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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.

The displayed matrices describe the orientation-preserving component $SE(2)$ of the full Euclidean group $E(2)=O(2)⋉{\mathbb{R}}^2$. Both have the same Lie algebra, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the Lie algebra of this Matrix Lie group asTaking matrix commutators givesAll remaining brackets follow from antisymmetry. The corresponding left-invariant vector fields in coordinates $(x,y,\psi )$ areTheir Lie brackets of vector fields obey the same relations. This also fixes which sign of the rotation generator is being used.

There is an important qualification to the printed connection prescription. For an arbitrary vector field $Z$ and smooth function $f$, its literal right-hand side obeysAn affine connection instead requires ${\nabla }_{L_i}(fZ)=L_i(f)Z+f{\nabla }_{L_i}Z$. Thus the formula for arbitrary $Z$ defines a connection only when $\lambda =1$. For example, at $\psi =0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the Leibniz rule requires $L_1$.

The intended one-parameter family is the bracket connection on a Lie group: prescribe the rule for left-invariant $Z$, then extend it by the Leibniz rule. If $[L_i,L_j]=c_{ij}{}^k{L}_k$, this meansExtend linearly over functions in the first slot. This is a well-defined affine connection for every real $\lambda$. The following Ricci tensor calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda =1$ member exists.

Use the curvature conventionFor left-invariant vector fields the structure constants of a Lie algebra are constant. The Jacobi identity givesWriting $A=\lambda (\lambda -1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, areFor instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\mathrm{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, soThis is the Killing form. Here ${\mathrm{ad}}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. ThereforeSince the coframe component dual to $L_3$ is $d\psi$, the same Ricci tensor is $2\lambda (1-\lambda )d\psi \otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.

At both $\lambda =0$ and $\lambda =1$, the full curvature vanishes. At $\lambda =0$, the left-invariant vector fields form a global parallel frame. At $\lambda =1$, every right-invariant vector field is parallel because left- and right-invariant vector fields commute; the connection is flat in that global parallel frame. Flatness does not imply zero torsion: for left-invariant arguments the torsion tensor isThus the two endpoint connections are flat, with opposite nonzero torsion. For comparison, $\lambda =1/2$ is torsion-free but has nonzero curvature, with ${\mathrm{Ric}}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a Levi-Civita connection.

A differential form of degree $k$ is integrated over an oriented $k$-dimensional smooth manifold by integrating its coefficient in orientation-preserving coordinates. For an oriented parametrization $F:U\to M$, this means integrating the pullback $F^{\ast }\alpha$ over $U$; a partition of unity combines charts. The change of variables formula makes the result independent of the charts. Reversing the orientation of a smooth manifold reverses the integral. The Generalized Stokes theorem says, for a compact oriented $k$-dimensional smooth manifold with boundary and a smooth $(k-1)$-form $\beta$,where the right-hand side includes the boundary pullback, and the boundary has the outward-normal-first boundary orientation. Compact support suffices on a noncompact manifold. This unifies the fundamental theorem of calculus, circulation and flux identities.

For example, take the unit disk $D$ oriented by $dx\wedge dy$ and the differential one-form $\beta =\frac{1}{2}(xdy-ydx)$. Its exterior derivative is $d\beta =dx\wedge dy$, so the area integral is $\pi$. Parametrizing the positively oriented circle by $(x,y)=(\cos t,\sin t)$ gives the pullback $\beta =\frac{1}{2}dt$ and the boundary integral ${\int }_0^{2\pi }\frac{1}{2}dt=\pi$, directly illustrating Generalized Stokes theorem.

For the distance calculation, write $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$, $x=u+v$ and $y=u-v$; these are scalar radii, not position vectors. The triangle inequality and its reverse give $r\le u+v=x$ and $|y|=|u-v|\le r$. Also,Thus $r\le x\le 2R-|y|$ and $|y|\le r$. These conditions, with $u,v\ge 0$, are also sufficient for a triangle with side lengths $u,v,r$. Endpoints correspond to collinear configurations.

Assume the two positions are independent random variables, each having constant volume probability density inside the radius-$R$ ball. Independence is needed: uniform marginal distributions alone do not determine the distance distribution. Put $C=3/(4\pi R^3)$ and $\mathbf{s}={\mathbf{r}}_2-{\mathbf{r}}_1$. The change of variables $({\mathbf{r}}_1,{\mathbf{r}}_2)\mapsto ({\mathbf{r}}_1,\mathbf{s})$ has unit Jacobian determinant, so the joint probability volume differential form is the product of the two normalized volume forms. In spherical coordinates for ${\mathbf{r}}_1$ and for $\mathbf{s}$ relative to its axis, it isThe polar coordinates fail on axes and at zero radii, which are sets of zero volume and do not affect the probability. Here $\theta$ is the angle from ${\mathbf{r}}_1$ to $\mathbf{s}$. The law of cosines becomesDifferentiating and taking the wedge product with $du\wedge dr$ eliminates the terms involving $du$ and $dr$:Consequently the positive integration density transforms by $|\sin \theta d\theta |=vdv/(ur)$ at fixed $u,r$. The minus sign is accounted for by reversal of limits: $v$ decreases as $\theta$ increases. The radial variable retained is $r$, while the polar angle is replaced by $v$; equivalently one can start with the angle between the two position vectors and replace that angle by $r$. The wording of the printed hint conflates these two coordinate choices, but its volume form gives the stated transformation directly.

Integrating $\sin {\theta }_{1}d{\theta }_{1}d{\varphi }_1$ over the first orientation gives $4\pi$, and integrating $d\chi$ gives $2\pi$. The probability density function of $r$ is thereforewhere $D_r$ has $0\le u,v\le R$ and $|u-v|\le r\le u+v$. There are no such configurations for $r>2R$.

To evaluate this integral explicitly, the Jacobian determinant of $(u,v)=((x+y)/2,(x-y)/2)$ has absolute value $1/2$. Hence $uvdudv=(x^2-y^2)dxdy/8$ as a positive density. For $0\le r\le 2R$, set $a=\min (r,2R-r)$. The domain is $-a\le y\le a$, $r\le x\le 2R-|y|$, and evenness in $y$ givesAn antiderivative, zero at $y=0$, yieldsSubstituting $a=r$ for $r\le R$ and $a=2R-r$ for $r\ge R$ gives the same polynomial in both intervals:Thus the distance between two uniform points in a three-dimensional ball has the final densityIt is zero outside this interval. Its nonnegativity also follows from $f_R(r)=3r^2(4R+r)(2R-r{)}^2/(16R^6)$, and direct integration gives ${\int }_0^{2R}{f}_R(r)dr=1$. The derivation uses the full six-dimensional probability volume differential form, rather than treating the three scalar distances as independent.