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On a star-shaped planar domain containing zero, let $\eta$ be a smooth closed differential one-form and set $f(x)={\int }_{[0,x]}\eta$. Green theorem on the triangle with vertices $0,x,x+h$ gives $f(x+h)-f(x)={\int }_{[x,x+h]}\eta$. Dividing by a displacement and taking its limit proves $df=\eta$. Thus every such form is an exact differential form, a degree-one version of the Poincaré lemma which uses only the planar integral theorem.

Take $\varphi$ in the identity component ${\mathrm{Symp}}_0(M,\omega )$ and choose a smooth symplectic isotopy ${\varphi }_t$ from the identity to $\varphi$. Let its generating vector field be $X_t={\dot{\varphi }}_t\circ {\varphi }_t^{-1}$. Differentiating ${\varphi }_t^{\ast }\omega =\omega$ givesThus ${\eta }_t={\iota }_{X_t}\omega$ is a closed differential one-form. By the vanishing of the first de Rham cohomology, it is exact, so choose $H_t$ with ${\eta }_t=-d{H}_t$.

The primitives can be chosen smoothly in both $t$ and the point. On each connected component fix a base point $x_0$ and define $H_t(x)=-{\int }_{x_0}^x{\eta }_t$. Exactness makes this independent of the chosen path. In a coordinate neighbourhood of any point, use a fixed path to the neighbourhood's centre followed by a smoothly varying short path; this shows smooth dependence locally, and hence globally. One may subtract a time-dependent mean without changing the Hamiltonian vector field. Therefore the original symplectic isotopy is Hamiltonian, and $\varphi \in \mathrm{Ham}(M,\omega )$.

Here the usual description of ${\mathrm{Symp}}_0$ by smooth isotopies agrees with the topological identity component. To see the local path-connectedness behind that description, use a Weinstein neighborhood of the diagonal in $(M\times M,-\omega \oplus \omega )$. The graph of a symplectomorphism close to the identity corresponds to the graph of a small closed one-form on the diagonal. Multiplying that one-form by $t\in [0,1]$ gives Lagrangian graphs; for sufficiently small forms their first projections remain diffeomorphisms, so they give symplectomorphisms connecting the original map to the identity. These local paths, translated in the group and concatenated with smooth time reparametrizations, identify its connected and smooth-path components.

Conversely every Hamiltonian isotopy is symplectic by part (a), and connects its time-one map to the identity. Hence

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

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.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

Put $d=\ell /(2\sqrt{3})$ and $a=e^{2\pi i/3}$. Consistent with the actual PDF vertices, parametrize the three oriented sides byThey run counterclockwise, with unit tangent $i{\beta }_j$ and outward normal ${\beta }_j$. Consequently the normal derivative and tangential derivative satisfyLet $\kappa =k{\beta }_j$. On this side the exponential factor becomesPulling back the closed differential one-form therefore givesDefine ${\Phi }_j(\kappa )$ and ${\Psi }_j(\kappa )$ by integrating the Dirichlet terms and normal term in this expression. The Generalized Stokes theorem gives ${\int }_{\partial D}W=0$, soThe factor $2i$ follows by solving $\sum E\Phi +(i/2)\sum E\Psi =0$. The bars on the side-2 rotation are essential; replacing them with $a$ would use the wrong side orientation. The converted TeX also reverses the sign in the upper vertex's exponential, whereas the PDF puts that vertex above the real axis; the parametrization here follows the PDF.

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the modified Helmholtz equation, including its Laplace equation case $\lambda =0$. Equivalently,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.

Fix counterclockwise arclength on each side and take the outward normal derivative. WriteThe unit tangent is $i{c}_j$ and the outward unit normal is $c_j$. On side $j$ this gives the chain-rule identityPut $w=c_{j}k$. The exponential in the closed differential one-form pulls back toAlso $-\lambda qd\bar{z}/(ik)=(\lambda /w)q^{(j)}ds$. Consequently the contribution of side $j$ is ${\rho }_j(c_{j}k)$, where the required local side spectra areThese are spectral integral transforms in the local coordinate of each side. Keeping the tangential-derivative term is important: premature integration by parts in a single side integral would leave corner values, which cannot simply be discarded. At the special samples used below, those endpoint terms do vanish for the rotationally invariant solution.

