I’m intrigued by the shapes shown in “The Elegant Universe” and other introductions to superstring theory but (no topologist I!) cannot see what they are supposed to mean, nor how they are constructed. Here’s some examples:
I suspect that the holes have a kind of Klein bottle weirdness about them and the colours are indicative of extent in a further dimension than 3, and I know that these are merely 3-D cross-sectional representations of 6-D shapes. All I seek is some help, however vague and dumbed down, in making sense of these things (The wiki page wasn’t much help in this respect!).
I’m amazed none of our resident theoretical physicists have piled on already. Figures it falls to a geometer to explain physics. Please note: this will twist your brain more than the C-Y manifolds are.
The first thing you have to grok is the notion of a manifold. Basically, it’s this: something that looks, close up, like regular n-dimensional space. Examples: regular n-dimensional space (duh), a sphere (the surface of a ball), a torus (the surface of a doughnut). Basically these are things that, in a small area around any point, look pretty close to n-dimensional space, just like the Earth appears to be flat since we only look at a tiny patch of it at once.
Next you have to go back and rework what you mean by “n-dimensional space”, since the dimensions we’re really concerned with here are complex dimensions. That means we’re not looking at n-dimensional space as n-tuples of real numbers, but as n-tuples of complex numbers. Alternately, these are 2n-tuples of real numbers, and the numbers are paired off in a way that mimics the way pairs of real numbers make up complex numbers. WARNING: not all 2n-real-dimensional manifolds carry what’s called a “complex structure” – there are real 2n-manifolds that don’t come from complex n-manifolds.
I should give a couple examples here of honest-to-God complex manifolds. Both will be 1-dimensional.
The Riemann sphere is constructed like this: take two copies of the complex plane, and call the coordinate on one z and on the other w. Now, cut out a circle from each centered at the origin with radius a little bigger than 1. Glue them together by pasting points with z and w coordinates such that z = 1/w. That is, the point z=1 matches up with w=1, z=i matches with w=-i, and so on. This is a complex 1-manifold, since every point on the surface has a small area around it completely contained in either the z patch or the w patch, and so looks (locally) like C[sup]1[/sup].
We can do the same thing with four squares of complex plane. First, paste the top of one square to the bottom of the second, and the top of the second to the bottom of the first (use a similar notion of identifying overlapping points as we did around the equator of the sphere). This gives a tube. Do the same with the third and fourth squares to get another tube. Now, glue the edges of these tubes together into a doughnut shape. That torus is another complex 1-manifold.
Now, a bit of a technical aside: If the surface in question isn’t orientable (like the Klein bottle you suggest has no “inside” vs. “outside”), it can’t be complex, so these things are not weird in the same way a Klein bottle is.
Second technical aside: “compact” basically means it has no bits of the surface running off to infinity. You can put it inside a (sufficiently big) ball. Also, we’ll assume there are no edges.
“Kähler” is a technical notion related to the metric, which is a defined way of measuring lengths and angles of tangent vectors. No, that doesn’t just come naturally. Manifolds as we’ve defined them up until now don’t come with a notion of lengths and angles. In fact, there are many manifolds that have more than one essentially different such way. This is where topology (the shapes) crosses into geometry (the sizes). Basically, the metric is a machine at every point that takes two tangent vectors at that point and spits out a number, similar to the dot product you may remember from physics.
There’s a different way of looking at the metric, though, which is basically as a sort of function, though in a sense that’s too hairy to get into right now. What Kähler means is that this function doesn’t change as you move around the surface. That is, in some sense, the notion of measuring lengths and angles is the same everywhere. This might sound weird, but remember that a lot of the predictions of general relativity come from the fact that mass-energy bends space in such a way that the metric of spacetime does change as you move around.
