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At a point in a normal orthonormal frame, torsion-freeness makes the frame brackets vanish there, hence the dual one-forms have zero exterior derivative there. Differentiating the interior product formula then leaves only ${\sum }_i{E}_i(f_i){\omega }_g$, the divergence of a Riemannian vector field. Since the point is arbitrary, the identity holds globally.

In a positive Riemannian orthonormal frame $X={\sum }_i{f}_i{E}_i$ with dual coframe ${\omega }_i$, the Riemannian volume form is ${\omega }_1\wedge \cdots \wedge {\omega }_n$. Inserting $E_i$ in its first slot omits ${\omega }_i$ and produces sign $(-1{)}^{i-1}$, giving the displayed interior product formula.

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 exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

Since the Riemannian orthonormal frame is positively oriented, its dual coframe satisfies ${\omega }_g={\omega }_1\wedge \cdots \wedge {\omega }_n$. The form $\nu$ is the interior product ${\iota }_X{\omega }_g$. The contraction rule for an exterior product givesOne can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X={\sum }_i{f}_i{E}_i$ proves the interior contraction of the Riemannian volume form formula:The exponents $i-1$ and $i+1$ have the same parity.

Use the standard convention that these compact smooth manifolds have no boundary. The interior product identity for a one-form gives $df\wedge {\iota }_X{\omega }_g=df(X){\omega }_g=X(f){\omega }_g$. For example, in the coframe from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded Leibniz rule and part (c) giveIntegrate and apply the Generalized Stokes theorem. Since $\partial M=\varnothing$, the integral of $d(f\nu )$ is zero. Rearranging proves integration by parts for Riemannian divergence:If a boundary were allowed, the exact general formula would instead have the additional term ${\int }_{\partial M}f\nu$ on the right; compactness alone does not remove that boundary contribution.