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The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincaré lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

Fix the convention ${\iota }_{X_f}\omega =-df$ for a Hamiltonian vector field. Nondegeneracy of the symplectic form gives one unique smooth $X_f$ for each smooth real-valued function $f$. Define the Poisson bracket byIn coordinates with $\omega =\sum d{q}^i\wedge d{p}_i$, this is $\sum (f_{q^i}{g}_{p_i}-f_{p_i}{g}_{q^i})$. Thus the sign convention is explicit and agrees with the usual coordinate Poisson bracket. By Cartan's magic formula, ${\mathcal{L}}_{X_f}\omega =d{\iota }_{X_f}\omega +{\iota }_{X_f}d\omega =0$. The contraction–Lie derivative commutator identity now givesHence the Hamiltonian Lie algebra homomorphism is $\boxed{\Phi (f)=X_f,[X_f,X_g]=X_{\{f,g\}}}$. It is linear and onto the space of Hamiltonian vector fields by definition. For completeness, the Jacobi identity for the bracket on functions follows from closure of $\omega$: evaluating $d\omega =0$ on $X_f,X_g,X_h$ and using the displayed commutator relation gives the cyclic Jacobi sum zero. Thus this really is a map of Lie algebras, not just a bracket-preserving notation.

Its kernel is determined by nondegeneracy:On a connected manifold these are precisely the real constants; on a disconnected manifold the constant may differ on each component. Therefore the quotient by these functions is isomorphic to the Lie algebra of Hamiltonian vector fields. The printed “homeomorphis” is interpreted as homomorphism: the map before taking this quotient is not an isomorphism because of its nontrivial kernel. The alternative convention ${\iota }_{X_f}\omega =df$ requires a corresponding bracket sign change to retain this homomorphism.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

A symplectic vector field satisfies ${\mathcal{L}}_X\omega =0$, so its local flow consists of symplectomorphisms. By Cartan's magic formula and $d\omega =0$, this is equivalent to the differential one-form ${\iota }_X\omega$ being closed. In the convention fixed above, a Hamiltonian vector field satisfies ${\iota }_X\omega =dH$ for a globally defined smooth function $H$. It is therefore a symplectic vector field, but the converse requires this closed one-form to be exact.

On the torus with $\omega =dx\wedge dy$,Thus ${\partial }_x$ is a symplectic vector field. The closed differential one-form $dy$ is not exact on the torus: its integral on the loop $\gamma (s)=[(0,s)]$, $0\le s\le 1$, equals one, while the integral of an exact one-form around any closed loop is zero. The vector field is symplectic but not Hamiltonian. The local candidate $H=y$ does not descend to a single-valued function on ${\mathbb{R}}^2/{\mathbb{Z}}^2$.

A Hamiltonian group action is a smooth Lie group action of $G$ on $(M,\omega )$ by symplectomorphisms, together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$. For $\xi \in \mathfrak{g}$, let ${\xi }_M(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ be its fundamental vector field. In our sign convention the defining identities areHere the left coadjoint action means $({\mathrm{Ad}}_g^{\ast }\nu )(\xi )=\nu ({\mathrm{Ad}}_{g^{-1}}\xi )$. Each component ${\mu }^{\xi }$ is therefore a Hamiltonian function for the corresponding infinitesimal action. Equivariance is part of the definition; merely requiring each infinitesimal generator to be a Hamiltonian vector field is the weaker condition of a weakly Hamiltonian action. Reversing the defining sign of Hamiltonian vector fields reverses the moment map sign as well.

Let $P=({\omega }^{ab})$ be a smooth antisymmetric matrix field. It defines a Poisson bivector throughThis Poisson bracket is bilinear, antisymmetric and a derivation in each argument. It defines a Poisson manifold when it also satisfies the Jacobi identity. Apply that identity in the form $\{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0$ to the coordinate functions. Since $\{x^a,x^b\}={\omega }^{ab}$, it givesThis is the coordinate Jacobi condition for a Poisson bivector. It is also sufficient: in the Jacobi identity for arbitrary functions the terms containing second derivatives cancel by antisymmetry, leaving $J^{abc}({\partial }_{a}f)({\partial }_{b}g)({\partial }_{c}h)$. Thus the coordinate condition captures the whole obstruction.

