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If $U$ is a $N\times N$ random unitary matrix uniformly distributed with respect to Haar measure, a $M\times M$ block $A$ from it has distribution given by $$ \det(1-AA^\dagger)^{N-2M}.$$

If $O$ is a $N\times N$ random real orthogonal matrix uniformly distributed with respect to Haar measure, a $M\times M$ block $A$ from it has distribution given by $$ \det(1-AA^T)^{(N-2M-1)/2}.$$

My question is what is the distribution of the $M\times M$ top left block from a $N\times N$ random unitary symmetric matrix from the Circular Orthogonal Ensemble.

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  • $\begingroup$ I think you need to assume $N-2M\geq 0$. $\endgroup$ Commented May 18, 2022 at 19:46

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The general formula in the circular ensembles follows from equation 2.10 of arXiv:cond-mat/9612179: $$P(H)\propto\text{det}\,(1-H)^{\tfrac{1}{2}\beta(N-2M+1-2/\beta)},\;\;N-2M\geq 0,\qquad\qquad(\ast)$$ where $H=AA^\dagger$ and $A$ is an $M\times M$ principal submatrix of an $N\times N$ unitary matrix $U$ (circular unitary ensemble, $\beta=2$) or unitary symmetric matrix (circular orthogonal ensemble, $\beta=1$) or unitary selfdual matrix (circular symplectic ensemble, $\beta=4$).

The case in the OP is $\beta=1$, in that case $H=AA^\dagger=AA^\ast$ (in the physics notation that $\ast$ is complex conjugation and $\dagger$ is Hermitian conjugate). The assumption that $N-2M\geq 0$ is needed, because if $M>N/2$ there will be an additional set of $M-N/2$ eigenvalues of $H$ that are pinned to unity.

The case that $U$ is real orthogonal (circular real ensemble, $H=AA^\dagger=AA^\top$) has an additional factor, $$P(H)\propto\text{det}\,H^{-1/2}\,\text{det}\,(1-H)^{\tfrac{1}{2}(N-2M-1)}.\qquad\qquad(\ast\ast)$$ That case is considered in an earlier MO posting. Note that $(\ast\ast)$ agrees with the corresponding result in the OP (the second equation), which gives $P(A)$ instead of $P(H)$.


To make contact with the physics literature: $U$ represents a scattering matrix, $A$ is the reflection matrix, and the eigenvalues of $1-H=1-AA^\dagger$ are the transmission eigenvalues $T_1,T_2,\ldots T_M$. Equation (2.10) in the cited paper gives the joint probability distribution of the $T_i$'s, which can equivalently be expressed as the equation $(\ast)$ above.

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  • $\begingroup$ So the exponent is the same as for the orthogonal group $\endgroup$
    – Marcel
    Commented May 18, 2022 at 20:01
  • $\begingroup$ not quite, if the distribution of $AA^\dagger$ is compared, there is an additional factor for the orthogonal group that is missing for the COE. $\endgroup$ Commented May 18, 2022 at 20:48
  • $\begingroup$ Which additional factor? I see $(N-2M-1)/2$ in both cases $\endgroup$
    – Marcel
    Commented May 18, 2022 at 22:50
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    $\begingroup$ I meant the additional factor $\text{det}\,(AA^\dagger)^{-1/2}$, comparing $(\ast)$ and $(\ast\ast)$ $\endgroup$ Commented May 19, 2022 at 6:11

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