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Let $U \in \mathbb{C}^{n \times n}$ be an arbitrary $n \times n$ unitary matrix, $U_1 = \operatorname{Re} U$ (real part of $U$), and $U_2 = \operatorname{Im} U$ (imaginary part of $U$). The construct the $2n \times 2n$ matrix

$$ J = \begin{pmatrix} U_1 & -U_2 \\ U_2 & U_1 \end{pmatrix} $$

One can show that $J$ is an orthogonal matrix (thanks to the comment of @Salcio): calculate

$$ \begin{align*} J^\mathsf{T} J &= \begin{bmatrix} U_1^\mathsf{T} U_1 + U_2^\mathsf{T} U_2 & -U_1^\mathsf{T} U_2 + U_2^\mathsf{T} U_1 \\ U_1^\mathsf{T} U_2 - U_2^\mathsf{T} U_1 & U_1^\mathsf{T} U_1 + U_2^\mathsf{T} U_2 \end{bmatrix} \end{align*} $$

Since $U$ is unitary, we get

$$ \begin{align*} U^\dagger U &= (U_1^\mathsf{T} - iU_2^\mathsf{T}) (U_1 + iU_2) \\ &= (U_1^\mathsf{T} U_1 + U_2^\mathsf{T} U_2) + i(U_1^\mathsf{T} U_2 - U_2^\mathsf{T} U_1) = 1 \end{align*} $$

The real and the imaginary parts of this equation gives

$$ U_1^\mathsf{T} U_1 + U_2^\mathsf{T} U_2 = 1, \quad U_1^\mathsf{T} U_2 - U_2^\mathsf{T} U_1 = 0 $$

Therefore $J^\mathsf{T} J = 1$, and $\det J = \pm 1$. Now the problem is that in my numerical test I always get $\det J = +1$. The code for the test is shown below (in Python, using the unitary matrix generator from SciPy):

import numpy as np
# generator of random unitary matrices
from scipy.stats import unitary_group
# dimension of the unitary matrix
n = 3
u = unitary_group.rvs(n)
u1, u2 = u.real, u.imag
j = np.block([[u1, -u2], [u2, u1]])
print(np.linalg.det(j))

How to further prove (or disprove) that $\det J = +1$ (ruling out the possibility $-1$)?

Motivation of the definition of $J$: consider two complex vectors $z, w \in \mathbb{C}^n$ related by $w = Uz$. Let

$$ \begin{align*} z &= z_1 + iz_2 && (z_1,z_2 \in \mathbb{R}^n) \\ w &= w_1 + iw_2 && (w_1,w_2 \in \mathbb{R}^n) \end{align*} $$

Then the real vectors $z_1,z_2,w_1,w_2$ are related by

$$ \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = J \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} $$

Zhengyuan Yue
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  • From Hadamard's inequality you get $|detJ| <=1$. At the same time (still from the Hadamard's inequality, this time its proof or geometric interpretation) equality would imply that rows of $J$ are orthogonal which is ... unlikely. – Salcio Jul 19 '22 at 01:49
  • Have you tried numerical examples when $n=2$? – Greg Martin Jul 19 '22 at 02:12
  • I updated the question now that I can show that $J$ is orthogonal, but still do not know why I always get $\det J = +1$ in numerical tests. – Zhengyuan Yue Jul 19 '22 at 03:04
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    The map $U\mapsto J$ is a homomorphism $U(n)\to O(2n)$. (You pick a unitary matrix, interpret it as an $\Bbb R$-linear transformation of $\Bbb C^n$ with $\Bbb R$-basis ${e_1,\cdots,e_n,ie_1,\cdots, ie_n}$.) Since $U(n)$ is connected and the homomorphism is continuous, the image must be in the identity component of $O(2n)$, namely $SO(2n)$. Or equivalently, compose $U\mapsto J$ with the homomorphism $\det:O(2n)\to{\pm1}$, and apply the same logic with the codomain being discrete. That's why $\det J$ must be $1$. – anon Jul 19 '22 at 03:13
  • @runway44 Maybe you can make your comment an answer. But somehow I believe that there must exist a more elementary proof... – Zhengyuan Yue Jul 19 '22 at 03:21
  • Your block matrix is a special case of symplectic matrix. Every symplectic matrix (over any field) has determinant $1$. You may see here for some elementary proofs. – user1551 Jul 19 '22 at 09:23

0 Answers0