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A smooth time-dependent vector field on a compact manifold without boundary has flow maps throughout each compact time interval on which the field is defined. Finite-time escape is impossible on the compact state space. Uniqueness and the backward-time equation supply smooth inverses, so the flow is a smooth isotopy.

A smooth interval family of contact forms on a compact manifold without boundary is carried back to its initial contact distribution by a smooth isotopy. The pullback of each form is a positive smooth multiple of the initial form. Solve for the horizontal generator of contact stability and integrate the resulting scalar equation for the contact conformal factor along an isotopy.

On an oriented even-dimensional Riemannian manifold of strictly positive sectional curvature, a nonconstant closed geodesic has a length-decreasing smooth variation. Its parallel transport fixes the tangent and acts on the odd-dimensional normal space by a special orthogonal group element. An odd-dimensional special orthogonal transformation has a fixed vector, so there is a nonzero periodic parallel normal field $V$. The second variation of geodesic energy is then strictly negative. The Cauchy-Schwarz inequality converts lower energy into strictly lower length because the original geodesic has constant speed. For an embedded geodesic the small variation is a smooth isotopy; otherwise it is a deformation through immersed loops.

On a compact connected smooth manifold without boundary, two positive volume forms for the same orientation are related by an orientation-preserving diffeomorphism exactly when their total integrals agree. A positive interpolation and Cartan's magic formula give the smooth isotopy realizing the equality.

Put ${\xi }_t=\ker {\alpha }_t$, the contact distribution. If $\dim M=2r+1$, the contact form condition is ${\alpha }_t\wedge (d{\alpha }_t{)}^r\ne 0$. It implies that $d{\alpha }_t{|}_{{\xi }_t}$ is a symplectic form on each contact hyperplane: inserting a vector transverse to ${\xi }_t$ in that volume form leaves the nonzero top exterior power $(d{\alpha }_t{|}_{{\xi }_t}{)}^r$. Hence there is a unique smooth time-dependent vector field $Y_t\in {\xi }_t$ satisfyingIts smoothness follows by inverting the smoothly varying nondegenerate matrix of this contact distribution symplectic form. This is the horizontal generator of contact stability.

Let $R_t$ be the Reeb vector field of ${\alpha }_t$. The one-form ${\dot{\alpha }}_t+{\iota }_{Y_t}d{\alpha }_t$ vanishes on ${\xi }_t$, so it is $h_t{\alpha }_t$. Evaluating on $R_t$, using ${\alpha }_t(R_t)=1$ and ${\iota }_{R_t}d{\alpha }_t=0$, identifies the coefficient:By Cartan's magic formula, the Lie derivative of a differential form isbecause $Y_t$ lies in the contact hyperplane.

Let ${\rho }_t$ solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0={\mathrm{id}}_M$. Since $M$ is compact without boundary and $Y_t$ is smooth on the closed time interval, its solutions cannot escape and exist for the entire interval. Reversing the time-dependent equation supplies a smooth inverse for each ${\rho }_t$. Thus these flow maps give a smooth isotopy. Differentiating the pullback of a differential form along this isotopy yieldsFor each base point this is a scalar linear equation for a covector, initially ${\alpha }_0$. Its solution isBy finite-time flow completeness on a compact manifold, this construction gives the whole time interval. The contact conformal factor along an isotopy is smooth in $(t,x)$ and nowhere zero, proving Gray stability theorem. The precise domains here are ${\rho }_t:M\to M$ and $\rho :M\times [0,1]\to M$; the extra time factor attached to the already indexed ${\rho }_t$ in the source is a notational slip. The proof produces the required family on its whole specified interval. If a parameter domain of all $\mathbb{R}$ is intended, extend the smooth field slightly beyond $[0,1]$, multiply it by a time cutoff, and use compactness to obtain the extended isotopy $\rho :M\times \mathbb{R}\to M$. Its restriction to $[0,1]$ is unchanged.

Use the energy of a curve normalizationFor a variation $F(s,t)$ put $T={\partial }_{t}F$ and $V={\partial }_{s}F$. The Levi-Civita connection is torsion free, so ${\nabla }_{s}T={\nabla }_{t}V$. Differentiating energy once gives $E^{\prime }=[⟨V,T⟩{]}_a^b-\int ⟨V,{\nabla }_{t}T⟩$. Along a geodesic, ${\nabla }_{t}T=0$. Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, consistent with the specified positive sectional curvature. Indeed ${\nabla }_s{\nabla }_{t}V={\nabla }_t{\nabla }_{s}V+R(V,T)V$, and $⟨R(V,T)V,T⟩=-⟨R(V,T)T,V⟩$. The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form $I(V,V)$.

Let $\dim M=2m$. Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent $T$, and its restriction to $T^{\perp }$ is an orientation-preserving orthogonal map of odd dimension $2m-1$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue $+1$. Choose a nonzero fixed vector $v$ normal to $T$ and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field $V$ with ${\nabla }_{t}V=0$.

For the exponential variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$, the endpoints match periodically. Strictly positive sectional curvature givesThus $E({\gamma }_s)<E(\gamma )$ for small nonzero $s$. By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.

For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.

Moser theorem says that if $M$ is a compact smooth manifold without boundary and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms with constant de Rham cohomology class, then there is a smooth isotopy ${\psi }_t$, with ${\psi }_0=\mathrm{id}$, such thatThus the endpoints are symplectomorphic. The hypothesis is a whole path of symplectic forms, not merely equality of the endpoint de Rham cohomology classes.

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.

Take the usual convention that the smooth isotopy starts at ${\rho }_0=\mathrm{id}$. The exterior derivative commutes with a pullback of a differential form, soConsequentlyThe right-hand condition says precisely that ${\rho }_t$ is a symplectic isotopy.

Here an exact symplectic manifold is necessarily noncompact in positive dimension: if $\dim M=2n$ and $\omega =-d\lambda$, then ${\omega }^n=-d(\lambda \wedge {\omega }^{n-1})$, contradicting the positive symplectic volume and the Generalized Stokes theorem on a compact smooth manifold without boundary. Thus the paper's introductory word “closed” must here be read as “without boundary”, as its parenthesis suggests, rather than as including compactness.

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

Let $M=S^2$ with an area form and let $L$ be an equator dividing the sphere into two open hemispheres of equal area. Every curve in a symplectic surface is Lagrangian. A small normal push moves $L$ to a nearby latitude, so it is displaceable by a smooth isotopy.

Suppose a symplectic isotopy had final image $L^{\prime }$ disjoint from $L$. The curve $L^{\prime }$ must lie in one hemisphere. Of the two discs bounded by $L^{\prime }$, the one contained in that hemisphere has area strictly below half the total area, and the other has area strictly above half. On the other hand, a symplectomorphism maps the original two hemispheres to the two discs bounded by $L^{\prime }$ and preserves their areas, so both would have half the total area. This contradiction is the symplectic non-displaceability of an area bisector.

Take $L=S^2$. For any Lagrangian embedding of $S^2$ into a symplectic four-manifold, the symplectic form identifies the normal bundle with $T^{\ast }{S}^2$. The self-intersection formula and the Euler characteristic giveup to the harmless orientation sign. If a smooth isotopy displaced $L$, its final image would represent the same homology class but have intersection number zero with $L$, a contradiction. This is the smooth non-displaceability from self-intersection.

For a compact ambient example, equip $S^2\times S^2$ with $\omega \oplus (-\omega )$. Its diagonalis Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.