Nearby Lagrangian conjecture

Nearby Lagrangian conjecture

OpenAI's claim

Nearby Lagrangian counterexample: For some sufficiently large even $N$, the paper claims a closed embedded exact Lagrangian in $T^*(S^9\times S^{N-1})$ that cannot be taken to the zero section by a Hamiltonian isotopy, despite sharing the base’s diffeomorphism type. Thus the unrestricted nearby-Lagrangian assertion is contradicted.

Mathematical statement of the claimed result

Mathematical claim 1.

For some sufficiently large even $N$, let $Q=S^9\times S^{N-1}$. The manuscript claims a closed exact smoothly embedded Lagrangian $L\subset T^*Q$ with $L\cong Q$ as smooth manifolds, but with no compactly supported Hamiltonian isotopy sending the zero section onto $L$. Exactness means that the canonical cotangent one-form restricts to $dF$ on $L$ for some function $F$.

Outline of the argument

Nearby Lagrangian counterexample: Generating families are organized by their homogeneous quadratic behavior at large vertical radius. Stable tube data determine spherical fibrations, and a nontrivial obstruction is selected from the tube-to-fibration comparison. A regular vertical-gradient construction produces an exact immersion, followed by a separate injectivity argument; the retained stable obstruction prevents a Hamiltonian isotopy to the zero section.

Formal verification

Nearby Lagrangian counterexample: The pinned catalogue lists no Lean formalization for this result.

This site has not run the Lean development. These links describe OpenAI’s selected formal statements and should be read with their stated scope.

Status caveat

OpenAI reports that an internal model produced the manuscripts. Their arguments have not been independently verified as proofs by this site. Any portion outside the documented formal scope may contain mistakes, and the catalogue label reports the announced mathematical claim.

Sources

• OpenAI, A counterexample to the nearby Lagrangian conjecture, 2026-09-23; citation key OAI:A-counterexample-to-the-nearby-Lagrangian-conjecture-September-23-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• OpenAI catalogue entry.

Nearby Lagrangian conjecture

Definitions and mathematical statement

For the smooth base manifold $N$ in the standard closed-base cotangent formulation, $T^*N$ has canonical one-form $\lambda$ and symplectic form ${\,\mathrm{d}}\lambda$. A closed embedded submanifold $L$ is Lagrangian when $\dim L=\dim N$ and $({\,\mathrm{d}}\lambda)|_L=0$, and exact when $\lambda|_L={\,\mathrm{d}} f$ for some function $f:L\to{\mathbb{R}}$. The conjecture asserts that there is a Hamiltonian isotopy $\Phi_t$ of $T^*N$ with \[ \Phi_1(L)=\text{the zero section }N. \] The isotopy is generated by a time-dependent Hamiltonian in the source's support convention. Homotopy equivalence of the projection $L\to N$, or abstract diffeomorphism of $L$ and $N$, is weaker than this ambient Hamiltonian-isotopy conclusion.

Formulation and concept references: [S1].

Short English statement

Is every closed exact Lagrangian in a cotangent bundle obtainable from its zero section by a Hamiltonian isotopy?

Research attempt

Scope and source check.

GPT-6 Astra Ultra prepared this unresolved attempt on 27 September 2026 by reviewing the formulation source and checking explicit symplectic calculations. The base is closed; for the proof below take a connected component of the base at a time. The source distinguishes a homotopy classification from the requested Hamiltonian isotopy. Its surrounding lecture notes contain some explicitly tentative arguments, so those arguments are not adopted as theorem inputs here. The elementary route chosen below solves the case in which the cotangent projection has no critical points, and identifies the exact additional geometric task in the general case. All Hamiltonians constructed below can be compactly supported, which meets the usual closed-base support convention.

Exact graphs admit an explicit isotopy.

