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MadMax
MadMax
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Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where and $d$ is exterior derivative., thus $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1} = F$ ) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1} = F$ ) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$ and $d$ is exterior derivative, thus $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1} = F$ ) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.
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MadMax
MadMax
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Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1}$$F \to g F g^{-1} = F$ ) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant (meaning $F$ transforms as $F \to g F g^{-1}$) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant/invariant (meaning $F$ transforms as $F \to g F g^{-1} = F$ ) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.
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MadMax
MadMax
  • 4.8k
  • 11
  • 38

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A$ is $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant (meaning $F$ transforms as $F \to g F g^{-1}$) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where the gauge field one-form $A$ is $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant.

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$, thus, $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\ = dg \wedge A g^{-1} + g dA g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1} \\ = g dA g^{-1} + (dg \wedge A g^{-1} - g A \wedge dg^{-1} + i dg \wedge dg^{-1}) $$ where $d$ is exterior derivative.

  • For Abelian case, the part in $(\dots)$ drops out, therefore $F=dA$ is covariant (meaning $F$ transforms as $F \to g F g^{-1}$) as in Electromagnetism.
  • For non-Abelian case, the part in $(\dots)$ is non-zero, therefore you need the extra $A\wedge A$ as in the non-Abelian $F = dA + A\wedge A$ to make $F$ covariant. This is THE major conclusion of Yang and Mills in their 1959 paper.
Source Link
MadMax
MadMax
  • 4.8k
  • 11
  • 38