Appendix 1: Derivation of Relation (29)
In this appendix, we discuss the relation (29) that appears in the proof of Lemma 2. Recall the term \({\mathcal M}_k\):
$$\begin{aligned} \mathcal {M}_k= & {} \text{ tr }\Big [ U_k U_j^{\dagger }-U_i U_k^{\dagger } U_i U_j^{\dagger }+U_i U_k^{\dagger }-U_i U_j^{\dagger } U_k U_j^{\dagger } +U_k U_i^{\dagger }\\&\qquad -U_j U_k^{\dagger } U_j U_i^{\dagger }+U_j U_k^{\dagger }-U_j U_i^{\dagger } U_k U_i^{\dagger }\Big ]. \end{aligned}$$
Our goal is to establish the following relation:
$$\begin{aligned} \mathcal {M}_k= & {} 4\Vert U_i-U_j\Vert ^2 -\text{ tr }\Big [ (U_k-U_j)(U_k^{\dagger }-U_j^{\dagger })(U_i-U_j)(U_i^{\dagger }-U_j^{\dagger }) \\&+\, (U_k-U_i)(U_k^{\dagger }-U_i^{\dagger })(U_i-U_j)(U_i^{\dagger }-U_j^{\dagger })\Big ]. \end{aligned}$$
Rearranging the terms inside the bracket in \({\mathcal M}_k\) and using the properties \(\text{ tr }(AB) = \text{ tr }(BA)\) and \(U_j, U_i \in {\mathcal U}_d\) yields
$$\begin{aligned} {\mathcal M}_k&= \text{ tr } \Big [ (I_d - U_i U_j^{\dagger } ) U_k U_j^{\dagger } + U_i U_k^{\dagger } (I_d - U_i U_j^{\dagger }) + (I_d - U_j U_i^{\dagger }) U_k U_i^{\dagger } + U_j U_k^{\dagger } (I_d - U_j U_i^{\dagger }) \Big ] \nonumber \\&= \text{ tr } \Big [ (U_k U_j^{\dagger } + U_i U_k^{\dagger }) (I_d - U_i U_j^{\dagger } ) + (U_k U_i^{\dagger } + U_j U_k^{\dagger } )(I_d - U_j U_i^{\dagger }) \Big ] \nonumber \\&= \text{ tr } \Big [ (U_k U_j^{\dagger } + U_i U_k^{\dagger }) U_i U_i^{\dagger } (I_d - U_i U_j^{\dagger } ) + (U_k U_i^{\dagger } + U_j U_k^{\dagger } ) U_j U_j^{\dagger } (I_d - U_j U_i^{\dagger }) \Big ] \nonumber \\&= \text{ tr } \Big [ (U_k U_j^{\dagger } + U_i U_k^{\dagger }) U_i (U_i^{\dagger } - U_j^{\dagger }) - (U_k U_i^{\dagger } + U_j U_k^{\dagger } ) U_j (U_i^{\dagger } - U_j^{\dagger }) \Big ] \nonumber \\&= \text{ tr } \Big [ \underbrace{\Big ( (U_k U_j^{\dagger } U_i + U_i U_k^{\dagger } U_i - U_k U_i^{\dagger } U_j - U_j U_k^{\dagger } U_j \Big ) (U^{\dagger }_i - U_j^{\dagger })}_{{\mathcal J}_1} \Big ]. \end{aligned}$$
(47)
Note that the terms \({\mathcal J}_1\) can be simplified as follows:
$$\begin{aligned} {\mathcal J}_1&= U_k U_j^{\dagger } (U_i - U_j) (U^{\dagger }_i - U_j^{\dagger }) + U_i U_k^{\dagger } (U_i - U_j) (U^{\dagger }_i - U_j^{\dagger }) + U_i U_k^{\dagger } U_j (U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&\quad - U_k U_i^{\dagger } (U_j - U_i) (U^{\dagger }_i - U_j^{\dagger }) - U_j U_k^{\dagger } (U_j - U_i) (U^{\dagger }_i - U_j^{\dagger }) - U_j U_k^{\dagger } U_i(U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&= \Big ( U_k U_j^{\dagger } + U_i U_k^{\dagger } + U_k U_i^{\dagger } + U_j U_k^{\dagger } \Big ) (U_i - U_j)(U^{\dagger }_i - U_j^{\dagger }) + \Big ( U_i U_k^{\dagger } U_j - U_j U_k^{\dagger } U_i \Big ) (U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&=: {\mathcal J}_{11} + {\mathcal J}_{12}, \end{aligned}$$