Now, that’s not to say these things aren’t curved. Kähler manifolds can be curved plenty, but the Kähler condition provides a check on how it can curve. Curvature is basically described by a matrix-valued function. If you take a tangent vector and move it around a loop in the manifold, in changes…
Okay, have to explain that. The whole idea of putting a vector anywhere you want like you were told in school is bogus and only works in “flat” space. The whole point of curvature is that as you move, directions change. Here’s the clearest example I can give:
Stand at the North pole, holding an arrow pointing towards Greenwich (England, not CT). Walk down the prime meridian, never turning your arrow at all until you hit the equator. Now, turn East and walk forward until you’re a quarter of the way around the Earth. Don’t turn the arrow, though, so it stays pointing south. Now, turn North again, the arrow now pointing behind you. When you get back to the North pole and check the arrow against the copy you left behind when you started, the arrow you carried is turned by 90 degrees.
Okay, so for every path that comes back to its starting point, there’s a matrix that tells how vectors transform when you move around that path. Now (hold on tight), on an n-dimensional manifold, you can take certain combinations of nth powers of these matrices, take the determinant of the resulting matrices, and get an integrand, suitable for integration over the manifold just like in multivariable calculus. The Calabi-Yau condition is that this integrand (and, a forteriori, its integral over the manifold) is zero.
Now that I’ve totally lost you, what does it matter? What Calabi conjectured – and Yau proved – is that given these insanely abstruse conditions, the manifold will be sort of flat. Still not totally flat, but flat in a way that just happens to correspond to Einstein’s empty space (no cosmological constant) equations. That is, if you think of this as a sort of spacetime, the equation of general relativity tells you that there’s no mass-energy in it. It’s all just curved, empty space.
A final n.b.: This doesn’t mean there’s no curvature, since “gravitational waves” are solutions of the GR equations in our regular everyday spacetime with no mass-energy present.
Mathochist, you are a credit to these boards. Thanks for the time and effort you put in there.
I hereby cast aspersions on the manhood and/or fortitude of the SDMB’s theoretical physicists, in the hope of goading them into stepping up to the plate also.
I’m a physicist, more computational than theoretical, and this stuff is mostly beyond me. Great explanation, Mathochist!
One note: I believe “‘gravitational waves’ are solutions of the GR equations in our regular everyday spacetime with no mass-energy present” is incorrect. Gravitational waves have both energy and momentum, but no associated matter (which is probably what you were thinking of).
Gravitational waves are vacuum solutions to the GR equation.
The Einstein tensor is on one side of the equation, and the stress-energy tensor is on the other. Gravitational waves are solutions of the equation you get when you set the stress-energy equal to zero, so the Einstein tensor must vanish as well. Equivalently (to the vanishing of the Einstein tensor), the Ricci tensor vanishes, and the space is “Ricci flat”, which is the conclusion of the C-Y theorem.
I’ll defer to your mathematics, Mathochist, but I can’t understand how a gravitational wave could not have energy and momentum. How else could they be detectable? Wikipedia’s Gravitational Radiation entry (a non-authoritave cite) indicates that they do.
I understand and agree that gravitational waves are vacuum solutions. After all, they do not require matter. Does the stress-energy tensor only apply to rest mass/energy?
An analogy to electromagnetics: e-m waves are vacuum solutions to Maxwell’s equations, yet they have both energy and momentum.
Ah, looking at Wikipedia’s Stress-energy Tensor, I believe the answer is the tensor represents non-gravitational energy. It is clearly true that a gravitational wave has no non-gravitational mass/energy/momentum. But it does have gravitational energy/momentum. Phew, I feel better now.
More or less. Basically, when physicists talk about “gravitational waves”, they’re really talking about a first-order perturbational solution about some background metric, usually flat spacetime. You can try to define an effective second-order perturbational stress-energy tensor for a gravitational wave, but it’s a rather subtle procedure and I won’t pretend that I thoroughly understand it.
(If perturbation theory isn’t your cup of tea, BTW, let me know and I’ll try to explain it more intuitively.)
Right, but they don’t have charge, which would be a better analogy. Just as Maxwell’s equations can be written as (derivatives of the EM field) = (currents and charges), Einstein’s equations can be written as (derivatives of the metric) = (stress-energy tensor).
