I'm using the reference "Differential Geometry, Gauge Theories and Gravity" by M. Göckeler and T. Schücker and I am having trouble to vary correctly the lagrangian

$$ \mathcal{L}_M=\dfrac{1}{2g^2}F \wedge *F $$

with respect to the vierbein $e^a$ in order to find the Energy-momentum of the Maxwell action.

By doing $e\rightarrow e+f$ in the lagragian above I find $$ \mathcal{L[e+f]}_M-\mathcal{L[e]}_M=\dfrac{1}{g^2}f^c\left(\frac{1}{4} F^{ab}\epsilon_{abcd} e^d \wedge F+\frac{1}{2}F_{cb}e^b \wedge *|_eF\right). $$

However, the right answer, present in the foregoing reference is

$$ \mathcal{L[e+f]}_M-\mathcal{L[e]}_M=\dfrac{1}{g^2}f^c\left(\frac{1}{4} F^{ab}\epsilon_{abcd} e^d \wedge F-\frac{1}{2}F_{cb}e^b \wedge *|_eF\right). $$

(the only difference is the minus sign in the expression inside the brackets)

I worked a lot, but couldn't identify my mistake. So, does anyone know anything that is not trivial in the treatment with the forms in this case?

(In this case the signature used is lorentzian)

The whole calculation that I did was the following:


$$ \mathcal{L}_M[e]=\dfrac{1}{2g^2}F \wedge *F=\frac{1}{2g^2}\left[ \frac{1}{2}F_{ab} e^a \wedge e^b \wedge \left(\frac{1}{4}F^{\alpha \beta}\epsilon_{\alpha \beta cd}e^c \wedge e^d\right) \right] =\frac{1}{16g^2}\left( F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} e^a \wedge e^b \wedge e^c \wedge e^d \right) $$

Then, by doing $e \rightarrow e+f$, and neglecting terms quadratic in $f$, we are lead to

$$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{16g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(f^a \wedge e^b \wedge e^c \wedge e^d + e^a \wedge f^b \wedge e^c \wedge e^d + e^a \wedge e^b \wedge f^c \wedge e^d + e^a \wedge e^b \wedge e^c \wedge f^d \right) =\frac{1}{16g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(2f^a \wedge e^b \wedge e^c \wedge e^d + 2e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{8g^2} F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd} \left(f^a \wedge e^b \wedge e^c \wedge e^d + e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{2g^2} \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}f^a \wedge e^b \wedge e^c \wedge e^d + \frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}e^a \wedge e^b \wedge f^c \wedge e^d \right) =\frac{1}{2g^2} \left[f^a \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + f^c \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^a \wedge e^b \wedge e^d \right] =\frac{1}{2g^2} \left[f^a \wedge \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + f^a \wedge \left(\frac{1}{4}F_{cb}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^c \wedge e^b \wedge e^d \right] $$

And then, we have $$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{2g^2} f^a \wedge \left[ \left(\frac{1}{4}F_{ab}F^{\alpha \beta}\epsilon_{\alpha \beta cd}\right) e^b \wedge e^c \wedge e^d + \left(\frac{1}{4}F_{cb}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^c \wedge e^b \wedge e^d \right] =\frac{1}{2g^2} f^a \wedge \left[ F_{ab}e^b \wedge *|_e F + \left(\frac{1}{2}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) F \wedge e^d \right] =\frac{1}{2g^2} f^a \wedge \left[ F_{ab}e^b \wedge *|_e F + \left(\frac{1}{2}F^{\alpha \beta}\epsilon_{\alpha \beta ad}\right) e^d \wedge F \right] $$

Relabeling the indices, we finally have

$$ \mathcal{L}_M[e+f]-\mathcal{L}_M[e] =\frac{1}{g^2} f^c \wedge \left[\frac{1}{2} F_{cb}e^b \wedge *|_e F + \left(\frac{1}{4}F^{ab}\epsilon_{abcd}\right) e^d \wedge F \right] =\frac{1}{g^2} f^c \wedge \left[\left(\frac{1}{4}F^{ab}\epsilon_{abcd}\right) e^d \wedge F +\frac{1}{2} F_{cb}e^b \wedge *|_e F \right] $$

This is not in accordance with the reference due to the plus sign (it should be a minus) and, again, I could not identify where I have done something wrong.

