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I'm reading chapter 11.2 of the Cheng and Li textbook 'Gauge theory of elementary particle physics'. It says that $T_+$, $T_-$ and $Q$ do not form a closed algebra. In order to fix this problem the theory is extended to contain extra fermions.

As Georgi and Glashow did, a heavy lepton $E$ is added to form a triplet together with a electron neutrino and electron. My question is about the commutator relation of $T_+$ with $T_-$. Can someone explain how they get to this commutator relation? In the book they say this relation is straighforward but I can't see how they get it. See the pictures for clarity.

from Cheng and Li textbook

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    $\begingroup$ On the first line of (11.23), the symbol is supposed to be $E^\dagger$, and not $E^+$, in case that is what is confusing you. The operator destroys neutral particles and creates positive charged new leptons. $\endgroup$ – Cosmas Zachos Jun 1 at 0:18
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If you are reading this text and have advanced to that section, especially past its (11.20,21,22), you must be proficient in the standard QFT current algebra application of the Jordan map for fermions, and whence appreciate how $$ J_{ +}=\sqrt{2} \begin{pmatrix} 0&1&0\\ 0& 0& 1\\ 0&0&0&\end{pmatrix} ,\qquad J_+^\dagger =J_-, \quad J_3=\operatorname{diag} (1,0,1),\\ [J_+,J_-]=2J_3. $$

Your Jordan map fermion sandwich (11.23) then follows directly for each of your $P_L$ and $P_R$ fermion isotriplet of (ii); note using the fermion anticommutation law between (11.20) & (11.21) keeps the triplets apart, as cross terms die off by the orthogonal projection operators $P_L \& P_R$.

You then have inside the obvious respective integrals, $$ [T_+, T_+^\dagger]\mapsto 2\left ( (E,N\sin\alpha+\nu \cos\alpha ,e))^\dagger J_3 P_L(E,N\sin\alpha+\nu \cos\alpha ,e)^T + (E,N,e)^\dagger J_3 P_R(E,N,e)^T \right) = 2 (E^\dagger (P_L+P_R)E - e^\dagger (P_L+P_R)e)= 2 (E^\dagger E - e^\dagger e) \mapsto 2Q. $$ Note how the difference in the middle entry of the two isotriplets drops out by being multiplied by the vanishing eigenvalue of $J_3$!

Neat, huh?

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