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For the symplectic form $\omega =idz\wedge d\bar{z}/(1+|z{|}^2{)}^2$, the circle group rotates $z$ by $e^{i\theta }$. Its real generator is $K=iz{\partial }_z-i\bar{z}{\partial }_{\bar{z}}$. The contraction is ${\iota }_K\omega =-dH$, so $H$ is a Hamiltonian function in this sign convention. In the other chart $w=1/z$, $H=1/(1+|w{|}^2)$, proving smoothness at infinity. Its image is $[0,1]$ with the endpoints at the fixed poles. The symplectic area is $2\pi$; rescaling to the round area $4\pi$ would double this moment function.

On the circle group $\mathbb{T}$, normalized Haar measure is the pushforward measure of uniform angle on $[0,2\pi )$. For a continuous function $f:\mathbb{T}\to \mathbb{C}$, defineThe factor $1/(2\pi )$ makes the integral of the constant function $1$ equal to $1$. For $g=e^{i\varphi }$, a change of angle givesThe last equality uses the $2\pi$-periodic function $u\mapsto f(e^{iu})$. Since the circle group is an abelian group, this also verifies right invariance. An arc of angular length $L$ has normalized Haar measure $L/(2\pi )$. In particular, the arc $A=\{e^{i\theta }:\pi /6\le \theta \le \pi /2\}$ and its quarter-turn $iA$ both have measure $(\pi /2-\pi /6)/(2\pi )=1/6$.

The continuous unitary characters ${\chi }_n(e^{i\theta })=e^{in\theta }$, indexed by integers, give a fully computed family of integrals:Here $n$ must be an integer, so $z^n$ is well defined and $e^{2\pi in}=1$. Every nonconstant mode cancels exactly over a full turn. For example, the integral of $f(e^{i\theta })=3+2\cos \theta -\sin (2\theta )$ is $3$, and the integral of ${\cos }^2\theta =(1+\cos (2\theta ))/2$ is $1/2$. More generally,This is the orthogonality of complex exponentials underlying the Fourier basis and Fourier series: invariant averaging separates distinct frequencies and retains the constant component.

Let $G$ be connected and compact with nonzero irreducible real Adjoint representation. Its Lie algebra has no proper invariant Lie algebra ideal. If it is Abelian, irreducibility forces dimension one, so $G$ is the circle group. Otherwise its centre has zero-dimensional Lie algebra and is finite. A proper closed normal subgroup likewise has zero-dimensional Lie algebra, hence is finite; connectedness of $G$ makes conjugation on that finite subgroup trivial, so the subgroup is central. A Lie group homomorphism onto a nontrivial connected target with surjective differential consequently has finite central group kernel. The qualifications “proper” and “nontrivial” are essential.

Conservation of particle number corresponds to global circle group phase rotations of a nonrelativistic Bose field. A positive-density uniform mean-field saddle chooses a phase, with degenerate saddle manifold a circle. The associated Goldstone boson is the long-wavelength phase mode. This is a thermodynamic symmetry-selection statement; an exact finite-volume number eigenstate does not itself choose a phase.

A symplectic manifold is a smooth even-dimensional manifold $M$ with a differential two-form $\omega$ that is closed, $d\omega =0$, and nondegenerate at every point. Nondegeneracy identifies vector fields with differential one-forms by the interior product $X\mapsto {\iota }_X\omega$. In this solution use the explicit Hamiltonian vector field conventionThis contraction sign differs from the negative-sign convention sometimes used for Hamiltonian vector fields; all subsequent signs are fixed by the displayed choice. In Darboux coordinates with $\omega ={\sum }_{a}d{x}_a\wedge d{y}_a$ it givesThe Poisson bracket is bilinear and antisymmetric, and the ordinary product rule gives $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Thus it remains to check the Jacobi identity for the Poisson bracket.

