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An anti-symplectic involution is an anti-symplectic map whose square is the identity. Its nonempty fixed-point set is a Lagrangian submanifold; the linear version has complementary $+1$ and $-1$ Lagrangian subspaces.

On a compact smooth manifold without boundary, every smooth self-map sufficiently $C^1$-close to a diffeomorphism is a diffeomorphism. Reducing to a map near the identity, invertibility of its differential follows uniformly by compactness. It is then a proper local diffeomorphism, hence a finite covering map. Sufficient $C^0$ closeness gives a homotopy to the identity on each component; its induced map on the fundamental group is surjective. The covering therefore has one sheet on each component, proving the assertion. This lets a nearby Lagrangian submanifold transverse to cotangent fibers be represented by a single global differential one-form graph.

If $[\sigma ]\ne 0$ in de Rham cohomology, a twisted cotangent symplectic form has no Lagrangian submanifold whose projection is a diffeomorphism onto the entire base. Such a submanifold would be a global one-form graph, and the twisted Lagrangian graph criterion would make $\sigma$ exact.

A compact Lagrangian submanifold $L$ with vanishing first de Rham cohomology cannot be a fibre of a smooth positive-dimensional fiber bundle whose nearby fibres are also Lagrangian submanifolds. In a Weinstein neighborhood, nearby fibres are one-form graphs of closed forms. They are exact by the cohomology assumption, so meet the zero section at a critical point of their potential, contradicting disjointness of distinct fibres.

The cotangent bundle has the canonical one-form ${\lambda }_{(q,p)}(v)=p(d\pi (v))$ and the canonical symplectic form $d\lambda$ up to a conventional sign. Its zero section is a Lagrangian submanifold.

Geometric quantization associates quantum states and observables with a symplectic manifold. A first step uses a complex line bundle with a connection on a vector bundle whose curvature is proportional to the symplectic form divided by $\hslash$; its existence imposes an integrality condition. A real polarization is an integrable half-dimensional distribution with Lagrangian submanifolds as leaves. Requiring sections to be covariantly constant along the leaves reduces the variables to a configuration-space description. Restricting the curvature to a Lagrangian leaf gives zero, permitting local flat sections; global sections can require trivial holonomy. Exact Lagrangian graphs also describe local semiclassical phases through their generating functions.

For a smooth nonconstant holomorphic disk $h$ in ${\mathbb{C}}^n$ with boundary on a Lagrangian submanifold $L$, its symplectic area is ${\int }_{D^2}|{\partial }_{s}h{|}^{2}dsdt>0$. The Cauchy-Riemann equations give ${\partial }_{t}h=J{\partial }_{s}h$, so this is the pullback of the standard symplectic form. The Stokes theorem identifies the area with the period of a primitive along the boundary loop. Thus the Liouville class of a Lagrangian submanifold evaluates nontrivially on this loop and cannot vanish.

An isotropic submanifold has every tangent space an isotropic subspace of a symplectic vector space. Its dimension is at most half the ambient dimension, and equality makes it a Lagrangian submanifold.

A Lagrangian boundary condition for a J-holomorphic curve requires specified boundary components of the source to map into specified Lagrangian submanifolds. For the strip, this is $u(s,0)\in L_0$ and $u(s,1)\in L_1$. For a compatible almost complex structure these target submanifolds are totally real. The condition accompanies the interior equation $u_s+J{u}_t=0$.

A Lagrangian embedding is a smooth embedding whose image is a Lagrangian submanifold.

A nonempty fixed-point set of an anti-symplectic involution is a Lagrangian submanifold. Local linearization identifies its tangent space with the $+1$ eigenspace of the derivative.

The graph of a one-form $\alpha$ on a smooth manifold $Q$ is a Lagrangian submanifold of its cotangent bundle exactly when $d\alpha =0$. Indeed the graph pulls the tautological one-form back to $\alpha$, and therefore pulls the canonical symplectic form back to $-d\alpha$. Locally the Poincaré lemma writes $\alpha =dS$, giving $p_i={\partial }_{i}S$. Globally a closed one-form may have nonzero periods and no single generating function.

A leaf of a Lagrangian foliation is an immersed Lagrangian submanifold tangent to its integrable distribution. A prequantum line bundle restricts to a flat connection along such a leaf, because its curvature is proportional to the symplectic form. Nonzero global parallel sections require trivial holonomy along every leaf loop.

At a regular free circle group level of a moment map, the inverse image of a Lagrangian submanifold of the reduced space is a Lagrangian submanifold of the original space. Its restricted symplectic form is the pullback of the reduced form and hence vanishes. Its dimension is one more than that of the reduced Lagrangian, exactly half the ambient dimension. It is a circle bundle over the original Lagrangian.

Choose a reference path between Lagrangian submanifolds $L_0,L_1$ in a symplectic manifold. With a connecting homotopy strip oriented by $ds\wedge dt$, define $\mathcal{A}(\gamma )=-\int u^{\ast }\omega$. If the path space is connected and $H_2(M,L_0\cup L_1)=0$, relative symplectic-area independence makes this single-valued up to a global constant. Its derivative is $d\mathcal{A}(\xi )=\int \omega (\dot{\gamma },\xi )dt$, so its critical points are the constant paths in $L_0\cap L_1$. For a compatible almost complex structure, its formal downward $L^2$ flow is the J-holomorphic curve equation $u_s+J{u}_t=0$.

