[proofplan] The proof constructs the desired isotopy as the flow of a time-dependent vector field. Constancy of the de Rham class implies that the time derivative $\dot\omega_t$ is exact, and compactness lets us choose primitives $\sigma_t$ smoothly in $t$. We then solve the algebraic equation $\iota_{X_t}\omega_t=-\sigma_t$ using nondegeneracy of $\omega_t$. The pullback differentiation formula and Cartan's formula show that $\frac{d}{dt}(\psi_t^*\omega_t)=0$, so the pulled-back form stays equal to its initial value $\omega_0$. [/proofplan] [step:Choose smooth primitives for the exact derivatives of the symplectic forms] Because $(\omega_t)_{t\in[0,1]}$ is a smooth family of closed $2$-forms, its time derivative is a smooth family \begin{align*} \dot\omega_t:=\frac{d}{dt}\omega_t\in\Omega^2(M), \end{align*} where $\Omega^2(M)$ denotes the space of smooth differential $2$-forms on $M$. Since the cohomology class $[\omega_t]\in H^2_{\mathrm{dR}}(M)$ is independent of $t$, differentiating the curve $t\mapsto[\omega_t]$ in the [finite-dimensional vector space](/page/Finite-Dimensional%20Vector%20Space) $H^2_{\mathrm{dR}}(M)$ gives \begin{align*} [\dot\omega_t]=0 \end{align*} for every $t\in[0,1]$. Hence each $\dot\omega_t$ is exact. We use the standard parametric de Rham primitive lemma on compact manifolds: if $(\alpha_t)_{t\in[0,1]}$ is a smooth family of exact $k$-forms on a compact [smooth manifold](/page/Smooth%20Manifold), then there is a smooth family $(\beta_t)_{t\in[0,1]}$ of $(k-1)$-forms such that $d\beta_t=\alpha_t$ for all $t$. Its hypotheses hold here because $M$ is compact and smooth and $(\dot\omega_t)_{t\in[0,1]}$ is a smooth family of exact $2$-forms. Applying this lemma to $\alpha_t:=\dot\omega_t$, we obtain a smooth family \begin{align*} \sigma_t\in\Omega^1(M) \end{align*} where $\Omega^1(M)$ denotes the space of smooth differential $1$-forms on $M$ such that \begin{align*} \dot\omega_t=d\sigma_t \end{align*} for every $t\in[0,1]$. [guided] The first point is to turn the cohomological hypothesis into an equation of differential forms. For each $t\in[0,1]$, define \begin{align*} \dot\omega_t:=\frac{d}{dt}\omega_t\in\Omega^2(M), \end{align*} where $\Omega^2(M)$ denotes the space of smooth differential $2$-forms on $M$. The family $(\omega_t)$ is smooth, so $(\dot\omega_t)$ is a smooth family of $2$-forms. Since every $\omega_t$ is symplectic, every $\omega_t$ is closed: \begin{align*} d\omega_t=0. \end{align*} Differentiating this identity with respect to $t$ gives \begin{align*} d\dot\omega_t=0. \end{align*} Now use the hypothesis that the de Rham cohomology class $[\omega_t]\in H^2_{\mathrm{dR}}(M)$ is independent of $t$. The map \begin{align*} [0,1]&\to H^2_{\mathrm{dR}}(M) \end{align*} \begin{align*} t&\mapsto[\omega_t] \end{align*} is constant, so its derivative is zero. The derivative of the class is the class of the derivative, hence \begin{align*} [\dot\omega_t]=0 \end{align*} for every $t\in[0,1]$. This means precisely that $\dot\omega_t$ is exact for every $t$. Pointwise exactness is not enough for the flow argument, because the vector field we construct must depend smoothly on $t$. We therefore use the standard parametric de Rham primitive lemma on compact manifolds: if $(\alpha_t)_{t\in[0,1]}$ is a smooth family of exact $k$-forms on a compact smooth manifold, then there is a smooth family $(\beta_t)_{t\in[0,1]}$ of $(k-1)$-forms satisfying $d\beta_t=\alpha_t$ for every $t$. The lemma applies because $M$ is compact and smooth and because $(\dot\omega_t)_{t\in[0,1]}$ is a smooth family of exact $2$-forms. Applying