The Generalized Stokes theorem applied to the closed differential one-form givesEquivalently, the global relation isThe identical outward Neumann boundary data make the three known transforms equal:To reduce the unknown traces to a single function, use rotation invariance and uniqueness. For $\lambda >0$ the difference $u$ of two solutions has zero outward derivative andby Green's first identity. Thus $u=0$. Rotation by $a$ preserves the equation, the domain and the oriented boundary data, so it preserves $q$; hence $q^{(1)}=q^{(2)}=q^{(3)}=q$ and ${\Phi }_j=\Phi$. The global relation becomes the single equation with three spectral unknownsThe three unknowns are evaluations of the same boundary transform, not three freely specified boundary data.

The printed allowance of all real $\lambda$ requires a qualification. At $\lambda =0$, existence requires $3{\int }_{-l/2}^{l/2}f(s)ds=0$ by the divergence theorem, and solutions differ by a constant. Fixing a rotation-invariant normalization gives uniqueness and the same rotational conclusion. In fact the unnormalized solution is already invariant: its rotation difference is constant, and three successive rotations make three times that constant zero. For $\lambda <0$, uniqueness fails precisely when $-4\lambda$ is an eigenvalue of the Neumann Laplacian. At these parameters identical data do not force an arbitrary solution to be rotationally invariant. Averaging any solution over the three rotations produces a solution with the common trace used here; the difference is a homogeneous Neumann eigenfunction, or a sum of them. Part (e) describes both solvability and this remaining freedom rather than assuming uniqueness for every real parameter.

We use the following precise degree-zero/one kernel of the magnetic X-ray transform in degrees zero and one theorem, as permitted here: for a smooth Anosov magnetic flow on a closed oriented surface, ifwith $q$ a smooth base function and $\xi$ a smooth one-form, then $q=0$ and $\xi$ is exact. Exact one-forms give the converse, because $F(w\circ \pi )=dw(v)$. No assumption of pointwise negative Gaussian curvature is needed for this theorem. We now derive the contact obstruction from it rather than assume the desired rigidity conclusion.

Use the usual connected-surface convention. Since ${\int }_{M}f{\Omega }_a=0$, the top-degree de Rham cohomology class of $f{\Omega }_a$ vanishes: integration identifies $H_{\mathrm{dR}}^2(M)$ with $\mathbb{R}$. Choose a smooth one-form $\eta$ on $M$ satisfying $d\eta =f{\Omega }_a$, and setThe structure equations and the previous volume calculation giveThis primitive exists because of the zero-flux hypothesis, and need not itself be contact.

A useful elementary fact is that every smooth invariant function of an Anosov flow is constant on a connected manifold. If $Fh=0$, then $d{h}_x(v_s)=d{h}_{{\varphi }_{t}x}(D{\varphi }_t{v}_s)$, and boundedness of $dh$ with exponential contraction makes this zero as $t\to \infty$. Negative time does the same for an unstable vector. Also $dh(F)=0$. The Anosov splitting spans the tangent space, so $dh=0$. This avoids any extra ergodicity assumption.

Suppose a smooth contact form $\tau$ is preserved. Invariance and contraction imply $\tau (v_s)=\tau (D{\varphi }_t{v}_s)=0$ in the limit; negative time gives the same on unstable vectors. Thus $\tau$ annihilates both transverse bundles. Its value $\tau (F)$ cannot vanish anywhere, since otherwise $\tau$ would be zero at that point, contradicting the contact condition. It is a smooth invariant function and therefore a nonzero constant. Divide $\tau$ by that constant to arrange $\tau (F)=1$.

The Cartan formula for the Lie derivative now gives ${\iota }_{F}d\tau =0$. In dimension three, all two-forms annihilating $F$ are scalar multiples of ${\iota }_F\mu$, soThe scalar $k$ is smooth and nowhere zero by the contact condition. Both sides without $k$ are flow-invariant; hence $Fk=0$ and the preceding Anosov argument makes $k$ constant. It follows thatis a closed differential one-form.

