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I'm practicing some differential forms stuff and got a bit stuck on something. I'd type it out but the action is very long so it's easier if I just link to where I'm getting it from, this paper http://gesalerico.ft.uam.es/tesis/pablo_camara.pdf [Broken]

Equation (4.20) (pdf page 51) is the IIA action with particular forms (4.21)-(4.23). (4.25) is derived from varying the metric, (4.26) by varying the dilaton. (4.27) to (4.29) are the ones I have problem with. I know they are from varying B2, C3 and A1 (respectively), using the equations in the footnote on page 50. (4.29) I can derive. It's (4.28) I'm stuck on. Here's what I've done :

[tex]\frac{\partial \mathcal{L}}{\partial C_{3}} = d \frac{\partial \mathcal{L}}{\partial(dC_{3})}[/tex] and [tex](F_{n})^{2} = n! F_{n} \wedge \ast F_{n}[/tex]

so

[tex]0 = d \left( \frac{\partial}{\partial C_{3}} \left\{ -\frac{1}{48}e^{\frac{\phi}{2}}(F_{4})^{2} + \frac{1}{2}B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{6}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( \frac{\partial}{\partial C_{3}} \left\{ -\frac{1}{2}e^{\frac{\phi}{2}}F_{4}\wedge \ast F_{4} + \frac{1}{2}B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{6}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( - e^{\frac{\phi}{2}} \ast F_{4} \right) + d \left( \frac{\partial}{\partial C_{3}} \left\{ B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{3}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( - e^{\frac{\phi}{2}} \ast F_{4} \right) + d \left( 2 B_{2}\wedge dC_{3} + \frac{m}{3}B_{2}\wedge B_{2}\wedge B_{2} \right) [/tex]

[tex]d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = 2 dB_{2}\wedge dC_{3} + \frac{m}{3} 3 dB_{2}\wedge B_{2}\wedge B_{2} \right\} \right)[/tex]

[tex]\frac{1}{2}d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = dB_{2}\wedge dC_{3} + \frac{m}{2} dB_{2}\wedge B_{2}\wedge B_{2} \right\} \right)= H_{3} \wedge \left(dC_{3} + \frac{m}{2}B_{2}\wedge B_{2} \right) [/tex]

So

[tex]\frac{1}{2}d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = H_{3} \wedge \left( F_{4} + H_{3} \wedge A_{1} \right)[/tex]

The factor of 1/2 is wrong on the left hand side and I've the extra [tex]H_{3}\wedge H_{3} \wedge A_{1}[/tex] on the right hand side. I tried messing around with various things like integrating by parts etc but couldn't get it to work. I'm using the same methods I used to get (4.29) but it's not working here. Am I missing something obvious or is there a subtle trick?

Thanks for any help you can provide :)

Equation (4.20) (pdf page 51) is the IIA action with particular forms (4.21)-(4.23). (4.25) is derived from varying the metric, (4.26) by varying the dilaton. (4.27) to (4.29) are the ones I have problem with. I know they are from varying B2, C3 and A1 (respectively), using the equations in the footnote on page 50. (4.29) I can derive. It's (4.28) I'm stuck on. Here's what I've done :

[tex]\frac{\partial \mathcal{L}}{\partial C_{3}} = d \frac{\partial \mathcal{L}}{\partial(dC_{3})}[/tex] and [tex](F_{n})^{2} = n! F_{n} \wedge \ast F_{n}[/tex]

so

[tex]0 = d \left( \frac{\partial}{\partial C_{3}} \left\{ -\frac{1}{48}e^{\frac{\phi}{2}}(F_{4})^{2} + \frac{1}{2}B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{6}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( \frac{\partial}{\partial C_{3}} \left\{ -\frac{1}{2}e^{\frac{\phi}{2}}F_{4}\wedge \ast F_{4} + \frac{1}{2}B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{6}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( - e^{\frac{\phi}{2}} \ast F_{4} \right) + d \left( \frac{\partial}{\partial C_{3}} \left\{ B_{2}\wedge dC_{3} \wedge dC_{3} + \frac{m}{3}B_{2}\wedge B_{2}\wedge B_{2}\wedge dC_{3} \right\} \right)[/tex]

[tex]0 = d \left( - e^{\frac{\phi}{2}} \ast F_{4} \right) + d \left( 2 B_{2}\wedge dC_{3} + \frac{m}{3}B_{2}\wedge B_{2}\wedge B_{2} \right) [/tex]

[tex]d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = 2 dB_{2}\wedge dC_{3} + \frac{m}{3} 3 dB_{2}\wedge B_{2}\wedge B_{2} \right\} \right)[/tex]

[tex]\frac{1}{2}d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = dB_{2}\wedge dC_{3} + \frac{m}{2} dB_{2}\wedge B_{2}\wedge B_{2} \right\} \right)= H_{3} \wedge \left(dC_{3} + \frac{m}{2}B_{2}\wedge B_{2} \right) [/tex]

So

[tex]\frac{1}{2}d \left( e^{\frac{\phi}{2}} \ast F_{4} \right) = H_{3} \wedge \left( F_{4} + H_{3} \wedge A_{1} \right)[/tex]

The factor of 1/2 is wrong on the left hand side and I've the extra [tex]H_{3}\wedge H_{3} \wedge A_{1}[/tex] on the right hand side. I tried messing around with various things like integrating by parts etc but couldn't get it to work. I'm using the same methods I used to get (4.29) but it's not working here. Am I missing something obvious or is there a subtle trick?

Thanks for any help you can provide :)

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