Conformal Equivalence: Help with Penrose's Road to Reality

[harshant]

Hi fellas

I have been reading Road to Reality by Roger Penrose, but can't go beyond chapter 8. I do not understand why topological equivalence does not imply conformal equivalence. In particular I cannot really make sense of his argument as to why a thin torus is not conformally the same as a fat one something he refers to as "pretty clear". Another source of confusion is that as far as I know any two parallelograms can be transformed into each other by a holomorphic transformation (Riemann mapping theorem). Doesnt this imply that the tori generated by these parallelograms should be conformally equivalent? Please help.

 

[maze]

I don't think RMT applies to closed regions. I think it would be very unlikely that you could, for example, conformally map a closed square region to a closed triangular region. What would happen to the 4th corner?

 

[harshant]

RMT only applies to open regions but we can always exclude the boundary of the parallelograms and treat them as open subsets of complex plane. Then, can we find a holomorphic mapping between these regions? And if we can why are the generated tori not conformally equivalent? As for the 4th corner we really cannot talk about it since we are excluding the boundary.

 

[maze]

In 2D, for open regions, I think you're right. But not necessairily for closed regions, and in 3D and higher there is no equivalent of RMT so all that goes out the window.

As for the tori, I haven't completely thought it through, but perhaps something like the construction in this picture could be used:
http://img155.imageshack.us/img155/8638/toriconformallymc1.png

 

[harshant]

Thanks for the image maze, that really cleared things up. Also for a closed region RMT does not work. The open region which is enclosed by the boundary can be mapped conformally to any other open region but the boundaries of the two will not map to each other using this mapping in general. But now the question is that for making a torus do we necessarily need to include the boundary of the parallelogram or can we only consider the interior of the parallelogram?

 

[Doodle Bob]

But now the question is that for making a torus do we necessarily need to include the boundary of the parallelogram or can we only consider the interior of the parallelogram?

Yes, you do. In order for two tori given as the space of equivalence classes of two lattices on the complex plane to be holomorphic (and hence conformal -- since conformal implies holomorphic in 2 dimensions), the lattices need to be holomorphic to each other, which is equivalent to the existence of a holomorphic mapping that maps the generators of one lattice onto the generators of the other.

A little experimentation with this idea will reveal to you that there are indeed quite a few (infinitely many) conformal structures on the topological 2-torus. Look at the generators of the lattice for the square torus versus a torus with a fundamental domain given as a parallelogram with one side being vertical (parallel to the y-axis) and the other sides having an irrational slope.

This is another way to understand what Penrose is getting at.

 

[harshant]

Hey thanks Doodle Bob, the argument was a bit advanced for me and it could only sink in after some research on lattices etc, but I think I have more or less got the gist of your argument. I have some clarifications though :-

1. Let us assume that we define a torus using identification mappings or by gluing opposite edges of a parallelogram. Now what do we get if instead of gluing parallel straight lines, we glue together arbitrary curves (see attachment)? My guess is a torus-like figure with changing cross-section. Can this figure ever be equivalent to a torus (excluding the case where the curves are parallel)?

2. Let us say that we have the conformal mapping from the interior of one parallelogram to another. Now in general the boundary of the original parallelogram will map to some arbitrary closed curve with 4 corners. Now if we try to construct a figure by gluing together opposite edges of the new boundary as above, will we get the original torus formed by the old boundary? Will we get a torus at all? Another way to ask this question would be whether the interior of the parallelogram really matters when constructing a torus (exact opposite of what I asked before!) or do all conformal images of the boundary give the same torus irrespective of whether or not the image of the interior lies within the new boundary or not?

 

[Doodle Bob]

It occurs to me that there is a subtle point that is being allusive here: namely, that there is a difference between a topological torus and tori with a given geometric structure. Topologically, all 2-dimensional tori are equivalent, i.e., each is homeomorphic to a the standard "donut". This includes your clarification question #1, in which the gluing is over vaguely arbitrary continuous curves. The only issue for the topological torus is to ensure that the bijection from the donut to the proposed torus is that it is continuous.

Things change when there is specific measurements, i.e., geometry, that you want to be the same from one torus to another. Now, you have to worry that the bijection from one torus to another preserves some aspect of the differential geometry of each. For example, the Riemannian metrics or the complex structures.

A very deep result of Riemann surface theory tells us that every torus (as a complex manifold) is biholomorphic to the complex plane modded out by the lattice formed as the integer linear combinations of some basis (over R) of C. The square lattice given by the points m+in, where m and n are integers, is one such lattice. A parallelogram for this is the 1x1 square bounded by vertices (0,0), (0,1), (1,0), and (1,1).

Another lattice is given {(m+n)+ sqrt(2)m i: m,n are integers}. A parallelogram for this one would be given with vertices (0,0), (1,0), (2,sqrt(2)), (1,sqrt(2)).

There *is* a conformal mapping from one of these parallelograms to the other; that's the Riemann Mapping Theorem. BUT it will not be an affine map and hence not preserve the sides of the parallelograms nicely. That is, it will not map one side of the domain parallelogram linearly onto one side of the target parallelogram. In fact, the mapping will totally screw up the sides. Hence, this map cannot be extended to the *closed* parallelograms in such a way that it becomes a conformal map on the tori.

In order for the mapping given by RMT to extend to a conformal map to the entire tori, it has to preserve the lattices that created the parallelograms in the first place. So, it has to be a conformal linear map, i.e., the composition of a point similarity about the origin and some rotation matrix.

I hope that helps. I can't tell if I addressed your clarification question #2.

 

[Doodle Bob]

There *is* a conformal mapping from one of these parallelograms to the other; that's the Riemann Mapping Theorem. BUT it will not be an affine map and hence not preserve the sides of the parallelograms nicely. That is, it will not map one side of the domain parallelogram linearly onto one side of the target parallelogram. In fact, the mapping will totally screw up the sides. Hence, this map cannot be extended to the *closed* parallelograms in such a way that it becomes a conformal map on the tori.

Let me clarify this paragraph, since it addresses your question #2. The conformal mapping from one open parallelogram to another will extend to its boundary but in a strange way, ie, non-linearly. It is still possible to derive equivalence classes on the boundary of the target parallelogram via this mapping (i.e., gluing) and get something holomorphic to the original domain torus, but it will not be the same torus as that given by the target parallelogram with its line segments glued together in the usual way.

 

[harshant]

Thanks again Doodle Bob, I was thinking on similar lines as your last post.