The Cartan formula for the Lie derivative and closedness of the symplectic form give ${\mathcal{L}}_{X_f}\omega =d({\iota }_{X_f}\omega )+{\iota }_{X_f}d\omega =d^{2}f=0$. Henceand nondegeneracy givesApplying both sides to an arbitrary smooth function $h$ gives $\{f,\{g,h\}\}-\{g,\{f,h\}\}=\{\{f,g\},h\}$, which is exactly the Jacobi identity for the Poisson bracket. Consequently the smooth functions, with their pointwise product and this Poisson bracket, form a Poisson algebra. Constants have zero Hamiltonian vector field. If $f,g$ are components of moment maps, the same displayed relation relates the brackets of their Hamiltonian vector fields to the Poisson bracket of those components. In the opposite contraction convention the corresponding relation is the Hamiltonian Lie algebra homomorphism with the compatible bracket sign.

For the point vortices, remove the collision diagonals from $({\mathbb{R}}^2{)}^N$, since the logarithmic Hamiltonian function is singular there. Set $s_{ab}=(x_a-x_b{)}^2+(y_a-y_b{)}^2>0$ for $a\ne b$. Differentiating each pair contribution givesThe Hamiltonian vector field therefore gives the equations of motionEach contribution is perpendicular to the displacement from vortex $b$ to vortex $a$ and has magnitude $2/|{\mathbf{r}}_a-{\mathbf{r}}_b|$. This is the velocity induced by a point vortex of positive circulation, with the normalization $\Gamma /(2\pi )=2$. Each vortex is advected by the velocity of the others, without a self-induced singular velocity. Pairwise contributions cancel in the total velocity, so ${\sum }_a{x}_a$ and ${\sum }_a{y}_a$ are conserved. For two vortices the separation is constant and they rotate counterclockwise about their midpoint, as expected for identical positive point vortices.

For $L=\frac{1}{2}{\sum }_a(x_a^2+y_a^2)$, compute its Poisson bracket with $H$ directly:The last sum vanishes by pairing $(a,b)$ with $(b,a)$, since their numerators are opposite and denominators equal. Thus $L$ Poisson commutes with $H$ and is conserved. Equivalently, the Hamiltonian function depends only on pairwise distances and is invariant under simultaneous rotations.

The standard positive generator of the simultaneous circle group $SO(2)$ action isIts interior product with the symplectic form isUse the moment map convention $d⟨\mu ,\xi ⟩=-{\iota }_{Y_{\xi }}\omega$ for the usual positive rotation generator. Identifying $\mathfrak{so}(2{)}^{\ast }$ with $\mathbb{R}$, this proves $L$ is the moment map for the standard rotation action. It is invariant under the action, hence equivariant because circle group $SO(2)$ is abelian. Additive constants are possible; the natural normalization is $L=0$ at the origin of the ambient space. On the collision-free space this polynomial is its restriction. With the solution's positive contraction convention, the rotation generator is $Y=-X_L$. Confusing these two signs would reverse the final augmented Hamiltonian.

Let ${\mathcal{R}}_{\phi }$ denote simultaneous counterclockwise rotation through $\phi$, and introduce the co-rotating coordinates by $\mathbf{r}(t)={\mathcal{R}}_{\Omega t}{\mathbf{r}}^{\prime }(t)$. Both the symplectic form and $H$ are rotation-invariant, so pulling back the inertial Hamiltonian vector field leaves $X_H$ unchanged. Differentiating the rotation gives the co-rotating equationThe co-rotating positions are constant precisely when this Hamiltonian vector field vanishes. By nondegeneracy of the symplectic form, that is equivalent toConversely a collision-free critical point of this augmented Hamiltonian function produces a rigidly rotating solution. This is a rotating point-vortex relative equilibrium.

As a sign and scaling check, scaling every position by $s>0$ changes $H$ by $-N(N-1)\log s$. Therefore ${\sum }_a{\mathbf{r}}_a\cdot {\nabla }_{a}H=-N(N-1)$. At a rotating point-vortex relative equilibrium, ${\nabla }_{a}H+\Omega {\mathbf{r}}_a=0$, so $2\Omega L=N(N-1)$. For $N\ge 2$ the angular velocity is positive in this circulation convention. An equilateral three-vortex configuration of circumradius $\rho$ has $L=3{\rho }^2/2$ and $\Omega =2/{\rho }^2$, agreeing directly with the equations of motion.

