# B Base space Minkowski, fiber is S^1, total spaces?

1. Jul 31, 2016

### Spinnor

If our base space, B, is Minkowski spacetime and our fibers are circles S^1 are the following constructions ways to put together B and S^1 and have a total space, T, that is considered a fiber bundle.

Remove from Minkowski spacetime, M, a timelike line, L. At the remaining points of Minkowski spacetime attach a S^1 fiber. If possible I would like to "twist" the fibers such that if we make one 360 degree orbit around line L we advance or go back one turn in the S^1 space? Is there such a total space T?

Again let our base space be Minkowski spacetime, and now let our S^1 fibers be cut. At each point of Minkowski spacetime attach one end of a fiber, the other end is attached a small distance dx away from the first end. So if you make one complete orbit in any fiber S^1 you move in spacetime some small distance dx. Is that something that could be considered a fiber bundle? Could dx be a function of spacetime?

Thanks!

2. Aug 3, 2016

### Ben Niehoff

Topologically, Minkowski space (I assume you mean the 4-dimensional one) is just $\mathbb{R}^4$, so it may simplify your thinking just to imagine $\mathbb{R}^4$.

If you do mean 4-dimensional Minkowski space (and hence $\mathbb{R}^4$), then this description doesn't quite make sense, since a line in $\mathbb{R}^4$ is linked by an $S^2$, not an $S^1$. So the notion of a "360 degree orbit around line L" doesn't quite work. However, since the line is linked by an $S^2$, it is perfectly fine to fiber your $S^1$ over that $S^2$ such that it is twisted. One twist will give you the Hopf fibration. So in this case, your total space would be $S^3 \times \mathbb{R}_+ \times \mathbb{R}$, where $\mathbb{R}_+$ is the radial direction and $\mathbb{R}$ is from the line you removed.

You can also put in more twists, in which case I think you get lens spaces in place of the $S^3$.

No, this does not describe a way to make a fiber bundle. A fiber should be "attached" to only one point of the base.

3. Aug 3, 2016

### Spinnor

Thanks Ben! I wanted to know if Kaluza Klein theory could accommodate magnetic charge. After some Googling I came up with a slide from a talk by professor Gary Horowitz which seems to indicate that magnetic charge can be added to the theory,

What I am trying to understand in what exactly is a "topological twisting" of S^1. I thought it had something to do with fiber bundles?

Thanks for you time!

4. Aug 4, 2016

### Ben Niehoff

"Topological twisting" is exactly what I just described. The $S^1$ describes a Hopf fibration over the $S^2$ which links the particle's worldline. The magnetic charge is the degree of the twisting, and its sign is given by the handedness of the twisting.

5. Aug 4, 2016

### Spinnor

Great! Will study this. Thank you.

In classical electrodynamics it seems that you can redefine electric charge so that it is made up of both electric and magnetic charge.and Maxwell's equations do not change I guess if you redefine the current also. But in Kaluza Klein theory magnetic charge and electric charge seem like completely different animals, one is momentum in S^1 and the other is a topological twisting of S^1. Is this really the case, are electric and magnetic charge fundamentally different in Kaluza Klein theory?

Thanks!

6. Aug 5, 2016

### Ben Niehoff

It's not quite that simple. You can (using "duality rotations") define electric charge to be some combination of electric charge and magnetic charge. However, all particles must still have the same ratio of magnetic to electric charge, so you still only have one type of charge.

Not sure if you've heard of the term "charge lattice" before. The charge lattice (of an Abelien gauge theory) is the lattice of possible quantized charges (in the classical theory, this lattice would become a continuous vector space). In standard Maxwell theory, the charge lattice is one-dimensional: electric charges. If you allow magnetic monopoles, then the lattice becomes two-dimensional. But if you just start from the standard Maxwell theory and use duality rotations, then the charge lattice is still one-dimensional; it just has a different orientation.

