What Makes a $W$-Boson a Gauge Boson?

I understand that in beta decay, a quark changes its flavor and emits a $W$-Boson, and this $W$ (is the W Boson virtual, by the way?) quickly decays into an electron and an electron antineutrino or positron and neutrino. What makes the $W$-Boson a gauge boson? I don't see any fundamental force being exchanged, and the W seems to act as an intermediate particle.

• In $\beta$-decay we have $n\rightarrow p + \overline\nu_e + e^-$, in QFT (Feynman diagrams) the anti-neutrino is considered as an incoming (time backwards running, because it's an antiparticle) neutrino which upon absorption of the $W^-$ turns into an electron. Gauge freedom expresses the observation that potentials (electric magnetic e.g.) can be varied to some extent without changing the fields. If you have further questions, please create another post on SE. – Frederic Thomas Jan 27 '18 at 21:39
• @Indigo2003 Before Feynman diagrams, there were time-ordered diagrams, where one diagram has the $q$ emit a W that then decayed, while another had a W spawn a virtual $e\nu$ pair and then get absorbed by the quark. All this was possible because energy wasn't conserved at vertices (uncertainty principle). Feynman unified these into 1 diagram that conserved 4-momentum at all vertices, and here the qW vertex and We$\nu$ vertex have space like separation (the virtual W has negative invariant mass squared). The point: don't fixate on time ordering, it's just an "interaction". – JEB Feb 14 '18 at 15:21