In the language of differential forms, the field strength $F$ for the Yang-Mills theory is given by

$$F={\rm d}A+A\wedge A,$$

where $A$ is a matrix of one-forms.

In the language of Ricci calculus, the the field strength $F_{\mu\nu}^{a}$ for the Yang-Mills theory is given by


where $\mu$ and $\nu$ label the indices of the field strength and $a$ labels the $a^{\text{th}}$ component of the matrix $A.$

I understand that $F=\frac{1}{2}F_{\mu\nu}\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu}=(\partial_{\mu}\partial_{\nu}-\partial_{\nu}\partial_{\mu})\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu},$

but how do you write the $A\wedge A$ in tensor notation?


I understand that $F=\frac{1}{2}F_{\mu\nu}\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu}+\cdots=\frac{1}{2}(\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu})\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu}+\cdots,$

where the dots denote $A\wedge A$ in Ricci calculus.

How do you write the $A\wedge A$ in tensor notation?

  • 1
    $\begingroup$ Just in case, you made MANY misprints in your las equation $\endgroup$
    – OON
    Nov 1, 2016 at 7:40
  • $\begingroup$ You mean that I should have $F=\frac{1}{2}F_{\mu\nu}\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu}=\frac{1}{2}(\partial_{\mu}\ A_{\nu}-\partial_{\nu}\ A_{\mu})\; {\rm d}x^{\mu}\wedge {\rm d}x^{\nu}$? $\endgroup$ Nov 1, 2016 at 7:52
  • 1
    $\begingroup$ If you write that for $U(1)$, yes. If generally, also change $F$ to $dA$ $\endgroup$
    – OON
    Nov 1, 2016 at 7:58

1 Answer 1


$A$ should be decomposed with help of gauge group generators, $$A=A^aT^a$$ If we substitute, $$A\wedge A=\frac{1}{2}(A_\mu^a A_\nu^b-A_\nu^a A_\mu^b)T^a T^b dx^\mu\wedge dx^\nu = \frac{1}{2}A_\mu^a A_\nu^b [T^a,T^b] dx^\mu\wedge dx^\nu=\frac{1}{2}f^{abc}A_\mu^a A_\nu^b T^c dx^\mu\wedge dx^\nu$$

Then you combine it with $dA$ and decompose $F=F^a T^a$


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