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Why Is Abelian Gauge Theory So Special?

What is a “free” non-Abelian Yang-Mill's theory?(coordinates) Invariance/Covariance of Chern-Simons theory and Yang-Mills theoryEquations of motion for the Yang-Mills $SU(2)$ theoryEinstein-Yang-Mills ConnectionsLattice gauge theory + Chern-Simons TheoryHodge star operator in Yang-Mills theory and derivation of the YM equationsInterpretation of the field strength tensor in Yang-Mills TheoryMistake or Rewriting of Yang-Mills in NakaharaClassical Yang-Mills equation of motion with both electric and magnetic sources?Global Part of Non-Abelian Gauge Transformation

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1

1

$begingroup$

I have a perhaps stupid question about Maxwell equations.

Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as

$$Dstar F=star J tag1.1$$

$$DF=0 tag1.2$$

Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.

If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.

However, as we all know, the Maxwell equations are

$$dstar F=star J$$

$$dF=ddA=0$$

Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?

differential-geometry gauge-theory maxwell-equations yang-mills

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asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

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1

1

$begingroup$

I have a perhaps stupid question about Maxwell equations.

Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as

$$Dstar F=star J tag1.1$$

$$DF=0 tag1.2$$

Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.

If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.

However, as we all know, the Maxwell equations are

$$dstar F=star J$$

$$dF=ddA=0$$

Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?

differential-geometry gauge-theory maxwell-equations yang-mills

share|cite|improve this question

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

72422 silver badges1313 bronze badges

$endgroup$

add a comment | 

$begingroup$

I have a perhaps stupid question about Maxwell equations.

Let $G$ be a generic Lie group. We consider a $G$-gauge theory. Let $A$ be the associated connection $1$-form, and $F=dA+Awedge A$ be the corresponding curvature tensor. The Yang-Mills equations can be written as

$$Dstar F=star J tag1.1$$

$$DF=0 tag1.2$$

Equation (1.1) follows from variation of the Lagrangian, and equation (1.2) is the Bianchi identity, which holds universally.

If the gauge group $G$ is replaced by $U(1)$, then naively one would expect the Yang-Mills equations to stay the same except $F=dA$.

However, as we all know, the Maxwell equations are

$$dstar F=star J$$

$$dF=ddA=0$$

Is there any deep geometrical reasons why $U(1)$ gauge theory is so special that one has to replace the covariant differential operator $D$ by the de-Rham differential $d$?

differential-geometry gauge-theory maxwell-equations yang-mills

share|cite|improve this question

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

72422 silver badges1313 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

72422 silver badges1313 bronze badges

share|cite|improve this question

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

72422 silver badges1313 bronze badges

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

72422 silver badges1313 bronze badges

asked 9 hours ago

The Last Knight of Silk RoadThe Last Knight of Silk Road

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2 Answers
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Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).

For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.

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answered 8 hours ago

MannyCMannyC

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We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.

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answered 8 hours ago

ACuriousMind♦ACuriousMind

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1

$begingroup$

Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).

For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.

share|cite|improve this answer

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).

For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.

share|cite|improve this answer

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

Normally the $D$ would be the covariant derivative in the adjoint representation (because $F$ and $star F$ transform in the adjoint of the gauge group).

For abelian gauge theories the adjoint representation is the trivial one (the representation with charge zero). You can either see this because $f^abc = 0$ or because the photon is electrically neutral. Hence $D$ in the adjoint representation for an abelian gauge theory is just $d$.

share|cite|improve this answer

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

answered 8 hours ago

MannyCMannyC

1,56411 gold badge44 silver badges1515 bronze badges

3

$begingroup$

We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges

$endgroup$

add a comment | 

$begingroup$

We don't "replace" the gauge covariant derivative by the ordinary differential, it is simply that case that $DF = mathrmdF$, since the adjoint representation of $mathrmU(1)$ is trivial.

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges

answered 8 hours ago

ACuriousMind♦ACuriousMind

74.5k1818 gold badges137137 silver badges340340 bronze badges