(48)
where we used the cancellation:
$$\begin{aligned} U_k U_j^{\dagger } U_j (U^{\dagger }_i - U_j^{\dagger }) - U_k U_i^{\dagger } U_i (U^{\dagger }_i - U_j^{\dagger }) = 0. \end{aligned}$$
\(\bullet \) Case A (Estimate of \({\mathcal J}_{11}\)): For further simplification, we use the trick
$$\begin{aligned} U_k U_j^{\dagger } = (U_k U_j^{\dagger } - I_d) U_j U_j^{\dagger } + I_d = (U_k - U_j) U_j^{\dagger } + I_d \end{aligned}$$
to find
$$\begin{aligned} {\mathcal J}_{11}&= 4 (U_i - U_j)(U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&\quad + \Big ( (U_k - U_j) U_j^{\dagger } + (U_i - U_k) U_k^{\dagger } + (U_k - U_i) U_i^{\dagger } + (U_j - U_k) U_k^{\dagger } \Big ) (U_i - U_j)(U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&= 4 (U_i - U_j)(U^{\dagger }_i - U_j^{\dagger }) \nonumber \\&\quad - \Big ( (U_k - U_j) (U_k^{\dagger } - U_j^{\dagger } ) + (U_k - U_i) (U_k^{\dagger } - U_i^{\dagger } ) \Big )(U_i - U_j)(U^{\dagger }_i - U_j^{\dagger }). \end{aligned}$$
(49)
\(\bullet \) Case B (Estimate of \({\mathcal J}_{12}\)): Next, we show that
$$\begin{aligned} \text{ tr } {\mathcal J}_{12} = 0. \end{aligned}$$
By expanding \({\mathcal J}_{12}\) and using the linearity of the trace, the property \(\text{ tr }(AB) = \text{ tr }(BA)\) and the unitarity of \(U_i, U_j\), we obtain
$$\begin{aligned} \text{ tr }{\mathcal J}_{12}&= \text{ tr }(U_i U_k^{\dagger } U_j U_i^{\dagger } - U_j U_k^{\dagger } - U_i U_k^{\dagger } + U_j U_k^{\dagger } U_i U_j^{\dagger } ) \nonumber \\&= \text{ tr } \Big [ (U_i U_k^{\dagger }) (U_j U_i^{\dagger }) \Big ] - \text{ tr } ( U_j U_k^{\dagger } ) -\text{ tr }(U_i U_k^{\dagger }) + \text{ tr } \Big [ (U_j U_k^{\dagger }) ( U_i U_j^{\dagger }) \Big ] \nonumber \\&= \text{ tr } \Big [ (U_j U_i^{\dagger }) (U_i U_k^{\dagger }) \Big ]- \text{ tr } ( U_j U_k^{\dagger } ) -\text{ tr }(U_i U_k^{\dagger }) + \text{ tr } \Big [( U_i U_j^{\dagger }) (U_j U_k^{\dagger }) \Big ] \nonumber \\&= \text{ tr } (U_j U_k^{\dagger })- \text{ tr } ( U_j U_k^{\dagger } ) -\text{ tr }(U_i U_k^{\dagger }) + \text{ tr } (U_i U_k^{\dagger }) \nonumber \\&=0. \end{aligned}$$
(50)
Finally, in (47), we combine estimates (48), (49), and (50) to obtain the desired estimate. This completes the proof of (29).
Appendix 2: Derivation of Gronwall’s Inequalities in (38)
In this appendix, we provide the proof of claim (38) in the proof of Theorem 2.
