 Solution

The weak charged current creates a left-handed neutrino and a right-handed outgoing positron in the massless limit. In the parent's rest frame their momenta are opposite, so these two fixed helicities give total angular-momentum projection of magnitude one along their flight axis. A spin-one W boson can supply that projection, and its leptonic width remains nonzero:A spin-zero pion cannot produce the same helicity configuration. The forbidden component is rescued only by a charged-lepton mass insertion that reverses its helicity, giving helicity suppression.

This can also be seen directly from the pion decay constant. Lorentz symmetry gives a vacuum-to-pion axial current matrix element proportional to $f_{\pi }{p}_{\pi }^{\mu }$. Contracting $p_{\pi }=k_e+k_{\nu }$ with the leptonic current givesusing the massless neutrino equation and $(/k_e+m_e)v_e=0$. Hence ${\mathcal{M}}_{\pi }\propto m_e$ and ${\Gamma }_{\pi }\propto m_e^2\to 0$. The W boson polarization vector is not constrained to be proportional to the total momentum, so its amplitude has no analogous compulsory mass factor.