Put $e(t)=e^{2\pi it}$ and let $P(n)=\theta n^k+\cdots$ be a real polynomial of degree $k\ge 2$. A standard form of Weyl inequality is: if $(a,q)=1$ and $|\theta -a/q|\le q^{-2}$, then for every $\eta >0$The implied constant is uniform in all lower coefficients. For degree one the finite geometric series bound is $|\sum e(\theta n+\beta )|\le \min (N,(2\Vert \theta \Vert {)}^{-1})$, where $\Vert t\Vert$ is distance to the nearest integer.

Here is the Weyl differencing proof. Extend $z_n=e(P(n))$ by zero outside $[1,N]$. Expanding $|\sum z_n{|}^2$ by differences of indices gives a sum of correlations ${\sum }_n{z}_{n+h}\overline{z_n}$. Repeated Cauchy-Schwarz inequality and this correlation identity yield, after $k-1$ steps, with $p=2^{k-1}$,The interval $I(\mathbf{h})$ ensures that every shifted argument remains in $[1,N]$. For completeness, the induction is the following: at stage $j$, the factor is $N^{2^j-j-1}$ and there is a sum over $j$ shifts. Squaring, Cauchy-Schwarz over those $O(N^j)$ shifts supplies $N^j$; expanding the square of each remaining interval sum introduces the next shift. The new factor is $N^{2^{j+1}-j-2}$, as asserted for stage $j+1$.

The final differenced phase is linear, of slope $k!\theta h_1\cdots h_{k-1}$. Zero products contribute $O_k(N^{k-1})$ to the sum. For nonzero products the finite geometric series bound applies. Grouping by $r=|h_1\cdots h_{k-1}|\le N^{k-1}$, the subpower bound for the divisor function bounds the number of representations by $O_{k,\rho }(N^{\rho })$ for any chosen $\rho >0$. The divisor input is elementary: for $t=k-1$, the number of ordered factorizations of $r$ is ${\prod }_{p^v\parallel r}\left(\genfrac{}{}{0ex}{}{v+t-1}{t-1}\right)$. For all sufficiently large primes this is bounded factorwise by $p^{ϵv}$, and for the finitely many smaller primes the polynomial in $v$ is bounded by a constant times $p^{ϵv}$. Multiplication gives $O_{t,ϵ}(r^{ϵ})$. ThusThe reduced denominator of $k!a/q$ is between $q/k!$ and $q$. Its approximation error is bounded by a constant depending on $k$ times the inverse square of that denominator. Split the $r$-range into blocks of length a sufficiently small fixed multiple of $q$. Within a block the values $k!\theta r$ are separated modulo one by a constant multiple of $1/q$: rational differences have spacing at least $1/q$, while the approximation error in a block difference is smaller than half that spacing. At most a bounded number of values occupy each distance interval of width $1/q$. Summing reciprocals of their ordered distances proves the reciprocal fractional-part sum near a rational estimateFor $q>N^k$, the proposed inequality follows from $|S|\le N$. Otherwise the last display and the differencing estimate giveChoose $\rho$ sufficiently small, absorb the logarithm into a further arbitrarily small power of $N$, and take $p$th roots. This proves the stated Weyl inequality.

For the recurrence assertion, a rational $a=u/v$ gives an integer $a{m}^2$ with $m=v$. For irrational $a$, Weyl differencing implies that $a{m}^2$ is an equidistributed sequence modulo one. One way to avoid a rate issue for unusually good rational approximations is to keep a fixed differencing range $H$: the Van der Corput inequality for finite scalar sequences gives, for each nonzero integer $j$, a limsup bound ${\mathrm{limsup}}_N|N^{-1}{\sum }_{m\le N}e(ja{m}^2){|}^2\ll 1/H$, because every fixed nonzero difference has irrational linear slope and its geometric average tends to zero. Let $H\to \infty$ and apply the Weyl criterion. Hence some $m$ always enters the open $ϵ$-neighborhood of zero.

