Synchronization and Stability for Quantum Kuramoto

Abstract

We present and analyze a nonabelian version of the Kuramoto system, which we call the Quantum Kuramoto system. We study the stability of several classes of special solutions to this system, and show that for certain connection topologies the system supports multiple attractors. We also present estimates on the maximal possible heterogeneity in this system that can support an attractor, and study the effect of modifications analogous to phase-lag.

Appendices

A Details for the Linearization Computation

If we take Proposition 3.4 and plug in $A=B=I$, then we obtain Proposition 2.2. So in fact we will prove Proposition 3.4 directly and consider Proposition 2.2 as a corollary. First a lemma:

Lemma A.1

Let us assume that G has the Lohe property. Then the infinitesimal version of the Lohe property is: for any $Y,Z\in G$, $Q\in \mathfrak g$, we have $YQZ + Z^{-1}QY^{-1} \in \mathfrak g$.

Proof

Let us write $X = Ye^{\epsilon Q}Z$ with $Y,Z\in G$ and $Q\in \mathfrak g$. Then $X\in G$, and by the Lohe property, $X-X^{-1} \in \mathfrak g$. But then

$$\begin{aligned} \begin{aligned} X - X^{-1}&= Y(I+\epsilon Q)Z - Z^{-1}(I-\epsilon Q)Y^{-1} \\&=(YZ-Z^{-1}Y^{-1}) + \epsilon (YQZ + Z^{-1}QY^{-1}) + O(\epsilon ^2). \end{aligned} \end{aligned}$$

Since $X-X^{-1}\in \mathfrak g$ and $YZ-Z^{-1}Y^{-1} \in \mathfrak g$, we have $YQZ + Z^{-1}QY^{-1}\in \mathfrak g$. $\square$

Proof of Proposition 3.4

If we expand (23), then we have

$$\begin{aligned}&\frac{d}{dt}X_i \cdot X_i^{-1} \nonumber \\&\quad = \Omega _i + \frac{1}{2}\sum _{j=1}^n \gamma _{ij} \left( f(BX_jX_i^{-1}A^{-1}) - f(AX_iX_j^{-1}B^{-1}) - f(BA^{-1})-f(AB^{-1})\right) .\nonumber \\ \end{aligned}$$

(27)

Assume that $Y = \{Y_i\}$ is a fixed point, which means

$$\begin{aligned} \begin{aligned} 0&=\Omega _i + \sum _{j=1}^n \gamma _{ij} \left( f(BY_jY_i^{-1}A^{-1}) - f(AY_iY_j^{-1}B^{-1}) - f(BA^{-1})-f(AB^{-1})\right) . \end{aligned} \end{aligned}$$

(28)

Assume that $f(x) = x^p$, and the general case will follow by linearity. Writing

$$\begin{aligned} X_i(t) = Y_i \exp (\epsilon Q_i(t)) = Y_i(I + \epsilon Q_i(t)) + O(\epsilon ^2), \end{aligned}$$

then the left-hand side of (QK) (or (23)) is

$$\begin{aligned} Y_i(\epsilon Q_i^{\prime })(I - \epsilon Q_i)Y_i^{-1} = \epsilon Y_i Q_i^{\prime } Y_i^{-1} + O(\epsilon ^2). \end{aligned}$$

(29)

We compute

$$\begin{aligned} (BX_j X_i^{-1}A^{-1})^p&= (BY_j(I+\epsilon Q_j)(I-\epsilon Q_i)Y_i^{-1}A^{-1})^p \\&= (BY_j Y_i^{-1}A^{-1} + \epsilon BY_j (Q_j-Q_i) Y_i^{-1}A^{-1} + O(\epsilon ^2))^p\\&= (BY_jY_i^{-1}A^{-1})^p \\ {}&\quad + \epsilon \sum _{q=0}^{p-1} (BY_jY_i^{-1}A^{-1})^q (BY_j(Q_j-Q_i)Y_i^{-1}A^{-1})(BY_jY_i^{-1}A^{-1})^{p-q-1}. \end{aligned}$$

The last term comes from noting that we can ignore the term of $O(\epsilon ^2)$, and consider only the binomial. Then every term of $O(\epsilon )$ in the expansion comes from taking q powers of the O(1) term, one power of the $O(\epsilon )$ term, and then $p-q-1$ powers of the O(1) term. We can reorder these terms slightly, by peeling off one $BY_j$ from the front and one $Y_i^{-1}A^{-1}$ from the back, and we can write

$$\begin{aligned}&(BX_j X_i^{-1}A^{-1})^p = (BY_jY_i^{-1}A^{-1})^p \\&\quad + \epsilon \sum _{q=0}^{p-1} (BY_j)(Y_i^{-1}A^{-1}BY_j)^q (Q_j-Q_i)(BY_jY_i^{-1}A^{-1})^{p-q-1}(Y_i^{-1}A^{-1}). \end{aligned}$$

Similarly, we obtain

$$\begin{aligned}&(AX_i X_j^{-1}B^{-1})^p = (AY_iY_j^{-1}B^{-1})^p \\&\quad - \epsilon \sum _{q=0}^{p-1} (AY_i)(Y_j^{-1}B^{-1}AY_i)^q (Q_j-Q_i)(AY_iY_j^{-1}B^{-1})^{p-q-1}(Y_j^{-1}B^{-1}). \end{aligned}$$

