What is meant by two vectors transforming in the same way under SU(2)?

[silmaril89]

This question comes from section 2.3 of 'Quantum Field Theory' by Lewis Ryder. The discussion is on the Lie Group SU(2). He discusses the transformations of vectors under SU(2). Here it goes:

consider the basic spinor \xi = \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix};
\xi \to U \xi,
\xi^\dagger \to \xi^\dagger U^\dagger.

Then he says, we see that \xi and \xi^\dagger transform in different ways, but we may use the unitarity of U to show that \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix} and \begin{pmatrix} - \xi_2^* \\ \xi_1^* \end{pmatrix} transform in the same way under SU(2).

My question is, what is meant by 'they transform in the same way'? And what is meant by saying that \xi and \xi^\dagger don't transform in the same way?

 

[vanhees71]

Define
\bar{\xi}=\epsilon \xi^*,
where
\epsilon_{12}=-\epsilon_{21}=1, \quad \epsilon_{11}=\epsilon_{22}=0
is the Levi-Civita symbol in two dimensions.

Now let
\xi'=U \xi
with an arbitrary U \in \mathrm{SU}(2). Then obviously
\bar{\xi}'=\epsilon \xi'^*=\epsilon U^* \xi^*.
But now for SU(2) matrices you have
(U^{\dagger} \epsilon U^*)_{jk}=U_{lj}^* \epsilon_{lm} U^*{mk} = \epsilon_{jk} \det(U^*)=\epsilon_{jk}.
Now \epsilon^{-1}=\epsilon^{t}=\epsilon^{\dagger}=-\epsilon. Thus we can write
\epsilon^{-1} U^{\dagger} \epsilon U^{*}=1 \; \Rightarrow \; U^{*}=\epsilon^{-1} U \epsilon=\epsilon U \epsilon^{-1}.
This means that \bar{\xi} transforms under SU(2) by a representation that is equivalent to the fundamental representation, which means that \bar{\xi} does not define a new representation compared to the fundamental representation. Thus, for SU(2), the fundamental and the conjugate complex fundamental representation are equivalent and thus from the point of view of reprsentation theory the same.