Recall that a left order on a group is an order of the underlying set such that left multiplication by any element of the groups is monotonous (for very simple examples see Some orders on groups). Define an element of an ordered group to be positive if it is greater than the unit of the group. Define an element to be negative if its inverse is positive. Clearly an element of an ordered group is either positive, negative or it is the unit of the group.
We now restrict our atention to orders on the group . The goal is the following: Suppose given an order on , prove that there exists a straight line passing through such that the elements strictly above this line have all the same sign, and the elements strictly below this line have also the same sign. That is, the elements strictly above are either all positive or all negative, and the same holds for the elements strictly below the line.
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 be a triangulated locally finite space. Suppose given a right order in . Let be the universal cover of . Then there exists such that is an embedding.
In particular the universal cover of embeds in . Compare this with the classical drawing of the universal cover of !