Gauge transformations and Covariant derivatives commute

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I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on fields on which the algebra is closed off-shell"

which was taken from section 11.2.1 (page 223) of Supergravity by Freedman and Van Proeyen. In this text the authors support this statement by proving the following $$delta(epsilon)mathcalD_muphi-epsilon^AmathcalD_muT_Aphi=0 tag1$$
where $epsilon^A(x)$ is the gauge parameter for the symmetry transformation generated by the operator $T_A$, $delta(epsilon)$ is a gauge transformation, $mathcalD_mu=partial_mu-B_mu^AT_A$ is the covariant derivative and $B_mu^A(x)$ is the gauge field corresponding to each gauged symmetry.

I have no problem in deriving the result $(1)$, but I don't understand why the result $(1)$ is equivalent to the statement in the yellow box. In particular, why is the statement in the yellow box not equivalent to $$delta(epsilon)mathcalD_muphi-mathcalD_mudelta(epsilon)phi=0~?$$

symmetry gauge-theory lie-algebra covariance

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edited Aug 30 at 9:08

asked Aug 30 at 8:30

NormalsNotFar

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up vote
4
down vote

favorite

1

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on fields on which the algebra is closed off-shell"

which was taken from section 11.2.1 (page 223) of Supergravity by Freedman and Van Proeyen. In this text the authors support this statement by proving the following $$delta(epsilon)mathcalD_muphi-epsilon^AmathcalD_muT_Aphi=0 tag1$$
where $epsilon^A(x)$ is the gauge parameter for the symmetry transformation generated by the operator $T_A$, $delta(epsilon)$ is a gauge transformation, $mathcalD_mu=partial_mu-B_mu^AT_A$ is the covariant derivative and $B_mu^A(x)$ is the gauge field corresponding to each gauged symmetry.

I have no problem in deriving the result $(1)$, but I don't understand why the result $(1)$ is equivalent to the statement in the yellow box. In particular, why is the statement in the yellow box not equivalent to $$delta(epsilon)mathcalD_muphi-mathcalD_mudelta(epsilon)phi=0~?$$

symmetry gauge-theory lie-algebra covariance

share|cite|improve this question

edited Aug 30 at 9:08

asked Aug 30 at 8:30

NormalsNotFar

174218

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on fields on which the algebra is closed off-shell"

which was taken from section 11.2.1 (page 223) of Supergravity by Freedman and Van Proeyen. In this text the authors support this statement by proving the following $$delta(epsilon)mathcalD_muphi-epsilon^AmathcalD_muT_Aphi=0 tag1$$
where $epsilon^A(x)$ is the gauge parameter for the symmetry transformation generated by the operator $T_A$, $delta(epsilon)$ is a gauge transformation, $mathcalD_mu=partial_mu-B_mu^AT_A$ is the covariant derivative and $B_mu^A(x)$ is the gauge field corresponding to each gauged symmetry.

I have no problem in deriving the result $(1)$, but I don't understand why the result $(1)$ is equivalent to the statement in the yellow box. In particular, why is the statement in the yellow box not equivalent to $$delta(epsilon)mathcalD_muphi-mathcalD_mudelta(epsilon)phi=0~?$$

symmetry gauge-theory lie-algebra covariance

share|cite|improve this question

edited Aug 30 at 9:08

asked Aug 30 at 8:30

NormalsNotFar

174218

I would like to understand the statement

"Gauge transformations and Covariant derivatives commute on fields on which the algebra is closed off-shell"

which was taken from section 11.2.1 (page 223) of Supergravity by Freedman and Van Proeyen. In this text the authors support this statement by proving the following $$delta(epsilon)mathcalD_muphi-epsilon^AmathcalD_muT_Aphi=0 tag1$$
where $epsilon^A(x)$ is the gauge parameter for the symmetry transformation generated by the operator $T_A$, $delta(epsilon)$ is a gauge transformation, $mathcalD_mu=partial_mu-B_mu^AT_A$ is the covariant derivative and $B_mu^A(x)$ is the gauge field corresponding to each gauged symmetry.

I have no problem in deriving the result $(1)$, but I don't understand why the result $(1)$ is equivalent to the statement in the yellow box. In particular, why is the statement in the yellow box not equivalent to $$delta(epsilon)mathcalD_muphi-mathcalD_mudelta(epsilon)phi=0~?$$

symmetry gauge-theory lie-algebra covariance

share|cite|improve this question

edited Aug 30 at 9:08

asked Aug 30 at 8:30

NormalsNotFar

174218

share|cite|improve this question

share|cite|improve this question

edited Aug 30 at 9:08

edited Aug 30 at 9:08

edited Aug 30 at 9:08

asked Aug 30 at 8:30

NormalsNotFar

174218

asked Aug 30 at 8:30

NormalsNotFar

174218

asked Aug 30 at 8:30

NormalsNotFar

174218

174218

add a comment | 

add a comment | 

1 Answer
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accepted

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_mu$ that transforms to $D'_mu$ and a field $phi$ that transforms as $phi' = U(epsilon)phi$ under a finite gauge transformation $U(epsilon) = mathrme^epsilon^a T^a$, we require that the transform of $D_mu phi$ is equal to $U(epsilon) D_mu phi$. In formulae:
$$ D'_mu phi' = UD_muphi.$$