For a given time t, choose indices i and j depending on t such that
$$\begin{aligned} d(U(t),\tilde{U}(t))^2=\Vert U_i(t) U_j^{\dagger }(t)-\tilde{U}_i(t) \tilde{U}_j^{\dagger }(t)\Vert ^2. \end{aligned}$$
Then, we have
$$\begin{aligned} \frac{d}{dt}d(U,\tilde{U})^2&= \frac{d}{dt} \text{ tr } \Big [ (U_i U_j^{\dagger } - {\tilde{U}}_i {\tilde{U}}_j^{\dagger }) (U_j U_i^{\dagger } - {\tilde{U}}_j {\tilde{U}}_i^{\dagger } ) \Big ] \nonumber \\&=\frac{d}{dt} \text{ tr }\Big (2I_d -U_i U_j^{\dagger }\tilde{U}_j\tilde{U}_i^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger }U_j U_i^{\dagger } \Big ) \nonumber \\&= -\text{ tr } \Big [ \frac{d}{dt} \Big ( (U_i U_j^{\dagger })(\tilde{U}_j\tilde{U}_i^{\dagger }) + (\tilde{U}_i \tilde{U}_j^{\dagger })(U_j U_i^{\dagger } ) \Big ) \Big ]. \end{aligned}$$
(51)
On the other hand, it follows from (45) that
$$\begin{aligned}&\frac{d}{dt}U_i U_j^{\dagger } = -{\mathrm i} (H_i U_i U_j^{\dagger } - U_i U_j^{\dagger }H_j) +\frac{K}{2N}\sum _{k =1}^{N} \left[ U_k U_j^{\dagger }-U_i U_k^{\dagger } U_i U_j^{\dagger }+U_iU_k^{\dagger }-U_i U_j^{\dagger } U_k U_j^{\dagger }\right] , \nonumber \\&\frac{d}{dt}{\tilde{U}}_j {\tilde{U}}_i^{\dagger } = -{\mathrm i} (H_j {\tilde{U}}_j {\tilde{U}}_i^{\dagger } - {\tilde{U}}_j {\tilde{U}}_i^{\dagger }H_i) +\frac{K}{2N}\sum _{k =1}^{N} \left[ {\tilde{U}}_k {\tilde{U}}_i^{\dagger }-{\tilde{U}}_j {\tilde{U}}_k^{\dagger } {\tilde{U}}_j {\tilde{U}}_i^{\dagger }+ {\tilde{U}}_j {\tilde{U}}_k^{\dagger }-{\tilde{U}}_j {\tilde{U}}_i^{\dagger } {\tilde{U}}_k {\tilde{U}}_i^{\dagger }\right] . \end{aligned}$$
(52)
In (51), we use relation (52) to obtain
$$\begin{aligned} \frac{d}{dt}d((U(t),\tilde{U}(t))^2 = {\bar{\mathcal L}} - \frac{K}{2N}\sum _{k=1}^{N} {\bar{\mathcal M}}_k, \end{aligned}$$
(53)
where the terms \({\bar{\mathcal L}}\) and \({\bar{\mathcal M}}_k\) are given by the following relations:
$$\begin{aligned} {\bar{\mathcal L}} ={\mathrm i} \times \text{ tr }\Big [ H_i U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } -U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } H_i + H_i \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } -\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } H_i\Big ] \end{aligned}$$
(54)
and
$$\begin{aligned} {\bar{\mathcal M}}_k&= \text{ tr }\Big [ U_k U_j^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }-U_i U_k^{\dagger } U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }+U_i U_k^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }-U_i U_j^{\dagger } U_k U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }\\&\qquad +U_i U_j^{\dagger }\tilde{U}_k \tilde{U}_i^{\dagger }-U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_k^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }+U_i U_j^{\dagger }\tilde{U}_j \tilde{U}_k^{\dagger }-U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } \tilde{U}_k \tilde{U}_i^{\dagger }\\&\qquad +\tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }- \tilde{U}_i \tilde{U}_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }+\tilde{U}_i \tilde{U}_k^{\dagger } U_j U_i^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger } \tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\\&\qquad +\tilde{U}_i \tilde{U}_j^{\dagger } U_k U_i^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_k^{\dagger } U_j U_i^{\dagger }+\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_k^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } U_k U_i^{\dagger }\Big ]. \end{aligned}$$
Lemma 4
The terms \({\bar{\mathcal M}}_k\) can be decomposed into three parts:
$$\begin{aligned} {\bar{\mathcal M}}_k = {\bar{\mathcal A}}_k+ {\bar{\mathcal B}}_k+ {\bar{\mathcal C}}_k, \end{aligned}$$
(55)
where
