Some orders on groups

A left order on a group is a total order of the underlying set such that left multiplication by any element of the group is monotonous. Dually we have the definition of right order. A bi-order is an order that is both a left and a right order.

Example 1: The usual order on the integers is a bi-order for the usual group structure.

Example 2: The same is true for the real numbers.

Problem 1: Give a bi-order in the free group generated by an arbitrary set.

Now we will give a very strong consequence of having a right order. First we start with a nice excercise.

Problem 2: Find an infinite countable discrete subset of the real numbers such that the induced order in this set does not have maximum nor minumum and such that for any pair of distinct elements there is a third element in between.

We are ready to state the following very nice result:

Problem 3: Let B be a triangulated locally finite space. Suppose given a right order in \pi_1(B). Let p : E\to B be the universal cover of B. Then there exists h : E \to \mathbb{R} such that (p,h) : E \to B \times \mathbb{R} is an embedding.

In particular the universal cover of B embeds in B\times \mathbb{R}. Compare this with the classical drawing of the universal cover of S^1!

Some orders on groups

Covering Surfaces

Compact orientable surfaces are classified (up to homeomorphism) by its genus. The genus is just a natural number. Thus we can enumerate them: The genus zero surface is the sphere S^2, the genus one surface is the torus T^2, the genus two surface can be constructed by taking two toruses, cutting a little disc out of each torus, and then gluing the two boundaries that we created when cutting the holes. In this way we can construct the genus n compact orientable surface for each n. The operation of cutting a little disc out of two (connected) surfaces and gluing them by the boundary is called the connected sum. For two surfaces A and B their connected sum is usually denoted by A \# B. For n \geq 0 we denote the genus n surface by n T^2 = T^2 \# \ldots \# T^2. When n is zero we define the connected sum of zero surfaces to be the sphere S^2.

Problem 1: Prove that n T^2 covers 2 T^2 for any n>1.

The classification of compact non-orientable surfaces is similar. If P^2 is the projective plane then a compact non-orientable surface is homeomorphic to n P^2 for some positive natural number n.

Problem 2: Prove that (n-1)T^2 covers n P^2 for any positive natural n.

Problem 3: Notice that T^2 \# P^2 is a compact surface. Describe it.

Problem 4: Prove that n P^2 covers 3 P^2 for any n\geq 3.

A nice corollary of this exercise is that, since composition of finite coverings is again a covering, the fundamental group of any compact surface is a subgroup of the fundamental group of 3 P^2, except maybe for T^2 and 2 P^2.

Covering Surfaces