J-Hermitian determinantal point processes: balanced rigidity and balanced Palm equivalence

Abstract

We study Palm measures of determinantal point processes with J-Hermitian correlation kernels. A point process $\mathbb {P}$ on the punctured real line $\mathbb {R}^* = \mathbb {R}_{+} \sqcup \mathbb {R}_{-}$ is said to be balanced rigid if for any precompact subset $B\subset \mathbb {R}^*$, the difference between the numbers of particles of a configuration inside $B\cap \mathbb {R}_+$ and $B\cap \mathbb {R}_-$ is almost surely determined by the configuration outside B. The point process $\mathbb {P}$ is said to have the balanced Palm equivalence property if any reduced Palm measure conditioned at 2n distinct points, n in $\mathbb {R}_+$ and n in $\mathbb {R}_-$, is equivalent to the $\mathbb {P}$. We formulate general criteria for determinantal point processes with J-Hermitian correlation kernels to be balanced rigid and to have the balanced Palm equivalence property and prove, in particular, that the determinantal point processes with Whittaker kernels of Borodin and Olshanski are balanced rigid and have the balanced Palm equivalence property.

Appendix

Proof of Proposition 2.3

By homogenity, we may assume, without loss of generality, that $\Vert A\Vert _{\mathscr {L}_{1|2}} \le 1$ and $\Vert B\Vert _{\mathscr {L}_{1|2}} \le 1$. Write A and B in block forms:

$$\begin{aligned} A = \left[ \begin{array}{cc} a_1 &{} b_1 \\ c_1 &{} d_1 \end{array}\right] , \quad B = \left[ \begin{array}{cc} a_2 &{} b_2 \\ c_2 &{} d_2 \end{array}\right] , \end{aligned}$$

then we have

$$\begin{aligned} AB = \left[ \begin{array}{cc} a_1a_2 + b_1 c_2 &{} a_1 b_2 + b_1d_2 \\ c_1 a_2 + d_1 c_2 &{} c_1 b_2 + d_1d_2 \end{array}\right] . \end{aligned}$$

By applying the operator ideal property $\Vert a b\Vert _1 \le \Vert a\Vert _1 \Vert b\Vert$, $\Vert a b\Vert _1 \le \Vert a\Vert _2 \Vert b\Vert$ and the Hölder inequality $\Vert ab\Vert _1 \le \Vert a \Vert _2 \Vert b\Vert _2$, we get

$$\begin{aligned} \Vert AB\Vert _{\mathscr {L}_{1|2}} = \,&\Vert a_1a_2 + b_1 c_2 \Vert _1 + \Vert c_1 b_2 + d_1d_2\Vert _1 + \Vert a_1 b_2 + b_1d_2\Vert _2 + \Vert c_1 a_2 + d_1 c_2 \Vert _2 \\ \le \,&\Vert a_1\Vert _1 \Vert a_2\Vert + \Vert b_1\Vert _2 \Vert c_2\Vert _2 + \Vert c_1\Vert _2\Vert b_2\Vert _2 + \Vert d_1\Vert _1\Vert d_2\Vert \\&+\, \Vert a_1\Vert _1\Vert b_2\Vert + \Vert b_1\Vert _2 \Vert d_2\Vert + \Vert c_1\Vert _2 \Vert a_2\Vert + \Vert d_1\Vert \Vert c_2\Vert _2 \\ \le \,&\Vert a_1\Vert _1 + \Vert b_1\Vert _2 + \Vert c_1\Vert _2 + \Vert d_1\Vert _1 + \Vert a_1\Vert _1 + \Vert b_1\Vert _2 + \Vert c_1\Vert _2 + \Vert d_1\Vert \\ \le&\, 2 ( \Vert a_1\Vert _1 + \Vert b_1\Vert _2 + \Vert c_1\Vert _2 + \Vert d_1\Vert _1) \le 2. \end{aligned}$$

$\square$

Proof of Proposition 2.4

The proof is easy from the definition of $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$ and the ideal property of trace-class and Hilbert–Schmidt class. $\square$