Write $\pi:T^*N\to N$ and use the convention $\omega=d\lambda=\sum_i dp_i\wedge dq_i$ and $\iota_{X_H}\omega=-dH$. If $\alpha$ is a one-form on $N$, its graph $j(q)=(q,\alpha_q)$ satisfies \[ j^*\lambda=\alpha,\qquad j^*\omega=d\alpha. \] Thus the graph is Lagrangian precisely when $\alpha$ is closed, and exact precisely when $\alpha=df$ for a globally defined function. For $H(q,p)=f(q)$ the Hamilton equations in this convention are $\dot q=0$, $\dot p=-df(q)$. Its time-one map sends the graph of $df$ to the zero section.

The Hamiltonian just written is not compactly supported in the fibers. This is easily repaired without changing the prescribed isotopy. The swept set \[ \{(q,s\,df(q)):q\in N,\ 0\le s\le1\} \] is compact. Choose a smooth compactly supported cutoff $\chi$ equal to one on a neighborhood of it. For $\widehat H=\chi f$, both the Hamiltonian and its differential agree with those of $H$ on that neighborhood. Uniqueness for the Hamiltonian ordinary differential equation shows that the points on the moving graph follow exactly the same paths. The cutoff flow exists throughout the required interval because its vector field has compact support. This proves the full ambient conclusion for exact graphs, including the support condition.

A projection with no critical points must be a graph.

There is a stronger elementary subcase. Suppose $L$ is nonempty, closed, embedded and exact, and $\pi|_L$ is a local diffeomorphism. Compactness makes this map proper. Its image is both open and closed in connected $N$, hence is all of $N$; it is a finite covering. I claim that its degree is one. This conclusion uses exactness and embeddedness together, and does not assume that a covering of the base is automatically trivial.

Choose $f:L\to\mathbb R$ with $df=\lambda|_L$. If the covering had at least two sheets, the space \[ C=\{(x,y)\in L\times L:\pi(x)=\pi(y),\ x\ne y\} \] would be a nonempty compact smooth manifold. For a finite covering the diagonal is both open and closed in the fiber product, so removing it preserves compactness; there is no sequence of distinct sheets converging to the diagonal. The function $F(x,y)=f(x)-f(y)$ has a maximum on $C$. At such a maximum, vary the common base point in an arbitrary tangent direction $v\in T_qN$, lifting it to both sheets. Write $x=(q,p_x)$ and $y=(q,p_y)$. The differential of $F$ is \[ dF(v)=p_x(v)-p_y(v). \] It vanishes for every $v$, so $p_x=p_y$. Consequently $x=y$ as points of the embedded submanifold of $T^*N$, contradicting their membership in $C$. The cover therefore has degree one and is a diffeomorphism. Its inverse identifies $L$ with the graph of $d(f\circ(\pi|_L)^{-1})$. The preceding compactly supported Hamiltonian sends it to $N$.

This proof also excludes a disconnected exact multisection over a connected base under the same nonsingularity hypothesis: the fiber product argument does not require connectedness of $L$. It does not exclude disconnectedness or prove the conjecture for a general exact Lagrangian whose projection has critical points. The compact covering structure, and specifically compactness of $C$, was essential.

Why exactness cannot be removed.

On $T^*\mathbb T^m$, a constant nonzero covector $a$ defines a closed Lagrangian graph. Its Liouville periods on the coordinate circles are the components of $a$, with the corresponding circle-length convention. If one is nonzero, the graph is not exact and cannot be Hamiltonian isotopic to the zero section. Here is the invariance calculation. For a Hamiltonian flow $\Phi_t$, \[ \frac{d}{dt}\Phi_t^*\lambda =\Phi_t^*\left(\iota_{X_{H_t}}d\lambda+ d(\lambda(X_{H_t}))\right) =d\,\Phi_t^*\left(\lambda(X_{H_t})-H_t\right). \] Integrating in time shows that $\Phi_t^*\lambda-\lambda$ is exact. Its integral around every closed loop is zero. Thus a nonzero period cannot become a zero-section period under such a flow.

