= Solution
For $G=U(n)$ take the diagonal <maximal torus>
$$
T=\{\operatorname{diag}(z_1,\ldots,z_n):|z_i|=1\}.
$$
The <spectral theorem> for <unitary matrices> says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct <eigenvalues> has centralizer exactly $T$. Its conjugates therefore determine its <eigenvalues> up to <permutation>, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The <character lattice of a torus> $T$ is $\mathbb Z^n$, with $e^\lambda(z)=z_1^{\lambda_1}\cdots z_n^{\lambda_n}$. <Conjugation> on the <matrix unit> $E_{ij}$ has <weight> $e_i-e_j$, so these are the <roots>. Choose $e_i-e_j$ positive for $i<j$; then
$$
\rho=\frac12(n-1,n-3,\ldots,1-n).
$$
<Dominant integral weights> are precisely the integer tuples $\lambda_1\geq\cdots\geq\lambda_n$. Negative entries are allowed: they correspond to <determinant twists>, so <polynomial representations> alone do not exhaust the <irreducible representations> of $U(n)$.
Normalize <Haar measure> on $G$ and $T$ to have mass one, and put
$$
\Delta(z)=\det(z_i^{n-j})_{i,j=1}^n=\prod_{i<j}(z_i-z_j).
$$
The <Weyl integration formula for U(n)> for a <continuous> <class function> is
$$
\boxed{\int_{U(n)}f(g)\,dg
=\frac1{n!}\int_{[0,2\pi)^n} f(\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n}))
\prod_{i<j}|e^{i\theta_i}-e^{i\theta_j}|^2
\prod_{i=1}^n\frac{d\theta_i}{2\pi}.}
$$
For a general <continuous function>, replace $f(t)$ in the <torus> integral by $\int_{G/T}f(gtg^{-1})\,d(gT)$, using the normalized invariant quotient measure.
To prove the formula, consider the <conjugation> map $q:G/T\times T\to G$, $q(gT,t)=gtg^{-1}$. On regular <torus> elements its fibres have exactly $n!$ points: the possible orderings of the distinct <eigenvalues>. Equip the <group> with an invariant <inner product> on its <Lie algebra> and split $\mathfrak u(n)=\mathfrak t\oplus\mathfrak t^\perp$. After translating the derivative back to the identity, its <torus> directions contribute the identity, while its transverse directions contribute $\operatorname{Ad}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane $\operatorname{Ad}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real <determinant> of the transverse derivative is
$$
J(t)=\prod_{i<j}|z_i/z_j-1|^2=|\Delta(z)|^2.
$$
The change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized <invariant measures>. The nonregular locus has measure zero: repeated <eigenvalues> are the zero-set of the nonzero discriminant, and the <torus> coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct <torus characters>. <Fourier orthogonality> gives $\int_T|\Delta|^2\,dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.
We can now derive the <character> formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional <irreducible representation> has a unique <highest weight> $\lambda$ with a one-dimensional highest-weight space; all its other <weights> are $\lambda$ minus a nonnegative sum of <positive roots>. Its <character> is Weyl invariant. Also, <character orthogonality for compact groups> gives $\int_G|\chi_\lambda|^2=1$. This <orthogonality> follows by averaging the <representation> on $\operatorname{End}(V_\lambda)$: the integral projects onto <intertwining operators>, whose <dimension> is one by the <Schur lemma>.
Let $\delta=(n-1,n-2,\ldots,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_\eta(z)=\det(z_i^{\eta_j})$. The finite <Laurent polynomial> $\Delta\chi_\lambda$ is alternating. Every alternating <Laurent polynomial> is a <linear combination> of the $A_\eta$: collect its monomials by <permutation> orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_\eta$ are <orthogonal> on $T$ and each has squared norm $n!$.
The coefficient of $z^{\lambda+\delta}$ in $\Delta\chi_\lambda$ is exactly one. Indeed, a contribution from the term $z^{w\delta}$ would require a <weight> $\lambda+\delta-w\delta$ in $V_\lambda$. For $w\ne1$, the vector $\delta-w\delta$ is a nonzero nonnegative combination of the positive <simple roots>, so that <weight> lies above $\lambda$ and is impossible. The identity <permutation> supplies the highest-weight coefficient one. Consequently
$$
\Delta\chi_\lambda=A_{\lambda+\delta}+\sum_{\eta\ne\lambda+\delta}c_\eta A_\eta.
$$
Apply the integration formula to $|\chi_\lambda|^2$. <Torus> <orthogonality> gives
$$
1=\frac1{n!}\int_T|\Delta\chi_\lambda|^2\,dt
=1+\sum_{\eta\ne\lambda+\delta}|c_\eta|^2.
$$
All the other coefficients vanish, proving
$$
\boxed{\chi_\lambda(z)=\frac{\det(z_i^{\lambda_j+n-j})}{\det(z_i^{n-j})}.}
$$
The quotient extends across repeated <eigenvalues> because its product with the denominator is the already defined <continuous> <character>. Equivalently, alternant divisibility makes it a <Laurent polynomial> invariant under <permutations>; multiply by a <determinant> power if negative exponents must first be removed.
Finally, $\delta-\rho=\frac{n-1}{2}(1,\ldots,1)$ is fixed by the <Weyl group>, so its common factor cancels in the quotient. Since $A_\rho=e^\rho\prod_{\alpha>0}(1-e^{-\alpha})$, the <determinant> expression is exactly
$$
\chi_\lambda=\frac{\sum_{w\in W}\varepsilon(w)e^{w(\lambda+\rho)-\rho}}{\prod_{\alpha>0}(1-e^{-\alpha})}.
$$
Using $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual <character> of the <torus>. At the identity, taking the leading Vandermonde term of the two <determinants> gives the accompanying <dimension> formula
$$
\dim V_\lambda=\prod_{i<j}\frac{\lambda_i-\lambda_j+j-i}{j-i}.
$$
Back to article page