Infinitesimally, this means
$$ delta(epsilon)(D_mu phi) = (epsilon^a T^a) D_mu phi = epsilon^a D_mu(T^aphi),$$
since the infinitesimal version of $D'_mu phi'$ is $delta(epsilon) D_mu phi$ by definition and the infinitesimal action of $U$ on $D_muphi$ is $(epsilon^a T^a)D_mu phi)$. This is precisely what Freedman and Van Proeyen show.

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answered Aug 30 at 9:27

ACuriousMind♦

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_mu$ that transforms to $D'_mu$ and a field $phi$ that transforms as $phi' = U(epsilon)phi$ under a finite gauge transformation $U(epsilon) = mathrme^epsilon^a T^a$, we require that the transform of $D_mu phi$ is equal to $U(epsilon) D_mu phi$. In formulae:
$$ D'_mu phi' = UD_muphi.$$

Infinitesimally, this means
$$ delta(epsilon)(D_mu phi) = (epsilon^a T^a) D_mu phi = epsilon^a D_mu(T^aphi),$$
since the infinitesimal version of $D'_mu phi'$ is $delta(epsilon) D_mu phi$ by definition and the infinitesimal action of $U$ on $D_muphi$ is $(epsilon^a T^a)D_mu phi)$. This is precisely what Freedman and Van Proeyen show.

share|cite|improve this answer

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

add a comment | 

up vote
4
down vote



accepted

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_mu$ that transforms to $D'_mu$ and a field $phi$ that transforms as $phi' = U(epsilon)phi$ under a finite gauge transformation $U(epsilon) = mathrme^epsilon^a T^a$, we require that the transform of $D_mu phi$ is equal to $U(epsilon) D_mu phi$. In formulae:
$$ D'_mu phi' = UD_muphi.$$

Infinitesimally, this means
$$ delta(epsilon)(D_mu phi) = (epsilon^a T^a) D_mu phi = epsilon^a D_mu(T^aphi),$$
since the infinitesimal version of $D'_mu phi'$ is $delta(epsilon) D_mu phi$ by definition and the infinitesimal action of $U$ on $D_muphi$ is $(epsilon^a T^a)D_mu phi)$. This is precisely what Freedman and Van Proeyen show.

share|cite|improve this answer

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_mu$ that transforms to $D'_mu$ and a field $phi$ that transforms as $phi' = U(epsilon)phi$ under a finite gauge transformation $U(epsilon) = mathrme^epsilon^a T^a$, we require that the transform of $D_mu phi$ is equal to $U(epsilon) D_mu phi$. In formulae:
$$ D'_mu phi' = UD_muphi.$$

Infinitesimally, this means
$$ delta(epsilon)(D_mu phi) = (epsilon^a T^a) D_mu phi = epsilon^a D_mu(T^aphi),$$
since the infinitesimal version of $D'_mu phi'$ is $delta(epsilon) D_mu phi$ by definition and the infinitesimal action of $U$ on $D_muphi$ is $(epsilon^a T^a)D_mu phi)$. This is precisely what Freedman and Van Proeyen show.

share|cite|improve this answer

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

Saying that the covariant derivative and gauge transformations "commute" is a bit of a sloppy expression for the following precise statement:

Given a covariant derivative $D_mu$ that transforms to $D'_mu$ and a field $phi$ that transforms as $phi' = U(epsilon)phi$ under a finite gauge transformation $U(epsilon) = mathrme^epsilon^a T^a$, we require that the transform of $D_mu phi$ is equal to $U(epsilon) D_mu phi$. In formulae:
$$ D'_mu phi' = UD_muphi.$$

Infinitesimally, this means
$$ delta(epsilon)(D_mu phi) = (epsilon^a T^a) D_mu phi = epsilon^a D_mu(T^aphi),$$
since the infinitesimal version of $D'_mu phi'$ is $delta(epsilon) D_mu phi$ by definition and the infinitesimal action of $U$ on $D_muphi$ is $(epsilon^a T^a)D_mu phi)$. This is precisely what Freedman and Van Proeyen show.

share|cite|improve this answer

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

share|cite|improve this answer

share|cite|improve this answer

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

answered Aug 30 at 9:27

ACuriousMind♦

67.9k17118287

67.9k17118287

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add a comment | 

 

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