I am currently reading Applications of Group Theory by M.S. Dresselhaus

On page 18 irreducible representations are defined and an example is given as follows: enter image description here

The author claims that these representations are irreducible. However for E and A the $\Gamma_2$ 2D representation is in block form and it can be seen that it could be expressed in terms of the $\Gamma_1$ and $\Gamma_1'$ representations as follows $\begin{pmatrix} \Gamma_1 & 0\\ 0 & \Gamma_1' \end{pmatrix}$ , so why is $\Gamma_2$ considered irreducible? I am using the last line of the definition here.


1 Answer 1


The representation of the group (in this case) is given by a set of six matrices. Reducibility is a property of the representation as a whole, not the individual matrices that form it.

So, you can't say that the $E$ matrix on its own is (ir)reducible, or that $A$ on its own is (ir)reducible. You have to look at the set of six matrices as a whole, and whether they can all be simultaneously block-diagonalised. If they can, then the entire representation is reducible; if they can't, then the entire representation is irreducible.

In this case, $E$ and $A$ are block-diagonal, but the other matrices $B$, $C$, $D$, and $F$ are not. You could find some change of basis such that one or more of the latter become diagonal. However, it's not possible to find a change of basis which simultaneously diagonalises all six matrices. So, the representation $\Gamma_2$ is irreducible.

Another way of looking at it is that you are trying to decompose $\Gamma_2 = \Gamma_1 \oplus \Gamma_1'$. This is indeed satisfied for the $E$ and $A$ matrices, but not for the others. So, it's not possible to say that the representations add up, because the representation consists of all six matrices, and all six matrices must add up (technically, form the correct direct sum) for this to be the case.

The cited definition is actually very rigorous and does mention this (emphasis mine):

If by one and the same equivalence transformation, all the matrices in the representation of a group can be made to acquire the same block form, then the representation is said to be reducible [...]

  • $\begingroup$ Yet a bit more rigor might have been in order, as taking the quoted definition literally would seem to imply that $\Gamma_1$ would be reducible, all of its matrices being already in "the same block form" (and indeed identical). $\endgroup$ Jan 17 at 23:54
  • $\begingroup$ In fact, compared to, say, Wikipedia's definition, the one given in the book seems pretty muddled. I can't even tell whether or not it's mixing up reducibility and decomposability, since the book never seems to define what it means by "same block form". $\endgroup$ Jan 17 at 23:55

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