Holomorphic and antiholomorphic representation for $U(2)$ equivalent? Consider the following two irreducible $U(2)$-representations:
$$V_1:=(\Bbb C^2)^*,\quad V_2:=\Bbb C^2,$$
where $U(2)$ acts as follows (with $v\in\Bbb C^2$):
$$(\rho_1(A)f)(v):=f(A^{-1}v),\quad \rho_2(A)v:=Av.$$
Let us describe the highest weights of $V_j$. Note that the complexified Lie algebra of $U(2)$ decomposes into
$$\mathfrak u(2)_\Bbb C=\Bbb C\operatorname{id}_2\oplus\ \mathfrak{sl}(2,\Bbb C).$$
Thus, the roots of $\mathfrak{u}(2)_\Bbb C$ arise from the roots of $\mathfrak{sl}(2,\Bbb C)$ by extending them by $0$ on the center $\Bbb C\operatorname{id}_2$. These are given by $\pm (e_1-e_2)$ with positive system $e_1-e_2$. The weights of $V_1$ resp. $V_2$ are given by $\{-e_1,-e_2\}$ resp. $\{e_1,e_2\}$ with highest weights $-e_2$ resp. $e_1$. However, we have $e_1+e_2=0$ so that the two weights coincide. By the theorem of the highest weight, $V_1$ and $V_2$ have to be isomorphic, but I do not think that they are. What am I missing? Thanks in advance for any hint.
 A: These are the defining representation $V = \mathbb{C}^2$ and its dual, and they are not isomorphic; you can very straightforwardly check that they have different characters by computing the trace of the action of a diagonal matrix $\left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right], |a| = |b| = 1$, which in the defining representation is $a + b$ but which in the dual is $a^{-1} + b^{-1} = \overline{a + b}$.
The direct sum decomposition you write down is not a direct sum of Lie algebras, and your very last claim about weights is false; where weights of $U(2)$ are concerned it is just not true that $e_1 + e_2 = 0$. What is true is that if you restrict the action to $SU(2)$ then the two representations become isomorphic, because that's when it becomes true that $e_1 + e_2 = 0$. The weight lattice of $U(2)$ has rank $2$ because the maximal torus $T = U(1) \times U(1)$ (the diagonal matrices) has rank $2$.
Formally, on the Lie group level we have a short exact sequence
$$1 \to SU(2) \to U(2) \xrightarrow{\det} U(1) \to 1$$
which can be split by the map sending $z \in U(1)$ to the diagonal matrix $\left[ \begin{array}{cc} z & 0 \\ 0 & 1 \end{array} \right]$. This exhibits $U(2)$ as a nontrivial semidirect product
$$U(2) \cong SU(2) \rtimes U(1)$$
where the action of $SU(2)$ on $U(1)$ is given by conjugation by the diagonal matrix above. We have a corresponding nontrivial semidirect product decomposition
$$\mathfrak{u}(2) \cong \mathfrak{su}(2) \rtimes \mathfrak{u}(1).$$
