# Doubt on $SU(2)_{L} \times U(1)_{Y}$ covariant derivative and its action on a fermion

## I) Introduction

### I.1)

The mathematical structure is quite clear: given a spacetime $$M$$, and a Lie group $$G$$ (the gauge group), we can construct the Principal bundle $$P^{G}_{M}$$. The connection $$1$$-form in $$P^{G}_{M}$$ introduces the gauge $$\mathcal{A}$$ field in physics.

Given the spacetime and the principal bundle, matter fields are introduced in another manifold called the associated bundle $$A_{P}$$. Inside this space, we craft the gauge covariant derivative:

$$D_{\mu} \phi = \partial_{\mu} \phi+\rho(\mathcal{A}_{s})\phi, \tag{1}$$

where, $$\rho$$ is the representation $$\rho: \mathfrak{g} \to GL(V)$$ of a given Lie group $$G$$; $$\mathfrak{g}$$ is the lie algebra.

Now, the mathematics is imprecise, for example the $$\mathcal{A}$$ inside $$\rho(\mathcal{A})$$ in $$(1)$$, isn't just the connection, but rather the local connection $$1$$-form $$\mathcal{A}_{s}$$. But my point is: given the principal bundle and the associated bundle the connection is "just one".

### I.2)

In Weinberg-Salam model, the gauge group is $$SU(2)_{L} \times U(1)_{Y}$$. This group have this form because we didn't "measure" the right electron neutrino $$\nu_{{e}_{R}}$$. So, we need a space to acomodate a doublet:

$$\psi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ \end{pmatrix},$$

together with a singlet:

$$\phi = e_{R}.$$

So, the Weinberg-Salam-Glashow model deals with "triplets":

$$\xi = \begin{pmatrix} \nu_{e_{L}} \\ e_{L} \\ e_{R} \end{pmatrix} \implies \Psi = \begin{pmatrix} \psi \\ \phi \end{pmatrix} \equiv \begin{pmatrix} L \\ R \end{pmatrix}$$

## I.3)

The covariant derivative for Weinberg-Salam-Glashow model is:

$$D_{\mu} = \partial_{\mu} + \rho(\mathcal{A}) = \partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu} \tag{2}.$$

## II) Question

Most texts books says that when you act the covariant derivative in the triplet, $$D_{\mu}\Psi$$, some internal calculation "breaks" the single $$D_{\mu}$$ in $$(2)$$ into two different ones:

$$D_{\mu}\Psi =\Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)\Psi \implies \tag{3}$$

$$D_{\mu}L = \Big(\partial_{\mu} - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu}\Big)L \tag{4}$$

$$D_{\mu}R = \Big(\partial_{\mu} - \frac{ig'}{2}B_{\mu}\Big)R \tag{5}$$

My question is: what is the calculation that breakes $$(3)$$ into $$(4)$$ and $$(5)$$?

## III) My attempt

$$D_{\mu}\Psi = \Biggr(\partial_{\mu} \otimes \mathbb{I}_{3\times3} - \frac{ig}{2}\begin{pmatrix} W^{3}_{\mu} & W^{1}_{\mu} - iW^{2}_{\mu} & 0\\ W^{1}_{\mu} + iW^{2}_{\mu} & W^{3}_{\mu} & 0 \\ 0 & 0 & 0 \end{pmatrix} - \frac{ig'}{2}\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & B_{\mu} \end{pmatrix}\Biggr)\Psi$$

this matrix expression breaks the $$D_{\mu}$$ into different action for left particles and right particles, but is clear that won't give us the right set of equations.

So, I shall ask again: what is the calculation that breakes $$(3)$$ into $$(4)$$ and $$(5)$$? Please, consider to write a step-by-step calculation answer.

• You wrote a bland direct sum; what calculation beyond that are you seeking? Jul 18, 2022 at 16:52
• @CosmasZachos I want to know why the covariant derivative acts differently for left particles and right particles. But, I want to know the calculation for that. Jul 18, 2022 at 17:02
• "Why"? It's a just so statement: a fact. Jul 18, 2022 at 17:04
• @CosmasZachos I agree with you: experimental physics just tell us that. But exists a formal mathematical calculation where in some place the action of $W$ field is zero for a right particle. Jul 18, 2022 at 17:08
• No, the asymmetric isolation of right-handed particles is a mysterious experimental fact. Feynman called the structure "cockeyed" because of it, and ribbed Weinberg about it. He was unreasonable. Jul 18, 2022 at 18:26