Using (28), we see that when we take the $\gamma$ sum the O(1) term disappears, and then we obtain

$$\begin{aligned} \epsilon Y_i Q_i^{\prime } Y_i^{-1}&= \frac{\epsilon }{2}\sum _{j=1}^n \gamma _{ij}\sum _{q=0}^{p-1} (BY_j)(Y_i^{-1}A^{-1}BY_j)^q (Q_j-Q_i)(BY_jY_i^{-1}A^{-1})^{p-q-1}(Y_i^{-1}A^{-1})\\&\quad +\frac{\epsilon }{2}\sum _{j=1}^n \gamma _{ij}\sum _{q=0}^{p-1} (AY_i)(Y_j^{-1}B^{-1}AY_i)^q (Q_j-Q_i)(AY_iY_j^{-1}B^{-1})^{p-q-1}(Y_j^{-1}B^{-1}). \end{aligned}$$

Solving for $Q_i^{\prime }$ gives

$$\begin{aligned} Q_i^{\prime }&= \frac{1}{2}\sum _{j=1}^n \gamma _{ij}\sum _{q=0}^{p-1} (Y_i^{-1}BY_j)(Y_i^{-1}A^{-1}BY_j)^q (Q_j-Q_i)(BY_jY_i^{-1}A^{-1})^{p-q-1}(Y_i^{-1}A^{-1}Y_i)\\&\quad +\frac{1}{2}\sum _{j=1}^n \gamma _{ij}\sum _{q=0}^{p-1} (Y_i^{-1}AY_i)(Y_j^{-1}B^{-1}AY_i)^q (Q_j-Q_i)(AY_iY_j^{-1}B^{-1})^{p-q-1}(Y_j^{-1}B^{-1}Y_i). \end{aligned}$$

We now use Lemma A.1. If we pair the qth term in the first sum with the $p-q-1$st term in the second, they are of the form $\Upsilon (Q_j-Q_i)\Xi + \Xi ^{-1}(Q_j-Q_i)\Upsilon ^{-1}$, and this term is in $\mathfrak g$. Therefore the entire right-hand side of (24) is also in $\mathfrak g$ by linearity. This recovers (24). $\square$

B Details Regarding the Twist Solution and Its Stability

There are several results in the body of the paper whose proofs were promised here, namely: Propositions 2.16,  2.22, 2.24, and  2.25 and Theorems 2.17,  2.18,  2.21. Before we proceed with these proofs, we have three lemmas which we state and prove first, and then proceed with the proofs of the remaining results.

Lemma B.1

Let us assume that $0< \ell < n/K$. Define

$$\begin{aligned} f_{k, \ell }(x) = \left\{ \cos \left( \frac{2\pi }{n} {k(\ell +x)}\right) +\cos \left( \frac{2\pi }{n} {kx}\right) - \cos \left( \frac{2\pi }{n} {k\ell }\right) -1\right\} , \end{aligned}$$

and then $f_{k,\ell }(x)$ is periodic of period n / k. On the fundamental domain, it is zero at $x=0, n/k-\ell$, negative for $x\in (0,n/k-\ell )$, and positive for $x\in (n/k-\ell ,n/k)$.

Proof

The statement of periodicity is straightforward, and we can see the zeros of the function by plugging in. If we define $g_{k,\ell }(x)$ by

$$\begin{aligned} g_{k, \ell }(x) = \left( \cos \left( \frac{2\pi }{n} {k(\ell +x)}\right) + \cos \left( \frac{2\pi }{n} {kx}\right) \right) , \end{aligned}$$

then this function is (up to a negative multiplicative constant) its own second derivative, meaning that it is concave down when positive and concave up when negative. Moreover, we see that this function has exactly two roots in the fundamental domain, at $x = -\frac{\ell }{2} \pm \frac{n}{4k}$. Finally, note that $f_{k,\ell }(x)$ is just $g_{k,\ell }(x)$ shifted (down) by a constant, and therefore $f_{k,\ell }(x)$ has at most two roots in the fundamental domain. Thus the statement about the signs on the intervals follows. $\square$

Lemma B.2

Let A be real symmetric, and B real skew-symmetric, and define

$$\begin{aligned} C = \left( \begin{array}{cc}A&{}-B\\ B&{}A\end{array}\right) . \end{aligned}$$

(30)

Then the eigenvalues of C are the eigenvalues of $A+\mathrm {i}B$, each repeated twice. If v is an eigenvector of $A+\mathrm {i}B$ with eigenvalue $\omega$, then $(\mathfrak {R}(v),\mathfrak {I}(v))$ and $(\mathfrak {R}(v),-\mathfrak {I}(v))$ are eigenvectors of C with eigenvalue $\omega$.

Remark B.3

Under the assumptions, C is real symmetric and $A+\mathrm {i}B$ is Hermitian, so the eigenvalues are real. Moreover, the eigenvalues of $A-\mathrm {i}B$ are the same as $A+\mathrm {i}B$.

Proof

Let $(A+\mathrm {i}B) v = \omega v$. Taking adjoints also gives $(A-\mathrm {i}B)v=\omega v$. Expanding this last gives

$$\begin{aligned} A\mathfrak {R}v + B \mathfrak {I}v + \mathrm {i}(-B\mathfrak {R}v + A \mathfrak {I}v) = \omega \mathfrak {R}v + \mathrm {i}\omega \mathfrak {I}v. \end{aligned}$$

From this it follows that

$$\begin{aligned} \left( \begin{array}{cc}A&{}B\\ -B&{}A\end{array}\right) \left( \begin{array}{c}\mathfrak {R}v \\ \mathfrak {I}v\end{array}\right) = \left( \begin{array}{c}A\mathfrak {R}v + B \mathfrak {I}v\\ -B\mathfrak {R}v + A \mathfrak {I}v\end{array}\right) =\omega \left( \begin{array}{c}\mathfrak {R}v \\ \mathfrak {I}v\end{array}\right) , \end{aligned}$$

and using the adjoint equation gives the same result for $(\mathfrak {R}v,-\mathfrak {I}v)$. $\square$

Lemma B.4

Let A be real symmetric and B, C real skew-symmetric $n\times n$ matrices. Define

$$\begin{aligned} Q = \left( \begin{array}{cccc}A&{}\quad B&{}\quad C&{}\quad 0\\ -B&{}\quad A&{}\quad 0&{}\quad C\\ -C&{}\quad 0&{}\quad A&{}\quad B\\ 0&{}\quad -C&{}\quad -B&{}\quad A \end{array}\right) . \end{aligned}$$

Then the eigenvalues of Q are the eigenvalues of $A\pm \mathrm {i}(B\pm C)$, or, equivalently, the eigenvalues of $A+\mathrm {i}(B\pm C)$ repeated twice.

