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Suppress tildes in the normalized equation and let $X=x-vt$. A Gross–Pitaevskii solitary wave stationary in these translating coordinates obeysThe boundary fixes the bulk complex argument and describes a localized disturbance with zero net vortex winding number; a vortex pair is possible. No extra plane-wave factor is needed when only the coordinates are changed. A full Galilean transformation of the field would additionally change the bulk flow and its complex argument convention.

Subtract the uniform background from the grand potential. A convenient dimensionless energy and the renormalized momentum of a condensate areThe subtraction in $\mathbf{p}$ removes the troublesome background complex argument gradient. In particular, for the usual subsonic localized far field $\psi -1=O(r^{-1})$, $\nabla \psi =O(r^{-2})$, and $1-|\psi {|}^2=O(r^{-2})$, the energy density is $O(r^{-4})$ and the momentum integrand is $O(r^{-3})$, both integrable in two dimensions. Merely writing $\psi \to 1$ without sufficient decay is not by itself a convergence proof. The common solitary-wave branch is subsonic: linearizing this normalization gives ${\omega }^2=k^2/2+k^4/4$ and speed of sound $1/\sqrt{2}$.

Integration by parts, with decaying variations and their boundary terms vanishing, givesThus the translating equation is precisely $\delta (E-v{p}_x)=0$. Along a differentiable family ${\psi }_{\lambda }$ with fixed bulk normalization, evaluate this stationarity on ${\partial }_{\lambda }\psi$ to obtainwhere $d{p}_x/d\lambda \ne 0$. This energy–momentum slope of a solitary wave does not require differentiating the Lagrange multiplier $v$ inside the field variation. At a turning point the parametrized identity remains true, while $E$ need not be a single-valued function of $p_x$ locally. The chosen factors make $i{\psi }_t=\delta E/\delta {\psi }^{\ast }$, consistent with the normalized equation; changing the normalization of $E$ alone would change the claimed slope.

The axial angular momentum per unit volume is $mn(r)r{u}_{\theta }(r)=\hslash n(r)$ for a singly quantized positive quantum vortex. Thus every particle in the prescribed shell contributes $\hslash$:This yieldsThis is $\hslash$ times the shell particle number. For a wall-truncated cloud with $R=R_0<R_r$, the corresponding result is $L_z=\pi H_0\hslash n_0[R^2-a_0^2-(R^4-a_0^4)/(2R_r^2)]$. Reversing the quantized circulation reverses $L_z$ without changing the flow kinetic energy. For winding number $q$, $L_z$ is multiplied by $q$.

Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

Let $u(t)=\gamma (t)/|\gamma (t)|$, a continuous map into the unit circle. By uniform continuity, partition $[0,1]$ so finely that on each interval beginning at $t_j$, $|u(t)/u(t_j)-1|<1$. This ratio lies in the right half-plane and has a continuous complex argument with value zero at $t_j$. Starting from ${\theta }_0$, add these local arguments successively, matching endpoint values. This constructs a continuous real lift with $u(t)=e^{i\theta (t)}$.

Any two such lifts with the same initial value differ by a continuous function taking values in $2\pi \mathbb{Z}$, and hence coincide. Since the curve closes, $e^{i(\theta (1)-\theta (0))}=1$. Define its winding number byChanging the chosen initial argument adds a constant multiple of $2\pi$ to the whole lift, so the winding number is independent of that choice. No differentiability of the curve is required.

Choose continuous argument lifts ${\theta }_1,{\theta }_2$ for the two curves. Their sum is a lift for their pointwise product, becauseSubtract its initial value from its final value to obtainIn particular, winding numbers subtract under pointwise division of curves avoiding zero.

The appropriate homotopy invariance of winding number is: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for each $s$, then all loops $t\mapsto H(s,t)$ have the same winding number. The basepoint may move; every intermediate curve must close and avoid zero.

Compactness gives $m={\min }_{s,t}|H(s,t)|>0$. Uniform continuity gives $\delta >0$ such that $|s-s^{\prime }|<\delta$ implies $|H(s,t)-H(s^{\prime },t)|<m$ for every $t$. Part (d) then makes the winding numbers equal for nearby $s,s^{\prime }$. Subdivide $[0,1]$ into steps smaller than $\delta$ and chain these equalities. ThereforeThis proves the theorem directly for continuous loops, without requiring an integral formula or smooth homotopy.

