Last week, I sat down to compute the character table of for the first time in several years.

My first instinct, following what I learned five years ago, was to use the standard tricks for computing character tables: compute the characters of some easily constructed representations, use the fact that the sum of the squares of the dimensions of the irreducible characters is , and use the orthogonality of the rows and columns of the character table to finish it off.

This, however, is rather tedious for a group as large as . But I had recently learned that the irreducible representations of the symmetric group are completely classified, and can be constructed using group actions on standard Young tableaux. Was there a way to use this theory to compute each entry of the character table directly?

It turned out that it was possible to laboriously use this theory to compute the characters one at a time, by finding an explicit matrix representation for each irreducible representation. However, I still had a nagging feeling that there should be a faster way to compute the characters.

I dug through the standard references in search of an answer, and I found it: *the Murnaghan-Nakayama Rule* (thanks to Enumerative Combinatorics, Vol. 2, by Richard Stanley). This concise and elegant method gives a simple combinatorial way of computing the character table of any symmetric group . And it goes like this:

**Step 1. ** The conjugacy classes of are the permutations having a fixed number of cycles of each length, corresponding to a partition of called the *shape* of the permutation. For instance, the conjugacy classes for have representative elements corresponding to the seven partitions of . Since the characters of a group are constant on its conjugacy classes, we index the columns of our character table by these partitions:

**Step 2.** There are precisely as many irreducible characters as conjugacy classes, so we can also index the irreducible characters by the partitions of . We represent each partition as a Young diagram and write these down the left of the table:

**Step 3: The Murnaghan-Nakayama Rule. ** We can now calculate the entry in row and column . Let be the parts of in decreasing order. Drawing as a Young diagram, define a *filling of with content * to be a way of writing a number in each square of such that the numbers are weakly increasing along each row and column and there are exactly squares labeled for each .

Consider all fillings of with content such that for each label , the squares labeled form a *connected* skew tableaux that does not contain a square. (A skew tableau is connected if the graph formed by connecting horizontally or vertically adjacent squares is connected.) Such a tableaux is called a *border-strip tableaux*, with each label coloring a *border strip*. For instance, the following is a border-strip tableaux:

For each label in the tableau, define the *height* of the corresponding border strip to be one less than the number of rows of the border strip. We then weight the tableau by where is the sum of the heights of the border strips that compose the tableau. (For instance, the heights of the border strips in the above tableaux are , , , and , and so the weight of the tableau is .) Then, the entry in the character table is simply the sum of these weights.

For example, in the case , suppose we are trying to calculate the entry for and . Then the three possible border-strip tableaux are drawn below:

They have weights , , and , for a total weight of . So, the corresponding entry of the character table is .

That’s all! This gives a quick way to churn out all the entries in the character table:

Using the Murnaghan-Nakayama rule, I was able to compute this table by hand in about 30 minutes. What a gemstone!

Whoa! They really should teach this method in Rep. Theory. I’d like to see a proof of why this works, because it seems almost too good to be true. (btw the weight of the first example tableau is actually -1 because the first height is 0, not 1)

The only proof I’ve ever seen is kind of horrible, unfortunately. It feels like, if it’s true, there should be a nice natural reason why it’s true, and yet I’ve never found a proof that doesn’t require 30 pages of setup and technical lemmas first.

You might be interested in two cute facts I came across at some point.

Firstly: permuting the rows and/or columns of the character table will permute its entries, of course. It is a fact that every permutation of the rows/cols of the character table of S_n produces a different table (i.e. permutations of rows/cols permute the possible tables faithfully), unless n = 4 or n = 6 (in which cases there is a kernel of size 2). The latter case is because S_6 is the only symmetric group to have outer automorphisms, though I don’t know why the former is true. [A similar result is true for A_n unless n = 6 – though you have to exclude the obvious swaps where representations/conjugacy classes of S_n split in A_n, which correspond to conjugation by odd elements of S_n, i.e. outer automs.]

Secondly: given any column of the character table of S_n (unless n = 4), you can work out the order of an element in that conjugacy class. This seems to be a slightly stronger form of a much more general theorem, which I *think* goes as follows: given a column of a character table of any group, you can work out the primes that divide the order of an element in that conjugacy class (but not their multiplicity).

(I forget where I found these, annoyingly. But it’s rather surprising to see how incredibly asymmetric the symmetric group can be!)

Cool! That’s very strange about S_4. It is clear that you can use an outer automorphism to swap the columns and get the same table… but S_4 is just a coincidence I guess? And yet no other such coincidences come up? Wild.

Let me know if you come across any other gemstones like that, and if you have a resource for the proof. Very interesting stuff, thanks for sharing!

-Maria

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