Proof

Since Q is symmetric, its eigenvalues are all real. Since A is symmetric and B, C are antisymmetric, either of the two matrices $A+\mathrm {i}(B\pm C)$ are Hermitian. Note also that $A\pm \mathrm {i}(B+C)$ have the same eigenvalues, since if

$$\begin{aligned} (A+\mathrm {i}(B+C))z = \mu z, \end{aligned}$$

then

$$\begin{aligned} (A-\mathrm {i}(B+C)){\overline{z}} = \mu \overline{z}. \end{aligned}$$

(Also note that the eigenvalue counts match up; since there are two choices of sign in $A\pm \mathrm {i}(B+C)$, this gives 4 lists of length n, or 4n total eigenvalues.)

We compute:

$$\begin{aligned} (A+\mathrm {i}(B+C))(x+\mathrm {i}y) = (Ax-(B+C)y)+\mathrm {i}((B+C)x+Ay), \end{aligned}$$

and if $\mu$ is a real eigenvalue of $A+\mathrm {i}(B+C)$, then we can separate real and imaginary parts as

$$\begin{aligned} Ax - (B+C)y = \mu x,\quad (B+C)x + Ay = \mu y. \end{aligned}$$

We also can compute directly that

$$\begin{aligned} Q(x,-y,-y,-x)^t&= (Ax - (B+C)y,-(B+C)x-Ay,-(B+C)x\\ {}&\quad -Ay,-Ax+(B+C)y)^t\\&=\mu (x,-y,-y,-x)^t, \end{aligned}$$

and we have shown that $x+\mathrm {i}y$ is an eigenvector of $A+\mathrm {i}(B+C)$, iff $(x,-y,-y,-x)$ is an eigenvector of Q with the same eigenvalue. Similar computations show that

$$\begin{aligned} (A+\mathrm {i}(B+C))(x-\mathrm {i}y) = \mu (x-\mathrm {i}y)&\iff Q(-x,-y,-y,x)^t=\mu (-x,-y,-y,x)^t,\\ (A+\mathrm {i}(B-C))(x+\mathrm {i}y) = \mu (x+\mathrm {i}y)&\iff Q(x,-y,y,x)^t=\mu (x,-y,y,x)^t,\\ (A+\mathrm {i}(B-C))(x-\mathrm {i}y) = \mu (x-\mathrm {i}y)&\iff Q(-x,-y,y,-x)^t=\mu (-x,-y,y,-x)^t, \end{aligned}$$

and we are done. $\square$

Proof of Proposition 2.16

We prove this in the case where $f(x) = x^p$ for $p\ge 1$ and the rest follows by linearity. We have

$$\begin{aligned} X_i^{\prime }X_i^{-1} = \sum _{j=1}^n \gamma _{ij}((X_jX_i^{-1})^p - (X_iX_j^{-1})^p). \end{aligned}$$

(31)

We will show that if we plug in $Y_i = T^i$ into the right-hand side of (31), we obtain zero. Plugging in gives

$$\begin{aligned} \sum _{j=1}^n \gamma _{ij}(T^{p(j-i)} - T^{p(i-j)}). \end{aligned}$$

There are two cases: either n is odd or n is even. If n is odd, then we obtain

$$\begin{aligned} \sum _{k=0}^{\lfloor n/2 \rfloor } \gamma _{i,i+k}(T^{pk} - T^{-pk}) + \gamma _{i,i-k} (T^{-pk}-T^{pk}). \end{aligned}$$

(32)

If n is even, then we obtain

$$\begin{aligned} \sum _{k=0}^{ n/2-1} \gamma _{i,i+k}(T^{pk} - T^{-pk}) + \gamma _{i,i-k} (T^{-pk}-T^{pk}) + \gamma _{i,i+n/2}(T^{pn/2}-T^{-pn/2}). \end{aligned}$$

(33)

Since $T^{pn} = (T^n)^p = I^p = I$, we have $T^{pn/2}T^{pn/2} = I$ and thus $T^{pn/2} = T^{-pn/2}$, so that the last term of (33) is zero and thus (33) is the same as (32). Since $\gamma _{i,i+k} = \gamma _{i,i-k}$ for all i, k, then all terms in (32) cancel. $\square$

Proof of Theorem 2.17

Let us first consider a single $\ell$-twist with $\left| {\ell }\right| > 1$, and then we will show that $\mu _{\ell ,m}$ is positive for some m. Since $\mu _{\ell ,m}$ in (16) is invariant under the change $\ell \mapsto -\ell , m\mapsto -m$, we can assume without loss of generality that $\ell >0$.

Assume then that $\ell > 1$. From Lemma B.1, the function $f_{k,\ell }(x)$ is periodic with period n / k and positive on $(n/k-\ell ,n/k)$, and from this it implies that it is positive on the domain $(n-\ell ,n)$ for every k. This interval has width $\ell$ and thus contains some integers (e.g. $n-1$ is always in the interior of this interval). In summary, if $\ell > 1$, then $f_{k,\ell }(n-1) > 0$. But now notice that $f_{k,\ell }(m)$ is just $\mu _{\ell ,m}$ in (17) with $\gamma$ chosen where $\gamma _k = 1$ and other $\gamma _j = 0$. Since (16) is linear in the components of $\gamma$, and the argument above is independent of k, this implies that for any choice of $\gamma$, for a single rotation with $\left| {\ell }\right| >1$, there is a linearly unstable mode.