$$\begin{aligned} {\bar{\mathcal A}}_k&:= \text{ tr }\Big [ \tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_k (I_d -\tilde{U}_k^{\dagger } U_k U_i^{\dagger } \tilde{U}_i) (\tilde{U}_k^{\dagger }\tilde{U}_i-I_d)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j) \\&\qquad +(\tilde{U}_i^{\dagger }\tilde{U}_k-I_d) (I_d -\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \\&\qquad +(\tilde{U}_j^{\dagger }U_kU_j^{\dagger }\tilde{U}_j-I_d) (I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \\& +(I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (\tilde{U}_k^{\dagger }\tilde{U}_j-I_d) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \\&\qquad +(I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k) (\tilde{U}_k^{\dagger }\tilde{U}_i-I_d ) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i) \\&\qquad +(\tilde{U}_i^{\dagger }U_kU_i^{\dagger }\tilde{U}_i-I_d) (I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i) \\&\qquad +(\tilde{U}_j^{\dagger }\tilde{U}_k-I_d) (I_d-\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i) \\&\qquad +\tilde{U}_j^{\dagger } U_j U_k^{\dagger } \tilde{U}_k (I_d -\tilde{U}_k^{\dagger } U_k U_j^{\dagger } \tilde{U}_j) (\tilde{U}_k^{\dagger }\tilde{U}_j-I_d) (I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\Big ], \end{aligned}$$
$$\begin{aligned} {\bar{\mathcal B}}_k&:=\text{ tr } \Big [ (\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k-I_d) (I_d -\tilde{U}_k^{\dagger }U_k U_i^{\dagger }\tilde{U}_i) (I_d-\tilde{U}_i^{\dagger }U_i U_j^{\dagger }\tilde{U}_j)\\&\qquad +(\tilde{U}_j^{\dagger }U_j U_k^{\dagger }\tilde{U}_k-I_d) (I_d -\tilde{U}_k^{\dagger }U_k U_j^{\dagger }\tilde{U}_j) (I_d -\tilde{U}_j^{\dagger }U_j U_i^{\dagger }\tilde{U}_i)\Big ], \end{aligned}$$
and
$$\begin{aligned} {\bar{\mathcal C}}_k&:=2 \text{ tr }\Big [ (I_d-\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) +(I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j)\\&\qquad +(I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i) +(I_d -\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i)\Big ]. \end{aligned}$$
Proof
We verify relation (55) in several steps.
\(\bullet \) Step A: The terms in \({\bar{\mathcal M}}_k \) can be rearranged as follows:
$$\begin{aligned} {\bar{\mathcal M}}_k&=\text{ tr }\Big [-U_i U_k^{\dagger } U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } + U_i U_j^{\dagger }\tilde{U}_j \tilde{U}_k^{\dagger } - U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } \tilde{U}_k \tilde{U}_i^{\dagger }+ U_k U_j^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }\\&\qquad -U_i U_j^{\dagger } U_k U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }+U_i U_j^{\dagger }\tilde{U}_k \tilde{U}_i^{\dagger }-U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_k^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }+U_i U_k^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }\\&\qquad -\tilde{U}_i \tilde{U}_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }+\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_k^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } U_k U_i^{\dagger }+\tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\\&\qquad -\tilde{U}_i \tilde{U}_j^{\dagger } \tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }+\tilde{U}_i \tilde{U}_j^{\dagger } U_k U_i^{\dagger }-\tilde{U}_i \tilde{U}_j^{\dagger } U_j U_k^{\dagger } U_j U_i^{\dagger }+\tilde{U}_i \tilde{U}_k^{\dagger } U_j U_i^{\dagger }\Big ]. \end{aligned}$$
\(\bullet \) Step B: Using \(\text{ tr }(AB) = \text{ tr }(BA)\), it follows that
$$\begin{aligned} {\bar{\mathcal M}}_k= & {} \text{ tr }\Big [-\tilde{U}_i^{\dagger } U_i U_k^{\dagger } U_i U_j^{\dagger } \tilde{U}_j + \tilde{U}_k^{\dagger } U_i U_j^{\dagger }\tilde{U}_j - \tilde{U}_i^{\dagger } \tilde{U}_k \tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j+ \tilde{U}_i^{\dagger } U_k U_j^{\dagger }\tilde{U}_j \\&-\, U_k U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } U_i U_j^{\dagger }+\tilde{U}_k \tilde{U}_i^{\dagger } U_i U_j^{\dagger }- \tilde{U}_k^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }U_i U_j^{\dagger } \tilde{U}_j+ U_k^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }U_i\\&-\, \tilde{U}_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\tilde{U}_i +U_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j -U_k U_i^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } +\tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\\&-\, \tilde{U}_j^{\dagger } \tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\tilde{U}_i+ \tilde{U}_j^{\dagger } U_k U_i^{\dagger }\tilde{U}_i- \tilde{U}_j^{\dagger } U_j U_k^{\dagger } U_j U_i^{\dagger }\tilde{U}_i+ \tilde{U}_k^{\dagger } U_j U_i^{\dagger }\tilde{U}_i\Big ]. \end{aligned}$$