Finally, $U_m=\{a\in \mathbb{R}/\mathbb{Z}:\Vert a{m}^2\Vert <ϵ\}$ are open sets covering the compact circle group. A finite subcover has a largest index $N$. That single $N$ works for every real $a$. This is uniform square recurrence on the circle; the compactness step gives existence, not an explicit quantitative bound.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.

An isomorphism of Lie algebras identifies the local multiplication laws of the identity components. In exponential coordinates the local product is governed by the Baker--Campbell--Hausdorff formula,A bracket-preserving linear isomorphism carries this law to the corresponding law in the other Lie algebra. Thus the identity components are locally isomorphic. This statement is local; the exponential coordinates need not cover the whole group.

The stronger global description uses a universal covering Lie group. The universal cover of a connected Lie group acquires a group law by lifting multiplication with a chosen identity; uniqueness of lifts supplies associativity and inversion. The covering map is a local diffeomorphism and a Lie group homomorphism, with discrete kernel. That kernel is central: conjugation of one of its elements gives a continuous map from the connected covering group into a discrete set, so it is constantly that element.

By integration of the given Lie algebra isomorphism and its inverse on simply connected covers, these covers are isomorphic. Their two compositions have identity differential and therefore are identity maps by the preceding uniqueness argument. Consequently, for some connected simply connected Lie group $\tilde{G}$,where ${\Gamma }_1,{\Gamma }_2$ are discrete central subgroups. Different quotients need not be isomorphic. The additive group $\mathbb{R}$ and the circle group $\mathbb{R}/2\pi \mathbb{Z}$ have the same one-dimensional Lie algebra but different compactness and fundamental groups. Another familiar pair is the SU(2) group and the SO(3) group, with the latter the quotient of the former by $\{I,-I\}$.

Thus isomorphic Lie algebras give isomorphic connected simply connected groups, and identify arbitrary identity components up to discrete central quotients. For disconnected groups the Lie algebra also omits the group of components and its conjugation and extension data. For example $\mathbb{R}\times F$, for any finite group $F$, always has Lie algebra $\mathbb{R}$, while its component group is $F$.

Identify the circle group with $[0,1)$ modulo its endpoints, and let $\lambda$ be normalized Lebesgue measure. We first check invariance. On the $r$th inverse branch, $x=(y+r)/m$ for $0\le r<m$. For an integrable function $g$, substitution givesIn particular, testing indicator functions shows that the integer multiplication map on the circle preserves $\lambda$.

To prove ergodicity, let $B$ be a measurable set with $\lambda (E_m^{-1}B△B)=0$ and put $h=1_B$. Then $h(E_{m}x)=h(x)$ almost everywhere. Its Fourier coefficients areUsing invariance of $h$ and substituting separately on the same $m$ branches givesIn the last line, the finite geometric series of $m$th roots of unity is zero unless $m$ divides $k$, when it equals $m$. For any nonzero integer $k$, divide repeatedly by $m$ until the resulting integer is no longer divisible by $m$. The recurrence then implies $c_k=0$.

Since the Fourier basis is a complete orthonormal basis of $L^2([0,1),\lambda )$, the absence of every nonconstant Fourier coefficient implies $h=c_0=\lambda (B)$ almost everywhere. An indicator function takes only the values zero and one, so $\lambda (B)$ must be zero or one. This proves

A strongly continuous unitary representation of the circle group decomposes as $H={\widehat{⨁}}_{m\in \mathbb{Z}}{H}_m$, with $U_z\xi =z^m\xi$ on $H_m$. Its positive energy unitary representation of the circle condition is that the occurring integers are bounded below. Equivalently $U_{e^{it}}=e^{itL}$ with a self-adjoint generator bounded below. An integer character can shift the lowest weight to zero.

Let $M=S^{\prime \prime }$. Because $S=S^{\ast }$, this is a Von Neumann algebra, and rotation invariance makes $U_z{M}^{\prime }{U}_z^{\ast }=M^{\prime }$. Joint irreducibility implies $M^{\prime }\cap \{U_z{\}}^{\prime }=\mathbb{C}I$: a nonscalar self-adjoint element of that joint commutant would have a nontrivial invariant spectral projection gives a reducing subspace.

