# Question on gauge fields "acting on different representations"

-First of all, in the end of this question, unfortunatelly, I'll kind of request the solution of an exercise. But, this isn't for any kind of classroom evaluation. It is just that I kind of grasp the heuristic picture, but I don't know how to perform the technical calculations, due to my poor knowledge on lie algebra and lie group theory.

• I'm following the text book $$[1]$$.

• The question is written in section VII$$)$$, on equation $$(8)$$.

Now, standard model of particle physics is a (chiral) gauge theory. Therefore, it is a theory which is enconded in the language of Principal fiber bundles and associated fibre bundles. I will write a series of topics that explain better my problem here.

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

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

The map, $$\rho: G \to GL(V)$$ is the representation of the Lie group (gauge group) Therefore, using this map we produce matrices that acts on vectors of a vector space $$V$$.

Furthermore, given a representation $$\rho$$, it is possible to define its differential representation:

$$\rho_{*}: \mathfrak{g} \to \mathrm{End}(V),$$

where the operation $$\cdot_{*}$$ is the pushfoward.

The necessity of dealing with spinors as multiplets introduces a algebraic structure called: "Twisted Spinor Bundle" $$[1]$$:

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

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 $$[1]$$, this means that the whole twisted spinor bundles "slipts" in right part $$R$$ and left part $$L$$ as:

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

The spinor field notion is then defined with the one observes that first $$S$$ 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 $$\psi = \psi_R \oplus \psi_L$$

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

Due to this mathematical structure, and knowing that $$E_{L}$$ and $$E_{R}$$ depends on representations. There are two (possibly distinct) representations of $$G$$ on complex vector spaces $$V_{R}$$ and $$V_{L}$$, i.e., the whole formal bundle that I'm talking about is (with all structures explicitly showed):

$$TS_{\mathrm{chiral}} = S \otimes E = (S_{L} \otimes E_{L}) \oplus (S_{R} \otimes E_{R}) = (S_{L} \otimes (P\times_{\rho_{L}}V_{L})) \oplus (S_{R} \otimes (P\times_{\rho_{R}}V_{R})). \tag{6}$$

The map, $$\rho_{L}: G \to GL(V_{L})$$ is the representation of the Lie group (gauge group), that produce matrices that acts on vectors of a vector space $$V_{L}$$. Similarly, the map, $$\rho_{R}: G \to GL(V_{R})$$ is the representation of the Lie group (gauge group), that produce matrices that acts on vectors of a vector space $$V_{R}$$. Therefore, we have two distinct induced representations as well: $$\rho_{L*}: \mathfrak{g} \to \mathrm{End}(V_{L})$$ and $$\rho_{R*}: \mathfrak{g} \to \mathrm{End}(V_{R})$$.

## 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

Explicitly, I have the following data:

• $$G = SU(2)_{L} \times U(1)_{Y}$$
• $$\mathfrak{g} = \mathfrak{su}(2)_{L} \oplus \mathfrak{u}(1)_{Y}$$
• $$V_{L} = \mathbb{C}_{L}^2 \otimes \mathbb{C}_{Y}$$
• $$V_{R} = \mathbb{C}_{L} \otimes \mathbb{C}_{Y}$$

The subscripts are nothing but labels, the $$\mathbb{C}$$'s are your standard linear algebra complex vector spaces. Also, for your convenience:

• $$\rho_{L*}:\mathfrak{su}(2)_{L} \oplus \mathfrak{u}(1)_{Y} \to \mathrm{End}(\mathbb{C}_{L}^2 \otimes \mathbb{C}_{Y})$$
• $$\rho_{R*}:\mathfrak{su}(2)_{L} \oplus \mathfrak{u}(1)_{Y} \to \mathrm{End}(\mathbb{C}_{L} \otimes \mathbb{C}_{Y})$$

So my question is: how can I show the explicit calculation (i.e. could you please write the calculation) that:

$$\rho_{R*}(A_{s}(X)) = \rho_{R*}(W_{s}(X)+B_{s}(X)) = \rho_{R*}(W_{s}(X))+\rho_{R*}(B_{s}(X)) = \rho_{R*}(B_{s}(X))? \tag{8}$$

There are three different, but equivalent, ways to ask question:

$$1)$$ Exercise $$8.11.8$$ of $$[1]$$, page $$525$$.

$$2)$$ What happens, explicitly, with $$\rho_{R*}(W_{s}(X))$$? I mean, this term seems to simply "vanish".