Now consider a rotation around multiple axes, indexed by $\ell _a$. If we have $\left| {\ell _a}\right| >1$ for any $\ell _a$, then the argument above also applies to $\mu _{\ell _a,m}$ in (17). Thus the only remaining case is if we have multiple nonzero $\ell _a,\ell _b$ with $\left| {\ell _a}\right| = \left| {\ell _b}\right| = 1$. As before, since $\kappa ^\pm _{\ell _1,\ell _2,m}$ is invariant under the transformation $\ell _1\mapsto -\ell _1,\ell _2\mapsto -\ell _2,m\mapsto -m$, we can assume that $\ell _1 = 1$. Then simplifying (18) gives

$$\begin{aligned} \kappa ^{+}_{1,1,m} = \kappa ^{-}_{1,-1,m} = 2\left( \cos \left( \frac{2\pi }{n} {k(m+1)}\right) - \cos \left( \frac{2\pi }{n} {k}\right) \right) \end{aligned}$$

and this is clearly positive at $m=n-1$. $\square$

Proof of Theorem 2.18

We first prove that if $\gamma = \delta _K$ and K does not divide n, then one of the $\mu _{1,m}$ in (16) is positive. Consider again Lemma B.1: the function $f_{K,1}(x)$ is positive on any interval of the form $(qn/K-1,qn/K)$ with $q\in \mathbb {Z}$. If n / K is not an integer, then qn / K is not an integer for some $q\in \mathbb {Z}$, and therefore $f_{K,1}(x)$ is positive on an integer, meaning that one of the $\mu _{1,m} > 0$. Similarly, if K does divide n, then all of the $\mu _{1,m} \le 0$, since all of the intervals will have width one and integral endpoints.

Now assume $K=1$. In this case, we see that the function $f_{1,1}(x)$ is positive only on the interval $(n-1,1)$, and therefore $\mu _{1,m} \le 0$ for all m, and strictly negative for $m\ne 0,-1$.

Now let us consider a general choice of $\gamma _k$ with $K< n/4$, and consider the formula for $\lambda _{1,m}$ in (16). Let us write this in the form $\sum _{k=1}^K \gamma _k \alpha _{k,m,n}$. We see by inspection that $\alpha _{k,0,n} = 0$ and therefore $\lambda _{1,0} = 0$ independent of $\gamma _k$. However, if $k < n/4$ and $m \ne 0$, then $\alpha _{k,m,n}\le 0$, and moreover if $m\ne 0$, $\alpha _{1,m,n} < 0$. Therefore $\lambda _{1,m} < 0$ if $\gamma _1 > 0$.

Now consider the formula for $\mu _{1,m}$ in (16). Let us write this in the form $\sum _{k=1}^K \gamma _k \beta _{k,m,n}$. We see that $\beta _{k,m,n} = 0$ if $m=0,n-1$, so this sum is again zero. As mentioned above, it is possible for $\beta _{k,m,n}$ to be positive (in particular it will be for some m if k does not divide m). However, we know that $\beta _{1,m,n} < 0$ for any n and $m\ne 0,n-1$, so therefore this sum has a negative coefficient for $\gamma _1$. Therefore, if $\gamma _1$ is large enough, this guarantees a negative sum, which gives the result. In fact, we can say more: for each m, the condition on $\gamma _1$ in terms of the other $\gamma _k$ is linear, and therefore to guarantee stability we need to show that

$$\begin{aligned} \max _{m=1,\dots , n-1} \sum _{k=1}^K \gamma _k \beta _{k,m,n} <0, \quad \text{ or, }\quad \gamma _1 > \min _{m=1,\dots , n-1}\left( -\frac{1}{\beta _{1,m,n}} \sum _{k=2}^K \gamma _k \beta _{k,m,n}\right) . \end{aligned}$$

Since the right-hand side is a finite minimum of linear functions, it is piecewise linear. $\square$

Remark B.5

In the special case of $K=2$, this reduces to $\gamma _1 > \rho ^*(n) \gamma _2$, where

$$\begin{aligned} \rho ^*(n) = \max \left( 0,\min _{m=1,\dots , n-1} \left( -\frac{\beta _{2,m,n}}{\beta _{1,m,n}}\right) \right) \end{aligned}$$

In particular, when n is even this is exactly zero, but for n odd this is always positive (since at least one of the $\beta _{2,m,n}$ is negative). In particular, we can see that when $n = 2k+1$ is odd, this expression is maximized at $m=k$, and for k large we find that

$$\begin{aligned} -\frac{\beta _{2,k,2k+1}}{\beta _{1,k,2k+1}} = \frac{\pi ^2}{4k^2} - \frac{\pi ^3}{4k^3} + O(k^{-4}), \end{aligned}$$

so that $\rho ^*(n) \approx (\pi /n)^2$ for n odd.

Proof of Theorem 2.21

Let us define two integrals:

$$\begin{aligned} I_m(\alpha )&:= \int _0^{2\pi \alpha } \cos (x) (\cos (mx)-1)\,dx, \end{aligned}$$

(34)

$$\begin{aligned} J_m(\alpha )&:= \int _{0}^{2\pi \alpha } \left( \cos ((m+1)x) + \cos (mx)-\cos (x)-1\right) \,dx. \end{aligned}$$

(35)

If we discretize by $x_k = 2\pi k/n$, $\Delta x = 2\pi /n$, then $\lambda _{1,m}$ (resp. $\mu _{1,m}$) is a scalar multiple of the (left) Riemann sum of $I_m$ (resp. $J_m$). In particular, if $I_m(\alpha )<0$, then this implies that $\lambda _{1,m}$ is negative on the $\alpha$-sequence for n sufficiently large, and similarly for $\mu _{1,m}$. We will show that there exists $\alpha ^*$ such that for all $0<\alpha <\alpha ^*$, $I_m(\alpha ),J_m(\alpha )<0$ for all m, and this will establish the result.

