# Integration by parts on manifold with a boundary

+ 6 like - 0 dislike
360 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

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

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

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

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

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

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

edited Aug 10, 2015

+ 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 (1,105 points)
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!

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.