Reduce $SU(3)$ fundamental representation into irreps of $SU(2)$ subalgebra I'm working on problems in Georgi's book on Lie algebras and can't figure out what I'm supposed to do on this problem (7B):
Show that $T_1$, $T_2$, and $T_3$ generate an $SU(2)$ subalgebra of $SU(3)$. Every representation of $SU(3)$ must also be a representation of the subalgebra. However, the irreducible representations of $SU(3)$ are not necessarily irreducible under the subalgebra. How does the representation generated by the Gell-Mann matrices transform under this subalgebra? That is, reduce, if necessary, the three dimensional representation into representations which are irreducible under the subalgebra and state which irreducible representations appear in the reduction. Then answer the same question for the adjoint representation of $SU(3)$.
It's clear to me why there are $SU(2)$ subalgebras of $SU(3)$, and I suppose if you will answer this you also know that without me explicitly writing the $T_a$s. What I don't understand is what he means by "transforming under the subalgebra," nor can I find an explanation online. What does "irreducible under the subalgebra" mean?
 A: A triplet v transforming under SU(3) means 
$$
\mathbf {v} \mapsto \exp (i\theta^a \lambda_a)~~ \mathbf {v}.
$$
So you see the third component of this complex 3-vector is left alone under action of the SU(2) subgroup (has been reduced out), generated by the SU(2) subalgebra $\{\lambda_1\equiv T_1,\lambda_2\equiv T_2,\lambda_3\equiv T_3\}$.  That is, for $\theta^a=(a,b,c,0,0,0,0,0)\equiv\vec\phi$, a 3-vector parameter, 
$$
\mathbf {v} \mapsto \exp (i\vec\phi \cdot \vec T)~~ \mathbf {v}.
$$
The upper two components, though, transform into each other, so they are irreducible under this restricted, SU(2) subgroup transformation.
So v is really a 2  ⊕ 1 , the singlet being the 3rd component.
Now the 8-vectors $ W^a {\mathbf v}^\dagger \lambda^a {\mathbf v} $ transform in the adjoint of SU(3), 
$$
 W^a {\mathbf v}^\dagger \lambda^a {\mathbf v} \mapsto  W^a {\mathbf v}^\dagger \!\exp (-i\theta^c \lambda_c)~~ \lambda^a \exp (i\theta^b \lambda_b){\mathbf v},
$$
so you see how the 3 * transforms: by the hermitian conjugate generators.
So, again, if you restrict the 8 λs to the three Ts of the SU(2) subalgebra, 
the adjoint octet transforms under SU(2) as
$$
 W^a {\mathbf v}^\dagger \lambda^a {\mathbf v} \mapsto  W^a {\mathbf v}^\dagger \!\exp (-i\vec\phi\cdot \vec T)~~ \lambda^a \exp (i\vec\phi\cdot \vec T){\mathbf v}.
$$
Now, consider the transformations among the 8 components of $W^a$.
Again, you see that the upper 3 components of $W^a$, corresponding to the subalgebra of the Ts , are mixed into each other by the SU(2) subalgebra, and so are an irreducible triplet. The eighth component, corresponds to $\lambda_8$, which commutes with the Ts, so it is a singlet under SU(2).
The other four components split into two conjugate SU(2) doublets,($\lambda_4 +i\lambda_5, \lambda_6+i\lambda_7$), and the $i\to -i$ combination, ("the K and $\bar K$" of the meson octet)  as you may observe by the transformation properties of $\{ \lambda_4,\lambda_5,\lambda_6,\lambda_7  \}$ under the SU(2)--their commutators with the Ts.  The other singlet of the nonet reduction your teacher wrote is ${\mathbf v}^\dagger {\mathbf v}$ which is also a singlet under SU(3), so it never entered the discussion to start with.  