For a closed piecewise smooth path $\gamma$ avoiding $w$, its winding number isA meromorphic function has a zero of order $m>0$ at $z_0$ whenwith $g$ holomorphic; it has a pole of order $m$ at $w_0$ when $(z-w_0{)}^{m}f(z)$ extends holomorphically and nonvanishingly there.

The argument principle states that, if $f$ has no zero or pole on $\gamma$,For a positively oriented simple contour this is the number of enclosed zeros minus poles, counted with multiplicity. It follows from the residue theorem because the residue of $f^{\prime }/f$ at a zero of order $m$ is $m$, while at a pole of order $m$ it is $-m$.

LetOn the imaginary axis,Thus $p(iy)$ always lies in the open right half-plane, has no zero there, and its continuous argument has net change zero as the imaginary-axis part of a large right-half-disc contour is traversed. On the right semicircle, $p(z)/z^4\to 1$ uniformly, so the argument change tends to that of $z^4$, namely $4\pi$. The argument principle therefore givesroots in the open right half-plane, counted with multiplicity.

Label the edges so that barycentric coordinates on $T$ satisfySuppose for contradiction that $A\cap B\cap C=\varnothing$. The distance from a point to a closed set gives continuous nonnegative functions $d_A,d_B,d_C$. Their sum never vanishes, while at every point at least one of them vanishes because $A\cup B\cup C=T$. Henceis a continuous map from $T$ to its boundary in barycentric-coordinate space. Moreover, $F(I)\subseteq I$, $F(J)\subseteq J$, and $F(K)\subseteq K$. On each edge, the straight-line homotopy between $F$ and the identity remains in that edge, so $F{|}_{\partial T}$ has winding number one. On the other hand, a map from the whole triangle to its boundary makes its boundary restriction null-homotopic, and hence gives winding number zero. This contradiction proves the three-set covering lemma:

If a retraction $f:D\to \partial D$ fixed every boundary point, choose a homeomorphism from $T$ to $D$ taking its three edges to three consecutive closed arcs of $\partial D$ with empty triple intersection. The inverse images under $f$ of those arcs would be closed, would cover $D$, and would contain the corresponding boundary arcs. Transporting them to $T$ would contradict the result just proved. Therefore

For the final statement, use compactness of the three closed arcs inside the corresponding open sets. A finite open cover of a compact metric space admits a closed shrinking, and the shrinking can be chosen to preserve specified compact subsets already lying in the respective open sets. Thus there are closed setswhich cover $D$ and contain $\alpha ,\beta ,\gamma$, respectively. The disc version of the three-set covering lemma, obtained from the triangle by a homeomorphism taking its edges to the three arcs, gives a point in $A\cap B\cap C$. Since these sets lie in the original open sets,

Parametrize the unit circle by a loop $\gamma :[0,2\pi ]\to {\mathbb{C}}^{\ast }$. Part (d) and $f(\gamma (0))=f(\gamma (2\pi ))$ implyThe kernel of the complex exponential function is $2\pi i\mathbb{Z}$, sofor some $k\in \mathbb{Z}$. This integer is the winding number of $f{|}_{S^1}$.

Outside a defect core writeAround any closed curve on which $\mathbf{p}\ne 0$, its topological charge is the winding numberSingle-valued polar order requires $q\in \mathbb{Z}$. A nonzero value prevents the field inside the loop from being continuously deformed to a uniform nonvanishing field, so the core must contain a zero or singularity.

Choose a real lift $\Delta \theta ={\theta }_f-{\theta }_i$. Classical paths fall into winding number sectors $j\in \mathbb{Z}$ and areEach sector has the same fluctuation determinant, so the image-sum form of the propagator isSet $a=\hslash T/(2m{R}^2)$. Applying the Poisson summation formula to the Gaussian $e^{i{x}^2/(4a)}$ giveswhich is exactly the spectral propagator found in part i. Thus the angular-momentum sum is dual to a sum over homotopy classes of classical paths.

