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A Lagrangian complement to $L$ in a symplectic vector space is a Lagrangian subspace $K$ with $E=L\oplus K$. Such complements exist smoothly for Lagrangian subbundles. Identify any initial complement with $L^{\ast }$ using the symplectic cross-pairing. If its internal pairing is $b(\eta ,\zeta )$, correct its lift by a vector $a\eta \in L$ with $\zeta (a\eta )=-b(\eta ,\zeta )/2$. The corrected complement has zero internal pairing.

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.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

A Lagrangian complement to $TX$ in $TM{|}_X$ is identified with $T^{\ast }X$ by $v\mapsto \omega (\cdot ,v){|}_{TX}$. This produces a vector bundle isomorphism $TX\oplus T^{\ast }X\to TM{|}_X$ that is the identity on $TX$ and carries the canonical pairing $\zeta (u)-\eta (v)$ to $\omega$. A suitably chosen tubular neighborhood map therefore matches the ambient two-forms along $X$, enabling the relative Moser theorem.