# Group theory - beginner

1. Nov 15, 2014

### Faust90

Hey folks,

I'm trying to dip into group theory and got now some questions about irreducibility.

A representation D(G) is reducibel iff there is an invariant subspace.

Do this imply now that every representation (which is a matrix (GL(N,K)) is reducibel if it is diagonalizable?

Best regards

2. Nov 16, 2014

### ShayanJ

It should block-diagonalizable which is more general than only being diagonalizable. A block-diagonal matrix, is a square matrix that can be thought of as having square matrices in its main diagonal and all other elements being zero. Then each block in the main diagonal is itself a representation.

3. Nov 16, 2014

### Andrea M.

Shyan is right, when you want find the irreducible representation you should block diagonalize. Only when your group is abelian block diagonalize is equivalent to diagonalize because, as you can prove using the Schur's lemma, all the irreducible representations of any abelian group must be of dimension one.

4. Nov 16, 2014

### ShayanJ

Is there any non-abelian group with all of its representations being one dimensional? i.e. is the converse true?

5. Nov 16, 2014

### Faust90

Hey,

I'm not actually sure if I understand this right.

Does block-diagonalizability implies diagonalizability or is it the other way round?

Best regards :)

6. Nov 16, 2014

### ShayanJ

Being diagonal is a special case of being block-diagonal so all diagonal matrices are block-diagonal but not all block-diagonal matrices are diagonal!

7. Nov 16, 2014

### Faust90

8. Nov 16, 2014

### Andrea M.

I think the converse il also true. Indeed we know that the dimensionality parameters $n_{\mu}$ for the inequivalent irreducible representation satisfy
$$\sum_{\mu}n_{\mu}^2=n_{G}$$
where $n_{G}$ is the order of the group. This implies that, if all the representation is one dimensional, the number of inequivalent representation must be equal to the order of the group. But we also know that the number of inequivalent representation of any finite (or compact) group is equal to the number of distinct conjugate class of G, so each element of G must be conjugate to itself so the group is abelian(?).
I'm not shire about the last step, i will think about it. Any suggestion is welcome :)

9. Nov 16, 2014

### Andrea M.

Are you sure that you could diagonalize with the same basis all the matrix of the representation?

10. Nov 16, 2014

### ShayanJ

I was missing something. Representations of a group should provide entities associated to the group elements and a composition law on them, such that they implement the structure of the group. But one dimensional representations are numbers and all numbers we have, with the usual products, are commutative so non-abelian groups can't have one dimensional representations, at least not until we find a composition law on some kind of numbers that is non-abelian.
I should say that I don't understand what you mean by "dimensionality parameters" and "order of the group".(It seems by order of the group, you don't mean its set's cardinality and I know no other meaning!)

To Faust: diagonal means having non-zero elements only on the main diagonal so $\left( \begin{array}{cc} 0 \ \ a \\ b \ \ 0 \end{array} \right)$ is not diagonal.

Last edited: Nov 16, 2014
11. Nov 17, 2014

### Andrea M.

By "dimensionality parameters" and "order of the group" i mean respectively the dimension of the representation and of the group.
I think you are right. If all the representations are one-dimensional the commutativity of the group follows from the definition of representation.