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

Finite energy requires $|\varphi |\to 1$ and $D\varphi \to 0$ on the circle at spatial infinity. Writing $\varphi \sim e^{i\chi }$ there gives $A\sim d\chi$. The vortex number is the winding numberBy Stokes theorem, the magnetic flux is quantized:

For the rotationally symmetric Abelian Higgs vortex ansatz, $A=f(r)d\theta$ and $\varphi =h(r)e^{ik\theta }$ giveAfter the angular integration, the Abelian Higgs model energy becomesCompleting the square in the two pairs of terms givesRegularity at the origin and approach to the vacuum at infinity requireMore precisely, $h(r)=O(r^k)$ and $f(r)=O(r^2)$ near the origin. The boundary term is $k$, soThe bound is saturated exactly when both squares vanish, giving the radial Bogomolny vortex equationsThe asymptotic value $f(\infty )=k$ also makes the flux $2\pi k$, so the ansatz has vortex number $N=k$.

For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would forceContinuity on the connected interval $[0,1/2)$ makes $\varphi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\varphi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.

Now let $g:[0,1]\to \mathbb{C}\setminus \{0\}$ and put $h=g/|g|$. This is a continuous path in the unit circle. The exponential map $\theta \mapsto e^{i\theta }$ is a covering map, so the path lifting theorem supplies a continuous lift $\theta :[0,1]\to \mathbb{R}$ after one value of $\theta (0)$ is chosen. HenceOne can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.

If $\theta$ and $\tilde{\theta }$ are two such phases, thenThis integer-valued function is continuous and hence constant. Thereforedoes not depend on the chosen lift.

For $u(t)=g(t^2)$, the phase $\theta (t^2)$ givesFor $v(t)=g(t{)}^2$, the phase $2\theta (t)$ gives

Finally, takeThey have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the winding number of the closed path.

Away from a zero of the Higgs field, writeSeparating real and imaginary parts of $(D_x+i{D}_y)\varphi =0$ givesThus $B=-\Delta u/2$ away from zeros. Combining this with $B=(1-e^u)/2$ givesIf $\varphi$ has a zero of multiplicity $N$ at the origin, its vortex number is the winding number $N$ and $u=2N\log r+O(1)$. Since $\Delta \log r=2\pi {\delta }^{(2)}(\mathbf{x})$ as a distributional identity, the complete Taubes equation isThe boundary conditions for a single $N$-vortex areThe latter is the finite-energy condition $|\varphi |\to 1$.

For a closed piecewise smooth curve $\gamma$ avoiding $w$, its winding number about $w$ is

The argument principle says that if a positively oriented closed curve $\gamma$ bounds a domain $U$, and a meromorphic function $F$ has no zeros or poles on $\gamma$, thenwhere zeros and poles are counted with multiplicity.

Suppose $F$ and $G$ are holomorphic on a neighbourhood of $\overline{U}$ andFor $0\le t\le 1$, $F+tG$ has no boundary zero, because a zero would imply $|F|=t|G|<|F|$. Its argument-principle count is integer-valued and continuous in $t$, hence constant. Thus Rouché's theorem states that

Now let $w_0=f(z_0)$. Since $|w_0|<1\le |f|$ on the unit circle, Rouché's theorem shows that $f$ and $f-w_0$ have the same number of zeros in the unit disc. The latter has the zero $z_0$, so $f$ has at least one zero there. For any $w$ with $|w|<1$, the same boundary inequality shows that $f-w$ has the same positive number of zeros as $f$. ThereforeThis is the unit-disc image from a boundary modulus lower bound.

Finally, take a sufficiently small positively oriented circle $C$ around zero. The residueis the winding number of $g(C)$ around zero, so $k\in \mathbb{Z}$. The pole is simple, hence its residue is nonzero and $k\ne 0$. Forwe haveThe second term cancels the complete principal part at zero, so the integer residue of a logarithmic derivative proves that $h^{\prime }/h$ has a removable singularity there.

Use the continuous determinant mapIt induces a homomorphism on fundamental groups. Under the standard identificationthe class of $t\mapsto e^{2\pi it}$ corresponds to its winding number $1$. Sincewe obtainFor $k\ne 0$ this is not the identity, so $[\gamma {]}^k$ could not have been the identity in ${\pi }_1(U(2),I)$.