Let us first consider

$$\begin{aligned} I_1(\alpha ) = \pi \alpha -\sin (2 \pi \alpha )+\frac{1}{4} \sin (4 \pi \alpha ). \end{aligned}$$

The Taylor series expansion shows that $I_1(\alpha ) < 0$ for $\alpha$ small and positive, and numerically we find that the first positive root of $I_1(\cdot )$ is at $\alpha ^* \approx 0.340461$. Now consider $m>1$. We have

$$\begin{aligned} I_m(\alpha ) = -\sin (2\pi \alpha ) + \frac{m}{m^2-1} \cos (2 \pi \alpha ) \sin (2 \pi m \alpha )- \frac{1}{m^2-1}\sin (2 \pi \alpha ) \cos (2 \pi m \alpha ). \end{aligned}$$

We break this into two cases: m even and m odd. If m is even, we observe that $I_m(\alpha ) = I_m(1/2-\alpha )$. Since the integrand in (34) is negative for all $x\in (0,1/4)$, this means that $I_m(\alpha )$ is monotone decreasing on (0, 1 / 4). Using the symmetry, this means that $I_m(\alpha ) \le 0$ for all $\alpha (0,1/2)$, and certainly for all $\alpha \in (0,\alpha ^*)$.

If m is odd, it is slightly more complicated. Let us write $m=2k+1$. We still have that $I_{2k+1}(\alpha )$ is monotone decreasing on $\alpha \in (0,1/4)$, and we can check that $I_{2k+1}(1/4) = -1$. Note that $1/2(I_{2k+1}(\alpha )+I_{2k+1}(1/2-\alpha )) = -\sin (2\pi \alpha )$ and

$$\begin{aligned} \frac{1}{2} (I_{2k+1}(\alpha )-I_{2k+1}(1/2-\alpha )) = \frac{1}{4k} \sin (4k\pi \alpha ) + \frac{1}{4(k+1)}\sin (4(k+1)\pi \alpha ) \end{aligned}$$

and thus

$$\begin{aligned} \frac{1}{2} \left| {I_{2k+1}(\alpha )-I_{2k+1}(1/2-\alpha )}\right| \le \frac{2k+1}{4(k^2+k)}. \end{aligned}$$

Putting all of this together, we have

$$\begin{aligned}&I_{2k+1}(\alpha ) = \frac{1}{2}(I_{2k+1}(\alpha )+I_{2k+1}(1/2-\alpha )) \nonumber \\&\qquad \qquad \qquad + \frac{1}{2} (I_{2k+1}(\alpha )-I_{2k+1}(1/2-\alpha )) \le -\sin (2\pi \alpha )+\frac{2k+1}{4(k^2+k)}. \end{aligned}$$

This means that

$$\begin{aligned} I_{2k+1}(\alpha ^*) \le -\sin (2\pi \alpha ^*)+\frac{2k+1}{4(k^2+k)} = -0.844328+\frac{2k+1}{4(k^2+k)} < 0, \end{aligned}$$

and thus $I_{2k+1}(\alpha ) <0$ for all $0<\alpha <\alpha ^*$.

Now we consider $J_m(\alpha )$. Notice that if $m=0,-1$, then $J_m(\alpha ) \equiv 0$, and otherwise

$$\begin{aligned} J_m(\alpha ) = \frac{\sin (2\pi \alpha (m+1))}{m+1} + \frac{\sin (2\pi \alpha m)}{m} - \sin (2\pi \alpha ) - 2\pi \alpha , \quad m\ne 0,-1. \end{aligned}$$

Using the inequality

$$\begin{aligned} \sin (x) \le x, \quad \sin (x) < x, x\ne 0, \end{aligned}$$

gives

$$\begin{aligned} \frac{\sin (2\pi \alpha m)}{m} -2\pi \alpha < 0 \end{aligned}$$

(36)

for all $m\ne 0$. Let us now show that if $\left| {q}\right| >1$, and $\beta >0$ is small enough, then

$$\begin{aligned} \sin (q \beta ) \le q \sin (\beta ). \end{aligned}$$

(37)

Fix $\beta$ and consider each side as a function of q. (Since both sides are odd in q, we can consider only $q>0$.) The right-hand side of (37) is linear, and the left-hand side is concave down on the set $q\in [0,\pi /\beta ]$, but the functions are equal at $q=0,1$. Therefore (37) holds for $q\in [1,\pi /\beta ]$. Since the left-hand side is negative in $[\pi /\beta ,2\pi /\beta ]$, then the inequality also holds here. Finally, note that the left-hand side is uniformly bounded above by 1, so that if the right-hand side is larger than one for all $q>2\pi /\beta$, then the inequality holds there. This is equivalent to saying that $\sin (\beta ) > \beta /2\pi$, which holds for all $0< \beta < \beta ^* \approx 2.6978$.

If we write $q=m+1, \beta = 2\pi \alpha$, then (37) becomes

$$\begin{aligned} \sin (2\pi \alpha (m+1)) \le (m+1)\sin (2\pi \alpha ), \end{aligned}$$

(38)

which holds for all $0< \alpha < \beta ^*/(2\pi ) \approx 0.429368$ and $m\not \in (-2,0)$. Since $\beta ^*/(2\pi ) > \alpha ^*$, this expression is negative on $(0,\alpha ^*)$. $\square$

Proof of Proposition 2.22

Recall that $X_q = T^q$, and that we are considering only $f(x) =x$. Comparing to Proposition 2.2, we have

$$\begin{aligned} \mathcal L_{X,ij} Q = \frac{1}{2}(T^{j-i}Q + QT^{i-j}), \end{aligned}$$

(39)

so that the dependence of L on i, j depends solely on $i-j$. Moreover, exchanging i, j replaces T with $T^*=T^{-1}$ in (39).

