I was trying to define an intertwiner between the alternating square $\Lambda^2(\mathbb{C}^2)$ and $\mathbb{C}$ to show that if $\rho$ is the standard Lie group representation, then $\Lambda^2\rho$ is isomorphic to the determinant representation $\det: \textrm{GL}_2(\mathbb{C})>G \to \mathbb{C}^{\times}$.
My idea was to define $\varphi: \Lambda^2(\mathbb{C}^2) \to \mathbb{C}$ by
$$ \varphi(v \wedge v') = \det(v,v')$$ This seems to do the trick in all respects except one: I can't show that it is linear.
In general, we have
$$\varphi(v_1 \wedge v_1' + v_2 \wedge v_2') = \varphi((v_1 + v_2) \wedge (v_1' + v_2')) = \det(v_1+v_2, v_1' + v_2')$$ which can be expanded into
$$\det(v_1,v_1')+ \det(v_2,v_2') + \det(v_1,v_2') + \det(v_2,v_1')\ .$$
What we want to show, in order to demonstrate linearity, is that this in fact equals $$\det(v_1,v_1') + \det(v_2,v_2')\ ,$$ i.e. that \begin{equation} \det(v_1,v_2') + \det(v_2,v_1') = 0\ . \tag{$\dagger$} \end{equation}
This clearly does not hold in general, but hinges upon the relation $v \wedge v' = - v' \wedge v$.
Whichever way I turn the signs and try to expand $\det(v_1+v_2, v_1' + v_2')$, I can't seem to deduce ($\dagger$). I have hence resorted to
Many thanks to Jyrki Lahtonen for his help. Using his suggestions, we here show that $\varphi$ is indeed linear.
As the statement is trivial for $v_1$ and $v_1'$ linearly dependent, we assume they are linearly independent and hence span $\mathbb{C}^2$.
Thus there exist $a,b,c,d \in \mathbb{C}$ such that $v_2 = av_1 + bv_1'$ and $v_2' = cv_1 + dv_1'$. We hence have:
$$v_2 \wedge v_2' = (av_1 + bv_1') \wedge (cv_1 + dv_1') = (ad-bc)(v_1 \wedge v_1')$$
And thus we conclude that
$$\varphi(v_1 \wedge v_1' + v_2 \wedge v_2') = (1+ad-bc)\det(v_1 , v_1') = \det(v_1,v_1') + \det(v_2,v_2')$$
Thereby proving the linearity of $\varphi$.