# Representations of finite groups

1. May 19, 2010

### kylie14

I'm just having a little trouble getting my head around how representation theory works.

Say for example we are working with the dihedral group D8. Then the degrees of irreducible representations over C are 1,1,1,1,2.
So there are 4 (non-equivalent) irreduible representations of degree 1, and one of degree 2. But what does this mean exactly? Can you use each one separately, or do you need all together to have the full representation?

Sorry if what I'm asking is really unclear.
I understand why we need to use representations of groups, and I've even found them for a higher order dihedral group.
My problem is in understanding why you get several for each group and why some are 1D and some 2D in, say, the case of D8.

2. May 20, 2010

### CompuChip

In a sense, irreducible representations are the 'building blocks' of any representation of a group.

So a representation consists of a set of matrices, one linked to every group element, such that the matrices behave the same (under matrix multiplication) as the group elements (under the group operation). Since an arbitrary group can be rather abstract and we have good intuition and an extensive set of mathematical tools (aka linear algebra) for matrices, this is of course very useful.

Now we can make the dimension of the matrices in the representation as large as we want. For example, we can add to any n x m matrix rows and columns of zeroes, which is rather trivial. A more complicated way to make a "new" representation is to stack different matrices block diagonally in a larger matrix - you can easily check that if every "block" satisfies the required relations, so does the entire matrix. We can even scramble these a bit, by performing an arbitrary (one-to-one) basis transformation. Then technically, the matrix is still made up of the same representations, but it may look very complicated.

Now it turns out, that when you filter out this operation, there are actually only so many "distinct" representations of a group, the irreducible ones. Let me put that another way. If you take a matrix in an arbitrary representation, you know from linear algebra that you can choose some basis which puts it in block diagonal form. The blocks you can get, due to the way diagonal matrix multiplication works, must all be (smaller) representations of the group themselves, and these are limited in number according to the theory.

So suppose you want to study arbitrary representations of a group. This is useful, I repeat, because all the information about the group structure is also captured in the representations, and these consist of matrices which are - in general - far less abstract than your average group.
In general there are infinitely many, but they are all built up of a small (in some cases still infinite, but in any case more tractable) number of irreducible representations. Studying these also gives us all the information you need.

3. May 20, 2010

### kylie14

Thank you, that does clarify things for me. Say you had a 1D irreducible representation of a finite group, though, and then also a 2D one for the same group; is there any advantage in using the 2D one?
What I mean is, if there is a 1D representation possible, why bother with a 2D one even if both are irreducible?

4. May 20, 2010

### CompuChip

For several reasons:
* It might be that a higher dimensional representation is more natural (for example, the natural representation of SO(3) is - obviously - three dimensional, although lower-dimensional representations exist)

* There always exists a 1-dimensional representation (which sends every group element to the 1-dimensional identity matrix, aka the number 1) but it's not very helpful in general.

* More generally, representations are not necessarily faithful (faithful means that every group element is represented by a distinct matrix).

* Suppose that you happen to find some set of matrices and you compute all their products and see that they are a representation of your group. Then all irreducible representations may be present.

Saving the best reason for last:
* Suppose that you are given some set of matrices, and you want to check if they form a group. Of course, you can check the group axioms by multiplying them all out. If you write down the relations that you find in the process, you might even be able to deduce if it's a group you already know (more strictly speaking, if your unknown group is isomorphic to a group you already know). Or you can go the easy way: block diagonalize your matrices, extract the irreducible representations, and see which group they define.

5. May 20, 2010

### kylie14

That all makes sense to me; thanks for explaining it so clearly!