For $T\in M^{\prime }$, take its strong Fourier coefficients $T_k={\int }_{\mathbb{T}}{z}^{-k}{U}_{z}T{U}_z^{\ast }dz$. They lie in $M^{\prime }$ and satisfy $U_z{T}_k{U}_z^{\ast }=z^k{T}_k$. Both $T_k^{\ast }{T}_k$ and $T_k{T}_k^{\ast }$ are rotation-invariant elements of $M^{\prime }$, hence scalar. If $T_k\ne 0$, both scalars equal $\Vert T_k{\Vert }^2$, so $T_k/\Vert T_k\Vert$ is a unitary. It shifts energy by $k$. For $k\ne 0$, applying it or its adjoint to a nonzero lowest-energy vector produces a vector below the lowest energy, a contradiction. Thus $T_k=0$ for every $k\ne 0$, while $T_0$ is scalar.

Fejér sum averaging of the strongly continuous orbit $U_{z}T{U}_z^{\ast }$ converges strongly to $T$. Its Fourier coefficients have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular every bounded operator commuting with $S$ commutes with every $U_z$. In fact the hypotheses force the stronger conclusion $S^{\prime }=\mathbb{C}I$.

A nonempty compact dynamical system $(X,T)$ is a minimal dynamical system if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward orbit is dense: the orbit closure of any point is a nonempty closed forward-invariant set, and conversely a closed invariant set contains the orbit closure of each of its points.

Consider the collection of nonempty closed forward-invariant subsets of $X$, ordered by reverse inclusion. Every chain has an upper bound, its intersection: the chain has the finite-intersection property, compactness makes the intersection nonempty, and continuity is not even needed for its forward invariance. The Zorn lemma supplies a minimal member $Y$, so $(Y,T{|}_Y)$ is a minimal subsystem. Also $T(Y)$ is compact, nonempty and forward invariant, so minimality gives $T(Y)=Y$. Here the printed word “nontrivial” must mean nonempty; a singleton with its identity map shows that a subsystem with more than one point cannot always be required.

If $y\in Y$, every tail of its forward orbit is dense in $Y$, since it is the orbit of $T^{N}y\in Y$. Therefore every neighborhood of $y$ contains $T^{n}y$ for arbitrarily large $n$. Choosing these times successively for balls of radii $1/j$ gives $n_j\to \infty$ with $T^{n_j}y\to y$. This proves the Birkhoff recurrence theorem: every continuous map of a nonempty compact metric space has a recurrent point. In fact every point of the chosen minimal subsystem is recurrent.

Write the continuous circle extension as the skew productA sufficient condition, and in this setting also a necessary one, is the circle skew-product minimality criterion: for every integer $m\ne 0$ there is no continuous function $f:X\to {\mathbb{S}}^1$ such thatThe cocycle $\rho$ is assumed continuous; a merely measurable cocycle need not define this topological extension.

To prove sufficiency, choose a minimal subsystem $Z$ of the extension. Its projection is a nonempty closed forward-invariant subset of $X$, hence all of $X$. For $a\in \mathbb{R}/\mathbb{Z}$, vertical translation sends $Z$ to another minimal subsystem, since it commutes with $\tilde{T}$. Two such translates either are disjoint or coincide: a nonempty intersection is closed and forward invariant, so minimality forces equality. Define their common stabilizerThis is a closed subgroup of the circle group. Its closedness follows by taking limits of translations of points in the compact set $Z$, using both a translation and its inverse. Each fiber $Z_x$ is nonempty. If $y,z\in Z_x$, the sets $Z$ and $Z+(z-y)$ intersect at $(x,z)$, so $z-y\in H$. Conversely translating any fiber point by any member of $H$ stays in the fiber. Thus every fiber is exactly one coset of $H$.