Now suppose the Poisson bivector is nondegenerate. Write $S=P^{-1}$, so $S_{ab}{\omega }^{bc}={\delta }_a^c$, and define the 2-formAn overall minus sign in identifying the symplectic form depends on the convention for Hamiltonian vector fields; it does not affect the closure argument. Differentiating the inverse matrix givesContract the coordinate Jacobi identity with $S_{ia}{S}_{jb}{S}_{kc}$. Antisymmetry gives ${\omega }^{dc}{S}_{kc}=-{\delta }_k^d$, and similarly for the other terms, henceThe cyclic expression is precisely the coefficient of the exterior derivative $d\Omega$. ThereforeSince $S$ is antisymmetric and nondegenerate, $\Omega$ is a symplectic form. This proves the closure of the inverse of a nondegenerate Poisson bivector. The matrix entries ${\omega }^{ab}$ are scalar functions; the codomain in the source's matrix description should be read as the space of antisymmetric matrices, rather than a vector-valued individual entry.

Write $z=(x,v)$ and let $\Phi (t,s,z)$ denote the Hamiltonian flow from time $s$ to time $t$. The characteristic curves satisfy Hamilton's equations:The Hamiltonian Liouville equation then reduces along each curve toThe signs and derivative variables here are those in the PDF.

The global characteristic flow for a Hamiltonian with bounded Hessian follows, for example, from $H\in C^2(\mathbb{R}\times {\mathbb{R}}^{2d})$ and, for every finite $T$,Thus the Hamiltonian vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is globally Lipschitz continuous in $z$ on each finite time interval and satisfies $|b(t,z)|\le A_T+L_T|z|$. The Picard-Lindelöf theorem gives local existence and uniqueness, while the Gronwall inequality gives, for example,This excludes finite-time escape. There is a unique Hamiltonian flow for all finite forward and backward times, and $\Phi (s,t)$ is the inverse of $\Phi (t,s)$. These sufficient conditions are deliberately stronger than necessary.

For $S_t(z)=\Phi (t,0,z)$, differentiability with respect to initial data gives the variational equationThe Jacobi determinant derivative formula givesThe mixed derivatives in the Hamiltonian vector field cancel:Consequently $J^{\prime }(t)=0$, and $J(0)=1$ gives preservation of phase space volume:The same argument applies to $\Phi (t,s)$ for every starting time $s$. This is the Liouville theorem in Hamiltonian mechanics. Explicit time dependence of $H$ does not affect the cancellation.

A Lie point symmetry is a local one-parameter family of diffeomorphisms of $(t,\mathbf{x})$ that takes solution graphs of the differential equation to solution graphs wherever the transformed graph remains a graph over time. Its infinitesimal generator is $Y=\tau (t,\mathbf{x}){\partial }_t+{\sum }_i{\eta }_i(t,\mathbf{x}){\partial }_{x_i}$. Its first prolongation isFor a regular first-order equation the infinitesimal criterion is ${\mathrm{pr}}^{(1)}Y(\Delta )=0$ on $\Delta =0$; it says that the prolonged vector field is tangent to the equation submanifold.

With the constant nondegenerate canonical matrix $J$, define the Poisson bracket by $\{f,g\}=(\nabla f{)}^{T}J\nabla g$ and the Hamiltonian vector field by $V_f=J\nabla f$. Our convention gives $V_f[h]=\{h,f\}$. Applying the Jacobi identity to a test function $h$,soA first integral $F$ obeys $dF/dt=\{F,H\}=0$. Consequently $[V_F,V_H]=0$, and the local flows commute. The time-independent transformation generated by $V_F$, with $t$ unchanged, therefore maps integral curves of a vector field for $V_H$ to integral curves of a vector field: it is a Lie point symmetry.

Conversely, such a time-independent symmetry gives $[V_F,V_H]=0$, hence $V_{\{F,H\}}=0$. Nondegeneracy of $J$ makes $\{F,H\}$ constant on each connected component. At a fixed point ${\mathbf{x}}_0$, $J\nabla H({\mathbf{x}}_0)=0$ implies $\nabla H({\mathbf{x}}_0)=0$, so that constant is zero on its component. Thus the converse holds on a connected phase space containing a fixed point, or on each component that contains one.

The printed global converse needs this connectedness convention: a fixed point on one component does not control the constants on other components. For example take one component with $H=(q^2+p^2)/2$, $F=0$, and another with $H=p$, $F=q$. The Hamiltonian vector fields commute everywhere and there is a fixed point on the first component, but $F$ is not a first integral on the second. Even on a connected space the fixed-point condition matters: $H=p$, $F=q$ commute but $\dot{F}=1$.

