irreducibility of certain subspaces of the permutation group in quantum mechanics Let $P_j$, $j = 1, \dotsc, N!$ be a set of unitary operators constituting a representation of the symmetric group $S_N$, acting in a sub-Hilbert space $V_0 \subseteq H$ (of a separable Hilbert space $H$) spanned by the images of a single non-zero vector $\lvert \psi_0\rangle$ under all possible permutations $P_k$.
Depending on the choice of $\lvert \psi_0\rangle$, the dimension of $V_0$ will lie between $1$ and $N!$, inclusive. 
The "character operators" introduced by Paul Dirac are given by
$\chi(P_k) = \frac{1}{N!} \sum\limits_{j = 1}^{N!} P_j \,P_k \,P_j^{\dagger}$.
It is easy to see that these operators (i) act invariantly in $V_0$, (ii) are class-functions, (iii) can be re-expressed as linear combinations over the $P_j$, and (iv) are the only functions of the permutation operators $P_j$ which simultaneously commute with all $P_k$ in $S_N$.
The eigenvectors of the $\chi(P_j)$ are proportional to the group characters, hence their name.  These character operators all commute with each other, so can be jointly diagonalized.  The joint eigensubspaces of the $\chi(P_j)$ within $V_0$ are invariant under the action of all $P_k$ in $S_N$.
Dirac seemed to think it obvious that these eigensubspaces are irreducible with respect to $S_N$.  Is there an easy proof of this irreducibility?
 A: The images of the $P_{k}$ under these so-called character operators are (non-zero multiples of) the images of the conjugacy class sums from the centre of the group algebra $\mathbb{C}S_{N}$ under the given representation on $V_{0}.$ The (mutually orthogonal) primitive central idempotents of $\mathbb{C}S_{N}$ are linear combinations of the class sums since both those idempotents and the class sums are a basis for the centre of the group algebra. Hence the image of each such central idempotent acts as a scalar on each of the simultaneous eigenspaces of the class sums. But being idempotents, they can only act as scalar 1 or scalar zero on each such simultaneous eigenspace, and, being mutually orthogonal, at most one idempotent can have non-zero action on any given simultaneous eigenspace. If the central primitive idempotent $e_{\mu}$ corresponding the irreducible complex character $\mu$ acts as scalar $1$ on a simultaneous eigenspace, then the simultaneous eigenspace affords  a multiple of the associated irreducible representation. But that multiple need not be $1$ in general. For example, if $V_{0}$ really has dimension $N!$ (which can happen, as you say), then the multiplicity of the representation affording character $\mu$ is $\mu(1)$, and when $N >2,$ some of these $\mu(1)$'s are greater than $1$.
A: Thanks for your response G.R., I must be misreading Dirac, whose quantum mechanics textbook is sometimes terse, but rarely wrong.  Indeed, I see now that just from character orthogonality, it apparently follows that in any regular representation of the permutation group $S_N$ (which in the above notation, corresponds to a case where $\dim V_0 = N!$), each irreducible representation of dimension $K$ appears exactly $K$ times. So as you said if $N > 2$ and $\dim V_0 = N!$, then other than the trivial and alternating representations, all other irreducible representations will show up multiple times in the corresponding joint eigensubspace of Dirac's operators.  For instance, if $N = 3$ and $\dim V_0 = 6$, then there is one trivial representation, one alternating representation, and two copies of the $2$-dimensional standard representation, not just one.