A Lagrangian submanifold diffeomorphic to an $n$-dimensional flat torus in a $2n$-dimensional symplectic manifold. Compact connected regular levels of a completely integrable Hamiltonian system are examples, by the Arnold-Liouville theorem.

Fix a primitive $\lambda$ of the standard symplectic form on ${\mathbb{R}}^{2n}$. With the coordinates of Question 2 one convenient choice isFor a Lagrangian submanifold with inclusion $i:L\hookrightarrow {\mathbb{R}}^{2n}$, $d(i^{\ast }\lambda )=i^{\ast }{\omega }_0=0$. Its Liouville class of a Lagrangian submanifold isAny two primitives differ by a closed one-form on the contractible ambient space, hence by an exact form, so the class is independent of the primitive. The cotangent primitive convention $\theta ={\sum }_j{p}_{j}d{q}_j$, with ${\omega }_0=-d\theta$, gives the negative of this class; rationality, positive period size and nonvanishing do not depend on that sign.

The period homomorphism sends an integral one-cycle $c$ to ${\int }_c{i}^{\ast }\lambda$. Its image $\mathrm{Per}(L)\subset \mathbb{R}$ is an additive subgroup. A rational Lagrangian submanifold is one for which this subgroup is discrete. In the nonzero case it has a unique positive generator:Equivalently ${\ell }_L$ is a real multiple of an integral cohomology class. The invariant $\gamma (L)$ is its least positive Liouville period, also expressible as $\inf \{a>0:a\in \mathrm{Per}(L)\}$. If the period group is zero, that infimum is $+\infty$ by the empty-set convention; some treatments instead reserve “rational” for the nonzero discrete case. Because the ambient space is contractible, the Stokes theorem identifies these periods with signed symplectic areas of disks whose boundaries lie on $L$.

Let $\varphi \in {\mathrm{Ham}}_c({\mathbb{R}}^{2n})$ be nonidentity and choose $x$ with $\varphi (x)\ne x$. Choose disjoint open neighbourhoods $U,V$ of $x,\varphi (x)$, respectively. Continuity gives a small Euclidean ball $B=B(x,r_0)$ with $B\subset U$ and $\varphi (B)\subset V$. Therefore $\varphi (B)\cap B=\varnothing$.

Inside this ball construct an explicit closed rational Lagrangian submanifold. For $0<a<r_0/\sqrt{n}$, take the product Lagrangian torusIts distance from $x$ is $\sqrt{n}a<r_0$. Each coordinate circle is one-dimensional, so the pullback of $d{q}_j\wedge d{p}_j$ to it is zero; summing proves that this $n$-dimensional product is a Lagrangian submanifold. It is compact without boundary.

On the $j$th fundamental circle, $q_j=a\cos t$, $p_j=a\sin t$, and the primitive in part (a) restricts to $\frac{1}{2}{a}^{2}dt$. Its integral is $\pi a^2$. Translation of the torus adds only an exact constant-coefficient one-form to the primitive, so it does not change the periods. Since $H_1(T^n;\mathbb{Z})$ is generated by these circles,The given displacement-energy bound applies to this explicitly constructed torus. Because $L\subset B$, the map $\varphi$ displaces $L$, and henceThis is Hofer nondegeneracy from rational Lagrangian displacement. For two different maps $\varphi ,\psi$, apply it to ${\varphi }^{-1}\psi \ne \mathrm{id}$ and use the distance definition. Therefore ${\rho }_{\infty }(\varphi ,\psi )>0$. The equivalent oscillation Hofer metric is nondegenerate as well.

Only the stated energy inequality was used; its conclusion was not used to presume positive energy of an arbitrary open set. The rational torus and its positive period have been constructed inside the particular displaced ball.

Let $\pi :T^{\ast }N\to N$ be the projection and $s_{\alpha }(q)=(q,{\alpha }_q)$ the section with image ${\Gamma }_{\alpha }$. For the canonical one-form on a cotangent bundle,Since $\pi \circ s_{\alpha }=\mathrm{id}$, evaluation on a tangent vector $v\in T_{q}N$ givesUse the same position-first standard symplectic form as before, $\omega =-d\theta$. Pullback commutes with the exterior derivative, soThe graph of a differential one-form is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^{\ast }N$. Therefore the vanishing of the pulled-back symplectic form is exactly the Lagrangian submanifold condition, andThis proves that the graph of a closed one-form is Lagrangian using the intrinsic one-form rather than a coordinate guess.

Write the standard complex coordinates as $z_j=q_j+i{p}_j$ and let $J$ denote multiplication by $i$. Use the standard symplectic form ${\omega }_0={\sum }_{j}d{q}_j\wedge d{p}_j$, for which ${\omega }_0(v,Jv)=|v{|}^2$. In real coordinates $s,t$ on the disk, holomorphicity and the Cauchy-Riemann equations implyTherefore the pulled-back two-form isIf this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the symplectic area is consequently strictly positive:Take the primitive $\lambda =\frac{1}{2}{\sum }_j(q_{j}d{p}_j-p_{j}d{q}_j)$ from Question 3. Let $c=h{|}_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the Stokes theorem givesThe restriction $\lambda {|}_L$ is closed because $L$ is a Lagrangian submanifold. If its Liouville class of a Lagrangian submanifold were zero, it would be an exact differential form, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. The Liouville class is therefore nonzero. This proves that a holomorphic disk detects a nonzero Liouville class directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.