# Doubt on: $G = SU(2)_{L} \times U(1)_{Y}$ representations, the Chiral Spinor bundle and the "split" of covariant derivative for $G$

Firstly,

• I've made two other questions $$$$,$$$$ concerning the same situation, but I think that this one will clarify better what I'm trying to understand.

• I'm following the text book $$$$ and I have a poor knowledge of the physics of standard model and a rough introduction to clifford algebras and spinors.

• The question is written in section VII$$)$$.

## I) Connection $$1$$-$$\mathrm{forms}$$

Now, in Salam-Weinberg model, the gauge group of the theory is $$G = SU(2)_{L} \otimes U(1)_{Y}$$. The connection constructed in the principal bundle, the $$(SU(2)_{L} \otimes U(1)_{Y})$$-$$\mathrm{bundle}$$, is the connection $$1$$-$$\mathrm{form}$$:

$$A = W + B.\tag{1}$$

$$A$$ is not the electromagnetic gauge field, rather, the connection $$1$$-$$\mathrm{form}$$ of the principal bundle. $$W$$ is the weak gauge field and $$B$$ the hypercharge gauge field.

## II) Local Connection $$1$$-$$\mathrm{forms}$$

A local version of $$(1)$$ (the local gauge field) that "puts the gauge field on spacetime" is given by:

$$A_{s} = s^{*}A.\tag{2}$$

Where, $$s$$ is a section on the principal bundle (the local gauge choice), and the $$s^{*}$$ is the pullback of the section (when we apply this map on $$A$$, we bring the information of the gauge field for a region located on the base manifold $$\mathcal{M}$$). Furthermore, since our algebraic landscape deals with groups and lie groups, the action of $$A_{s}$$ on a vector field $$X \in T_{p}\mathcal{M}$$ lies on the lie algebra: $$A_{s}(X) \in \mathfrak{g}$$.

## III) Spinors and Multiplets

Now, the necessity of dealing with spinors as multiplets introduces a algebraic structure called: "Twisted Spinor Bundle" $$$$:

$$TS = S \otimes E = S \otimes (P\times_{\rho}V). \tag{3}$$

Where, S is the spinor bundle, and E the associated bundle (the $$P$$ is the principal bundle and $$\rho: G \to GL(V)$$ the representation).

The tensor structure $$(3)$$ tells us: "we have spinor fields in $$S$$ and the fact that we construct a tensor product with $$E$$ we construct the well-known multiplets $$\psi$$".

Actually, Twisted Spinor Bundles are also called Gauge Multiplet Spinor Bundles.

## IV) Covariant Derivatives 1

In $$(3)$$ we can construct the covariant derivative of the theory acting on multiplets (spinors) as:

$$D^{A}_{\mu}\psi = \partial_{\mu}\psi + \rho(A_{s}(X))\psi = \partial_{\mu}\psi - \frac{ig}{2}W^{a}_{\mu}\sigma_{a}\psi - \frac{ig'}{2}B_{\mu}\psi. \tag{4}$$

## V) Chirality

An important feature of the standard model is its chirality. Following $$$$, this means that the whole twisted spinor bundles "slipts" in right part $$R$$ and left part $$L$$ as:

$$S \otimes E = (S_{L} \otimes E_{L}) \oplus (S_{R} \otimes E_{R}). \tag{5}$$

Hence, the multiplet splits as:

$$\psi = \psi_{L} + \psi_{R}.\tag{6}$$

The structure $$(5)$$ is called "Twisted Chiral Spinor Bundle".

## VI) Covariant Derivatives 2

In same fashion, we can construct the covariant derivative of the theory acting on chiral multiplets (spinors) as:

$$D^{A}_{\mu}\psi= \partial_{\mu}\psi + \rho_{L}(A_{s}(X))\psi_{L} + \rho_{R}(A_{s}(X))\psi_{R} \implies$$

$$D^{A}_{\mu}\psi \equiv D^{A}_{\mu}(\psi_{L} + \psi_{R}) =\partial_{\mu}\psi_{L} + \rho_{L}(A_{s}(X))\psi_{L} + \partial_{\mu}\psi_{R} + \rho_{R}(A_{s}(X))\psi_{R}\tag{7}$$

## VII) My Question

Concerning the gauge group of Salam-Weinberg theory, how can I show that:

$$\rho_{L}(A_{s}(X)) = - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu} \tag{8}$$

and

$$\rho_{R}(A_{s}(X))= - \frac{ig'}{2}B_{\mu}?\tag{9}$$

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

$$$$ Doubt on the action of covariant derivative of $SU(2)_{L} \otimes U(1)_{Y}$ on a right-fermion

$$$$ Mark J.D. Hamilton Mathematical Gauge Theory, Springer, 2017.

There is nothing to show: The representations $$\rho_{L/R}$$ are part of the definition of the field $$\psi$$.
The idea that $$\psi$$ is part of some $$S\otimes E$$ for a fixed $$E$$ is wrong - the idea here is to observe first that $$S$$ splits as $$S_R\oplus S_L$$ and that we are hence free to define a field $$\psi_R$$ as a section of some $$S_R \otimes E_R$$ and a field $$\psi_L$$ as a section of some $$S_L \otimes E_L$$ where $$E_R$$ and $$E_L$$ have no a priori relation. Then we define $$\psi = \psi_R \oplus \psi_L$$. You choose $$E_R$$ so that your eq. (9) is true and $$E_L$$ so that your eq. (8) is true.
Unrelated nitpick: Stop using $$\otimes$$ to denote group direct products like $$\mathrm{SU}(2)\times \mathrm{U}(1)$$ - this is not a tensor product in the category of groups. See this answer by Qmechanic for more on this notational confusion.
• @M.N.Raia I don't understand what you mean. Part of the specification of the vector bundles $E_{L/R}$ is a representation $\rho_{R/L}$. You are free to choose a representation where $\rho_R(W) = 0$ (as long as that's a valid representation of the algebra, which it is in this case). The reason we choose different $\rho_L$ and $\rho_R$ in the Standard Model is because that's what matches experimental observation - there is no fundamental reason the SM would "have to be" chiral. Aug 12, 2022 at 20:51
• Well, so, when we act the $\rho_{R}(A) =\rho_{R}(W+B) = \rho_{R}(W)+\rho_{R}(B)$, plus the "experimental results" we need to state that $\rho_{R}(W) = 0$ (and therefore, $\rho_{R}(A) =\rho_{R}(B)$)? Aug 12, 2022 at 20:59
• You said that is possible to have $\rho_{R}(W) = 0$. So, the action of the representation on the whole local connection: $\rho_{R}(A)$ is $\rho_{R}(A) = \rho_{R}(B)$. And the action of $\rho_{L}(W) \neq 0$ and then, $\rho_{L}(A) = \rho_{L}(W) + \rho_{L}(B)$. Aug 12, 2022 at 21:03