Consider $N$ real random particles $x_1,\cdots, x_N$ distributed according to a density $\rho(x_1,\ldots,x_N)$ with respect to the Lebesgue measure on $\mathbb R^N$, which is assumed to be invariant under permutations : $$ \rho(x_{\sigma(1)},\ldots,x_{\sigma(N)})=\rho(x_1,\ldots,x_N),\qquad \sigma\in\mathfrak S_N. $$ We moreover assume the particles to interact as a determinantal point process with a kernel $K:\mathbb R \times \mathbb{R}\rightarrow\mathbb R$ which is, seen as a kernel operator on $L^2(dx)$, a projection operator.

This means roughly that the density distribution is given by

$$ \frac{1}{N!}\det\Big[ K(x_i,x_j) \Big] \prod_{i=1}^N dx_i. $$

Questions :

If we order the particles $x_1<\cdots< x_N$ and then extract a new system of particles $\{x_i\}_{i\in I}$, where $I\subset \{1,\ldots,N\}$, do we keep a determinantal structure (once forgetting the ordering on the $x_i$'s, $i\in I$ )?

And if Yes, what would be the new kernel ?

This question is motivated from the following particular case : Once the particles ordered, namely $x_1<\ldots< x_N$, it is well known that one has the Fredholm determinant representation : $$ \mathbb P(x_N\leq s)=\det(I-K)_{L^2(s,+\infty)}, \qquad x_1,\ldots,x_N\in\mathbb R. $$ I'm looking for a similar formula involving the operator $K$ for $ \mathbb P(x_k\leq s)$ when $1\leq k < N.$

  • 1
    $\begingroup$ For example, we take an $N \times N$ random Hermitian matrix, the eigenvalues $\lambda_1, \dots, \lambda_N$ form a determinantal process. We can put these real numbers in order $-2 \sqrt{N} \approx \lambda'_1 < \dots < \lambda'_N \approx 2 \sqrt{N}$. Do the $\lambda_{2k}$ form a determinantal process? I don't know. $\endgroup$ – john mangual Jul 27 '12 at 1:51

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