Not every weight vector is highest

Given: The Axiom of Choice, the Lie algebra ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$ with its standard basis $e,f,h$ (The special linear Lie algebra sl_2), the Cartan subalgebra $\mathfrak{h}=\mathbb{C}h$, the root $\alpha$ with $\alpha (h)=2$, the positive system $\{\alpha \}$, and the standard two-dimensional module $V={\mathbb{C}}^2$ with basis $u_1=\left(\genfrac{}{}{0px}{}{1}{0}\right)$, $u_2=\left(\genfrac{}{}{0px}{}{0}{1}\right)$, on which $e,f,h$ act by their matrices $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

The Axiom of Choice is assumed; it enters through the root-space and highest-weight theory used below (The Axiom of Choice).

For the chosen root and positive system the subalgebra ${\mathfrak{n}}^{+}$ is the root space ${\mathfrak{g}}_{\alpha }$, which for ${\mathfrak{s}\mathfrak{l}}_2$ equals $\mathbb{C}e$ because $[h,e]=2e$ (Positive and negative nilpotent subalgebras and the Borel, The special linear Lie algebra sl_2).

A weight vector $v\in V_{\mu }$ is a highest weight vector exactly when it is nonzero and ${\mathfrak{n}}^{+}\cdot v=0$ (Highest-weight vectors and modules, Weight and weight space).

The matrix action on the basis is $h\cdot u_1=u_1$, $h\cdot u_2=-u_2$, $e\cdot u_1=0$, $e\cdot u_2=u_1$, so $u_1$ has weight $\alpha \mathrm{/}2$ and $u_2$ has weight $-\alpha \mathrm{/}2$, and $V$ is a finite-dimensional module (The special linear Lie algebra sl_2, Finite-dimensional representations of sl_2).