# Representation theory of $SU(2)$: Left-Right symmetric models

My question concerns the following excerpt of the article A Fermionic bi-Doublet Effective Field Theory for Dark Matter: I do not understand how the action of $$SU_L(2) \times SU_R(2)$$ on $$Mat^{2 \times 2}(\mathbb{C})$$ is supposed to be $$(U_L,U_R).\mathscr{D} = U_L \mathscr{D} U_R$$ as this is neither a left nor a right action. I would instead expect the correct action to be something like $$(U_L,U_R).\mathscr{D} = U_L \mathscr{D} U_R^\dagger$$, which would be a left action. This would also be more consistent with the electric charges assigned to the components of $$\mathscr{D}$$ because then the third component of the spin $$I_3^L + I_3^R$$ would be given by:

$$S_3 \mathscr{D} = -i \partial_t|_{t=0} [(\exp(\frac{i}{2} \sigma_3) , \exp(\frac{i}{2} \sigma_3) ).\mathscr{D}] = \frac{1}{2} [\sigma_3,\mathscr{D}]$$

which leads to $$\left[ {\begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} } \right]$$ and $$\left[ {\begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} } \right]$$ having $$z$$-spin 0, $$\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$$ having spin 1 and $$\left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} } \right]$$ having spin -1 as indicated by the superscripts in equation 2.5 (see screen shot above).

As I am quite new to the whole topic, I am uncertain about the validity of my arguments and would be grateful if you could either confirm them or explain to me the notation used in the paper quoted.

• Have you studied and understood ref ? Sep 25, 2021 at 14:54
• Isn't ref  postulating exactly the transformation behaviour that I proposed i.e. $(U_L, U_R).D = U_L D U_R^\dagger$ instead of $(U_L, U_R).D = U_L D U_R$? Sep 25, 2021 at 18:38
• ? Are you talking about (3.3) of this? what is D? What are A,B? Above, $\mathscr{D}$ (2.5) has the same transformation properties and charge assignments as (2.3). Sep 25, 2021 at 19:03
• Sorry for the misleading notation, I removed the $A$,$B$ and $D$ now and replaced them with a notation more consistent with the one used in your answer. Sep 25, 2021 at 19:26

I cannot decipher your peculiar notes of your understanding, but the paper appears straightforward.

Defining $$U_L=\exp (i\vec \tau \cdot \vec \theta),\qquad U_R=\exp (i\vec \tau \cdot \vec \phi)$$, you may observe $$\mathcal{ H}= \begin{bmatrix} H^+\\ H^0 \\ \end{bmatrix}, ~~~\tilde{\mathcal{ H}}=-i\tau_2 \mathcal{ H}^*= \begin{bmatrix}- H^0\\ H^- \\ \end{bmatrix} ~~~\leadsto \\ \mathscr{H}=[\tilde{\mathcal{ H}} , \mathcal{ H} ],$$ so that, $$\mathcal{ H} \to U_L \mathcal{ H} , \qquad \tilde{\mathcal{ H}} \to U_L \tilde{\mathcal{ H}} ~~~\implies \\ \mathscr{H}\to U_L \mathscr{H}.$$

But also $$[ -H^{0~*}, H^+] \to [-H^{0~*}, H^+] U_R, \qquad [ H^{ -}, H^0]\to [ H^{ -}, H^0] U_R \implies \\ \mathscr{H}\to \mathscr{H} U_R.$$

You are told $$\mathscr{D}$$ transforms exactly the same way, so that $$\operatorname{Tr} \mathscr{H}^\dagger \mathscr{D}= \operatorname{Tr} \mathscr{D}\mathscr{H}^\dagger$$ is an invariant under both $$U_L$$ and $$U_R$$.

As you have discovered, people switch conventions at the drop of a hat, and so the paper you are discussing has very different ones (check the v.e.v. !) from those of the ur-reference . (Frankly, Longhitano's paper is superior.) So the right action need not have a dagger on it, since you never take the vector diagonal product of the the two isospins.

To help you review the conventional picture, in the SM, for ultimately degenerate u,d quarks, you'd have the Yukawa terms, L-R invariant, present as $$[u_L^\dagger, d_L^\dagger]\mathscr{H} \begin{bmatrix} -u_R\\ d_R \\ \end{bmatrix}.$$

So the rightmost, R-spinor doublet transforms through $$U_R^\dagger$$, not $$U_R$$ !

Note each of the four terms of the above dot product is electrically natural.

• Thanks for your answer! The assignments of the isospin is still puzzling me though. For example, I would think that if you do not have the dagger for $U_R$, then $M = \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} } \right]$ should have $I_3^L$ eigenvalue of $1/2$ as $\frac{1}{2} \sigma_3 M = \frac{1}{2}$ but also $I_3^R$ eigenvalue of $1/2$ since $M \frac{1}{2} \sigma_3 = \frac{1}{2}$. But then the charge of the particle would be $1$ instead of $0$. Sep 25, 2021 at 18:47
• (if on the other hand we act on the right with $U_R^\dagger$ then the sign of $I_3^R$ is flipped, leading to results consistent with the assertion that $M$ corresponds to a state with charge $0$.) Sep 25, 2021 at 18:53
• As I suggested, I have no idea what your mysterious symbols and implications are. I am just fixing and parsing out the basic facts of the L-R construction. Sep 25, 2021 at 19:05
• Well my problem is that in the end $H^{+,0,-}$ are supposed to have charge $Q = I_3^R + I_3^L$ of $1,0$ and $-1$. I think that I understand how this arises in  for example in the construction of $\Phi$ (eq. 2.2). But my reasoning doesn't work in the construction that I asked about i.e. with the transformation $H \rightarrow U_L H U_R$ instead of $U_L H U_R^\dagger$. So if you could explain to me how the correct charges arise here or recommend some literature to me, this would help a lot! Sep 25, 2021 at 19:17
• Oh, that's a convention! θ and φ are completely independent, so my and the paper's φ, for simplicity, are the opposite of 's φ. You may stick a minus sign in the definitions above. L and R actions are characterized by the location of the unitary matrices, not their angles! Sep 25, 2021 at 19:27