# Index Concordia

## Speculations on an eleven dimensional manifold

Posted in Classical Electrodynamics, Gauge Theory, Geometry by Index Guy on July 15, 2007

[This post is inspired by exercise 11 on chapter 2 of Spacetime and Geometry. I am not completely sure that my analysis is correct, so readers are encouraged to point out my mistakes.]

We consider an 11-dimensional spacetime $\mathcal{M}$ with 3-form gauge potential $A^{(3)}$ and an associated 4-form field strength $F^{(4)} = \text{d}A^{(3)}$. Let us first consider traditional, 4-dimensional Minkowski spacetime. In this spacetime, the gauge potential $A$ is a 1-form and it couples to point particles with an action term of the form

$S = \displaystyle\int_{\gamma}A,$

with $\gamma$ being the 1-dimensional trajectory of the 0-dimensional particle through spacetime, the worldline. Analogously we can say that our 3-form gauge potential on $\mathcal{M}$ can be integrated along a sort-off 3-dimensional spacetime trajectory called the worldvolume. The action term will look like

$S = \displaystyle \int_{\Gamma}A^{(3)}.$

This would imply that the object that the gauge potential will couple to will be 2-dimensional, a membrane.

Now let us go back to 4-d Minkowski spacetime. When we have an electron (a 0-d object) and want to measure its charge, we use Gauss’s Law and integrate the dual of the field strength $*F$ over a 2-sphere. This is because in 4-d, the field strength and its dual have the same dimension; both are 2-forms. Switching again to our manifold $\mathcal{M}$. In our case, the dual of the field strength is an 11 – 4 = 7-form. This would imply that the charge of our 2-brane can be measure by integrating the dual field strength over a 7-sphere $\Sigma$. The charge would be

$Q = \displaystyle \int_{\Sigma}*F^{(7)}.$

How does Stokes’s Theorem looks in the language of differential forms? The answer is beautiful:

$\displaystyle \int_{\Gamma}\text{d}B = \displaystyle \int_{\partial\Gamma}B.$

In this expression, $\Gamma$ is the region of integration and $\partial\Gamma$ its boundary. Awesome. Now what? Remember that an $(n-1)$-sphere is the boundary of an $n$-ball. Again, for Minkowski spacetime, the charge is

$Q = \displaystyle \int_{\partial\Sigma}*F = \displaystyle \int_{\Sigma}\text{d}*F = \displaystyle \int_{\Sigma}*J.$

The last expression is equivalent to the charge density. We used Maxwell’s equations to get the last step. Similarly we can argue that

$Q = \displaystyle \int_{\partial\Sigma}*F^{(7)} = \displaystyle \int_{\Sigma}(\text{d}*F)^{(8)}.$

Thus, having $\text{d}*F = 0$ will make the boundary integral to vanish, implying some sort of zero-net-flux. I guess we can take this as an interpretation of charge conservation, although the dual field strength does not represent the charge… This is the part I am not sure about, since in Minkowski spacetime charge conservation is expressed as

$\text{d}*J = 0.$

This follows again from Maxwell’s equations. I will continue forward with the other items of the exercise. Now assume that there exist some other gauge potential $B$ that defines the dual field strength like

$*F = \text{d}B.$

This gauge potential will be a 6-form. It would couple naturally with a 5-dimensional object.Now we consider the action term for the gauge potential itself. On Minkowski spacetime this is given by an integral over all the 4-d spacetime 4-volume and has terms like

$S = \displaystyle \int_{M} F \wedge *F + A \wedge *J,$

with the first term being gauge invariant under the transformation

$A \rightarrow A + \text{d}\alpha .$

For our 11-dimensional theory we must consider the transformations

$A \rightarrow A + \text{d}\alpha$ and $B \rightarrow B + \text{d} \beta .$

The question will be then: what terms can we write for the action of our gauge potential 3-form? Keep in mind that $\alpha$ is a 2-form and $\beta$ is a 5-form. Let’s assume that this forms vanish at infinity.First we note that the field strength 4-form $F$ is invariant under this gauge transformation. The same goes for the dual field strength. Hence we can readily include a term like

$F\wedge *F$

in our action since this is also a 11-form. We will limit ourselves with terms that are first, second and third order in $A$. What we need is to construct 11-forms using all the forms defined above, derivatives and duals of them. The first term that comes to my mind is the (4 + 4 + 3)-form

$F\wedge F \wedge A.$

Note that under the gauge transformation we have this same term plus an additional term,

$F\wedge F \wedge A \rightarrow F\wedge F \wedge A + F\wedge F \wedge \text{d}\alpha.$

Since by definition we have $\text{d}F = 0$, we can write this additional term as

$\text{d}(F\wedge F\wedge \alpha).$

The integral of this derivative over all space can be written as the integral of the argument over the boundary of this space, and since we are assuming that $\alpha$ vanishes at infinity we can take this term to be zero. So then my first guess is a gauge invariant term. It is of third order in $A$.

Other 11-forms we can construct look like

$\text{d}(F\wedge B)$ or $\text{d}(*F\wedge A).$

Both of these forms can be shown to be equivalent to $F\wedge *F$ by using the product law for the exterior derivative. One last possibility is the form

$B\wedge *B.$

Here a complication arises: the fact that the Hodge star and the exterior derivative are non-commuting operators. This makes it hard to compare terms like

$B\wedge \text{d}*\beta.$

I thought of defining the dual gauge transformations as

$*B \rightarrow *B + \text{d}\tilde{\beta},$

same for the other gauge potential. But this transformation is not require by anything, unlike the transformations of the gauge potentials that leave the field strengths unchanged. Also the usual Stokes-theorem-argument does not work with this one. In fact, it gives a term that explicitly depends on $\tilde{\beta}$. This is very bad. Unless we define some sort of covariant derivative… Maybe later. For now I am going to conclude that the action for the gauge potential looks like

$S_{A} = \displaystyle\int_{\Gamma} F\wedge *F + F \wedge F \wedge A.$

Of course, if we do not require $\text{d}*F = 0$, we can also say that this term is equal to some current-form $J$ and add a term $A\wedge *J$.

Update: After searching with Google “11d gauge theory” I came upon John Baez’s exposition on supergravity and M-theory, which can be found here. Professor Baez mentions the terms that I have so maybe the two terms that I guessed are the only ones one can write. I am still not convinced, but also I do not want to write down the indices notation explicitly.