if we take it into account, we'll get such an expression for subgroup H:
$$\begin{pmatrix}<br /> e^{i\phi} & 0 &0 \\ <br /> 0 & e^{i\psi}& 0\\ <br /> 0 & 0 & e^{-i\phi-i\psi}<br /> \end{pmatrix}$$

It seems to me that group of such matrices is isomorphic to U(1)xU(1). So, the desired manifold is M= SU(3)/(U(1)xU(1)). Firstly, I want to ask about some site or book where many of the relations using the factor groups are gathered (like $$SU(2)/Z_2 = SO(3)$$). They really make life easier sometimes; and, in particular, I want to know if SU(3)/(U(1)xU(1) is isomorphic to some more simple-written group. Moreover, I completely don't understand how to find homotopy groups like $$\pi_2$$ (SU(3)/(U(1)xU(1))). Give an idea, please. Probably SU(3)/(U(1)xU(1)) is homotopically equivalent to something easy?

Thank you.

 

[tuu-tikki]

The solution.

I found the answer to it by myself.
There's such a theorem that if there is a bundle (G, G/H, H), G is compact and simply-connected then $$\pi_2 (G/H) = \pi_1 (H)$$. Also if G is connected, $$\pi_1 (G/H) = \pi_0 (H)$$. So, the desired homotopy classes are $$\pi_2 (SU(3)/(U(1)\times U(1))) = \pi_1 (U(1) \times U(1)) = \pi_1 (U(1)) \times \pi_1 (U(1)) = Z \times Z ; \pi_1 (SU(3)/(U(1)\times U(1))) = \pi_0 (U(1) \times U(1)) = 0$$.