I am studying an undergraduate thesis, called A Fast Fourier Transform for the Symmetric Group, by Tristan Brand, one time a student at Harvey Mudd College. Source:


It contains many obvious errors. So, I don't trust it, but it is the only source that I have found that actually gives (or claims to give) the matrices for the representation of $S_4$ that correspond to the [22] partition. He gives the same matrices that I get for the (12) and (23) transpositions, but his matrix for the (34) element of $S_4$ is different than either of those for (12) and (23), while I get the same matrix for (34) as the one I get for (12).

His 3 distinct matrices generate a group of order 12, which I have identified as the dihedral group, $D_6$. My 2 distinct matrices generate a group of order 6, namely, $D_3=S_3$.

My question is: Which group is generated by the matrices of this representation of $S_4$?

If the author of this thesis is right, then my second question follows immediately:

Please explain how the following rules for constructing the seminormal representations of the symmetric group lead to different matrices for the (12) and (34) elements of the group.

Let $t_i^\lambda$,...,$t_n^\lambda$ be the standard Young tableaux on a given shape, $\lambda$, arranged in some standard order. Then the representation matrix, $R^\lambda$($j$-1,,$j$) for the ($j$-1,$j$) transposition is computed according to the following rules:

  1. If $t_i^\lambda$ contains $j$-1 and $j$ in the same row, then $R^\lambda$($j$-1,$j$)=+1.
  2. If $t_i^\lambda$ contains $j$-1 and $j$ in the same column, then $R^\lambda$($j$-1,$j$)=-1.
  3. If $j$-1 and $j$ are not in the same row or column, then there are two tableaux, $t_i^\lambda$ and $t_k^\lambda$, with $i<k$, that are the same except that $j$-1 and $j$ are interchanged. Then we have the following submatrix: (see NOTE below)
  4. All other entries are 0.

NOTE: I can't get the matrix to come out right. Even copying and pasting the example on your own help page doesn't work! Can somebody fix it? I figure that if you know enough to answer my question, you know what matrix should be there! In any case, you can find in the cited thesis. Moreover, the third rule doesn't apply in the case of the (12) and (34) permutations because (1,2) and (3,4) always occur in either the same row or same column in both standard tableaux for the [22] partition. So, it doesn't matter here.


It's just a mistake in the thesis. If you work out the character table for $S_4$, you find that the 2-dimensional irreducible representation, which corresponds to the [22] partition, has character 2 on the conjugacy class of (12)(34). Therefore (12)(34) and its conjugates are the identity map. So (12) and (34) must be the same map.

  • $\begingroup$ Thanks for the confirmation. Being self-taught, and seeing how many fools stubbornly think they are right - no matter how you try to talk sense into them, I never know whether to believe my own opinions or not. The last thing I want to be is like those fools! I always try to jump up and down on what I believe to see if it lets me down - knowing that the truth never will. Also, being self-taught, your theoretical explanation means nothing to me. Paraphrasing what Morris Kline once said of Wronski, I'm "more of an algorithmist than a mathmatician". $\endgroup$ – Koilon Sep 1 '11 at 18:13

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