# Elementary question about distributive property of variation operator on an exterior product

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.

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

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.

• The exterior derivative acts that way because it is an antiderivation. I see no reason why the Euler-Lagrange variation should be one. – Ryan Unger Jul 24 '15 at 20:35
• What do you mean by "antiderivation"? – leastaction Jul 24 '15 at 20:37
• en.wikipedia.org/wiki/… – Ryan Unger Jul 24 '15 at 20:42
• The variation is a linear operator on the action functional, not on the forms themselves. For example, for a functional $F[x(t)]$, $\delta F[x] := F[x(t)+\epsilon \eta(t)] - F[x(t)]$. – Ultima Jul 24 '15 at 21:29
• Thanks, that was rather silly of me. I am going to edit my original post and add the variational calculation, as there's a minor issue there as well, which I would like to discuss. – leastaction Jul 25 '15 at 0:28

1. The variational operator $\delta$ satisfies the product rule

$$\delta (A \wedge B) = (\delta A)\wedge B + A \wedge (\delta B)$$

1. To find the variation of $G \wedge \star G$ where $\star G$ is the Hodge dual of $G$, note that

$$G \wedge \star G = \frac{1}{p!}G_{\mu_1 \ldots \mu_p} G^{\mu_1 \ldots \mu_p} \sqrt{|g|}d^{D}x$$

From this, it is immediately obvious that under a variation of the form $C$ (where $G = dC$)

$$\delta(G \wedge \star G) = \frac{2}{p!}G_{\mu_1\ldots \mu_p}\delta G^{\mu_1\ldots\mu_p}\sqrt{|g|}d^{D}x = 2(\delta G) \wedge \star G = 2 G\wedge \star(\delta G)$$

(it is useful to be able to go back and forth between form notation and tensor notation)

Corollary: $$\delta (G \wedge \star G) = (\delta G) \wedge \star G + G \wedge \star (\delta G) = 2 (\delta G) \wedge \star G = 2 G \wedge \star (\delta G)$$

1. For the third term, it is indeed true that $$\delta(C\wedge G\wedge G) = 3 (\delta C) \wedge G \wedge G$$.