Our story continues with having reached the main technical idea according to Varadarajan of Weyl’s Character Formula, which is that one should consider the orthogonality relation between characters of irreducible representations and decompose along two different directions. On one hand, along the diagonal elements and on the other hand along the conjugacy classes each of which is going to be a submanifold of some sort in .
SINGULAR POINTS
The integration formula that Weyl uses for the orthogonality relation is valid for points that are ‘nonsingular’ in a sense. The condition is that the conjugacy class must have maximum possible dimension. At a point we want to understand the condition on the conjugacy class to which belongs, recall that belong to the conjugacy class if and only if there exists such that . We next want to characterize the conjugacy class as exactly the coset space where is the centralizer of , i.e. those that commute with . Let’s be elementary about this, so if the the condition is the same as which occurs if and only if . Therefore we can exactly identify the distinct elements of the conjugacy class of with . The two points of note is first that maps on to the conjugacy class of , and a different argument for distinctness of points of the conjugacy class is implies which implies that $latex y_2^{1} y_1 \in G_x$ and .
Now let us try to understand the dimension of this space . If are distinct eigenvalues of with multiplicities then we want to calculate the dimension of in terms of these multiplicities. Elements of preserve the eigenspaces for implies both $latex as well as . Therefore for a fixed we certainly have that at most components contribute to . Therefore . On the other hand, Varadarajan claims that this is an equality, i.e. that . In other words, that the dimension of matrices that commute with the matrix of the multiplicity eigenspace of eigenvalue has dimension . This is not clear to me yet why. This dimension, seems too large at first glance but I am quite confident that my worries are unnecessary so skipping the issue:
Our original question is when is maximized. We have the constraint that . Therefore .
Now and the two sides are equal when and . Actually this last part tells us that we needed only and not equality.
Since is maximum if and only if is minimum, above can be used to conclude is maximum if and only if has distinct eigenvalues, which is the same as the characteristic polynomial having distinct roots.
THE DISCRIMINANT FUNCTION
The characteristic polynomial with a variable . Now is invariant under permutations and by Newton’s theorem is a function of the symmetric elementary functions where for example and so on. The discriminant is such that latex x$ has distinct roots. We call is regular if has distinct eigenvalues and denote the set of regular values as and call the complement the singular set. The key point is that the singular set is small. More precisely, the Haar measure which is used for all the integration is absolutely continuous with respect to the Lebesgue measure in coordinates and since the singular set is defined as the zero set of an analytic function, it has Haar measure zero.
OUTLINE OF THE PROOF OF WEYL CHARACTER FORMULA
Recall that the Character Formula is that the irreducible characters of are in natural onetoone correspondence with where . The character
where is the character
Define
For an irreducible representation define . Then the following are relatively easily established:
(i) has finite Fourier series with integer coefficients
(ii)
(iii)
Since a finite Fourier series with integer coefficients which is skew symmetric with integer coefficients from the formula while is a symmetric finite Fourier series with integer coefficients (i) and (ii) are clear while (iii) is just the orthogonality relation.
The interesting and quite sophisticated part of the proof is to construct all possible that satisfy the three conditions.
I am going to come back to the pieces maybe tomorrow.
Ok now it is tomorrow, March 18 2018.
For any diagonal character the alternating sum
is a finite Fourier sum and if then . For the first part, we just inserted and then we changed the index of the sum over all the permutations in which shows (ii).
Any character satisfying (i) and (ii) must be of the form
where the coefficients are integers and
If and are in different orbits then . Recall that the characters are elements of and the torus. If and are in different orbits then there are no permutations that can equate . The orthogonality is actually the outcome of a separate result known as ‘the orthogonality relation for characters’: . If the then .
HERE IS ANOTHER ATTEMPT TO FOLLOW THE PROOF:

Let pi be an irreducible representation of U(n).
So the key issue in the Weyl Character Formula is to consider the three conditions (i) Phi is a finite Fourier series with integer coefficients, (ii) Phi_pi^w = sign(w) Phi_pi, (iii) C \int Phi_pi Phi_{pi’} = delta_{pi pi’}.
For any character chi whatsoever, we can construct a character
tilde{chi} = sum_{w in W} sign(w) chi^w
and this will satisfy (i) obviously and (ii) by taking for any permutation s and tilde{chi}^s = sum_w sign(s)^2 sign(w) chi^{sw} = sign(s) sum_w sign(sw) chi^{sw} = sign(s) sum_{sw} sign(sw) chi^{sw}. The general orthogonality relation for characters guarantees that (tilde{chi}_1}\tilde{chi}_2) = 0 when chi_1 and chi_2 are in different Worbits. If they are in the same orbit, chi_2 = chi_1^w then
tilde{chi_2} = tilde{ chi_1^ w } = sign(w) chi_1 by (ii)
We have D the ntorus in the diagonal of U(n) and the characters in the dual \hat{D}. So far we have produced a map \tilde: \hat{D} > { characters that satisfy (i), (ii) } and we analyzed what happens in the image of this map under the action of permutations W.
Ok, now we step back and recall that all the characters of U(n) can be written as diag(t_1,…,t_n)>t^m = t_1^{m_1} … t_n^{m_n} and we consider what this tildemap is doing in terms of these.
The tildemap will zero out those chi_m which have repeated values in m=(m_1,…,m_n) because if we take m_i=m_j then we can find permutation w swapping i and j indices leaving all else the same which has sign 1 so chi_m^w = sign(w) chi_m = – chi_m so chi_m=0. This imples the image of the tildemap only contains combinations of chi_m with distinct elements. Then we can find a permutation that reorders the elements of (m_1,…,m_n) in decreasing order.
So ANY Phi satisfying (i) and (ii) a combination with integer coefficients and finite sum
and furthermore  Phi ^2 = 1. This last condition is going to force only one of the coefficients to be nonzero because the coefficients are all integers. and that coefficient will be +/1.
So far we have shown any Phi satisfying (i) and (ii) will pick out an m with a coefficient +/ 1. We still have to show that all possible m occur as the image of the tilde map with Phi satisfying (i) and (ii).
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