The example is a stress test of the hypotheses, not a counterexample to the problem. It also distinguishes Hamiltonian translation by $df$ from symplectic translation by an arbitrary closed one-form. The latter preserves $d\lambda$ but need not have a globally defined Hamiltonian. Ignoring this distinction would incorrectly prove an assertion without the necessary exactness assumption.

Exact Lagrangian paths would be enough.

A useful reduction is that a smooth path of closed embedded exact Lagrangians can be extended, at the level of their images, to a Hamiltonian isotopy. To check the statement, let $j_t:L_0\to T^*N$ be a smooth family of Lagrangian embeddings and $V_t=\partial_tj_t$ its velocity along the image. Define \[ \beta_t=j_t^*(\iota_{V_t}\omega). \] Differentiating $j_t^*\omega=0$ gives $d\beta_t=0$. If $j_t^*\lambda=df_t$ with primitives chosen smoothly in $t$, Cartan's formula gives \[ \beta_t=d\left(\partial_tf_t-\lambda(V_t)\circ j_t\right). \] Write the last primitive as $h_t$. Smooth primitives can be obtained by fixing base values and integrating the smoothly varying exact forms; there is no changing period obstruction.

Choose a smooth Hamiltonian near the moving image with $H_t\circ j_t=-h_t$, and extend it with compact support. Such an extension exists using a tubular neighborhood of the compact family of embeddings. For every tangent vector $w$ to $L_t$, \[ \omega(X_{H_t}-V_t,w)=0. \] The symplectic orthogonal of a Lagrangian tangent space is that same tangent space. Hence $X_{H_t}-V_t$ is tangential: the Hamiltonian flow and $j_t$ move the same submanifold, differing only by a time-dependent reparametrization. This proves the reduction. It does not construct the required path from an arbitrary $L$ to the zero section.

A merely formal deformation of tangent planes, a homotopy of the projection, or a path through immersed Lagrangians does not meet this criterion. The path must consist of actual embedded exact Lagrangians. At a self-intersection there need not be one compatible value of the local Hamiltonian prescription on the ambient image. This is one reason that an immersion-level construction cannot simply be declared an ambient isotopy.

The fiber-contraction shortcut fails at its endpoint.

Consider $\delta_s(q,p)=(q,sp)$. For $s>0$ it is a diffeomorphism and $\delta_s^*\omega=s\omega$, so it takes exact Lagrangians to exact Lagrangians. But it is generally not symplectic and is not itself a Hamiltonian map. More decisively, at $s=0$ its restriction to $L$ is just the projection onto the zero section. Critical points of $\pi|_L$ can destroy the immersion condition, and several points can have the same image. Thus the family need not be a smooth path of embeddings including its endpoint. Applying the preceding extension lemma on $s\in[\epsilon,1]$ for every positive $\epsilon$ does not supply a limiting Hamiltonian diffeomorphism at zero.

Even uniform convergence of the images as sets would not fix this: invertibility and derivative bounds need not survive such a limit. The graph subcase avoids the problem because its projection is already a diffeomorphism and the explicit Hamiltonian supplies a nonsingular path all the way to the endpoint.

Verification, first gap, and outcome.

The sign of the Hamilton equations was checked against the source's $d\lambda$ convention, the cutoff preserves the entire swept graph, and the covering proof explicitly checks compactness after removing the diagonal. The period obstruction tests the necessity of exactness. The first missing step is an embedded exact deformation that removes the critical points of the cotangent projection, or another construction of the required Hamiltonian isotopy without doing so. The projection subcase and path-extension reduction are proved; neither supplies that step for arbitrary $L$. The nearby Lagrangian conjecture remains UNRESOLVED, and no novelty is claimed for these elementary restricted deductions.

Sources

• S1 Formal statement, Conjecture 1.5. https://www.math.stonybrook.edu/~fzheng/seminar-f24/notes/week-1-2.pdf. Formulation source.