Let us choose as a basis for $\mathfrak {so}(d)$ the basis vectors $M_{ij}$, $i<j$, where $M_{ij}$ has a 1 in position (i, j), a $-1$ in position (j, i), and is zero otherwise. Then a direct computation shows that

$$\begin{aligned} \frac{1}{2}(T M_{12} + M_{12}T^*)&= \cos \theta M_{12},\\ \frac{1}{2}(T M_{1q} + M_{1q}T^*)&= \frac{1}{2}(1+\cos \theta ) M_{1q} + \frac{1}{2}\sin \theta M_{2q},\quad q>2,\\ \frac{1}{2}(T M_{2q} + M_{2q}T^*)&= \frac{1}{2}(1+\cos \theta ) M_{2q} - \frac{1}{2}\sin \theta M_{1q},\quad q>2, \end{aligned}$$

and for all other $M_{ij}$ not listed above, we have $TM_{ij} + M_{ij}T^* = M_{ij}$. In particular, if we consider the operator $L:\mathfrak {so}(d)\rightarrow \mathfrak {so}(d)$ given by $LQ=(TQ+QT^*)/2$, and choose as ordered basis for $\mathfrak {so}(d)$ the matrices $(M_{12},M_{13},M_{23},M_{14},M_{24},\dots , M_{34}, \dots )$, then the matrix representation of L in this basis is given by the matrix

$$\begin{aligned} Q(\theta ) = (\cos \theta ) \oplus \left( \begin{array}{cc}\frac{1}{2}(1+\cos \theta )&{} - \frac{1}{2}\sin \theta \\ \frac{1}{2}\sin \theta &{}\frac{1}{2}(1+\cos \theta )\end{array}\right) ^{\oplus (d-2)}\oplus I_{(d-2)(d-3)/2}. \end{aligned}$$

(40)

Since T is a twist by angle $\theta$, replacing T with $T^p$ just replaces $\theta$ with $p\theta$ in (40). What this means is that if we look at the entry $Q_{\alpha \beta }(\theta )$ in (40), this will allow us to generate the $(\alpha ,\beta )$ block of $J_Y$ in (10) as follows: the (i, j)th entry of $J_{\alpha \beta }$ is given by

$$\begin{aligned} (J_{\alpha \beta })_{ij} = \gamma _{ij} Q_{\alpha \beta }((j-i)\theta ). \end{aligned}$$

Moreover, since $\Gamma$ is circulant, we have $\gamma _{ij} = \gamma _{\left| {i-j}\right| }$, and we can more simply write

$$\begin{aligned} (J_{\alpha \beta })_{i,i\pm k} = \gamma _k Q_{\alpha \beta }(\pm k\theta ). \end{aligned}$$

Note that if $Q_{\alpha \beta }$ is an even function of $\theta$, this implies that $J_{kl}$ is symmetric, and if $Q_{kl}$ is an odd function of $\theta$, then $J_{kl}$ is skew-symmetric.

More compactly, let us the matrix $\mathsf {A}_{\Gamma }({f}({\theta }))$ as the $n\times n$ matrix with entries:

$$\begin{aligned} \left( \mathsf {A}_{\Gamma }({f}({\theta }))\right) _{ij} = {\left\{ \begin{array}{ll} \gamma _k f(k\theta ), &{}\quad j=i+k,\\ -\sum _{\ell \ne 0}\gamma _\ell f(\ell \theta ),&{}\quad j=i.\end{array}\right. } \end{aligned}$$

and then it follows that the Jacobian at X is given by the matrix

$$\begin{aligned} J_{X} = \mathsf {A}_{\Gamma }(\cos \theta ) \oplus \frac{1}{2} \left( \begin{array}{cc}\mathsf {A}_{\Gamma }(1+\cos \theta )&{} - \mathsf {A}_{\Gamma }(\sin \theta )\\ \mathsf {A}_{\Gamma }(\sin \theta )&{}\mathsf {A}_{\Gamma }(1+\cos \theta )\end{array}\right) ^{\oplus (d-2)}\oplus \mathsf {A}_{\Gamma }(1)^{\oplus {(d-2)(d-3)/2}} \end{aligned}$$

(41)

Recalling (16), if we can show that $\lambda _\ell$, $\mu _\ell$, $\nu _\ell$ are the eigenvalues of the three matrices appearing in (41), then we are done. To establish this, we note that the first and third matrices are circulant, and while the second matrix is not quite “block circulant with circulant blocks” (BCCB), we can attack it in a similar fashion to that used for BCCB matrices [17, 32, 62]. Using Lemma B.2, we see that the eigenvalues of the middle matrix are the same as the eigenvalues of

$$\begin{aligned} \frac{1}{2}(\mathsf {A}_{\Gamma }(1+\cos \theta ) + \mathrm {i}(\mathsf {A}_{\Gamma }\sin \theta )) = \frac{1}{2} \mathsf {A}_{\Gamma }(1+e^{\mathrm {i}\theta }). \end{aligned}$$

