Do the first two columns of the character table of the permutation group $S_n$ (the conjugacy classes () and (1 2) - that's a fairly standard order) always suffice to identify an irrep? (E.g. take $S_6$: four irreps have dimension 5 but the second entry is 1,-1,3 or -3.) If yes, can this be generalized to other groups? How many entries can be equal for two different irreps anyway? (E.g. $S_6$ again, trivial and standard irrep have two equal entries out of eleven.)

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    $\begingroup$ If you tensor a representation with the sign representation, this will preserve the value of the character of the conjugacy classes of even permutations as well as odd permutations with character $0$. For example, the characters corresponding to $4+1$ and $2+1+1+1$ agree on $5$ out of $7$ conjugacy classes. $\endgroup$ – Douglas Zare Aug 19 '16 at 21:18
  • $\begingroup$ I added a couple of tags to indicate more clearly the scope of the question. Aside from that, you are asking several related questions, so it's difficult to see which of them an "answer" would apply to. Can you prioritize them and try to follow the usual guidelines for asking questions? $\endgroup$ – Jim Humphreys Aug 21 '16 at 22:50
  • $\begingroup$ Regarding the first question, are there not partitions that are not self-conjugate where the corresponding irreducible character is zero on transpositions? If so, this would give a pair of irreducible characters that are the same in the first two columns. $\endgroup$ – Jeremy Rickard Aug 22 '16 at 10:05
  • $\begingroup$ @Jeremy: This would indeed answer my question in the negative (could someone rev up the GAP program?). $\endgroup$ – Hauke Reddmann Aug 22 '16 at 11:06
  • $\begingroup$ @Jim: If Jeremy is right, the second question is void anyway, and since Douglas pointed out that the "similarity" between two even/odd-related irreps is easily above 50% (I forgot to take the zeroes in account), the third is only a minor question. So, yes, I priorize an actual example of a non-self-conjugate Sn irrep with transposition character 0. $\endgroup$ – Hauke Reddmann Aug 22 '16 at 11:11

For $S_{16}$ there appear (I asked GAP) to be two pairs of non-self-conjugate partitions where the value of the corresponding characters are zero on transpositions.

I haven't tried to figure out what the partitions are.

Edit. Suzuki ("The values of irreducible characters of the symmetric group", Arcata conference on representations of finite groups, Part II, Proceedings of symposia in pure mathematics 47, 1986) gives a simple criterion for an irreducible character of a symmetric group to take the value zero on transpositions. If the corresponding Young diagram has boxes with coordinates $(x,y)$ with the main diagonal $x=y$ then the criterion is that $\sum (x-y)=0$. This is obviously satisfied by self-conjugate partitions, but he also gives the example of $6+3+2+2+2$, which gives a pair of degree $112112$ characters for $S_{15}$ that I missed when asking GAP.


It is easy to evaluate the second character if the partition is self-conjugate, since then it must be $0$. It's not easy to construct pairs of non-conjugate partitions that correspond to representations with the same dimension, perhaps because there is a large spread of possible dimensions, but my guess was that there was no particular reason self-conjugate partitions had to correspond to representations with distinct dimensions.

Here are partitions of $57$ that are self-conjugate with the same dimension $701951320702493736151313458962540000$:

$\lambda_1 = 14+8+8+6+5+4+3+3+1+1+1+1+1+1$

$\lambda_2 = 13+8+8+6+6+5+3+3+1+1+1+1+1$

I used Mathematica to generate the partitions, test if they were self-conjugate, and compute the dimensions. This took a few minutes per symmetric group of about this size.

  • $\begingroup$ Too bad that only one answer can be checked :-) $\endgroup$ – Hauke Reddmann Aug 22 '16 at 18:42

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