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I am trying to understand the relationship between the Kostant partition function and the Haar measure. Both seem to involve the Vandermonde determinant:

$$ \Delta(\theta) = \prod_{i< j} |e^{i\theta_i} - e^{i\theta_j}|$$

According to question: Explicit computations using the Haar measure integrating over the Unitary group seems to involve integrals with Vandermonde determinant in the numerator,

$$ \langle f \rangle_1 = \int_{[0,2\pi]^n} d\theta_1\dots d\theta_n \;\Delta (\theta)^2 f(\theta)$$

while Kostant partition function seems to involve Vandermonde determinant in the demominator:

$$ \langle f \rangle_1 = \int_{[0,2\pi]^n} d\theta_1\dots d\theta_n \;\Delta (\theta)^{-2} f(\theta)$$

Is there a name for this second measure? These notes suggest the Weyl denominator and Kostant partition function are reciprocal.

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    $\begingroup$ Are you sure? Does not look integrable to me.... Unless you only integrate functions that vanish quadraticaly in $\theta_i-\theta_j$. $\endgroup$
    – ofer zeitouni
    Commented Dec 12, 2014 at 16:11
  • $\begingroup$ @oferzeitouni Here is an example by Doron Zeilberger, arxiv.org/abs/math/9811108. It looks like Selberg integral, but perhaps not quite? I am trying to find clarification. $\endgroup$
    – john mangual
    Commented Dec 12, 2014 at 16:28
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    $\begingroup$ The function $|e^{i\theta_1}-e^{i\theta_2}|^{-2}$ is not integrable. If you do not mean an integral but rather a formal power series expansion, that's a different story. $\endgroup$
    – ofer zeitouni
    Commented Dec 12, 2014 at 17:08
  • $\begingroup$ I would say "the Kostant partition function is the Fourier transform of the reciprocal of the Weyl denominator", in that the first is a function on the weight lattice of a torus and the second is a function on the torus itself. $\endgroup$
    – Allen Knutson
    Commented Dec 13, 2014 at 3:55

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