# Preferred directions on a torus?

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I would like to regard an intrinsically flat 2-torus. This is usually sketched as a square with the left and right edge identified and the upper and lower edge identified, respectively.

The four corners of the square represent the same point. Now I connect this point to itself via a loop on the torus, which shall be a straight line, and there shall be only one winding. There are only two possibilities: Either along one of the edges, which is the short distance; or along one of the diagonals, which is the long distance.

Does this mean that there are preferred directions on the torus or is this only an artefact of an inappropriate graphic representation? Sorry, if this is a dumb question and thanks for any answers.

andrewkirk
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It depends on whether we are talking about a torus as a bare topological space or as a Riemannian manifold. The difference is that the latter has more structure around the notion of 'direction', in particular it has the notion of geodesics, which are the equivalent of straight lines.

A Riemannian torus has preferred directions. If, given a point ##p## on the torus, for any direction (specified by a tangent vector ##\vec v_p##) we define ##d(\vec v_p)## to be the distance, along the geodesic that follows that vector, that one has to travel until one crosses one's past path, that distance will vary by direction, with the shortest distance being for the vector that points directly around the short radius of the torus

For a bare topological torus - which is any set homeomorphic to a Riemannian torus, but without the metric structure - there is no notion of direction, so there cannot be a preferred direction.

mathwonk
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I assume you are taking n ot about direcvtions but about closed loops, i,.e,. whether there are distinguished loops in the torus. I think not, and you can test this by cutting your rectangle along its diagonal and regluing it so that the old diagonal becomes an edge. Then your same construction yields different choices if loops.

algebraiclly this sems to coirrespond to choosing a fre basis of two vectors in the group of (homology) loops for the torus. I.e. consider the abelian group ZxZ, with the symplextic form (a,b).(c,d) = ad-bc, which corresponds to the intersection pairing on the homology of the torus. The two loops you chose have the property that each intersects itself in zero and intersects th other in 1. But in fact all loops (a,b) have wlf intersection zero, and I susopect you can always choose a second one to, no wait, maybe not. Some loops (a,b) seem to have the property that there is a second loop that meets it in intersection number 1, but others do not. (i guess (2,2) cannot meet any other vector in just 1.) maybe those where a,b are rel prime? anyway i have to go make dinner, so play with this.

later:

no wait, sorry now i think you did mean directions in the sense that your loops or lines all go through the given point represented by the corners of your rectangle, hence represent a finite number of distinguished directiosn from that point. my same answer still seems to apply. and maybe it links up with the riemannian answer by asking which geodesics passing through your point, both close up on thmselves and form a loop that is part of an algebraic basis for the homology group of 2 cycles.

If one of your edges is represented by (1,0) and the other by (0,1), then your diagonal is (1,1), and I am guessing thqt all others correspond to vectors (a,b) where gcd(a,b) = 1. just a guess, but an informed one.

so to me the problem is your statementthat there are only three straight lines that do what you ask. here is another candidate: draw a straight line from the upper right corner to the middle of the left vertical edge. then continue that by drawing a straight line from the middle of the right vertical edge to the lower left corner. that will correspond to the vector (2,1). Then the vector (3,2) represents a loop with oriented intersection number =m 1 with that one.

but hey maybe if we want not just priented interscetion number but physical iontersection, we get abck to your choices?

no hey wait, we can pair (2,1) with (1,0) AND SEEMINGLY GET A ONE POINT OHYSICL INTERSECTION. i.e. the last broken line i gav e above meets the bottom edge only once.

so you could do the same with every loop of form (n,1), pairing with (1,0).

Now in general it seems two vectors (a,b) and (c,d) in the plane work as a basis iff the oparallelogram they spane has area ad-bc = 1. But then maybe we can use that parallelogram as the representation for the torus, and get a parallelogram having (a,b) as one edge and (c,d) as thw other. and these edges sem only yo meet physically once? so maybe we are back to the original conjecture that (a,b) has such a partner iff gcd(a,b) = 1. ??? can you see this in your original rectangle? anyway we are getting an infinite collection of (distinguished) directions.

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Thank you very much. I was talking about a torus as a Riemannian manifold (which shall be intrinsically flat). Highly interesting that on such a torus there are preferred directions, while on a sphere there are not (or is the latter wrong?).

andrewkirk
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or is the latter wrong?
That's correct. The sphere ##S^2## is isotropic (the same in all directions), while a Riemannian torus - at least the versions we are most familiar with, that are embedded in 3D space - is not. I suspect that there is no metric that can be put on a torus that makes it isotropic. It would be interesting to try to prove that.

lavinia
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I suspect that there is no metric that can be put on a torus that makes it isotropic. It would be interesting to try to prove that.

No two dimensional torus can be isotropic. Its Euler characteristic is zero so unless it is flat it must have regions of positive and negative Gauss curvature( by the Gauss-Bonnet Theorem).

An isometry of a flat torus would lift to an isometry of the Euclidean plane that preserves the group of covering transformations and this group is a lattice. So the flat torus is not isotropic.

Note that a flat torus is homogeneous even though it is not isotropic.

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lavinia
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On a flat torus there is always a closed geodesic of minimal length - although it may not be unique. There are no closed geodesics of maximal length and most geodesics are not closed.

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Infrared
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Perhaps I'm missing something, but doesn't

Some loops (a,b) seem to have the property that there is a second loop that meets it in intersection number 1, but others do not. (i guess (2,2) cannot meet any other vector in just 1.) maybe those where a,b are rel prime?

the symplextic form (a,b).(c,d) = ad-bc, which corresponds to the intersection pairing on the homology of the torus.

lavinia
Gold Member
Continuing on Mathwonk's line of thought.

Start with the flat torus obtained from the plane by identifying points whose (x,y) coordinates differ by integers. That is: ##(x,y)## is identified with ##(z,w)## if ##x-z## and ##y-w## are both integers. By adding or subtracting some integer from each coordinate one sees that every point is identified with some point in the unit square so one can think of the torus as all points in the unit square with opposite edges identified.

But the unit square is not the only region like this. There are an infinite number of parallelograms in the plane that also work. Every point in the plane is identified with some point in the parallelogram and opposite edges of the parallelogram are identified.

The edges of these parallelograms may be thought of as distinguished directions - but they may not be the same directions as the sides of the original unit square.

A way to generate these parallelograms is to take a linear transformation of the plane that is area preserving and which preserves the lattice of points with integer coordinates. Such a linear map warps the unit square into a parallelogram of area one whose opposite sides are identified. Take for instance the map,

##(x,y) →(x + ny, y) ## This map takes the unit square into a parallelogram whose base is still the line segment ##[0,1]## on the x-axis but whose other edge is now a line segment connecting the origin to the point ##(n,1)##. This is a new distinguished edge.

So mathwonk's point is that depending on which parallelogram one uses one gets different distinguished edges and there are infinitely many possibilities.

Problem: For each fundamental parallelogram above, do the edges necessarily project to generators of the fundamental group of the torus?

Problem: Is there an area and lattice preserving linear transformation that maps the unit square onto a paralellogram whose sides are arbitrarily given generators of the fundamental group of the torus?

- Another way to find all of the distinguished lines is to draw all of the straight lines connecting two points in the lattice. This will give every closed geodesic starting and ending at the projection of the lattice points onto the flat torus.

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