Note that this last matrix is circulant. Choose $\zeta$ to be an nth root of unity, and define the vector v such that $v_i = \zeta ^i$. Writing $E = (1/2)\mathsf {A}_{\Gamma }(1+e^{\mathrm {i}\theta })$, we have

$$\begin{aligned} (Ev)_i&= \sum _{k=1}^K E_{i,i+k}v_{i+k} + \sum _{k=1}^K E_{i,i-k}v_{i-k} + E_{ii}v_i\\&= \sum _{k=1}^K\frac{\gamma _k}{2}(1+e^{\mathrm {i}k\theta })\zeta ^{i+k}+\sum _{k=1}^K\frac{\gamma _k}{2}(1+e^{-\mathrm {i}k\theta })\zeta ^{i-k} + \left( -\sum _{k\ne 0}\frac{\gamma _k}{2} (1+e^{\mathrm {i}k\theta })\right) \zeta ^i\\&= \zeta ^i\left( \sum _{k=1}^K\frac{\gamma _k}{2} (1+e^{\mathrm {i}k\theta })(\zeta ^{k}-1)+\sum _{k=1}^K\frac{\gamma _k}{2} (1+e^{-\mathrm {i}k\theta })(\zeta ^{-k}-1)\right) , \end{aligned}$$

and since $v_i =\zeta ^i$, the expression in parentheses is an eigenvalue of E. Let us now assume that $\theta = 2\pi l/n$, and choose $\zeta = \exp (\mathrm {i}2\pi m/n)$. (Note here that l is fixed by the solution X, but m ranges over $0,\dots ,n-1$.) Then we have

$$\begin{aligned}&\sum _{k=1}^K\frac{\gamma _k}{2} (1+e^{\mathrm {i}k\theta })(\zeta ^{k}-1)+\sum _{k=1}^K\frac{\gamma _k}{2} (1+e^{-\mathrm {i}k\theta })(\zeta ^{-k}-1)\\&\quad = \sum _{k=1}^K\frac{\gamma _k}{2}\left( e^{\mathrm {i}\frac{2\pi }{n} k(\ell +m)} + e^{\mathrm {i}\frac{2\pi }{n} km}- e^{\mathrm {i}\frac{2\pi }{n} k\ell } -1\right) + c.c.\\&\quad = \sum _{k\ne 0 }\gamma _k \left\{ \cos \left( \frac{2\pi }{n} {k(\ell +m)}\right) +\cos \left( \frac{2\pi }{n} {km}\right) - \cos \left( \frac{2\pi }{n} {k\ell }\right) -1\right\} . \end{aligned}$$

This is the definition of $\mu _{\ell ,m}$ in (16). By (41) there are $d-2$ copies of the inner matrix, and by Lemma B.2, these eigenvalues are doubled, so each $\mu _{\ell ,m}$ appears with multiplicity $2(d-2)$.

The formulas for $\lambda _{\ell ,m}, \nu _{\ell ,m}$ are similar but a bit simpler. Again noting that these are circulant, the same approach shows that the eigenvalues of $\mathsf {A}_{\Gamma }(\cos \theta )$ are $\lambda _{\ell ,m}$ and the eigenvalues of $\mathsf {A}_{\Gamma }(1)$ are $\nu _{\ell ,m}$. $\square$

Proof of Propositions 2.24 and 2.25

The proof will be similar to that given in Proposition 2.22. The idea is as follows: let us assume that $T = \mathsf {Tw}(\theta _1,\theta _2,0,0,\dots ,0)$. Let us write $L Q = \frac{1}{2}(TQ+QT^*)$ as above, then we have

$$\begin{aligned} \mathcal L M_{12}&= \cos \theta _1 M_{12},\quad \mathcal L M_{34} = \cos \theta _2 M_{34},\\ \mathcal L \left( \begin{array}{c}M_{1q}\\ M_{2q}\end{array}\right)&= \frac{1}{2}\left( \begin{array}{cc}1+\cos \theta _1&{} -\sin \theta _1\\ \sin \theta _1&{}1+\cos \theta _1\end{array}\right) \left( \begin{array}{c}M_{1q}\\ M_{2q}\end{array}\right) , \quad q>4,\\ \mathcal L \left( \begin{array}{c}M_{3q}\\ M_{4q}\end{array}\right)&= \frac{1}{2}\left( \begin{array}{cc}1+\cos \theta _2&{} - \sin \theta _2\\ \sin \theta _2&{}1+\cos \theta _2\end{array}\right) \left( \begin{array}{c}M_{3q}\\ M_{4q}\end{array}\right) , \quad q>4,\\ \mathcal L\left( \begin{array}{c}M_{13}\\ M_{24}\\ M_{23}\\ M_{24}\end{array}\right)&= \frac{1}{2}\left( \begin{array}{cccc}\cos \theta _1+\cos \theta _2&{} -\sin \theta _2&{}-\sin \theta _1&{}0\\ \sin \theta _2&{}\cos \theta _1+\cos \theta _2&{}0&{}-\sin \theta _1\\ \sin \theta _1&{}0&{}\cos \theta _1+\cos \theta _2 &{}-\sin \theta _2\\ 0&{}\sin \theta _1&{}\sin \theta _2&{}\cos \theta _1+\cos \theta _2 \end{array}\right) \left( \begin{array}{c}M_{13}\\ M_{24}\\ M_{23}\\ M_{24}\end{array}\right) , \end{aligned}$$

and for all $M_{ij}$ not listed above, $\mathcal L M_{ij} = M_{ij}$. Form this it follows that the Jacobian at X is given by the matrix

