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Posted on July 31, 2022July 31, 2022 by
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The two-edge theorem (Theorem [twisted_reduction]) also provides a criteria to determine if the joint distribution of a collection of independent unitaries follows a twisted Haar measure of the form. In fact, for each $n$, we can find a collection of unitaries in $cal U(2^n)$ such that their joint distribution follows a Haar measure which is not twisted. This leads to a natural question.

Does there exist an $n$ such that the Haar measure on $cal U(2^n)$ is not twisted?

In the following, we will show that the answer to this question is ‘yes’.

Let $p_m=frac{1}{2}(1+eta_m)$, where $eta_m$ is a Rademacher sequence. By Proposition [twisted_measure], the Haar measure on $cal U(2^n)$ is not twisted if and only if $p_m=1$ for infinitely many $m$’s.

We first provide an upper bound on $p_m$’s. In fact, we can show that $p_mle 2^{ -n}$. To see this, note that by Hölder’s inequality, $$begin{aligned}
Eleft(frac{1+eta_m}{2}right)&le left(Eleft(frac{1+eta_m}{2^n}right)right)^{2/n}.end{aligned}$$ By Lemma [Rademacher_approximation], there is a function $h(x)$ such that $$begin{aligned}
Eleft(frac{1+eta_m}{2^n}right)&=frac{1}{2}+frac{1}{2^{2n}}sum_{l=1}^nE(h(x_l)).

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