Equivalence relations

[2017-09-17 Sun]

Equivalence relations are an important concept in mathematics, but sometimes they are not given the emphasis they deserve in an undergraduate course. Having a good grasp of equivalence relations is very important in the course MATHM205 (Topology and Groups) which I'm teaching this term, so I have written this blog post to remind you what you need to know about them. I will kick off with a few examples, then give a more formal definition.


Clock arithmetic

The simplest interesting example of an equivalence relation is equivalence of integers mod 2. You consider two integers to be equivalent if they have the same parity (both even or both odd), otherwise you consider them to be inequivalent. You end up with two equivalence classes of integers: the odd and the even integers. You can do arithmetic/algebra just as well with these equivalence classes as you could with the integers themselves: even + even = even, odd + even = odd, odd + odd = even etc.

More generally, given a number \(n\) you can consider equivalence of integers modulo \(n\): we write \(x\sim_n y\) if and only if \(n\) divides \(y-x\). We now have \(n\) equivalence classes according to the remainder you get when you divide by \(n\).


Let's take a different example, this time from topology. One way to think about a particle living on the 2-dimensional torus is as follows. You let the particle move around on the unit square in the \((x,y)\)-plane. Every time it falls off the right-hand side, you pick it up and put it onto the left-hand side at the same \(y\)-coordinate (and vice versa). Every time it falls off the top, you pick it up and put it onto the bottom at the same \(x\)-coordinate. In other words, we want to consider the points \((x,0)\) and \((x,1)\) to be the same point on the torus, and we want to consider the points \((0,y)\) and \((1,y)\) to be the same point on the torus. To formalise this, we introduce an equivalence relation on the set of points in the square for which \((x,0)\sim (x,1)\) and \((0,y)\sim (1,y)\) and we consider the torus to be the set of equivalence classes of points in the square modulo this equivalence relation. This proves to be a very useful way of constructing spaces (and we will see in the course how to equip the set of equivalence classes with a topology).


More formally, an equivalence relation on \(X\) is a subset \(\sim\subset X\times X\) with the following properties:

This becomes less mysterious if we write \(x\sim y\) instead of \((x,y)\in\sim\). Then we see that an equivalence relation is a criterion for two points in \(X\) to be identified as ``equivalent'', in such a way that this equivalence is symmetric in \(x\) and \(y\), reflexive (in the sense that \(x\) is equivalent to itself) and transitive (so that if \(x\) is equivalent to \(y\) and \(y\) is equivalent to \(z\) then \(x\) is also equivalent to \(z\)).

Given a point \(x\in X\), the equivalence class of \(x\) modulo \(\sim\) is the subset \[ [x]=\{y\in X\ :\ x\sim y\}.\] The quotient \(X/\sim\) is defined to be the set of equivalence classes modulo \(\sim\).

Examples revisited

Let us revisit the earlier examples, along with some new ones:


We use this final example of quotienting a group by a subgroup to illustrate some of the subtleties with equivalence relations.

It's always easy to form the quotient \(X/\sim\) as a set, but usually we have some structure (a group law, topology, or something) on \(X\) and we want \(X/\sim\) to inherit the same kind of structure.

When we quotient out by a subgroup, we usually don't have a canonical way to write down a group structure on the quotient. If we try the most naive thing and define the product on cosets by: \g_2\] then we run in to the following problem. Suppose we pick different representatives \(g_1\) and \(g'_1\) of the same equivalence class. Then \(g_1=g_1'h\) for some \(h\in H\). Now there is no guarantee that \(g_1g_2\) and \(g'_1g_2=g_1hg_2\) represent the same coset, which they would need to if the product we wrote down above were to be well-defined. Indeed, this only reliably works if \(H\) is a /normal subgroup/, that is \(ghg^{-1}\in H\) for all \(h\in H\) and \(g\in G\), because then \[g_1hg_2=g_1g_2h'\] for some \(h'\in H\), so \(g'_1g_2\) and \(g_1g_2\) define the same coset of \(H\).

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.