= Solution
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> convention
$$
\iota_{X_f}\omega=df,\qquad \{f,g\}=\omega(X_f,X_g).
$$
This 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 $\omega=\sum_a dx_a\wedge dy_a$ it gives
$$
X_f=\sum_a\left(f_{y_a}\partial_{x_a}-f_{x_a}\partial_{y_a}\right),\qquad \{f,g\}=\sum_a\left(f_{x_a}g_{y_a}-f_{y_a}g_{x_a}\right),\qquad X_f(g)=-\{f,g\}.
$$
The <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 $\mathcal L_{X_f}\omega=d(\iota_{X_f}\omega)+\iota_{X_f}d\omega=d^2f=0$. Hence
$$
\begin{aligned}
\iota_{[X_f,X_g]}\omega&=\mathcal L_{X_f}(\iota_{X_g}\omega)-\iota_{X_g}(\mathcal L_{X_f}\omega)\\
&=\mathcal L_{X_f}(dg)=d(X_f g)=-d\{f,g\},
\end{aligned}
$$
and nondegeneracy gives
$$
\boxed{[X_f,X_g]=-X_{\{f,g\}}.}
$$
Applying 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 gives
$$
H_{x_a}=-2\sum_{b\ne a}\frac{x_a-x_b}{s_{ab}},\qquad H_{y_a}=-2\sum_{b\ne a}\frac{y_a-y_b}{s_{ab}}.
$$
The <Hamiltonian vector field> therefore gives the equations of motion
$$
\boxed{\dot x_a=-2\sum_{b\ne a}\frac{y_a-y_b}{s_{ab}},\qquad\dot y_a=2\sum_{b\ne a}\frac{x_a-x_b}{s_{ab}}.}
$$
Each 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 $L=\tfrac12\sum_a(x_a^2+y_a^2)$, compute its <Poisson bracket> with $H$ directly:
$$
\{L,H\}=\sum_a(x_aH_{y_a}-y_aH_{x_a})=-2\sum_a\sum_{b\ne a}\frac{y_ax_b-x_ay_b}{s_{ab}}=0.
$$
The last sum vanishes by pairing $(a,b)$ with $(b,a)$, since their numerators are opposite and denominators equal. Thus \b[$L$ 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 is
$$
Y=\sum_a\left(-y_a\partial_{x_a}+x_a\partial_{y_a}\right).
$$
Its <interior product> with the <symplectic form> is
$$
\iota_Y\omega=-\sum_a(y_a\,dy_a+x_a\,dx_a)=-dL.
$$
Use the <moment map> convention $d\langle\mu,\xi\rangle=-\iota_{Y_\xi}\omega$ for the usual positive rotation generator. Identifying $\mathfrak{so}(2)^*$ with $\mathbb R$, this proves \b[$L$ 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 $\mathcal R_\varphi$ denote simultaneous counterclockwise rotation through $\varphi$, and introduce the co-rotating coordinates by $\mathbf r(t)=\mathcal R_{\Omega t}\mathbf r'(t)$. 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 equation
$$
\dot{\mathbf r}'=X_H(\mathbf r')-\Omega Y(\mathbf r')=X_H(\mathbf r')+\Omega X_L(\mathbf r')=X_{H+\Omega L}(\mathbf r').
$$
The co-rotating positions are constant precisely when this <Hamiltonian vector field> vanishes. By nondegeneracy of the <symplectic form>, that is equivalent to
$$
\boxed{d(H+\Omega L)(\mathbf r')=0.}
$$
Conversely 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 $\sum_a\mathbf r_a\cdot\nabla_aH=-N(N-1)$. At a <rotating point-vortex relative equilibrium>, $\nabla_aH+\Omega\mathbf r_a=0$, so $2\Omega L=N(N-1)$. For $N\geq2$ 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.
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