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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 \begin{equation} e = e_{\alpha}^{A}T_Adx^{\alpha}. \end{equation} The $T_A$ are the generators of the Lie algebra SO(1,2) and obey the usual commutation relations \begin{equation} [T_A,T_B] = -\epsilon_{ABC}T^C, \end{equation} 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 \begin{equation} D = d+[e, \quad]\end{equation}

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

It is explicity given by

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

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 ?

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  • $\begingroup$ 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. $\endgroup$ Jan 27, 2018 at 20:36
  • $\begingroup$ Thanks @AccidentalFourierTransform, would it be possible for you to write an example on how it act ? $\endgroup$
    – Ezareth
    Jan 27, 2018 at 20:55
  • $\begingroup$ $D\omega=\mathrm d\omega+[e,\omega]$, where $\omega=\omega^A_\alpha T_A\mathrm dx^\alpha$. $\endgroup$ Jan 27, 2018 at 20:56
  • $\begingroup$ Thanks again @AccidentalFourierTransform. But then, consider $D_{\beta}w$, is it equal to $dw + [e_{\beta}^AT_Adx^{\beta},w_{\gamma}^BT_Bdx^{\gamma}]$ ? $\endgroup$
    – Ezareth
    Jan 27, 2018 at 21:00

2 Answers 2

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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.

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The covariant derivative is defined as $$ \begin{equation} D = d+[e, \quad]\end{equation} $$ 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 \begin{equation} DX = dX+AX-(-1)^n XA, \end{equation} 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 $$ \begin{equation} DX = dX+[A, X] = dX+AX- XA. \end{equation} $$ And for odd(1, 3, ...)-form $X$, the covariant derivative is $$ \begin{equation} DX = dX+[A, X]_+ = dX+AX+ XA. \end{equation} $$

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.

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    $\begingroup$ Hi. Could you point out where I can see a derivation of the form you have: $$ D_A X = d X + A \wedge X \pm X \wedge A$$ $\endgroup$ Dec 12, 2021 at 5:42

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