The Unapologetic Mathematician

Mathematics for the interested outsider

The Character Table of S4

Let’s use our inner tensor products to fill in the character table of S_4. We start by listing out the conjugacy classes along with their sizes:


Now we have the same three representations as in the character table of S_3: the trivial, the signum, and the complement of the signum in the defining representation. Let’s write what we have.


Just to check, we calculate


so again, \chi^\perp is irreducible.

But now we can calculate the inner tensor product of \mathrm{sgn} and \chi^\perp. This gives us a new line in the character table:


which we can easily check to be irreducible.

Next, we can form the tensor product \chi^\perp\otimes\chi^\perp, which has values


Now, this isn’t irreducible, but we can calculate inner products with the existing irreducible characters and decompose it as


where \chi^{(5)} is what’s left after subtracting the other three characters. This gives us one more line in the character table:


and we check that


so \chi^{(5)} is irreducible as well.

Now, we haven’t actually exhibited these representations explicitly, but there is no obstacle to carrying out the usual calculations. Matrix representations for V^\mathrm{triv} and V^\mathrm{sgn} are obvious. A matrix representation for V^\perp comes just as in the case of S_3 by finding a basis for the defining representation that separates out the copy of V^\mathrm{triv} inside it. Finally, we can calculate the Kronecker product of these matrices with themselves to get a representation corresponding to \chi^\perp\otimes\chi^\perp, and then find a basis that allows us to split off copies of V^\mathrm{triv}, V^\perp, and V^\mathrm{sgn}\otimes V^\perp.

November 8, 2010 - Posted by | Algebra, Group theory, Representation Theory


  1. Is there a general procedure for filling out the character table, or is it a matter of trying different combinations until you find a new irreducible character orthogonal to the ones you’ve already got (and iterating until you have a full basis)?

    Comment by Joe English | November 9, 2010 | Reply

  2. We’ll come up with a general straightforward position for symmetric groups eventually.

    Comment by John Armstrong | November 9, 2010 | Reply

  3. […] We can check this in the case of and , since we have complete character tables for both of them: […]

    Pingback by Decomposing the Left Regular Representation « The Unapologetic Mathematician | November 17, 2010 | Reply

  4. Small error, it’s the complement of the trivial rep, not the signum….

    Comment by Charles Waldman | May 2, 2013 | Reply

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