If $Z$ is proper, $H$ is proper and therefore finite, say $H=\{0,1/m,\dots ,(m-1)/m\}$ for some positive integer $m$. The character $y\mapsto e^{2\pi imy}$ is constant on each fiber coset, giving a well-defined function $f(x)$ from $Z_x$. This function is continuous: its graph is the continuous image of $Z$ in $X\times {\mathbb{S}}^1$, and projection of that compact graph onto the Hausdorff space $X$ is a continuous bijection and hence a homeomorphism. The relation $\tilde{T}(Z)=Z$ gives $f(Tx)=e^{2\pi im\rho (x)}f(x)$, contrary to the criterion. Hence $Z=\tilde{X}$ and the extension is minimal.

Conversely, a solution of the displayed equation makes $F(x,y)=e^{2\pi imy}/f(x)$ a nonconstant continuous invariant function on the extension. Its nonempty level sets are proper closed invariant subsets, contradicting minimality. This also proves necessity of the circle skew-product minimality criterion.

Apply it with $\alpha =\sqrt{2}$, $X=\mathbb{R}/\mathbb{Z}$, $Tx=x+\alpha$, and $\rho (x)=2x+\alpha$. The base is a minimal dynamical system: the closure of the cyclic subgroup generated by an irrational angle is the whole circle group, and its positive forward orbit has the same closure. To see the subgroup assertion, a proper closed subgroup of the circle is finite; if it contained an irrational angle it would be infinite. The elementary classification follows because a closed subgroup with arbitrarily small nonzero elements is dense, while otherwise its least positive representative generates a finite subgroup.

There can be no forbidden $f$ for any $m\ne 0$. Indeed, the winding number of a continuous closed path of $x\mapsto f(x+\alpha )$ equals that of $x\mapsto f(x)$, by translation and homotopy invariance of winding number. Consequently their quotient has winding number zero. But the proposed quotient $e^{2\pi im(2x+\alpha )}$ has winding number $2m\ne 0$. Thus the skew product is minimal. Starting at $(0,0)$, induction givesbecause $n^2\alpha +2n\alpha +\alpha =(n+1{)}^2\alpha$. Its orbit is dense in the two-dimensional torus, so its second coordinate is dense in the circle. Removing the initial point does not change density, since every orbit tail in a minimal dynamical system is dense. Therefore

The circle group is $U(1)=\{z\in \mathbb{C}:|z|=1\}$ under multiplication. For continuous finite-dimensional complex representations, its irreducibles are the one-dimensional charactersA group representation is a continuous homomorphism to the invertible linear maps; it is irreducible when it has no proper nonzero invariant subspace. Compactness permits an invariant Hermitian inner product by averaging. Since the resulting unitary operators commute for $U(1)$, simultaneous diagonalization reduces an irreducible to a character, and the continuous characters of the circle have the stated integer winding number.

The group $SU(2)$ consists of unitary two-by-two complex matrices of determinant one. Each has the unique formidentifying it homeomorphically with the unit 3-sphere in ${\mathbb{R}}^4$. This is the Spin group $\mathrm{Spin}(3)$. A unitary matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$, and its conjugacy class is determined by $0\le \theta \le \pi$, or equivalently its trace $2\cos \theta$.

The irreducible representations of SU2 are $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ for integers $n\ge 0$, with dimension $n+1$. The symmetric power can be realized as homogeneous degree-$n$ polynomials in two variables, with the induced linear action. On the diagonal torus its eigenvalues are $e^{i(n-2j)\theta }$ for $j=0,\dots ,n$, so its character, the trace of the representing operator, isAt the endpoints use the continuous values ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The highest-weight classification says this list is complete.

For the final action, multiplication of block matrices gives $(AB{)}_1=A_1{B}_1$, so conjugation is a linear group representation. Decompose the underlying three-dimensional space as $V_1\oplus V_0$. Its endomorphism space isThe two off-diagonal blocks are each $V_1$, since the invariant alternating form identifies $V_1^{\ast }$ with $V_1$. The scalar part of $\mathrm{End}(V_1)$ is $V_0$ and its traceless part is $V_2$: the same alternating form identifies traceless endomorphisms with symmetric tensors of degree two. The lower scalar block is a second $V_0$. ThusThe dimensions add to $2+4+3=9$, and its character is $2{\chi }_0+2{\chi }_1+{\chi }_2$.