\(\bullet \) Step C: Again, we use \(\text{ tr }(UAU^{\dagger }) = \text{ tr }(A)\) for unitary U to find
$$\begin{aligned} {\bar{\mathcal M}}_k= & {} \text{ tr }\Big [-\tilde{U}_i^{\dagger } U_i U_k^{\dagger } U_i U_j^{\dagger } \tilde{U}_j + \tilde{U}_k^{\dagger } U_i U_j^{\dagger }\tilde{U}_j - \tilde{U}_i^{\dagger } \tilde{U}_k \tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j+ \tilde{U}_i^{\dagger } U_k U_j^{\dagger }\tilde{U}_j \\&-\, (\tilde{U}_j^{\dagger })U_k U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } U_i U_j^{\dagger }(\tilde{U}_j)+(\tilde{U}_j^{\dagger })\tilde{U}_k \tilde{U}_i^{\dagger } U_i U_j^{\dagger }(\tilde{U}_j)\\&-\, \tilde{U}_k^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }U_i U_j^{\dagger } \tilde{U}_j+ (\tilde{U}_j^{\dagger } U_j)U_k^{\dagger }\tilde{U}_j \tilde{U}_i^{\dagger }U_i(U_j^{\dagger }\tilde{U}_j)\\&-\, \tilde{U}_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\tilde{U}_i +(\tilde{U}_i^{\dagger } U_i)U_k^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j(U_i^{\dagger }\tilde{U}_i)\\&-\,(\tilde{U}_i^{\dagger })U_k U_i^{\dagger } \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }(\tilde{U}_i) +(\tilde{U}_i^{\dagger })\tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }(\tilde{U}_i)\\&-\, \tilde{U}_j^{\dagger } \tilde{U}_k \tilde{U}_j^{\dagger } U_j U_i^{\dagger }\tilde{U}_i+ \tilde{U}_j^{\dagger } U_k U_i^{\dagger }\tilde{U}_i- \tilde{U}_j^{\dagger } U_j U_k^{\dagger } U_j U_i^{\dagger }\tilde{U}_i+ \tilde{U}_k^{\dagger } U_j U_i^{\dagger }\tilde{U}_i\Big ], \end{aligned}$$
which factors into
$$\begin{aligned} {\bar{\mathcal M}}_k= & {} \text{ tr }\Big [(\tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_i-\tilde{U}_k^{\dagger }\tilde{U}_i)(-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)+ (\tilde{U}_i^{\dagger } \tilde{U}_k-\tilde{U}_i^{\dagger } U_k U_i^{\dagger }\tilde{U}_i)(-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)\\&+\, (\tilde{U}_j^{\dagger }U_k U_j^{\dagger } \tilde{U}_j-\tilde{U}_j^{\dagger }\tilde{U}_k)(-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)+(\tilde{U}_k^{\dagger } \tilde{U}_j-\tilde{U}_j^{\dagger } U_jU_k^{\dagger }\tilde{U}_j)(-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)\\&+\, (\tilde{U}_k^{\dagger }\tilde{U}_i-\tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_i)(-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)+ (\tilde{U}_i^{\dagger } U_k U_i^{\dagger }\tilde{U}_i-\tilde{U}_i^{\dagger } \tilde{U}_k)(-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\\&+\, (\tilde{U}_j^{\dagger }\tilde{U}_k-\tilde{U}_j^{\dagger }U_k U_j^{\dagger } \tilde{U}_j)(-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)+(\tilde{U}_j^{\dagger } U_jU_k^{\dagger }\tilde{U}_j-\tilde{U}_k^{\dagger } \tilde{U}_j)(-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\Big ].\\ \end{aligned}$$
\(\bullet \) Step D: Each of these eight summands is the product of two terms, and adding the first multiplicative terms of all eight summands results in zero. Thus, we have