Use the Hamiltonian vector field convention ${\iota }_{X_t}\omega =d{H}_t$, which is forced by the sign of the formula in this part. This is the opposite of the default sign in the linked general article. Since $d\lambda =-\omega$, Cartan's magic formula givesDifferentiate the pullback of a differential form along the smooth isotopy and integrate from $0$ to $t$:Thus the required exact differential form has the explicit primitive

Write the exact differential form as $\beta =dh$ and take the cotangent fiber translation family ${\rho }_t(x,\xi )=(x,\xi +td{h}_x)$. Its generating vector field is vertical and in local coordinates isWith $\omega ={\sum }_{j}d{x}_j\wedge d{\xi }_j$ and the Hamiltonian vector field convention from Question 3,Thus ${\rho }_t$ is a Hamiltonian isotopy and $f={\rho }_1$ is a Hamiltonian diffeomorphism. The local flow is explicitly defined for every real $t$, regardless of compactness. With the alternative convention ${\iota }_{X_H}\omega =-dH$, the same family is generated by $H=h\circ \pi$.

The definition in the question does not require compact support for the Hamiltonian function; the displayed Hamiltonian diffeomorphism uses that definition.

Use the paper's Fubini-Study form normalization ${\tilde{\omega }}_{FS}=\frac{i}{2}\partial \overline{\partial }\log (1+|z{|}^2)$, the Hamiltonian vector field convention ${\iota }_{X_H}\omega =dH$, and the Lie algebra generator whose circle action is $e^{i\theta }$. Write $D=|z_0{|}^2+|z_1{|}^2+|z_2{|}^2$. The normalized moment map for the torus action isThe moment map image is the filled triangleIndeed, the three quantities $|z_j{|}^2/D$ are nonnegative and add to one, and any such triple is realized by choosing their square roots as homogeneous coordinates.

For the diagonal circle action, a point in the fixed-point set satisfies $[z_0,t{z}_1,t{z}_2]=[z_0,z_1,z_2]$ for every $t\in S^1$. If $z_0\ne 0$, the projective scaling must be one, forcing $z_1=z_2=0$. If $z_0=0$, all coordinates are scaled together. ThereforeBy part (b), the diagonal moment map isThe factor $1/2$ follows from the specified Fubini-Study form, not the integral normalization in the linked general article. For example, in one affine complex coordinate $w=r{e}^{i\theta }$ the symplectic form is $r(1+r^2{)}^{-2}dr\wedge d\theta$, whose interior product of a differential form with ${\partial }_{\theta }$ is $-r(1+r^2{)}^{-2}dr=d[-r^2/(2(1+r^2))]$. Choosing the generator $e^{2\pi it}$ rescales all moment maps by $2\pi$; reversing the defining sign reverses their signs. The negative level in parts (d) and (e) uses the convention displayed here.

Fix $p\in X$. A linear change of coordinates first identifies ${\omega }_p$ with the standard symplectic form ${\omega }_0$. After shrinking to a star-shaped neighborhood, everyis nondegenerate. The Poincaré lemma gives $\omega -{\omega }_0=d\sigma$, with $\sigma (p)=0$. Define $X_t$ by ${\iota }_{X_t}{\omega }_t=-\sigma$ and let ${\varphi }_t$ be its local flow. Cartan's magic formula givesThus ${\varphi }_1^{\ast }\omega ={\omega }_0$, proving the Darboux theorem.

For a smooth function $H$ on a closed symplectic manifold, its Hamiltonian vector field is defined by ${\iota }_{X_H}\omega =-dH$. The same formula gives ${\mathcal{L}}_{X_H}\omega =-d^{2}H=0$, so the Hamiltonian flow preserves the symplectic form. To move one point to another in a connected $X$, join them by a path, cover the path by finitely many Darboux charts, and in each chart use a cutoff linear Hamiltonian to perform a small translation. Composing these compactly supported Hamiltonian diffeomorphisms proves that symplectomorphisms act transitively on each connected component.

In ${\mathbb{R}}^2$, every embedded curve is a Lagrangian submanifold. Let $L_1,L_2$ be circles enclosing different Euclidean areas. A plane symplectomorphism preserves area and carries the bounded complementary component of one circle to that of its image, so no symplectomorphism maps $L_1$ to $L_2$.

The same phenomenon exists in every ${\mathbb{R}}^{2n}$. Withtake the product toriThe Liouville class of a Lagrangian submanifold has respective period vectors $\pi (1,\dots ,1)$ and $2\pi (1,\dots ,1)$ on these tori. Every symplectomorphism of ${\mathbb{R}}^{2n}$ preserves the Liouville class up to the induced integral change of basis on $H_1(T^n)$, because its pullback changes $\lambda$ only by an exact form. An integral automorphism sends a primitive vector to a primitive vector and therefore cannot send the first period vector to the second. Hence

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.