Note that there is a little bit of a catch here, which is that charge quantization easily follows if you have magnetic charges (which are independent of electric charges), but has no obvious justification if you have only electric charges.

Yes, KK theory is not standard Maxwell theory; it is Maxwell theory plus magnetic charges (plus a scalar field, which you've been ignoring). The magnetic charges are independent of the electric ones, giving a 2-dimensional charge lattice.

Note that there is a bit of a catch again: KK theory is often described as 5d gravity reduced on a circle; the 4d fields after reduction then include a Maxwell 2-form and a massless uncharged scalar. However, there is a gauge choice involved here: the circle on which you reduce! The following is the metric of a "KK monopole", which is a single magnetic charge in KK theory:

$$ds^2 = - dt^2 + \frac{1}{r} (d\psi + \cos \theta \, d\phi)^2 + r \big( dr^2 + r^2 \, d\theta^2 + r^2 \sin^2 \theta \, d\phi^2 \big)$$
However, this is also the metric for 5-dimensional flat Minkowski space! The reason you get something non-trivial when reducing to 4 dimensions is because of the choice of circle on which to reduce, which in this case is the circle described by the coordinate $\psi$.

There is another thing that makes KK theory a little unrealistic, or at least restricts its validity to low energies, which is: In order to make the reduction to 4d, we assume that all fields are independent of the circle coordinate on which we're reducing. In a true 5d gravity theory, there is no natural reason to impose this restriction. The simplest way to approximate dependence on the "internal" direction is by Fourier modes around the circle, also called "higher KK modes". These modes give you an infinite tower of massive fields in the lower-dimensional theory, just like you have in string theory.

7. Aug 5, 2016

### Spinnor

So I guess a magnetic monopole and its opposite can annihilate. So can a Hopf Fibration of one twist and its opposite somehow come together and annihilate?

If so is that something that is straight forward for a mathematician to depict graphically?

Thanks!

8. Aug 6, 2016

### Spinnor

Ben, is it possible to make contact between the twisted space you describe and the vector potential fields for a Dirac monopole,

Thanks!

9. Aug 8, 2016

### Ben Niehoff

Take a look at the metric again:

$$ds^2 = - dt^2 + \frac{1}{r} (d\psi + \cos \theta \, d\phi)^2 + r \big( dr^2 + r^2 \, d\theta^2 + r^2 \sin^2 \theta \, d\phi^2 \big)$$
You can see the vector potential in the $d\psi$ term:

$$\frac{1}{r} (d\psi + A)^2, \qquad A = \cos \theta \, d\phi$$
You should be able to check that this is the same vector potential that you posted, up to a gauge transformation. If we change coordinates $\psi \to \psi \pm \phi$, then we get

$$A = (\pm 1 + \cos \theta) \, d\phi$$
which matches your example (except that this has magnetic charge +1, whereas your example has magnetic charge -1).

10. Aug 9, 2016

### Spinnor

Ben thanks for all your help! It is great when experts try and make simple that which is complicated. I also appreciate having misconceptions in my understanding corrected.

If I study and understand what is above should I be able to write down a metric similar to the one you wrote down but for a monopole-anti monopole pair separated by some distance?

Thanks!

11. Aug 10, 2016

### Ben Niehoff

A monopole-antimonopole pair is more difficult, because they will attract each other, and so the metric will depend on time. In fact, I'm not aware of anyone having written down such a solution in pure 5d gravity.

In 5d supergravity, however, one can have additional fields beyond just the metric. In N=2 SUGRA, you can have a number of Maxwell 2-forms and scalars (in 5 dimensions!). When you KK reduce to 4 dimensions, these additional fields can give you extra things. But anyway, using the extra fields, it is easy to write down a supersymmetric solution that contains a monopole-antimonopole pair, as well as a 5d electromagnetic field which holds them apart and thus allows the metric to be static. In fact, one can write down solutions for any number of monopoles and antimonopoles arranged in any way you like, together with a 5d Maxwell field of just the right form to hold all of the (anti)-monopoles in place.