$$\begin{aligned} {\bar{\mathcal M}}_k= & {} \text{ tr }\Big [(\tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_i-\tilde{U}_k^{\dagger }\tilde{U}_i)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)+ (\tilde{U}_i^{\dagger } \tilde{U}_k-\tilde{U}_i^{\dagger } U_k U_i^{\dagger }\tilde{U}_i)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)\\&+\, (\tilde{U}_j^{\dagger }U_k U_j^{\dagger } \tilde{U}_j-\tilde{U}_j^{\dagger }\tilde{U}_k)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)+(\tilde{U}_k^{\dagger } \tilde{U}_j-\tilde{U}_j^{\dagger } U_jU_k^{\dagger }\tilde{U}_j)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j)\\&+\, (\tilde{U}_k^{\dagger }\tilde{U}_i-\tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_i)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)+ (\tilde{U}_i^{\dagger } U_k U_i^{\dagger }\tilde{U}_i-\tilde{U}_i^{\dagger } \tilde{U}_k)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\\&+\, (\tilde{U}_j^{\dagger }\tilde{U}_k-\tilde{U}_j^{\dagger }U_k U_j^{\dagger } \tilde{U}_j)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)+(\tilde{U}_j^{\dagger } U_jU_k^{\dagger }\tilde{U}_j-\tilde{U}_k^{\dagger } \tilde{U}_j)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\Big ]. \end{aligned}$$
\(\bullet \) Step E: By the unitarity of \(U_i\),
$$\begin{aligned} {\bar{\mathcal M}}_k&=\text{ tr }\Big [\tilde{U}_i^{\dagger } U_i U_k^{\dagger }\tilde{U}_k(I_d-\tilde{U}_k^{\dagger }U_k U_i^{\dagger }\tilde{U}_i)\tilde{U}_k^{\dagger }\tilde{U}_i(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j) \nonumber \\& +\tilde{U}_i^{\dagger } \tilde{U}_k(I_d-\tilde{U}_k^{\dagger } U_k U_i^{\dagger }\tilde{U}_i)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger }\tilde{U}_j) \nonumber \\& +\tilde{U}_j^{\dagger }U_k U_j^{\dagger } \tilde{U}_j(I_d-\tilde{U}_j^{\dagger } U_j U_k^{\dagger }\tilde{U}_k)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j) \nonumber \\&+(I_d-\tilde{U}_j^{\dagger } U_jU_k^{\dagger }\tilde{U}_k)\tilde{U}_k^{\dagger } \tilde{U}_j(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j) \nonumber \\& +(I_d-\tilde{U}_i^{\dagger } U_iU_k^{\dagger }\tilde{U}_k)\tilde{U}_k^{\dagger } \tilde{U}_i(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i) \nonumber \\&+ \tilde{U}_i^{\dagger }U_k U_i^{\dagger } \tilde{U}_i(I_d-\tilde{U}_i^{\dagger } U_i U_k^{\dagger }\tilde{U}_k)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\nonumber \\& +\tilde{U}_j^{\dagger } \tilde{U}_k(I_d-\tilde{U}_k^{\dagger } U_k U_j^{\dagger }\tilde{U}_j)(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i) \nonumber \\&+\tilde{U}_j^{\dagger } U_j U_k^{\dagger }\tilde{U}_k(I_d-\tilde{U}_k^{\dagger }U_k U_j^{\dagger }\tilde{U}_j)\tilde{U}_k^{\dagger }\tilde{U}_j(I_d-\tilde{U}_j^{\dagger } U_j U_i^{\dagger } \tilde{U}_i)\Big ]. \end{aligned}$$
(56)
Note that there is a factor of two in \({\bar{\mathcal C}}_k\), and that matrix multiplication is distributive. In (56), the first line is the sum of the first term of \({\bar{\mathcal A}}_k\), the first term of \({\bar{\mathcal B}}_k\), and half of the first term of \({\bar{\mathcal C}}_k\). The second line is the sum of the second term of \({\bar{\mathcal A}}_k\) and half of the first term of \({\bar{\mathcal C}}_k\). The third and fourth lines are the sum of the third and fourth terms of \({\bar{\mathcal A}}_k\), respectively, with each half of the second term of \({\bar{\mathcal C}}_k\). The fifth and sixth lines are the sum of the fifth and sixth terms of \({\bar{\mathcal A}}_k\), respectively, with each half of the third term of \({\bar{\mathcal C}}_k\). The seventh line is the sum of the seventh term of \({\bar{\mathcal A}}_k\) and half of the fourth term of \({\bar{\mathcal C}}_k\). The eighth line is the sum of the eighth term of \({\bar{\mathcal A}}_k\), the second term of \({\bar{\mathcal B}}_k\), and half of the fourth term of \({\bar{\mathcal C}}_k\). \(\square \)
Proposition 2
Suppose that the coupling strength K and initial data \(U^0\) and \({\tilde{U}}^0\) satisfy
$$\begin{aligned} K>K_e>\frac{54}{17}D(H)\approx 3.1765D(H),\qquad U^0,~{\tilde{U}}^0 \in {\mathcal S}(\alpha _1). \end{aligned}$$
Then for any two Lohe flows \(\{U_i\}\) and \(\{\tilde{U}_i\}\),
$$\begin{aligned} -2K(1+3\alpha _1)d(U,\tilde{U})^2\le \frac{d}{dt}d(U,\tilde{U})^2 \le -2K(1-3\alpha _1)d(U,\tilde{U})^2. \end{aligned}$$
Proof
It follows from (53) that
$$\begin{aligned} \frac{d}{dt}d((U(t),\tilde{U}(t))^2 = {\bar{\mathcal L}} - \frac{K}{2N}\sum _{k=1}^{N} {\bar{\mathcal M}}_k. \end{aligned}$$
Below, we estimate \({\bar{\mathcal L}}\) and \({\bar{\mathcal M}}_k\) separately.
