= Solution
The charged <pion> has <spin> zero. In its <rest frame>, the outgoing <muon> or <Positron> and the <neutrino> travel in opposite directions. With a nearly massless left-helicity <neutrino>, total <angular momentum> zero requires the positive charged <antilepton> to have negative <helicity>. The <weak charged current> instead creates a massless positive <antilepton> with positive <helicity>. A nonzero charged-lepton <mass> supplies the required opposite-helicity component.
Thus the <decay amplitude> is proportional to $m_\ell$: this is <helicity suppression>. Explicitly, contracting the pion's axial-current matrix element, proportional to its four-momentum $p_\pi^\mu$, with the leptonic current and applying the <Dirac equation> leaves a factor $m_\ell$. The <leptonic pseudoscalar decay width> is, in the $f_\pi$ convention,
$$
\Gamma(\pi^+\to\ell^+\nu_\ell)=
\frac{G_F^2|V_{ud}|^2f_\pi^2m_\pi m_\ell^2}{8\pi}
\left(1-\frac{m_\ell^2}{m_\pi^2}\right)^2.
$$
The <phase space> advantage of the lighter <Positron> cannot offset its much stronger <helicity suppression>. The common factors cancel in the <pion electron-to-muon decay ratio>:
$$
\boxed{\frac{\Gamma_e}{\Gamma_\mu}
=\frac{m_e^2}{m_\mu^2}
\left[\frac{1-m_e^2/m_\pi^2}{1-m_\mu^2/m_\pi^2}\right]^2
\simeq1.28\times10^{-4}.}
$$
This tree-level estimate explains why the muonic branch dominates by roughly four orders of magnitude. The paper's table gives an electronic <branching fraction> $1.230\times10^{-4}$; electromagnetic corrections change the simple ratio by a few percent without changing the mechanism.
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