We require the Liouville mean of such a closed differential form, and derive it explicitly. Let $R_s$ be fibre rotation and average $\zeta$ over $0\le s\le 2\pi$. Since $d\zeta =0$, the Cartan formula for the Lie derivative givesConsequently its rotational average differs from $\zeta$ by an exact form. For a closed rotationally invariant one-form $\overline{\zeta }$, the identity $d(\overline{\zeta }(V))={\mathcal{L}}_V\overline{\zeta }-{\iota }_{V}d\overline{\zeta }=0$ makes its vertical value a constant $a$. Subtracting $a\omega$ leaves a rotationally invariant horizontal form, which descends to a base form $\xi$. We have therefore proved the averaging closed one-forms on a surface unit tangent bundle decompositionThe last relation follows from closedness and $d\omega =-K{\pi }^{\ast }{\Omega }_a$.

Evaluation at $F$ gives $\zeta (F)={\xi }_x(v)+af+Fh$. The average of the linear expression ${\xi }_x(v)$ around each unit circle is zero. The second term has total integral $2\pi a{\int }_{M}f{\Omega }_a=0$. Finally ${\int }_{SM}Fhd\mu =0$ by volume preservation and Stokes theorem. Thusfor every closed differential one-form $\zeta$ in this zero-flux system. Apply this to $\tau +k\sigma$: the directional average of ${\eta }_x(v)$ also vanishes, so its integral is $(1+k){\mathrm{Vol}}_{\mu }(SM)$. Positivity of the volume forces $k=-1$. Hence $d\tau =d\sigma$ and $\zeta =\tau -\sigma$ is closed.

Using its proved decomposition and $\tau (F)=1$, we obtainThe stated magnetic transform kernel theorem gives $af=0$ and exactness of $\eta -\xi$, hence $d\eta =d\xi$. If $a=0$, then $d\xi =0$ and $f{\Omega }_a=d\eta =0$, so $f=0$. If $a\ne 0$, the equation $af=0$ already gives $f=0$. Necessity is proved.

Conversely, if $f=0$, then $F=X$. The canonical form $\alpha$ is contact because $\alpha \wedge d\alpha =-\mu$ never vanishes, andThus the flow preserves $\alpha$, provingIf disconnected surfaces are admitted, the flux condition must hold on each component for the exactness step; a single cancelling total flux does not imply componentwise exactness. The result above uses the standard connected interpretation of a surface.

Let $\omega ={\theta }^{-1}(\alpha )$. The preceding invariant formula for the exterior derivative identifies the expression in the question withIts vanishing for every pair of global vector fields is equivalent to $d\omega =0$: global fields with arbitrary prescribed tangent values are available by bump-function extension. Thus it is exactly the condition that $\omega$ be a closed differential one-form.

By definition, the first de Rham cohomology is closed one-forms modulo exact differential forms. The hypothesis $H_{\mathrm{dR}}^1(M)=0$ makes every closed one-form exact. Hence $d\omega =0$ gives a globally smooth function $g$ with $\omega =dg$, andConversely, if $\alpha (X)=X(g)$, thenby the definition of the Lie bracket of vector fields. Equivalently, $d(dg)=0$. The bracket compatibility condition is necessary and, when $H_{\mathrm{dR}}^1(M)=0$, sufficient for a global potential. The potential is unique up to a constant on each connected component, since the difference of two potentials has zero differential.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

For $u_{z\bar{z}}={\beta }^{2}u$, pull back its spectral closed differential one-form along each straight side $z=m_j+s{h}_j$. Under counterclockwise traversal and outward normal derivatives $q_j$, the side density is $i{e}^{-i\beta (\lambda z-\bar{z}/\lambda )}[|h_j|q_j+\beta (\lambda h_j+{\bar{h}}_j/\lambda )g_j]$, where $g_j$ is the Dirichlet boundary data. The Generalized Stokes theorem gives the displayed global relation. Clockwise traversal reverses the normal-derivative coefficient; all orientations must be changed consistently.

If $x=\chi (x^{\prime })$ is an angular diffeomorphism and $\eta \cdot dx$ is a closed differential one-form, the map $y=\eta (\chi (x^{\prime }))+D\chi (x^{\prime }{)}^{-T}{y}^{\prime }$ is a symplectomorphism. It combines a cotangent lift of a diffeomorphism with translation by a closed differential one-form. In particular $\eta =du+a$ is allowed on a torus even when the constant one-form $a\cdot dx$ is not exact.