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Let $A,D\in \mathbb{C}^{n \times n}$ be diagonal matrices. I need to calculate $$\int_{U(n)}\det{(A-HDH^\dagger)}\,\mathrm{d}H$$ where $dH$ is the unit invariant Haar measure on the group of unitary matrices and $H^\dagger$ is the conjugate transpose of $H$. (If $A=I$ this is very easy to solve, but I want the answer for $A\neq I$ in terms of $A$ and $D$.)

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    $\begingroup$ First of all, the matrices should probably be in $\mathbb{C}^{n^2}$ not in $\mathbb{C}^n$. Second, what role does $D$ play? I don't see it in the integral, perhaps $D = L$? Third, what does $H'$ mean? Is it the transpose of $H$? Fourth, are the words "deterministic" really relevant here? $\endgroup$ – Vít Tuček Jan 28 '15 at 12:08
  • $\begingroup$ Oops, Sorry for these typos! I edited them. Does it make sense now? Thanks. $\endgroup$ – Peter Jan 28 '15 at 17:22
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    $\begingroup$ Wlog one can assume that $A$ is diagonal because the Haar measure is invariant under left- and rightmultiplication. $\endgroup$ – Johannes Hahn Jan 28 '15 at 17:28
  • $\begingroup$ Do you need an exact answer or would large $n$ asymptotics suffice? In the second situation, in the case where $A$ and $D$ are diagonal with free entries, one might be able to use an approximation of the spectral distribution of $A-HDH^*$ ($H$ chosen uniformly at random) by the free convolution of $\mu_A$ with $\mu_D$, see e.g. the first few pages of arxiv.org/abs/math/9809193 $\endgroup$ – Yemon Choi Jan 28 '15 at 18:17
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    $\begingroup$ @Peter: Just a follow up on Carlo's suggestion. I don't think you want to compute the integral of exp sum of traces as in your last comment. $\int\exp(t\ {\rm tr}(A^{-1}HDH^{\dagger}))dH$ is enough, then take derivatives in $t$. That's the famous Harish-Chandra-Itzykson-Zuber integral. See e.g. terrytao.wordpress.com/2013/02/08/… $\endgroup$ – Abdelmalek Abdesselam Jan 28 '15 at 23:04

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