Sunday 9 March 2008

Schur's Lemma

Schur's Lemma is a nice piece of group theory apparently needed to prove the orghogonality relations which are in turn another nice piece of group theory. I did not find the whole proof anywhere, so I will show it here. The nice thing is: Once you put things into correct mathematical notation, you're almost done.

We have a group G whose elements we call R. And we may think of the elements as symmetry operations.

We have two representations α and β of this group. So what is a representation of a group? In most cases you may think of it as a vector space V or its basis, and let the group operations work on it. To be precise you have to define how the group operations work on this vector space. Then a representation becomes a mapping (a homomorphism) between the group G and the group of invertible linear transformations on V (instead of linear transformations you may also think of matrices).

Typically β=α but for this prove it can be different and may work on a different vector space W

For Schur's Lemma we also need that α is irreducible. That means every subspace U of V that is invariant with all the group operations contains either just the null vector or is the whole V.

Finally we have another linear transformation (or matrix) f

which has the following property

In other words the two mappings are the same for all the group operations.

Then Schur's Lemma says that f has to be either the 0 mapping or it has to be invertible. In the second case are the dimensions of the representations the same, dim(V)=dim(W):

If you have read this far, you may be wondering why that is.


First we can assume that

without loss of generality because we could just relabel things if they aren't.

You have to consider the kernel of f. First we make the following observation

In other words:

But because α is irreducible, we know that ker(f), a subspace of V, has to look like:

In the second case obviously because everything is mapped onto the null vector. In the first case f is injective and it has to be bijective because W has at most the same dimension as V (it turns out that dim(V)=dim(W)).

That's what we wanted to show.

The applications of this are also kind of interesting. Maybe later.

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