this with $\alpha_t:=\dot\omega_t$ gives a smooth family \begin{align*} \sigma_t\in\Omega^1(M) \end{align*} where $\Omega^1(M)$ denotes the space of smooth differential $1$-forms on $M$ such that \begin{align*} \dot\omega_t=d\sigma_t \end{align*} for every $t\in[0,1]$. [/guided] [/step] [step:Define the time-dependent vector field by the Moser equation] For each $t\in[0,1]$ and $x\in M$, the [bilinear form](/page/Bilinear%20Form) \begin{align*} (\omega_t)_x:T_xM\times T_xM\to\mathbb R \end{align*} is nondegenerate. Therefore the [linear map](/page/Linear%20Map) \begin{align*} T_xM&\to T_x^*M \end{align*} \begin{align*} v&\mapsto \iota_v(\omega_t)_x \end{align*} is an isomorphism. Hence there is a unique vector $X_t(x)\in T_xM$ satisfying \begin{align*} \iota_{X_t(x)}(\omega_t)_x=-(\sigma_t)_x. \end{align*} The smooth dependence of $\omega_t$ and $\sigma_t$ on $(t,x)$, together with the smooth inversion of the bundle isomorphism induced by $\omega_t$, shows that these vectors define a smooth time-dependent vector field \begin{align*} X:[0,1]\times M\to TM \end{align*} with $X_t\in\mathfrak X(M)$, where $\mathfrak X(M)$ denotes the space of smooth vector fields on $M$, and \begin{align*} \iota_{X_t}\omega_t=-\sigma_t \end{align*} for every $t\in[0,1]$. [/step] [step:Integrate the vector field to a global isotopy] Since $M$ is compact and $X:[0,1]\times M\to TM$ is a smooth time-dependent vector field, the standard global existence theorem for smooth time-dependent vector fields on compact manifolds applies on the whole interval $[0,1]$ and gives a smooth flow \begin{align*} \psi:[0,1]\times M\to M \end{align*} satisfying \begin{align*} \psi_0=\operatorname{id}_M \end{align*} and \begin{align*} \frac{d}{dt}\psi_t(x)=X_t(\psi_t(x)) \end{align*} for every $x\in M$ and $t\in[0,1]$. For each fixed $t\in[0,1]$, global existence for the reversed time-dependent vector field on the compact interval from $t$ back to $0$ gives a smooth inverse to $\psi_t$. Hence each map $\psi_t:M\to M$ is a diffeomorphism of $M$, and $\psi$ is a smooth isotopy on the full interval $[0,1]$. [/step] [step:Differentiate the pulled-back forms along the isotopy] We apply the differentiation formula for pullbacks along a time-dependent flow. Since $\psi_t$ is generated by $X_t$, for the smooth family of $2$-forms $(\omega_t)$ we have \begin{align*} \frac{d}{dt}(\psi_t^*\omega_t)=\psi_t^*(\dot\omega_t+\mathcal L_{X_t}\omega_t). \end{align*} Cartan's formula for the Lie derivative gives \begin{align*} \mathcal L_{X_t}\omega_t=d(\iota_{X_t}\omega_t)+\iota_{X_t}(d\omega_t). \end{align*} Because $\omega_t$ is symplectic, $d\omega_t=0$. Therefore \begin{align*} \mathcal L_{X_t}\omega_t=d(\iota_{X_t}\omega_t). \end{align*} Using $\dot\omega_t=d\sigma_t$ and $\iota_{X_t}\omega_t=-\sigma_t$, we obtain \begin{align*} \dot\omega_t+\mathcal L_{X_t}\omega_t=d\sigma_t+d(\iota_{X_t}\omega_t). \end{align*} Substituting the Moser equation gives \begin{align*} \dot\omega_t+\mathcal L_{X_t}\omega_t=d\sigma_t+d(-\sigma_t)=0. \end{align*} Hence \begin{align*} \frac{d}{dt}(\psi_t^*\omega_t)=0 \end{align*} for every $t\in[0,1]$. [/step] [step:Conclude that the pulled-back form is constant in time] Since \begin{align*} \frac{d}{dt}(\psi_t^*\omega_t)=0, \end{align*} the family of $2$-forms $(\psi_t^*\omega_t)_{t\in[0,1]}$ is constant in $t$. Evaluating at $t=0$ and using $\psi_0=\operatorname{id}_M$, we get \begin{align*} \psi_t^*\omega_t=\psi_0^*\omega_0=\operatorname{id}_M^*\omega_0=\omega_0 \end{align*} for every $t\in[0,1]$. Thus $\psi$ is the required isotopy, and the theorem follows. [/step]