Differentiate the circle group action at angle zero. The resulting real vector field isIn real coordinates $z=x+iy$, this is $K=-y{\partial }_x+x{\partial }_y$; the complex coefficients are conjugates, so it is indeed real. Its integral curves satisfy $\dot{z}=iz$ and hence $z(t)=e^{it}z(0)$, with period $2\pi$ away from the fixed points.

The affine coordinate does not cover the whole complex projective line, so check the other chart. With $w=1/z$, the same vector field is $K=-iw{\partial }_w+i\bar{w}{\partial }_{\bar{w}}$, smooth at $w=0$. It vanishes at $z=0$ and $w=0$, the two fixed poles. It therefore generates the stated global rotation of $S^2$, not merely a flow in one chart.

Write $e(t)=\exp (2\pi it)$. A sequence in the circle group is an equidistributed sequence if, for every interval $I\subseteq \mathbb{R}/\mathbb{Z}$,where $|I|$ is its normalized length. Equivalently, averages of every continuous function along the sequence tend to its circle integral. Trigonometric polynomials approximate continuous functions, and interval indicators can be squeezed between continuous functions with arbitrarily close integrals. The nonconstant additive characters have integral zero. These facts give the Weyl criterion:This is the link between equidistribution and cancellation in exponential sums.

Suppose every positive-shift difference sequence were equidistributed. Fix $m\ne 0$ and set $z_n=e(m\varphi (n))$. For every fixed $h\ge 1$, the Weyl criterion would giveThe omitted final $h$ terms change a normalized average by at most $h/N$.

Here is the needed Van der Corput inequality for finite scalar sequences. Extend $z_n$ by zero outside $[1,N]$ and average $H$ consecutive translates of the sum. Cauchy-Schwarz givesIndeed, apply Cauchy-Schwarz to $H{\sum }_n{z}_n={\sum }_t{\sum }_{j=0}^{H-1}{z}_{t-j}$ and expand the squared inner sum. Taking $N\to \infty$ first leaves a bound $1/H$; then let $H\to \infty$. Every nonzero Fourier average of $\varphi$ vanishes, so the Weyl criterion makes $\varphi$ equidistributed. This is the differencing obstruction to equidistribution. By contraposition, a non-equidistributed sequence has a non-equidistributed difference for some positive $h$, hence for some $h\ne 0$ as requested.

For $\varphi (n)=\sqrt{2}{n}^2$, the difference is ${\Delta }_h\varphi (n)=-2h\sqrt{2}n-h^2\sqrt{2}$. For any nonzero integer $m$, its exponential sum is a constant phase times a geometric progression with ratio $e(-2mh\sqrt{2})\ne 1$. Its normalized magnitude is at most $2/(N|1-e(-2mh\sqrt{2})|)$, which tends to zero. Thus every positive-shift difference is equidistributed, and the contraposition just established proves

A homogeneous static saddle minimizes $f(\rho )=-\mu \rho +g{\rho }^2/2$ for $\rho =|\psi {|}^2\ge 0$. With $g>0$ and $\mu >0$,The mean-field Bose-Einstein condensate has a macroscopic coherent occupation of the uniform mode. Its continuous particle-number phase symmetry is $\psi \mapsto e^{i\chi }\psi$, the circle group $U(1)$ generated by particle number. Choosing one phase breaks it, and the saddle manifold is a circle. For $\mu \le 0$, the constrained minimum is the vacuum ${\rho }_0=0$, so the condensed saddle requires the stated positive-density regime.

Spontaneous symmetry breaking is understood through a phase-selected thermodynamic description: an exact finite-volume number eigenstate has vanishing field expectation. The resulting Goldstone boson is a gapless phase/sound mode, while phase gradients carry the superfluid velocity $\mathbf{v}=\nabla \varphi /m$. This is the saddle-point conclusion requested, not a proof of true condensate order in every dimension or temperature. Long-wavelength phase fluctuations can invalidate the assumed order, as quantified below.