  • $\begingroup$ I don't know which step you are missing, From (5.74) to (5.75) it is straightforward. Do you obtained (5.74) as same as in the book. $\endgroup$ Mar 11, 2017 at 7:51
  • $\begingroup$ I think $\mathcal{L[e+f]}_M-\mathcal{L[e]}_M=\dfrac{1}{g^2}f^c\wedge \left(\frac{1}{4} F^{ab}\epsilon_{abcd} e^d \wedge F-\frac{1}{2}F_{cb}e^b \wedge *|_eF\right).$ is more correct. $\endgroup$ Mar 11, 2017 at 7:53
  • $\begingroup$ I have edited the post to include the calculation that I did (I found it easier not follow strictly the Identity and calculation of book). $\endgroup$ Mar 11, 2017 at 16:23

2 Answers 2


I understand the notation in the book this way \begin{eqnarray} e^a \wedge e^b \wedge \star |_e F &\equiv& e^a \wedge e^b \wedge \star F\;\\ &=& F \wedge \star (e^a \wedge e^b)\;\\ &=& \star (e^a \wedge e^b) \wedge F\;. \end{eqnarray} Now in the more accurate definition of Hodge dual (in the proof from definition there is a step of flipping the indices for more symmetric looking result, but we not flip them here, at first) we have \begin{equation}\boxed{ \star (f^a \wedge e^b) =\frac 1 2 \epsilon^{abcd} f_c \wedge e_d }\;. \end{equation} In the case of $\star(e \wedge e)$, there are no problems for flipping the indices \begin{equation} \frac 1 2 \epsilon^{abcd} e_c \wedge e_d = \frac 1 2 \epsilon^{ab}{}_{cd}\, e^c \wedge e^d \;. \end{equation} But in the case of $\star(f \wedge e)$ we have \begin{eqnarray} \star (f^a \wedge e^b) &=& \frac 1 2 \epsilon^{abcd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} \eta^{fd} f_c \wedge e_d \\ &=& \frac 1 2 \epsilon^{ab}{}_{ef} \eta^{ce} f_c \wedge e^f \\ &=& - \frac 1 2 \epsilon^{ab}{}_{ef} f^e \wedge e^f\;. \end{eqnarray} This because $e$ obeys the transformation laws \begin{eqnarray} e^a &\longmapsto& e^a+f^a \\ e_a &\longmapsto& e_a-f_a \\ && \eta^{ab}f_a=-f^b\;. \end{eqnarray} With \begin{equation} f^a \wedge e^b \wedge \star |_e F = \star(f^a \wedge e^b) \wedge F =- \frac 1 2 \epsilon^{ab}{}_{cd} f^c \wedge e^d \wedge F \end{equation} we can obtained the same result as shown in the reference.


Sorry I have to dredge this thread up, but I've got the same problem as the author had. Let's summarize: user148471 was trying to say, that

  1. We have to make the substitution $e^a \mapsto e^a + f^a $, and THEN compute the Hodge dual, not conversely.
  2. We've got $$\star (e^a \wedge e^b)= \frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d$$ BUT $$\star (f^a \wedge e^b) = -\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d.$$
  3. We define $f_a$ as $$f_a:=-g_{ab}f^b=-\eta_{ab}f^b,$$ hence $$\eta^{ab}f_a=-\eta^{ab}\eta_{ac}f^c=-\delta ^b _{\ c}f^c=-f^b$$

Concerning 2. The formula $$\star (f^a \wedge e^b) = -\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d.$$ would solve the problem. But is it correct? The formula seems not to make sense since $f^a,\ a=0,1,2,3$ are any variations of 1-forms of the basis. What if we take $$f^a :=e^a,\ a=0,1,2,3?$$ Then we'll have $$\star (f^a \wedge e^b)=\star (e^a \wedge e^b)$$ and $$-\frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d=-\frac{1}{2} \epsilon ^{ab}_{\quad cd}f^c \wedge e^d=\star(f^a \wedge e^b)=\star (e^a \wedge e^b)=\frac{1}{2} \epsilon ^{ab}_{\quad cd}e^c \wedge e^d.$$ This tells us that $$\star (e^a \wedge e^b)=0,$$ but why should it equal 0? When $a=b$ then $$\star (e^a \wedge e^b)=0,$$ but $a$ and $b$ don't have to be equal.


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