# Equivalence relations and equivalence classes

1. Feb 22, 2010

### Rezaderex

Hey!

Hoping you guys could help me with a small issue. No matter how hard I try, I don't seem to fully understand the notion of an equivalence relation, and henceforth an equivalence class. What I do understand that, in order to have and equivalence relation, it is defined to satisfy three rules, symmetry, transitivity....forgot the name of the last one...the one where A~B, B~C, implies A~B. I have read that the equivalence class is the set of those elements which satisfy those definition rules.....can someone please elaborate, none of the literature I have helps which a less abstract full definition (or at least my interpretation of one).

The books I do have are James Munkres, Topology, and Nakaharas Geometry Topology and physics. Both books arent helping. ALSO, if a kind soul has a little more time, the main reason why I question my understanding of equivalence relations is due to an example in Nakaharas p94, where he gives some x-y belonging to a subgroup of G....then defines a relation on them....im confused by this.

Thanks for any help
Cheers
Reza

2. Feb 22, 2010

### Ben Niehoff

Say we define an equivalence class by the relation

$$A = B \mod 2$$

where A and B are integers. What this means is that we consider any two numbers "equivalent" if their remainders when diving by 2 are equal. That is, we divide the integers into two sets: the odd numbers, and the even numbers. Each of these sets constitutes an equivalence class. All the odd numbers are equivalent to each other (under the relation above), and all the even numbers are equivalent to each other.

Basically, an equivalence class is taking the set of all objects which are equivalent under some (reflexive, transitive, commutative) relation, and referring to this entire set as one object.

For example, one can consider modulo-N arithmetic as an arithmetic on equivalence classes, instead of numbers, where any two numbers are said to be equivalent if

$$A = B \mod N$$

This splits the integers into exactly N equivalence classes.

When defining equivalence classes, one often says that we "mod out by" some given operation. For example, if we take the x-y plane and mod out by rotations by $\pi/2$, then we are left with the quarter-plane of points with both x and y positive. That is, we split the plane into equivalence classes where two points (x, y) and (x', y') are said to be equivalent if they can be rotated into each other by any multiple of $\pi/2$.

A slightly more tricky example is that we can consider the 2-sphere to be the quotient

$$S^2 = \frac{SO(3)}{SO(2)}$$

That is, if we take the full rotation group in 3 dimensions, and mod out by rotations around one specific axis, we are left with the 2-sphere. To see what is happening, consider the Euler angles. One must specify three angles: $\theta$ and $\phi$ locate a point on the unit sphere, which determines a line through the origin; the third angle $\psi$ then determines an angle of rotation about this line. If we consider equivalence classes of all points that can be related by rotations in $\psi$ (that is, all points related by the action of SO(2)), then what we are left with is just the two remaining Euler angles, $\theta$ and $\phi$. And as already mentioned, these two angles simply locate a point on the unit sphere; and thus, what we are left with is precisely the 2-sphere.

On p.94 of Nakahara, he is basically doing the same thing as this last example I gave you. The elements of a quotient group G/H are equivalence classes of elements in G, where the equivalence relation is defined by being related by an element of H. In my example above, G is SO(3), and H is SO(2). G/H is then the 2-sphere.

The - in "x - y" is not subtraction; it means "x + -y" where + is the group operation and -y is the inverse element to y. Abelian groups are often written using addition instead of multiplication.

Last edited: Feb 22, 2010
3. Feb 22, 2010

### Werg22

Let X be any set equipped with an equivalence relation ~. Let C(x) be the set of all elements in X that are equivalent to x. This is called the equivalence class corresponding to x. Given two elements a and b in X, you can easily show that either C(a) = C(b), or C(a) and C(b) are disjoint. Note that x is always in C(x), so X = U C(x), x is in X. Since the equivalence classes of the elements of X are either disjoint or coinciding, you can see that we may partition X into equivalence classes, i.e. express as the union of disjoint sets, which are all equivalence classes.

Also, note that if a and b are in a same equivalence class C(x), then a ~ b. If a is in C(x), b is in C(y), with C(x) not equal to C(y), then a is not equivalent to b. So in addition to partitioning X, the equivalence classes of the relation ~ group equivalent elements together.

That's all there is to them.

4. Feb 25, 2010

### Rezaderex

Thanks peeps, I was looking at the Nakaharas thing all wrong.... Either way thanks for the help!!!

5. Feb 25, 2010

### Studiot

Have a care where you fit the zero integer in these classes.

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