# Gauge covariant derivative on form

Let $e$ be a one-form gauge field that belongs to the adjoint representation of the gauge group, that is SO(1,2). It is defined as $$e = e_{\alpha}^{A}T_Adx^{\alpha}.$$ The $T_A$ are the generators of the Lie algebra SO(1,2) and obey the usual commutation relations $$[T_A,T_B] = -\epsilon_{ABC}T^C,$$ with $A,B,C=0,1,2$ and $\epsilon_{012}=1$. The group indices $A,B,C$ are raised and lowered with the flat metric $\eta_{A,B}=diag(1,-1,-1)$.

The covariant derivative is defined as $$D = d+[e, \quad]$$

The field strengh is defined in terms of the commutator and it yields $$[D_{\alpha},D_{\beta}] = F_{\alpha \beta}^{A}T_{A}$$

It is explicity given by

$$F_{\alpha \beta} = \partial_{\alpha}e_{\beta}^{A}-\partial_{\beta}e_{\alpha}^{A}-\epsilon_{BC}^{A}e_{\alpha}^Be_{\beta}^C$$

Question I am used to the usual notation in term of coordinates but I am lost here. What shall I put in the commutator ? A random 1-form ? How to explicitely get the last result by evaluating the commutator ?

• The form of $D$ depends on what it acts on. What you wrote is only valid when $D$ acts on one-forms that live in the adjoint representation. – AccidentalFourierTransform Jan 27 '18 at 20:36
• Thanks @AccidentalFourierTransform, would it be possible for you to write an example on how it act ? – Ezareth Jan 27 '18 at 20:55
• $D\omega=\mathrm d\omega+[e,\omega]$, where $\omega=\omega^A_\alpha T_A\mathrm dx^\alpha$. – AccidentalFourierTransform Jan 27 '18 at 20:56
• Thanks again @AccidentalFourierTransform. But then, consider $D_{\beta}w$, is it equal to $dw + [e_{\beta}^AT_Adx^{\beta},w_{\gamma}^BT_Bdx^{\gamma}]$ ? – Ezareth Jan 27 '18 at 21:00

Using the Lie algebra valued function:

$$e_{\alpha} = e_{\alpha}^AT_A$$ We can write the covariant derivative components

$$D_{\alpha} = \partial_{\alpha} + \mathrm{ad}(e_{\alpha})$$

where, $\mathrm{ad}(X) = [X, .]$ is the adjoint representation. Please notice that it is linear in the components of $e_{\alpha}$

Thus

\begin{align*} [D_{\alpha}, D_{\beta}] &= \mathrm{ad}(e_{\alpha}) \partial_{\beta}- \partial_{\beta} \mathrm{ad}(e_{\alpha}) - \mathrm{ad}(e_{\beta}) \partial_{\alpha}+ \partial_{\alpha} \mathrm{ad}(e_{\beta}) + [\mathrm{ad}(e_{\alpha}) , \mathrm{ad}(e_{\beta}) \\ &= -\mathrm{ad}(\partial_{\beta}e_{\alpha}) + \mathrm{ad}(\partial_{\alpha}e_{\beta}) + \mathrm{ad}([e_{\alpha} , e_{\beta}]) \\ &= \mathrm{ad}(\partial_{\alpha}e_{\beta} -\partial_{\beta}e_{\alpha} + [e_{\alpha} , e_{\beta}]) \end{align*} where the linearity of the adjoint representation of the Lie algebra was used in the application of Leibniz rules and the fact that it is a representation: $[\mathrm{ad}(X), \mathrm{ad}(Y)] = \mathrm{ad} ([X,Y])$

The result is evident from the last expression.

The covariant derivative is defined as $$$$D = d+[e, \quad]$$$$ What shall I put in the commutator ? A random 1-form ?

For the sake of generosity, let's look at an n-form $$X$$ which transforms covariantly under some gauge transformation $$g$$ $$X \rightarrow gXg^{-1}.$$

The covariant derivative of $$X$$ should be $$$$DX = dX+AX-(-1)^n XA,$$$$ where $$A$$ is the gauge field 1-form corresponding to the gauge transform $$g$$ and wedge $$\wedge$$ product between forms is assumed. The first $$-$$ sign in the above expression comes from $$g^{-1}$$ in $$gXg^{-1}$$. The second $$(-1)^n$$ sign has to do with the wedge $$\wedge$$ product (anti)commuting properties between 1-form $$d$$ and n-form $$X$$.

So for even(0, 2, ...)-form $$X$$, the covariant derivative is $$$$DX = dX+[A, X] = dX+AX- XA.$$$$ And for odd(1, 3, ...)-form $$X$$, the covariant derivative is $$$$DX = dX+[A, X]_+ = dX+AX+ XA.$$$$

For example, the covariant derivative of the tetrad/vielbein/vierbein 1-form $$e$$ is $$T= D_\omega e = de+[w, e]_+ = de + \omega e + e\omega,$$ where $$\omega$$ is the spin connection 1-form, which is the gauge field corresponding to the local Lorentz gauge transformation $$g_{Lorentz}$$. Incidentally, $$T$$ is the torsion 2-form.