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Solution to Representative Characters

by Brian Hamrick

The key to solving the set of equations given is to understand that the 24 letters that are present correspond to the 24 irreducible representations of the alternating group A10. This is further confirmed because 1, 240, 630, 4725, 6048, 8400, 18900, 18900, 22400, 25200, 25200, 56700, 72576, 86400, 86400, 86400, 90720, 120960, 151200, 151200, 151200, 201600, 201600, and 226800 are the sizes of the 24 conjugacy classes of A10.

The equations then have a natural meaning, where concatenation of letters is the direct sum of two representations, corresponding to addition of the characters, and  ⊗  is the tensor product, which corresponds to pointwise multiplication of characters. See here for more information about characters of representations.

We can get the character table of A10 from any number of sources, including Sage. The entire character table is a 24 × 24 matrix, so for brevity we’re only going to list the dimensions of each representation here.

 Character Dimension χ1 1 χ2 9 χ3 35 χ4 36 χ5 42 χ6 75 χ7 84 χ8 90 χ9 126 χ10 160 χ11 210 χ12 224 χ13 224 χ14 225 χ15 252 χ16 288 χ17 300 χ18 315 χ19 350 χ20 384 χ21 384 χ22 450 χ23 525 χ24 567

To compute the decomposition of a character into irreducible characters, we can use the inner product on characters: χ, χ'〉 = 1/|A10| ΣC|C|χ(C)χ'(C), where C ranges over all the conjugacy classes of A10. When χ is irreducible, this formula gives us the number of times it appears in the decomposition of χʹ into irreducibles, because the irreducible characters are orthogonal with respect to this inner product.

With this knowledge, we can compute the decompositions of the tensor product of each pair of irreducible representation. From here it is reasonably straightforward to associate each letter with an irreducible representation.

For example, the only pair of irreducibles whose product decomposes into the sum of two irreducibles is χ2 ⊗ χ5 = χ8 + χ16, so B and M are χ2 and χ5 in some order, and I and T are χ8 and χ16 in some order. By looking at the rest of the equations, we quickly see that in fact M is χ2, and so B is χ5.

After more work along these lines, we get the following key. The letters A and N are indistinguishable, as are the letters W and F. The representations corresponding to W and F are complex conjugates, so distinguishing them isn’t possible with equations of this form, and A and N are similarly difficult to distinguish. Additionally, one representation never appears, and so it could be any of the letters Q, R, or X. However, the answer can be extracted in its entirety without resolving these ambiguities.

 Character Dimension Letter χ1 1 J χ2 9 M χ3 35 U χ4 36 P χ5 42 B χ6 75 Z χ7 84 E χ8 90 T χ9 126 O χ10 160 K χ11 210 C χ12 224 W/F χ13 224 F/W χ14 225 Q/R/X χ15 252 H χ16 288 I χ17 300 S χ18 315 L χ19 350 D χ20 384 A/N χ21 384 N/A χ22 450 Y χ23 525 V χ24 567 G

Now, looking at the table at the bottom of the puzzle, we see that it gives the values of a character on A10. Since some of the conjugacy classes have the same size, we need to figure out which value goes with which one, and after that we get that it decomposes into the representations MUPELIISV.

At this point, one could simply anagram for the answer, but the ordering is also given in the inequality chain at the bottom. It says that the second letter of the answer has the smallest dimension, followed by the fourth letter, then the third letter, and so on. Ordering the letters in this way gives the answer IMPULSIVE.