Proof of Proposition 2.5

By the relation (2.6), under the hypothesis of Proposition 2.5 on A, B, the two operators A, B are both in $\mathscr {L}_2(L^2(\mathbb {R}))$, hence $AB\in \mathscr {L}_1(L^2(\mathbb {R}))$. By the ideal property of $\mathscr {L}_1(L^2(\mathbb {R}))$, the operator $(1 + A)^{-1}AB$ belongs to $\mathscr {L}_1(L^2(\mathbb {R}))$ and hence belongs to $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$. We can write

$$\begin{aligned} (1 + A)^{-1} B = (1 + A)^{-1} ( (1 + A) B - AB) = B - (1 + A)^{-1}AB, \end{aligned}$$

hence the operator $(1 + A)^{-1} B$ belongs to $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$. Similar argument yields the fact that the operator $B (1 + A)^{-1}$ also belongs to $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$. $\square$

Proof of Proposition 2.7

Fix a pair of operators A, B in $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$. Note first that by Proposition 2.3, the operator $A+ B + AB$ is in the space $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$, hence the extended Fredholm determinant $\det ((1 + A)(1+B)) = \det (1+A+B+AB)$ is well-defined. By the multiplicativity property of the usual Fredholm determinant, the desired identity holds whenever $A, B \in \mathscr {L}_{1}(L^2(\mathbb {R}))$, see, e.g. [32, Thm. 3.8]. Thus by the continuity of the function $A\mapsto \det (1+A)$ on $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$, for proving the desired identity, it suffices to show that there exist two sequences $(A_n)_{n\in \mathbb {N}}$ and $(B_n)_{n\in \mathbb {N}}$ in $\mathscr {L}_{1}(L^2(\mathbb {R}))$ such that we have the following convergences in the space $\mathscr {L}_{1|2}(L^2(\mathbb {R}))$:

$$\begin{aligned} A_n \xrightarrow {n\rightarrow \infty } A, \, B_n \xrightarrow {n\rightarrow \infty } B \,\,\text {and}\,\,A_nB_n \xrightarrow {n\rightarrow \infty } AB. \end{aligned}$$

(4.76)

To this end, take any two sequences $(P_n)_{n\in \mathbb {N}}$ and $(Q_n)_{n\in \mathbb {N}}$ of finite rank orthogonal projections on $L^2(\mathbb {R}_{+})$ and $L^2(\mathbb {R}_{-})$, respectively, assume that $P_n$ and $Q_n$ converge in the strong operator topology to the orthogonal projections $P_{+}$ and $P_{-}$, respectively. Now we may set

$$\begin{aligned}A_n = (P_n + Q_n)A, \quad B = B(P_n + Q_n).\end{aligned}$$

Then it is clear that the finite rank operators $A_n$ and $B_n$ satisfy all the desired conditions in (4.76). Note that we intentionally obtain $A_n$ and $B_n$ by multiplying $P_n + Q_n$ on the left side of A and on the right side of B, so that the third condition in (4.76) is satisfied. $\square$

Proof of Proposition 2.8

From Grothendieck’s definition of Fredholm determinant:

$$\begin{aligned} \det (1 + T) = \sum _{k=0}^\infty {\mathrm {tr}}(\wedge ^k (T)), \quad T \in \mathscr {L}_1(L^2(\mathbb {R})), \end{aligned}$$

and the fact that, once $A\in \mathscr {L}_1(L^2(\mathbb {R}))$ and f is a bounded function, then

$$\begin{aligned} {\mathrm {tr}}(\wedge ^k (fA)) = {\mathrm {tr}}( \wedge ^k (M_f) \circ \wedge ^k (A)) = {\mathrm {tr}}( \wedge ^k (A) \circ \wedge ^k (M_f)) = {\mathrm {tr}}(\wedge ^k (Af)), \end{aligned}$$

we see that the identity (2.9) holds when $A\in \mathscr {L}_1(L^2(\mathbb {R}))$. For $A\in \mathscr {L}_{1|2}(L^2(\mathbb {R}))$, we may argue similarly as in the proof of Proposition 2.7. See also [11] for the proof in more general case. $\square$