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Integration by parts on manifold with a boundary

+ 6 like - 0 dislike
238 views

Suppose $C$ is a 3-form, and $G$ is a 4-form defined by $G = dC$. Also, $M_{11}$ is an 11-dimensional manifold (without a boundary), $W_{6}$ is a 6-dimensional submanifold of $M_{11}$ and $D_{\epsilon}W_6 = -S_{\epsilon}W_{6}$ is the 4-sphere bundle over $W_6$.

Further, suppose $\rho$ is a 0-form and $e_{2}^{1}$ is a 2-form. Under a variation,

$$\delta C = -d(\rho e_{2}^{1})$$

I want to compute the variation $\delta S_{CS}$ in the Chern-Simons integral

$$S_{CS} = -\lim_{\epsilon\rightarrow 0}\int\limits_{M_{11}\backslash D_{\epsilon}W_6} C \wedge G \wedge G$$

Apparently, the correct answer is

$$\delta S_{CS} = -\lim_{\epsilon \rightarrow}\int\limits_{S_{\epsilon}W_6}\rho e_{2}^{1}\wedge G \wedge G$$

But what I get is something else. Here is my detailed derivation.

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[\int\limits_{M_{11}\backslash D_\epsilon W_6}\delta C \wedge G \wedge G + 2 \int\limits_{M_{11}\backslash D_\epsilon W_6}\delta G \wedge C \wedge G\right]$$

From integration by parts

$$\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G = \int\limits_{M_{11}\backslash D_\epsilon W_6}d(\delta C \wedge C \wedge G) = \int\limits_{M_{11}\backslash D_\epsilon W_6} \delta dC \wedge C \wedge G - \int\limits_{M_{11}\backslash D_\epsilon W_6} \delta C \wedge G \wedge G$$

So it should follow that

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G + 3\int\limits_{M_{11}\backslash D_\epsilon W_6}\delta C \wedge G \wedge G\right]$$

As $M_{11}\backslash D_{\epsilon}W_6$, for finite $\epsilon$ cannot support an 11-form, the second integral vanishes inside the limit, and we are left with

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G\right]$$

Substituting $\delta C = -d(\rho e_2^1)$ we get

$$\delta S_{CS} = +\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)} d(\rho e_2^1) \wedge C \wedge G\right]$$

Now,

$$d(\rho e_2^1 \wedge (C\wedge G)) = d(\rho e_2^1) \wedge C\wedge G + \rho e_2^1 \wedge d(C\wedge G)$$

and $\partial(\partial(M_{11}\backslash D_\epsilon W_6)) \equiv 0$, so

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\rho e_2^1 \wedge G \wedge G\right]$$

As a last step, using $\partial(M_{11}\backslash D_\epsilon W_6) = -S_{\epsilon} W_6$, one gets the final expression

$$\delta S_{CS} = +\lim_{\epsilon\rightarrow 0}\left[2\int\limits_{S_\epsilon W_6}\rho e_2^1 \wedge G \wedge G\right]$$

This is off by a sign and a factor of 2.

What seems to be wrong in the derivation here?

Physics background: These algebraic manipulations are inspired by a calculation of the M5-brane anomaly in M-theory, perhaps discussed first in a paper by Freed, Minasian, Harvey and Moore (http://arxiv.org/abs/hep-th/9803205). It has been pointed out in some follow-up papers that there are an odd number of minus signs involved, and that the original paper may have overlooked one sign. (Of course the M5-brane anomaly is seen to cancel, but the cancellation is a bit involved and requires a careful understanding of minus signs and factors. Hence this question.)


This post imported from StackExchange Mathematics at 2015-08-09 22:38 (UTC), posted by SE-user leastaction

asked Aug 9, 2015 in Mathematics by leastaction (425 points) [ revision history ]
edited Aug 10, 2015 by leastaction

1 Answer

+ 3 like - 0 dislike

You have that

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[\int\limits_{M_{11}\backslash D_\epsilon W_6}\delta C \wedge G \wedge G + 2 \int\limits_{M_{11}\backslash D_\epsilon W_6}\delta G \wedge C \wedge G\right]  $$

and

$$\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G = \int\limits_{M_{11}\backslash D_\epsilon W_6}d(\delta C \wedge C \wedge G) =$$

$$\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta dC \wedge C \wedge G - \int\limits_{M_{11}\backslash D_\epsilon W_6} \delta C \wedge G \wedge G $$

From the last equation we derive that

$$\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta C \wedge G \wedge G =-\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G +\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta dC \wedge C \wedge G $$

it is to say

$$\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta C \wedge G \wedge G =-\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G +\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta G \wedge C \wedge G $$

Replacing the last equation in the first equation we obtain that

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[-\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G +\int\limits_{M_{11}\backslash D_\epsilon W_6} \delta G \wedge C \wedge G +\\\\ 2 \int\limits_{M_{11}\backslash D_\epsilon W_6}\delta G \wedge C \wedge G\right]  $$

which is reduced to

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0}\left[-\int\limits_{\partial(M_{11}\backslash D_\epsilon W_6)}\delta C \wedge C \wedge G  + 3 \int\limits_{M_{11}\backslash D_\epsilon W_6}\delta G \wedge C \wedge G\right]  $$

From this last result we derive that

$$\delta S_{CS} = -\lim_{\epsilon \rightarrow 0}\int\limits_{S_{\epsilon}W_6}\rho e_{2}^{1}\wedge G \wedge G$$

Do you agree?

answered Aug 10, 2015 by juancho (860 points) [ revision history ]
edited Aug 10, 2015 by juancho

Ah, so

$$\delta S_{CS} = \lim_{\epsilon\rightarrow 0} \int\limits_{\partial(M_{11}\backslash D_{\epsilon}W_6)}\delta C \wedge C \wedge G$$

Now, you write $\delta C = -d(\rho e_2^1)$ to get 

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0} \int\limits_{\partial(M_{11}\backslash D_{\epsilon}W_6)}d(\rho e_2^1) \wedge C \wedge G$$

At this point, one has to use the following identity

$$d(\rho e_2^1 \wedge C \wedge G) = d(\rho e_2^1) \wedge C \wedge G + (-1)^{2) \rho e_2^1 \wedge d(C \wedge G) = d(\rho e_2^1) \wedge C \wedge G + \rho e_2^1 \wedge G \wedge G$$

(note the sign)

So when you do integration by parts, you get an extra minus sign

$$\delta S_{CS} = +\lim_{\epsilon\rightarrow 0} \int\limits_{\partial(M_{11}\backslash D_{\epsilon}W_6)}\rho e_2^1 \wedge G \wedge G$$

Finally, you have one more minus sign 

$$\delta S_{CS} = -\lim_{\epsilon\rightarrow 0} \int\limits_{S_\epsilon W_6}\rho e_2^1 \wedge G \wedge G$$

Thanks @juancho!

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