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I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

EDIT: Solution appears below.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$ [WRONG!]

A similar manipulation is required on the third term.

EDIT (post comments):

$$\delta(G \wedge \star G) = \delta G \wedge \star G + G \wedge \delta \star G$$

Writing $G = dC$ and integrating the first term by parts,

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + dC \wedge \delta \star dC$$

Another integration by parts yields

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + C \wedge \delta (d\star dC)$$

But I'm not sure how helpful this is...am I missing something with $G \wedge \delta \star G$?

Apparently what one needs from the first term is $2 \delta C \wedge d\star G$, to get the correct equation of motion.

If $\chi$ is a $p$-form in $D$ dimensions, then $d\chi$ is a $(p+1)$-form and $\star d\chi$ is a $D-(p+1) = D-p-1$ form, whereas $d\star\chi$ is a $D-p+1$ form. So the exterior derivative does not commute with Hodge dualization. Does this statement have some deeper meaning?

So, because of this, I don't see how I can pull the d out of $\star G$ using integration by parts because to me, $\star d \neq d\star$. How does one get the extra factor of 2 then?

As for the third term

$$\delta(C \wedge G \wedge G) = 3 \delta C \wedge G \wedge G$$

This seems okay.

EDIT: Solution appears below.

I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

EDIT: Solution appears below.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$ [WRONG!]

A similar manipulation is required on the third term.

EDIT (post comments):

$$\delta(G \wedge \star G) = \delta G \wedge \star G + G \wedge \delta \star G$$

Writing $G = dC$ and integrating the first term by parts,

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + dC \wedge \delta \star dC$$

Another integration by parts yields

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + C \wedge \delta (d\star dC)$$

But I'm not sure how helpful this is...am I missing something with $G \wedge \delta \star G$?

Apparently what one needs from the first term is $2 \delta C \wedge d\star G$, to get the correct equation of motion.

If $\chi$ is a $p$-form in $D$ dimensions, then $d\chi$ is a $(p+1)$-form and $\star d\chi$ is a $D-(p+1) = D-p-1$ form, whereas $d\star\chi$ is a $D-p+1$ form. So the exterior derivative does not commute with Hodge dualization.

So, because of this, I don't see how I can pull the d out of $\star G$ using integration by parts because to me, $\star d \neq d\star$. How does one get the extra factor of 2 then?

As for the third term

$$\delta(C \wedge G \wedge G) = 3 \delta C \wedge G \wedge G$$

This seems okay.

EDIT: Solution appears below.

I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$ [WRONG!]

A similar manipulation is required on the third term.

EDIT (post comments):

$$\delta(G \wedge \star G) = \delta G \wedge \star G + G \wedge \delta \star G$$

Writing $G = dC$ and integrating the first term by parts,

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + dC \wedge \delta \star dC$$

Another integration by parts yields

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + C \wedge \delta (d\star dC)$$

But I'm not sure how helpful this is...am I missing something with $G \wedge \delta \star G$?

Apparently what one needs from the first term is $2 \delta C \wedge d\star G$, to get the correct equation of motion.

If $\chi$ is a $p$-form in $D$ dimensions, then $d\chi$ is a $(p+1)$-form and $\star d\chi$ is a $D-(p+1) = D-p-1$ form, whereas $d\star\chi$ is a $D-p+1$ form. So the exterior derivative does not commute with Hodge dualization. Does this statement have some deeper meaning?

So, because of this, I don't see how I can pull the d out of $\star G$ using integration by parts because to me, $\star d \neq d\star$. How does one get the extra factor of 2 then?

As for the third term

$$\delta(C \wedge G \wedge G) = 3 \delta C \wedge G \wedge G$$

This seems okay.

I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

EDIT: Solution appears below.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$ [WRONG!]

A similar manipulation is required on the third term.

EDIT (post comments):

$$\delta(G \wedge \star G) = \delta G \wedge \star G + G \wedge \delta \star G$$

Writing $G = dC$ and integrating the first term by parts,

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + dC \wedge \delta \star dC$$

Another integration by parts yields

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + C \wedge \delta (d\star dC)$$

But I'm not sure how helpful this is...am I missing something with $G \wedge \delta \star G$?

Apparently what one needs from the first term is $2 \delta C \wedge d\star G$, to get the correct equation of motion.

If $\chi$ is a $p$-form in $D$ dimensions, then $d\chi$ is a $(p+1)$-form and $\star d\chi$ is a $D-(p+1) = D-p-1$ form, whereas $d\star\chi$ is a $D-p+1$ form. So the exterior derivative does not commute with Hodge dualization. Does this statement have some deeper meaning?

So, because of this, I don't see how I can pull the d out of $\star G$ using integration by parts because to me, $\star d \neq d\star$. How does one get the extra factor of 2 then?

As for the third term

$$\delta(C \wedge G \wedge G) = 3 \delta C \wedge G \wedge G$$

This seems okay.

EDIT: Solution appears below.

I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

I am inclined to think so, because the variation is a linear operator acting from the left, just like the exterior derivative.

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$

A similar manipulation is required on the third term.

I am trying to work out the equations of motion of a 11-dimensional supergravity action

$$S = \frac{1}{2\kappa^2}\left(\gamma\int d^{11}x\sqrt{|g|}\mathcal{R} - \frac{\alpha}{2}\int G \wedge \star G - \frac{\beta}{6}\int C \wedge G \wedge G\right) $$

where $\mathcal{R}$ is the scalar curvature, $\alpha$, $\beta$, $\gamma$ and $\kappa$ are constants, $C$ is a 3-form and $G = dC$ is the corresponding 4-form field strength.

My very elementary question is that if $A_{(p)}$ and $B_{(q)}$ are p- and q- forms respectively, then we know that

$$d(A_{(p)}\wedge B_{(q)}) = dA_{(p)} \wedge B_{(q)} + (-1)^{p} A_{(p)} \wedge dB_{(q)}$$

Does this also hold for the variation operator $\delta$?

If yes,

$$\delta(G \wedge \star G) = \delta G \wedge \star G + (-1)^{4} G \wedge G \wedge \delta\star G$$ [WRONG!]

A similar manipulation is required on the third term.

EDIT (post comments):

$$\delta(G \wedge \star G) = \delta G \wedge \star G + G \wedge \delta \star G$$

Writing $G = dC$ and integrating the first term by parts,

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + dC \wedge \delta \star dC$$

Another integration by parts yields

$$\delta(G \wedge \star G) = \delta C \wedge d\star G + G \wedge \delta \star G\\ = \delta C \wedge d\star G + C \wedge \delta (d\star dC)$$

But I'm not sure how helpful this is...am I missing something with $G \wedge \delta \star G$?

Apparently what one needs from the first term is $2 \delta C \wedge d\star G$, to get the correct equation of motion.

If $\chi$ is a $p$-form in $D$ dimensions, then $d\chi$ is a $(p+1)$-form and $\star d\chi$ is a $D-(p+1) = D-p-1$ form, whereas $d\star\chi$ is a $D-p+1$ form. So the exterior derivative does not commute with Hodge dualization. Does this statement have some deeper meaning?

So, because of this, I don't see how I can pull the d out of $\star G$ using integration by parts because to me, $\star d \neq d\star$. How does one get the extra factor of 2 then?

As for the third term

$$\delta(C \wedge G \wedge G) = 3 \delta C \wedge G \wedge G$$

This seems okay.