$$3)$$ In physics (local chart) language: concerning the gauge group of Salam-Weinberg theory, how can I show that:

$$\rho_{L*}(A_{s}(X)) = - \frac{ig}{2}W^{a}_{\mu}\sigma_{a} - \frac{ig'}{2}B_{\mu} \tag{9}$$

and

$$\rho_{R*}(A_{s}(X))= - \frac{ig'}{2}B_{\mu}?\tag{10}$$

Where $$\sigma_{a}$$ are the Pauli Matrices.

$$[1]$$ Mark J.D. Hamilton Mathematical Gauge Theory, Springer, 2017.

To answer via the second suggestion, $$\rho_{R^*}$$ is $$0$$ on $$\mathfrak{su}(2)$$ because it is effectively just the trivial action on $$\mathbb{C}$$: $$\rho_{R^*}(W+B)(X\otimes Y) = \rho_1(W)X \otimes Y + X\otimes \rho_2(B)Y,$$ where $$\rho_1, \rho_2$$ are the actions of $$\mathfrak{su}(2)$$ on $$\mathbb{C}_L$$ and $$\mathfrak{u}(1)$$ on $$\mathbb{C}_Y$$ respectively. (This is simply the way that any direct sum of Lie algebras acts on a tensor product of their representations)

There are many possibilities for the action of $$\rho_2$$ (I assume one is picked implictly earlier on) but $$\rho_1$$ must be trivial as $$\mathfrak{su}(2)$$ has no non-trivial representations on a 1-dimensional space (the same is true of any semisimple Lie algebra).

Note: this is entirely based on the representation theory of Lie algebras (especially semisimple ones) and doesn't engage with the larger context you have mentioned.

Edit: I will admit, it is unclear to me why the constructions are not symmetric and for example why $$V_L$$ and $$V_R$$ have different dimensions. The same fact will not hold for $$\rho_{L^*}$$ as defined since we cannot then assume the corresponding $$\rho_1$$ is trivial. Perhaps there is some motivating physics reason for making these choices but, the way you have described it, it feels like $$S \otimes E$$ should split into $$S_R\otimes E \oplus S_L\otimes E$$ and I don't see where $$E_R, E_L$$ come from

• The group $SU(2)_{L} \times U(1)_{Y}$ is the way it is due to the physics of Electroweak unification. In this theory, a chiral theory, the electron is "divided" into "electron left $e_{L}$ and electron right $e_{R}$". Also, the "electron companion" the neutrino is divided into "Neutrino left $\nu_{e_{L}}$ and neutrino right $\nu_{e_{R}}$. Sep 11, 2022 at 7:58
• But, we didn't observe the $\nu_{e_{R}}$; then the whole theory had to be constructed with doublets of $SU(2)_{L}$: $$\psi_{L} = \begin{pmatrix}\nu_{e_{L}} \\ e_{L} \end{pmatrix}$$ and singlets of $U(1)_{Y}$: $$\psi_{R} = e_{R}$$. Sep 11, 2022 at 8:05
• Also, keep in mind that the group of my question is suitable for a high energy scale when electromagnetism and the weak force are "united". The universe have passed this phase and now weak force and electromagnetism are different. Furthermore, the group of electromagnetism is $U(1)_{\mathrm{em}}$ and therefore do not requires chiral bundles: the electromagnetism do not "sees" the difference between left or righ particles. The field $B$ acts on both $e_{L}$ and $e_{R}$, because after a phenomenon called symmetry breaking, we recover the electromagnetism out from the group of this question. Sep 11, 2022 at 8:16
• Then I stand by my main answer. I think then that the statement $S\otimes E = S_L\otimes E_L \oplus S_R\otimes E_R$ is incorrect/misleading mathematically. instead you are finding a subbundle of $S\otimes E$ via symmetry breaking. I thought that involved a reduction of some principal bundle (maybe that is to $U(1)$ here) but my actual physics knowledge here is very limited Sep 11, 2022 at 8:31
• I think your answer and edits are the very thing that I was looking for, actually. I was looking for some calculation/mathematical occurence that showed me some sort of vanishing of $\rho_{R*}(W_{s}(X))$. This occurs because the map $\rho_{R*}$ goes to $0$, since we are looking for $1$-dimensional representation of $\mathfrak{su}(2)_{L}$. Sep 11, 2022 at 8:37