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I am suffering from a question I encountered from the lecture notes of gauge theory by David Tong. The problem comes from page 67 on the gauge fixing in back-ground gauge method. In David Tong's lecture notes, the gauge field $A$ is decomposed into a background gauge field $\bar{A}$ plus a perturbation part $a$ (In David Tong's lecture notes, the perturbative part is denoted as $\delta A$, I would rather denote it as $a$, instead.). i.e.

$$A=\bar{A}+a$$

Then David Tong claimed that under a gauge transformation $U$, the two parts transform as

$$\bar{A}^{U}=U^{-1}dU+U^{-1}\bar{A}U,\quad\quad a^{U}=U^{-1}aU.$$

Or, infinitesimally,

$$\delta\bar{A}=\bar{D}\omega,\quad\quad \delta a=[a,\omega],$$

where $\bar{D}=d+[\bar{A},\,\,\,]$ is the covariant derivative with respect to the background gauge field $\bar{A}$.

My question is the following:

The above definition of gauge transformation seems different from the one used in Peskin. In the book "An Introduction to Quantum Field Theory" by Peskin, on page 534, Chapter 16, Peskin claims that the background gauge field $\bar{A}$ should be regarded as fixed, the Lagrangian has a local gauge redundancy implemented by transformations on $a$:

$$a^{U}=U^{-1}aU+U^{-1}\bar{D}U,$$

or infinitesimally,

$$\delta a=\bar{D}\omega+[a,\omega]\equiv D_{A}\omega.$$

On page 535, Peskin (equation 16.99) said that the transformation (16.99), which is equivalent to the "gauge transformation" in Tong's lecture notes, is just a local symmetry, which shouldn't be a gauge transformation at all.

Is David Tong wrong here?

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  • $\begingroup$ When using background fields, there are many possible choices for how gauge fields should transform. You only know how the total field transforms. You are free to choose how the different parts transform. Tong is using a different choice from Pesking -- they are both fine, and there are many other possible choices too. $\endgroup$ Commented Mar 2, 2019 at 15:43
  • $\begingroup$ Maybe post this as an answer? $\endgroup$
    – MannyC
    Commented Mar 2, 2019 at 15:57
  • $\begingroup$ @AccidentalFourierTransform Thank you for your answer. Would you answer another related question I encountered from Chern-Simons theory? The choice that Tong made here does not work for Chern-Simons theory. physics.stackexchange.com/q/419746/185558 $\endgroup$
    – Xenomorph
    Commented Mar 2, 2019 at 15:58

1 Answer 1

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Transformation of affine subvector spaces

Before I discuss the behaviour of gauge fields, let's start with a simpler example. Consider an affine sub vector space $$ B \subset \mathbb R^3 $$ Elements $a \in B$ transform under Galilean transformation $(v, R) \in \mathbb R^3 \rtimes SO(3)$ as $$ a \mapsto v + R a\,. $$ Once we fix a base point $a_0\in B$ we can write $B = a_0 + V$ where $V$ is a subvector space. I.e, any element $a \in B$ can be written as $a = a_0 + b$. If we write the transformed affine subvectorspace as $\tilde B = (v, R)\circ B =(v + R a_0)+ R V$, the transformation of $a$, $a_0$, and $b$ are given by \begin{align} a &\mapsto v + R a \\ a_0 &\mapsto v + R a_0\\ b &\mapsto R b\,. \tag{2} \end{align}

Of course, the choice of the base point $\tilde B = \tilde a_0 + R V$ is not unique. If we restrict to Galilean tranformations, such that $\tilde B = B$ we can either choose $\tilde a_0 := v + R a_0 \in \tilde B$ or $\tilde a_0 := R a_0 \tilde B$ as base point. In the former case, the transformation of $a$, $a_0$, and $b$ are given by $(2)$. In the latter case, the transformations are given by \begin{align} a &\mapsto v + R a \\ a_0 &\mapsto \tilde a_0 = R a_0\\ b &\mapsto v + R b\,. \tag{3} \end{align}

Note that the transformation of $a \in B$ is always the same.

Neither P&S nor Tong a wrong.

First things first: Both reference agree that the gauge transformation of the full gauge transformation reads as $$ A_\mu \mapsto U^{-1} A_\mu U + U^{-1} \partial_\mu U\,.\tag{4} $$ (Some authors also write the transformation in terms of $\tilde U := U^{-1}$.)

The $A_\mu$ do not live in a vector bundle but rather in an affine vector bundle. I.e., once you fix a base point $A^0_\mu$ any other connection can be written as $$ A_\mu = A^0_\mu + B_\mu $$ where $A^0_\mu$ transforms according to $(4)$ and $B_\mu$ tranforms in the adjoint representation, i.e., $$ B_\mu \mapsto U^{-1} B_\mu U\,. \tag{5} $$

As in the example with affine subvector spaces, instead of $\tilde A^0_\mu := U^{-1} A^0_\mu U + U^{-1} \partial_\mu U$, we can also choose $\tilde A^0_\mu := U^{-1} A^0_\mu U$ as base point of the transformed gauge connections.

  • In the former case $B_\mu$ transforms according to $(5)$, which is the transformation used by David Tong.

  • In the latter case $B_\mu$ transforms according to $$ \tilde B_\mu := U^{-1} B_\mu U + U^{-1} \partial_\mu U\,, $$ which is the transformation used by P&S.

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$
    – Xenomorph
    Commented Oct 5 at 10:17

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