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I am currently examining the symmetric group $S_4$, and I was tasked with finding a $2$-D, a $3$-D, and a $4$-D representation of the group. The $4$-D representation is reducible, so I first found it and then found that it was the direct sum of a $3$-D irreducible representation plus the fully symmetric representation. Consider one element:

enter image description here

This is not in block diagonal form. In python, there is a function available for a matrix m, m.jordan_form(), which applies a symmetry operation to the input matrix and returns that matrix in Jordan form. The result for this matrix was

enter image description here

I assume that $I$ is $i$, the imaginary number. Please correct me if I am wrong on this.

The issue with this result is we know that the above matrix is the direct sum of the fully symmetric representation and a $3$-D representation; but the corner elements are $-1$ and $I$, neither of which are $1$. The fully symmetric representation is always just $1$, so I expect that the block form result should either have a $1$ in the top left corner; and then the bottom right block is the $3$-D representation; or I expect the bottom right corner to be $1$ and then the top left block is the $3$-D representation. To clarify, if the fully symmetric representation is represented by $\Gamma^{(1)}$, then

$$\Gamma^{(1)}[(1342)]=1$$

and for the general group element $R$

$$\Gamma^{(1)}[R]=1$$

My question is then, what am I doing wrong? Is there a different block form other than Jordan form which I need to use in this situation? Thanks for your help!

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  • $\begingroup$ In the 3D representation, $S_4$ acts by symmetries of the cube, which includes $90^{\circ}$ rotations around axes through faces. These are order four elements corresponding to $4$-cycles in $S_4$. This clearly acts as $1$ in the fixed axis and the usual $90^{\circ}$ 2D rotation matrix for the other block. However, when complexified that 2D block diagonalizes to ${\rm diag}(i,-i)$, since these are conjugate the fourth roots of unity. $\endgroup$ – arctic tern Jun 2 at 23:00
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The problem is you need to decompose all of the matrices corresponding to the representation of $G$ at the same time, that is, find a basis $v_1, \ldots, v_4$ such that every matrix of the representation becomes block-diagonal in that basis. In the language of matrices, you need to find a single conjugating matrix $P$ such that $P \Gamma[\sigma] P^{-1}$ is block diagonal for all $\sigma \in S_4$. Once you have done this, the 1-dimensional and 3-dimensional subrepresentations will be clear.

For any finite group representation, the Jordan form of a matrix will be diagonal, with roots of unity down the diagonal. This is because the matrix of a group element has finite order.

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  • $\begingroup$ do you know of any procedure (other than simply generating a set of equations) that would help me to find that $P$ matrix? $\endgroup$ – Kraig Jun 3 at 12:15
  • $\begingroup$ @Kraig Well, it's the same problem as breaking a representation into a sum of irreducibles. In this simple case, you have the projector $P(v) = \frac{1}{4!} \sum_{\sigma \in S_4} \Gamma[\sigma] v$ onto the one-dimensional representation, and so $1 - P$ will project onto the leftover three-dimensional representation. $\endgroup$ – Joppy Jun 3 at 15:05

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