$$\begin{aligned} \begin{aligned} J_{X} = \mathsf {A}_{\Gamma }(\cos \theta _1) \oplus \mathsf {A}_{\Gamma }(\cos \theta _2) \oplus \frac{1}{2} \left( \begin{array}{cc}\mathsf {A}_{\Gamma }(1+\cos \theta _1)&{} - \mathsf {A}_{\Gamma }(\sin \theta _1)\\ \mathsf {A}_{\Gamma }(\sin \theta _1)&{}\mathsf {A}_{\Gamma }(1+\cos \theta _1)\end{array}\right) ^{\oplus (d-4)}\\\oplus \frac{1}{2} \left( \begin{array}{cc}\mathsf {A}_{\Gamma }(1+\cos \theta _2)&{} - \mathsf {A}_{\Gamma }(\sin \theta _2)\\ \mathsf {A}_{\Gamma }(\sin \theta _2)&{}\mathsf {A}_{\Gamma }(1+\cos \theta _2)\end{array}\right) ^{\oplus (d-4)}\oplus \mathsf {A}_{\Gamma }(1)^{\oplus {(d-4)(d-5)/2}}\oplus \\\oplus \frac{1}{2}\left( \begin{array}{cccc}\mathsf {A}_{\Gamma }(\cos \theta _1+\cos \theta _2)&{} -\mathsf {A}_{\Gamma }(\sin \theta _2)&{}-\mathsf {A}_{\Gamma }(\sin \theta _1)&{}0\\ \mathsf {A}_{\Gamma }(\sin \theta _2)&{}\mathsf {A}_{\Gamma }(\cos \theta _1+\cos \theta _2)&{}0&{}-\mathsf {A}_{\Gamma }(\sin \theta _1)\\ \mathsf {A}_{\Gamma }(\sin \theta _1)&{}0&{}\mathsf {A}_{\Gamma }(\cos \theta _1+\cos \theta _2) &{}-\mathsf {A}_{\Gamma }(\sin \theta _2)\\ 0&{}\mathsf {A}_{\Gamma }(\sin \theta _1)&{}\mathsf {A}_{\Gamma }(\sin \theta _2)&{}\mathsf {A}_{\Gamma }(\cos \theta _1+\cos \theta _2) \end{array}\right) . \end{aligned} \end{aligned}$$

(42)

All of the formulas for the eigenvalues except for $\kappa$ are derived exactly the way as they are in the proof of Proposition 2.22—the only difference in all of these is that instead of a single $\theta$ we now have $\theta _1$ or $\theta _2$. The matrix that is new is the last one. Putting Lemma B.4 together with the last matrix in (42), we see that the remaining $\kappa _{\ell _1,\ell _2,m}^\pm$ are the eigenvalues of

$$\begin{aligned} \frac{1}{2}\mathsf {A}_{\Gamma }(e^{\mathrm {i}\theta _1} + e^{\pm \mathrm {i}\theta _2}). \end{aligned}$$

Note that this last matrix is circulant. Choose $\zeta$ to be an nth root of unity, and define the vector v such that $v_i = \zeta ^i$. Writing $E = (1/2)\mathsf {A}_{\Gamma }(e^{\mathrm {i}\theta _1} + e^{\pm \mathrm {i}\theta _2})$, we have

$$\begin{aligned} (Ev)_i&= \sum _{k=1}^K E_{i,i+k}v_{i+k} + \sum _{k=1}^K E_{i,i-k}v_{i-k} + E_{ii}v_i\\&= \sum _{k=1}^K\frac{\gamma _k}{2}(e^{\mathrm {i}k\theta _1}+e^{\pm \mathrm {i}k\theta _2})\zeta ^{i+k}+\sum _{k=1}^K\frac{\gamma _k}{2}(e^{-\mathrm {i}k\theta _1}+e^{\mp \mathrm {i}k\theta _2})\zeta ^{i-k} + \left( -\sum _{k\ne 0}\frac{\gamma _k}{2} (e^{\mathrm {i}k\theta _1}+e^{\pm \mathrm {i}k\theta _2})\right) \zeta ^i\\&= \zeta ^i\left( \sum _{k=1}^K\frac{\gamma _k}{2} (e^{\mathrm {i}k\theta _1}+e^{\pm \mathrm {i}k\theta _2})(\zeta ^{k}-1)+c.c.\right) , \end{aligned}$$

and since $v_i =\zeta ^i$, the expression in parentheses is an eigenvalue of E. If we write $\theta _a = 2\pi \ell _a/n$, and choose $\zeta = \exp (\mathrm {i}2\pi m/n)$, then

$$\begin{aligned}&\sum _{k=1}^K\frac{\gamma _k}{2} (e^{\mathrm {i}k\theta _1}+e^{\pm \mathrm {i}k\theta _2})(\zeta ^{k}-1)+c.c.)\\&\quad = \sum _{k=1}^K\frac{\gamma _k}{2}\left( e^{\mathrm {i}\frac{2\pi }{n} k(\ell _1+m)} + e^{\mathrm {i}\frac{2\pi }{n} k(\pm \ell _2+m)}- e^{\mathrm {i}\frac{2\pi }{n} k\ell _1} -e^{\mathrm {i}\frac{2\pi }{n} k\ell _2}\right) + c.c.\\&\quad = \sum _{k=1}^K \gamma _k \left\{ \cos \left( \frac{2\pi }{n} {k(\ell _1+m)}\right) +\cos \left( \frac{2\pi }{n} {k(\pm \ell _2+m)}\right) - \cos \left( \frac{2\pi }{n} {k \ell _1}\right) - \cos \left( \frac{2\pi }{n} {\pm k\ell _2}\right) \right\} . \end{aligned}$$

This is the definition of $\kappa ^\pm _{\ell _1,\ell _2,m}$ in (17). By (42) there are $d-2$ copies of the inner matrix, and by Lemma B.2, these eigenvalues are doubled, so each $\mu _{\ell _a,m}$ appears with multiplicity $2(d-2)$.

More generally, let us assume that $T = \mathsf {Tw}(\theta _1,\theta _2,\dots )$ where the unwritten angles may or may not be zero. The proof above implies that the $4\times 4$ matrix that appears in (42) also appears as a term in the direct sum of the Jacobian for the higher-order rotation, and and such contains the eigenvalues $\kappa ^\pm _{\ell _1,\ell _2,m}$ as well. $\square$