\(\bullet \) Case A (Estimate of \({\bar{\mathcal L}}\)): We use the property \(\text{ tr }(AB) = \text{ tr }(BA)\) and (54) to find
$$\begin{aligned} {\bar{\mathcal L}} = {\mathrm i} \times \Big [ \text{ tr }(H_i U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger }) - \text{ tr }(U_i U_j^{\dagger } \tilde{U}_j \tilde{U}_i^{\dagger } H_i) + \text{ tr }(H_i \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger }) - \text{ tr }( \tilde{U}_i \tilde{U}_j^{\dagger } U_j U_i^{\dagger } H_i) \Big ] = 0. \end{aligned}$$
\(\bullet \) Case B (Estimate of \({\bar{\mathcal M}}_k\)): We use the decomposition of \( {\bar{\mathcal M}}_k\) in Lemma 4 to derive the estimate
$$\begin{aligned} {\bar{\mathcal M}}_k = 4d(U, {\tilde{U}})^2 ( 1 + \text{ small } \text{ quantities }). \end{aligned}$$
\(\bullet \) Case B.1 (Estimate of \({\bar{\mathcal A}}_k\)): First, we estimate the four terms below; the remaining terms can be estimated similarly. It follows from (16) and (17) that
$$\begin{aligned}&\diamond ~\Big | \text{ tr } \Big (\tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_k (I_d -\tilde{U}_k^{\dagger } U_k U_i^{\dagger } \tilde{U}_i) (\tilde{U}_k^{\dagger }\tilde{U}_i-I_d)(I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j) \Big ) \Big | \nonumber \\&\qquad \qquad \le \Vert \tilde{U}_i^{\dagger } U_i U_k^{\dagger } \tilde{U}_k \Vert \cdot \Vert I_d -\tilde{U}_k^{\dagger } U_k U_i^{\dagger } \tilde{U}_i \Vert \cdot \Vert \tilde{U}_k^{\dagger }\tilde{U}_i-I_d \Vert \cdot \Vert I_d-\tilde{U}_i^{\dagger } U_i U_j^{\dagger } \tilde{U}_j \Vert \nonumber \\&\qquad \qquad \le D({\tilde{U}}) d(U, {\tilde{U}})^2, \nonumber \\&\diamond ~ \Big | \text{ tr }\Big ((\tilde{U}_i^{\dagger }\tilde{U}_k-I_d) (I_d -\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \Big ) \Big | \nonumber \\&\qquad \qquad \le \Vert \tilde{U}_i^{\dagger }\tilde{U}_k-I_d \Vert \cdot \Vert I_d -\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i \Vert \cdot \Vert I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j \Vert \nonumber \\&\qquad \qquad \le D({\tilde{U}}) d(U, {\tilde{U}})^2, \nonumber \\&\diamond ~\Big | \text{ tr } \Big ( (\tilde{U}_j^{\dagger }U_kU_j^{\dagger }\tilde{U}_j-I_d) (I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \Big ) \Big | \nonumber \\&\qquad \qquad \le \Vert \tilde{U}_j^{\dagger }U_kU_j^{\dagger }\tilde{U}_j-I_d \Vert \cdot \Vert I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k \Vert \cdot \Vert I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j \Vert \nonumber \\ \end{aligned}$$
$$\begin{aligned}&\qquad \qquad \le D(U) d(U, {\tilde{U}})^2, \nonumber \\&\diamond ~\Big | \text{ tr } \Big ( (I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (\tilde{U}_k^{\dagger }\tilde{U}_j-I_d) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) \Big ) \Big | \le D({\tilde{U}}) d(U, {\tilde{U}})^2. \end{aligned}$$
(57)
In the third estimate of (57), we used the relation:
$$\begin{aligned} \Vert {\tilde{U}}_j^{\dagger } U_k U_j^{\dagger } {\tilde{U}}_j - I_d \Vert = \Vert {\tilde{U}}_j ({\tilde{U}}_j^{\dagger } U_k U_j^{\dagger } {\tilde{U}}_j - I_d ) {\tilde{U}}_j^{\dagger } \Vert = \Vert U_k U_j^{\dagger } - I_d \Vert = \Vert U_k - U_j\Vert \le D(U). \end{aligned}$$
