# Irreducible representation. I'm confused.

1. Jan 6, 2014

### LagrangeEuler

All commutative groups have one dimensional representation
$D(g_i)=1, \forall i$
I understand what is representation. Also I know what is one, two... dimensional representation. But what is irreducible representation I do not understand. How you could have two dimensional irreducible representation of commutative group when you have always one dimensional one. Also you could have two two-dimensional representation of which one is reducible and one is irreducible. I'm confused with the concept of irreducible representation. If you have group $G$ of order $ordG$ with $K$ classes then if
$\sum_{i=1}^{K}n_i|\chi_i|^2=ordG$
group is irreducible. Otherwise is reducible. $n_i$ is number of elements in every class.

2. Jan 6, 2014

### Office_Shredder

Staff Emeritus
A representation is a map from G to GL(n,F) such that in Fn (where F is your field, usually R or C) there is no stable subspace. Even restricting ourselves simply to the group Zn, if Zn is acting on R2 for it to have a one dimensional subspace which is stable then the image of the representation must have a common eigenvector. Since Zn is generated by one element this is equivalent to saying that D(g) has an eigenvector, but that is non-trivial for a general real matrix in R2 to have an eigenvector. For example rotation matrices have no eigenvectors over the reals (which can be used to create irreducible representations).

Intuitively, the point is that even though there exists a one dimensional representation, it is unlikely that my two dimensional representation looks like that one dimensional representation on any one dimensional subspace.

3. Jan 6, 2014

### LagrangeEuler

This is a problem for me. What I have from knowing that some representation is irreducible. If I have two two-dimensional representation, why is important for me that one of them is irreducible? So rotational matrix with elements $a_{11}=\cos \theta, a_{12}=\sin \theta,a_{21}=-\sin \theta,a_{22}=\cos \theta$ do not have real eigen-values? I'm not sure how to find is this representation of rotation irreducible and why is that important?

4. Jan 6, 2014

### Office_Shredder

Staff Emeritus
You don't show the representation of a rotation is irreducible, because "representation of a rotation" doesn't have meaning. The group Z4 has a representation in which the elements are mapped to rotation matrices; I recommend first that you calculate such a representation. Then tell me what your definition of an irreducible representation is and we'll work from there.

As far as importance, typically from my experience you don't care about whether a specific representation is irreducible or not. What is useful is that in the complex setting, any representation can be split into a sum of irreducible representations, and theorems which are hard to prove become easier by proving it for an irreducible representation only. There are probably theorems about irreducible representations that are important over the reals as well but I don't know them off the top of my head.

5. Jan 6, 2014

### George Jones

Staff Emeritus
In physics, irreducible representations are extremely important, and often have physical meaning. For example,

The idea is that invariant subspaces in some sense characterize "elementary" physical systems whose states only transform amongst themselves under the action of the group representation.

With respect representations of the rotation group, in quantum theory, if the rotation group acts on two irreducible subsystems of the total system, then the action of the rotation group on the of the total system is the tensor product of the two irreducible representations, which is usually reducible. The Clebsch-Gordan coefficients seen in quantum mechanics arise by writing this tensor product representation as a direct sum of irreducible representations.

Last edited: Jan 6, 2014
6. Jan 7, 2014

### LagrangeEuler

I need to put $\theta=0,\theta=\frac{\pi}{2},\theta=\pi,\theta=\frac{3\pi}{2}$. I get four different matrices. Characters are
$\chi_1=2, \chi_2=-2,\chi_3=0$
and $n_1=1,n_2=1,n_3=2$
$\sum_in_i|\chi_i|^2=4+4=8$
Representation is reducible. Right?

7. Jan 7, 2014

### Office_Shredder

Staff Emeritus
Is your definition of irreducible "the sum of the squares of characters equals the order of the group" (which is a terrible way to define it) or have you just proven that to be equivalent to the definition of irreducible.

8. Jan 7, 2014

9. Jan 7, 2014

### Office_Shredder

Staff Emeritus
That pdf doesn't define what an irreducible representation is, it assumes you already know. It simply talks about what the characters of irreducible representations are. Go back and look up what an irreducible representation is. (hint: it's in chapter 3)

10. Jan 8, 2014

### LagrangeEuler

I read chapter 3. I'm not sure what you want me to do. It's obvious in case $Z_4$ that the representation with rotational matrices is reducible.

11. Jan 8, 2014

### Office_Shredder

Staff Emeritus
It's not obvious because it depends on what field you are working over, the reals or complex numbers. All I want you to do is tell me what the definition of an irreducible representation is. You should also think about what that definition means geometrically.

12. Jan 9, 2014

### LagrangeEuler

I find eigen-values of rotational matrices which form group $Z_4$ and I got that some of them has complex eigen-values $i,-i$. All of them has real entries. Is that mean that group is reducible? And why?

13. Jan 9, 2014

### Office_Shredder

Staff Emeritus
You still haven't said what the definition of an irreducible representation is. It's going to be impossible to answer the question of 'is this representation reducible' if you don't know what the definition of reducible is!

14. Jan 10, 2014

### LagrangeEuler

In it is some linear subspace $W$ of space $V$ representation is called $G$ invariant if for any element $w \in W$ and every element $g$ in $G$.
$w\cdot g$ is in $W$. The restriction of representation to $W$ is known as subrepresentation. If group has only trivial subrepresentation it is irreducible. I'm not sure how to find in some cases invariant subspaces.

Last edited: Jan 10, 2014
15. Jan 10, 2014

### Office_Shredder

Staff Emeritus
Yes, very good. OK, now let's look at our rotation example. Here is where it is very important that we be clear what field we are working over. If we have a representation
$$Z_4 \to GL(\mathbb{R},2)$$
that means we are considering real matrices only, and they are operating on $\mathbb{R}^2$. Consider the rotation by $\pi/2$. Is there a subspace W of $\mathbb{R}^2$ which is stable under a rotation by pi/2?

The answer is no, and it should be clear geometrically that this is impossible. Mathematically, any subspace of $\mathbb{R}^2$ is a one dimensional subspace, and if a one dimensional subspace is stable under a matrix acting on it then the spanning vector of that subspace is an eigenvector. So you can confirm that the rotation by pi/2 has no eigenvectors when it is acting on $\mathbb{R}^2$ and hence the representation under the reals is a two dimensional irreducible representation.

On the other hand, we can consider the representation
$$\phi: Z_4 \to \mathbb{C}^2$$.

Let g be a generator of $Z_4$ such that $\phi(g)$ is the rotation by pi/2 matrix. As you observed this matrix has eigenvalues i and -i, and two eigenvectors to go with them. Those eigenvectors span stable one dimensional subspaces of $\mathbb{C}^2$. Furthermore, the only other matrices we have to consider are $\phi(g^2) = \phi(g)^2$ and $\phi(g^3) = \phi(g)^3$. In general if v is an eigenvector of A, it's an eigenvector of A2 and A3 as well, so those eigenvectors we found are eigenvectors of $\phi(g^2),\phi(g^3)$ and trivially as well $\phi(g^4)$ which is the identity matrix. Therefore they span stable one dimensional subspaces for the whole representation, and as a representation over the complex numbers the rotation matrix representation is reducible.