Is every diagonalizable representation of a group reducible?

[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

 

[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.

 

[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.

 

[ShayanJ]

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.

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

 

[Faust90]

Hey,

thanks for your answers! :)
I'm not actually sure if I understand this right.

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

Best regards :)

 

[ShayanJ]

Hey,

thanks for your answers! :)
I'm not actually sure if I understand this right.

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

Best regards :)

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!

 

[Faust90]

Thanks! :)

but then I'm a bit confused. For example, when I have a look at the D_n group and the representation of it.

http://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups

I know that this representation is irreducible but I could diagonalize all of these matrices?

Best regards!

 

[Andrea M.]

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

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 :)

 

[Andrea M.]

Thanks! :)

but then I'm a bit confused. For example, when I have a look at the D_n group and the representation of it.

http://groupprops.subwiki.org/wiki/Linear_representation_theory_of_dihedral_groups

I know that this representation is irreducible but I could diagonalize all of these matrices?

Best regards!

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

 

[ShayanJ]

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 :)

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.

 

[Andrea M.]

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!)

By "dimensionality parameters" and "order of the group" i mean respectively the dimension of the representation and of the group.

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 think you are right. If all the representations are one-dimensional the commutativity of the group follows from the definition of representation.