Today we prove the assertion that we made last time: that irreducible characters are orthogonal. That is, if and are -modules with characters and , respectively, then their inner product is if and are equivalent and otherwise. Strap in, ’cause it’s a bit of a long one.
Let’s pick a basis of each of and to get matrix representations and of degrees and , respectively. Further, let be any matrix with entries . Now we can construct the matrix
Now I claim that intertwines the matrix representations and . Indeed, for any we calculate
At this point, Schur’s lemma kicks in to tell us that if then is the zero matrix, while if then is a scalar times the identity matrix.
First we consider the case where (equivalently, ). Since is the zero matrix, each entry must be zero. In particular, we get the equations
But the left side isn’t just any expression, it’s a linear function of the . Since this equation must hold no matter what the are, the coefficients of the function must all be zero! That is, we have the equations
But now we can recognize the left hand side as our alternate expression for the inner product of characters of . If the functions and were characters, this would be an inner product, but in general we’ll write
Okay, but we actually do have some characters floating around: and . And we can write them out in terms of these matrix elements as
And now we can use the fact that for characters our two bilinear forms are the same to calculate
So there: if and are inequivalent irreps, then their characters are orthogonal!
Now if we can pick bases so that the matrix representations are both . Schur’s lemma tells us that there is some so that . Our argument above goes through just the same as before to show that
so long as . To handle the case where , we consider our equation
We take the trace of both sides:
and thus we conclude that . And so we can write
Equating coefficients on both sides we find
And finally we can calculate
exactly as we asserted.
Incidentally, this establishes what we suspected when setting up the character table: if and are inequivalent irreps then their characters and must be unequal. Indeed, since they’re inequivalent we must have . But if the characters were the same we would have to have . So since inequivalent irreps have unequal characters we can replace all the irreps labeling rows in the character table by their corresponding irreducible characters.