The Character Table of S4
Let’s use our inner tensor products to fill in the character table of 
. 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 
: 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, 
is irreducible.
But now we can calculate the inner tensor product of 
and . 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 
, which has values
Now, this isn’t irreducible, but we can calculate inner products with the existing irreducible characters and decompose it as
where 
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 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 
and 
are obvious. A matrix representation for 
comes just as in the case of by finding a basis for the defining representation that separates out the copy of inside it. Finally, we can calculate the Kronecker product of these matrices with themselves to get a representation corresponding to , and then find a basis that allows us to split off copies of , , and 
.
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November 8, 2010 -
Posted by John Armstrong |
Algebra, Group theory, Representation Theory
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)?
We’ll come up with a general straightforward position for symmetric groups eventually.
[…] We can check this in the case of and , since we have complete character tables for both of them: […]
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Small error, it’s the complement of the trivial rep, not the signum….