Decomposition of representation that has multiple copies of isomorphic irreducible representations, of a finite group.

264 Views Asked by At

To better show where the weightlifting point is, I'll use $D_{3d}$ group as an example. The problem is that a 24 dimensional reducible representation of $D_{3d}$ is given, which is a matrix form $r_{\mu\nu}(g)$, $\mu, \nu = 1,\ldots, 24$. By canonical decomposition, $r = 2 A_{1 g} \oplus 2 A_{2 g} \oplus 4 E_{g} \oplus A_{1 u} \oplus 3 A_{2 u} \oplus 4 E_{u}$. Here, $4 E_{u}$ means 4 copies of $E_u$ irreducible representations. The question is how to explicitly construct a decomposition of $4E_u$ into a direct sum of subrepresentations isomorphic to $E_u$. In other words, currently the basis of $4E_u$ representation is $(b_1, \ldots, b_8)$, under which the $4E_u$ representation is not $2\times2$ block diagonalized. How to obtain a set of basis $(e_1, \ldots, e_8)$ under which $4E_u$ representation becomes $2\times2$ block diagonal?

In Serre's $\it Linear\ Representations\ of\ Finite\ Groups$, the following projector is given: $$ p_{\alpha \beta}=\frac{n}{g} \sum_{t \in G} r_{\beta \alpha}\left(t^{-1}\right) \rho_{t} $$ But here representation $r_{\beta\alpha}$ is already the irreducible representation, whose basis is unkonw yet. Therefore I don't know how to use this formula to tackle this problem.

The background is in the supplimentary material of paper "Two-phonon Absorption Spectra in the Layered Honeycomb Compound α-RuCl3", where the answer, the basis of the irreducible representation, is given. But I don't know how to obtain it or $E_g$ representations.

Does any one know how to tacle it?


Update:

I've now better understood the projector defined above. This projector is defined for arbitrary representations of G, thus can be used to decompose any representations. However, the remaining question is how to obtain $r_{\alpha\beta}(g)$, which is a known matrix form of an irreducible representation of G? Looks like chicken and eggs problem again.

1

There are 1 best solutions below

0
On

It seems that to use the Serre's formula shown above, at least one irreducible representation has to be known. And typically they are known. See the comments in https://mathoverflow.net/questions/393186/decomposition-of-representation-that-has-multiple-copies-of-isomorphic-irreducib