• I've made two other questions $[1]$,$[2]$ 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 $[3]$ 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" $[3]$:

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

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 $[3]$, 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:

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


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

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

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

$[3]$ Mark J.D. Hamilton Mathematical Gauge Theory, Springer, 2017.


1 Answer 1


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.

  • $\begingroup$ I see. But, why the covariant derivative acts differently in left particles and right particles? In a pure mathematical sense. I'm trying to understand where is the part of the calculations that we see the field W "vanishing" when acting on a right particle. $\endgroup$
    – M.N.Raia
    Aug 12, 2022 at 20:49
  • $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Aug 12, 2022 at 20:51
  • $\begingroup$ 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, $\rho_{R}(A) =\rho_{R}(B)$)? $\endgroup$
    – M.N.Raia
    Aug 12, 2022 at 20:59
  • $\begingroup$ @M.N.Raia I'm afraid I do not understand the question. $\endgroup$
    – ACuriousMind
    Aug 12, 2022 at 21:01
  • $\begingroup$ 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, $\rho_{L}(A) = \rho_{L}(W) + \rho_{L}(B)$. $\endgroup$
    – M.N.Raia
    Aug 12, 2022 at 21:03

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