Do electrons have third component of isospin? From what I've read, it seems like there are two types of isospin: the strong and the weak. I guess this means that leptons have no strong isospin, but they do have weak isospin. Am I right?
I am a bit confused because my particle physics lecturer mentioned that the weak interaction does not conserve the third component of isospin. But he did not say that he was referring to the strong one specifically. As an example, he used beta decay, where from -1/2 (neutron) it goes to 1/2 (proton) (He assumed that the electron and the antineutrino have 0). I guess he was speaking about the strong isospin, right?
Since, although I'm struggling to find somewhere where this is stated (I would very much appreciate it if someone could link me to a source), I had written in a loose sheet of paper long ago that the electron, the muon and the tau have -1/2 value of the third component of the weak isospin. And the corresponding neutrinos have +1/2. This means that beta decay conserves the third component of weak isospin, since -1/2 -> 1/2 -1/2 -1/2 = -1/2.
Could someone tidy up my loose ideas into a more stable general statement of the above?
 A: The word "isospin" is overloaded. 


*

*Strong isospin: From what I've seen, the word "isospin" is most often used in this sense. It refers to the approximate symmetry between up and down quarks, which would be an exact symmetry if the electromagnetic and weak interactions were ignored and if their mass terms were equal. This isospin is an approximate global symmetry (not a gauge symmetry) of QCD. I don't know if the term "strong isospin" is standard, but I'll follow the OP's lead and use it here to help distinguish it from the other usage of the word "isospin"...

*Weak isospin: In the Standard Model (ignoring neutrino masses here), the $SU(2)_L\times U(1)_Y$ electroweak gauge group acts differently on the left-handed and right-handed parts of the spinor fields. Only the left-handed parts carry non-zero $SU(2)_L$ charge. The left-handed parts come in $SU(2)_L$ doublets. One example is 
$$
 \{\text{up quark},\text{ down quark}\}_L,
\tag{1}
$$
and another example is
$$
 \{\text{electron},\text{ electron neutrino}\}_L.
\tag{2}
$$
I'm using the subscript $L$ as a reminder that only the left-handed parts of these fields are involved in these doublets. In contrast, the right-handed parts are all singlets under $SU(2)_L$, with various charges ("hypercharges") under $U(1)_Y$. The term "weak isospin" is sometimes used in reference to how things are grouped according to the $SU(2)_L$ gauge symmetry.
The fact that both usages of the word "isospin" involve an $\{$up quark, down quark$\}$ pair can be confusing. To help underscore the difference between them, note that if the s-quark had the same mass as the u- and d-quarks, and if we again ignore the EM and weak interactions and mass-differences, then "strong isospin" would be an $SU(3)$ global symmetry rather than $SU(2)$. In contrast, the $SU(2)_L$ gauge interaction remains $SU(2)$ and acts only on the left-handed components of the fermion fields.
Beta decay does not conserve strong isospin: a neutron ($udd$) changes to a proton ($uud$), and the emerging electron and neutrino do not involve quarks. Strong isospin is an approximate symmetry of quark flavors, by definition, so the electron and neutrino have zero isospin by this definition. Strong isospin is only an approximate symmetry, ignoring the weak interaction (among other things).
In contrast, "weak isospin" is a gauge symmetry and is therefore exact by definition. (A gauge symmetry is just a redundancy in the relationship between field operators and observables.) Since an up quark (or a down quark, or an electron) is a superposition of both left- and right-handed components, it would not be strictly correct to talk about the "weak isospin" of an up quark (or down quark, or electron). For that to make sense, we need to specify whether we're talking about the left-handed or the right-handed component. Roughly speaking, the decay $udd\rightarrow uud + e + \nu$ involves only the left-handed component of the decaying $d$ quark. 
