The following is an about a Left-Right Symmetric model.
$SU(2)\otimes SU(2)$ $(2\otimes 2=3\oplus 1)$ will generate a triplet, which in Left-Right Symmetric model is $$\vec{\Delta}=\begin{pmatrix}\delta_{1}\\ \delta_{2} \\ \delta_{3} \end{pmatrix}$$. The $SU(2)$ quantum number/charge on above is $$\begin{pmatrix} 1\\0\\-1 \end{pmatrix}$$. We can write this in $2\times 2$ representation as $\Delta=\frac{1}{2}\vec{\tau}\cdot \vec{\Delta}$, where $\vec{\tau}$ is vector made out of pauli matrices.
What is the complete mathematical explanation of $2\otimes 2=3\oplus 1$?
How to make a scalar triplet from a scalar doublet?
Answer. There is some procedure like $H^{T} i \tau_{2} \tau H$. (PS. I don't know exact form of this at this point of time).
What is the exact form? and why it is correct or how to make a cross product out of two doublet scalar fields?
How to calculate the charges of the components $\Delta$?
Answer. There is some mechanism to calculate the $T_{3L}+T_{3R}$ using commutator as $T_{3L}+T_{3R}=\frac{1}{2}[\tau_{3},\Delta]$. Or we can say the formula to calculate $Q$ is $Q=T_{3L}+T_{3R}+\frac{1}{2}(B-L)=\frac{1}{2}[\tau_{3},\Delta]+ \frac{1}{2}(B-L)$.
Main Question. What is the explanation to this? I tried to solve this but fails. There is also some charge distribution which is unexplained to me and canbe written as: $$\begin{pmatrix}\delta_{1}^{++}\\ \delta_{2}^{+}\\ \delta_{3}^{0}\\ \end{pmatrix}$$
Why is the $B-L$ charge for $\Delta_{L}$ is not $0$ like higgs doublet?
Answer. We need to break the $U(1)_{B-L}$ symmetry so we need $B-L$ charge in Triplets.
What is the other explanation to this? We can have this charge on Bi-doublet too, but we are not choosing that for some reason.
Answer. Bi-Doublet is responsible for symmetry breaking of Intermediate Salam-Weinberg Model gauge group which needs zero charge of $B-L$.
How come Bi-doublet is balancing left-handed and right-handed scalar fields i.e., We need a doublet in minimal Standard model now we should need another doublet instead we are defining a bidouble. Doesn't the left handed part will disturb the right handed fields, If not, how?
How to distinguish between $\Delta_{L}$ & $\Delta_{R}$?
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