The other terms can be estimated similarly; thus, we obtain
$$\begin{aligned} |{\bar{\mathcal A}}_k| \le (2D(U) + 6 D({\tilde{U}})) d(U, {\tilde{U}})^2 \le 8 \alpha _1 d(U, {\tilde{U}})^2. \end{aligned}$$
\(\diamond \) Case B.2 (Estimate of \({\bar{\mathcal B}}_k\)): Similar to Case A, we obtain using (37) that
$$\begin{aligned} |{\bar{\mathcal B}}_k| \le 2 d(U, {\tilde{U}})^3 \le 2(D(U) + D({\tilde{U}})) d(U, {\tilde{U}})^2 \le 4 \alpha _1 d(U, {\tilde{U}})^2. \end{aligned}$$
\(\diamond \) Case B.3 (Estimate of \({\bar{\mathcal C}}_k\)): Recall that
$$\begin{aligned} {\bar{\mathcal C}}_k&:=2 \text{ tr }\Big [ (I_d-\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j) +(I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j)\\& +(I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i) +(I_d -\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j) (I_d -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i)\Big ]. \end{aligned}$$
Directly expanding the expression for \({\bar{\mathcal C}}_k\) yields
$$\begin{aligned} {\bar{\mathcal C}}_k&=2 \text{ tr }\Big [ (I_d-\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j+\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j)\\& +(I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j+\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j)\\& +(I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i+\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i)\\&+(I_d -\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i+\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i)\Big ]. \end{aligned}$$
Using the relation \(\text{ tr }(AB) = \text{ tr }(BA)\) and \(\text{ tr }(UAU^{\dagger }) = \text{ tr }(A)\) for unitary U to cancel terms with index k yields
$$\begin{aligned} {\bar{\mathcal C}}_k&=2 \text{ tr }\Big [ (I_d-\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i -\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j+\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j)\\&\quad +(I_d -\tilde{U}_j^{\dagger }U_jU_k^{\dagger }\tilde{U}_k-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j+U_k^{\dagger }\tilde{U}_k\tilde{U}_i^{\dagger }U_i)\\&\quad +(I_d -\tilde{U}_i^{\dagger }U_iU_k^{\dagger }\tilde{U}_k -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i+U_k^{\dagger }\tilde{U}_k\tilde{U}_j^{\dagger }U_j)\\&\quad +(I_d -\tilde{U}_k^{\dagger }U_kU_j^{\dagger }\tilde{U}_j -\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i+\tilde{U}_k^{\dagger }U_kU_i^{\dagger }\tilde{U}_i)\Big ]\\&=4 \text{ tr }\Big [2I_d-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j-\tilde{U}_j^{\dagger }U_jU_i^{\dagger }\tilde{U}_i\Big ]\\&=4 \text{ tr }\Big [(I_d-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j)(I_d-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j)^{\dagger }\Big ]\\&=4\Vert I_d-\tilde{U}_i^{\dagger }U_iU_j^{\dagger }\tilde{U}_j\Vert ^2\\&=4d(U(t),\tilde{U}(t))^2. \end{aligned}$$
Note that indices i and j are chosen so that the last equality holds. Finally, in (53), we combine all estimates in Case A and Case B to obtain
$$\begin{aligned} \left| \frac{d}{dt} d(U,\tilde{U})^2+2 K d(U,\tilde{U})^2\right| \le 6 \alpha _1 K d(U,\tilde{U})^2. \end{aligned}$$
This completes the proof. \(\square \)
