Phantom_Ghost

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  1. 2月前
    2018-11-13 06:10:03
    Phantom_Ghost 更新于 Symplectic Geometry

    Contactization of exact symplecticmanifolds, wave fronts

    Let $(M,\omega=d\lambda)$ be an exact symplectic manifold

    Def. The contactization of $(M,d\lambda)$ is $\mathbb{R}\times M$ ($\text{dim}\,M=2n$) with the contact sturcture defined by $dz-\text{pr}_2^*\lambda$ (nowhere vanishing 1-form)
    $(dz-\text{pr}_2^*\lambda)\wedge d(dz-\text{pr}_2^*\lambda)^n=dz\wedge(\text{pr}_2^*\omega)^n$ is a volume form.

    If $f:L\to M$ is an exact Lagrangian immersion and fixing a primitive $H$ of $f^*\lambda$ one obtains a lift $F:L\to\mathbb{R}\times M$ (i.e. $dH=f^*\lambda$) via $F(p)=(H(p),f(p))$
    \begin{array}[c]{ccc}
    & & \mathbb{R}\times M\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\
    &\underset{\underset{\;}{\;}}{\stackrel{F\;\;\;\;\;}{\nearrow}} & \;\Bigg\downarrow\scriptstyle{\text{pr}_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\\
    L &\underset{\;\;\;f\;\;\;\;}{\xrightarrow{\;\;\;\;\;\;\;}} & M\;\;\;\;\;\;\;\;\;\;\;\;\;\;
    \end{array}
    Then $F^*(dz-\text{pr}_2^*\lambda)=dH-(\underset{f}{\underbrace{\text{pr}_2\circ F}})^*\lambda=0$.

    Def.: Let $(N,\xi)$ be a contact manifold. An immersion $f:L\to N$ is Legendrian if $f_*(TL)\subset\xi$.

    $\Rightarrow$ Exact Lagrangian immersion lifted to Legendrian immersion in the contactization.

    Apply this to cotangent bundle $(T^*M,\omega_\text{st.}=d\lambda_\text{st.})$. Let $\pi:T^*M\to M$ denote the bundle projection.

    Def.: Let $f:L\to T^*M$ be an exact Lagrangian immersion and $dH=f^*\lambda_\text{st.}$. The wave front of $f$ is the image of the following composition:
    \[
    L\xrightarrow{F}\mathbb{R}\times T^*M\xrightarrow{\pi}\mathbb{R}\times M
    \]
    The wave front is not an immersion everywhere.

    Remark: A wave front has no vertical tangencies.
    If $p\in L$ so that $\pi\circ F$ is an immersion, then one can construct $F$ respectively $f$ from the (reparameterized) wave front. From $(\pi\circ F)_*(T_pL)$ one can read off $dH$ $\Rightarrow$ the unique 1-form $\alpha$ in $T_p^*M$ so that $\text{ker}(dz-\alpha)=(\pi\circ F)_*(T_pL)$.

    [attachment:5bef5a55a88c3]

    Since $f^*\lambda_\text{st.}=dH$, we can read off $f^*\lambda_\text{st.}|_p$ from the wave front. However $f^*\lambda_\text{st.}|_p=f(p)$, $f(p)\in T_x^*M$ where $x=(\pi\circ f)(p)$ (reparameterized)
    \[
    L\xrightarrow{F}\mathbb{R}\times T^*M\xrightarrow{\text{pr}_2}T^*M\xrightarrow{\pi}M
    \]
    Let $X\in T_xM$, $\hat{X}\in f_*(T_pL)\subset T_xM$, then $X=\pi_*\hat{X}$.
    $\Rightarrow$ $f(p)(\pi_*\hat{X})=\lambda_\text{st.}(\hat{X})$
    $(dH)|_p(Y)=(f^*\lambda_\text{st.})|_p(Y)=\lambda_\text{st.}(f_* Y)=f(p)(\pi_*f_*Y)$, $Y\in T_pL$.
    [attachment:5bef5a55d047c]

    This allows us to 'draw' a Lagrangian immersion.

    Question: Which manifold admits Lagrangian immersion in $(T^*\mathbb{R}^n,\omega,J)\cong(\mathbb{C}^n,\omega,\mathfrak{i})$

    Lemma: A necessary and sufficient condition for the existence of a Lagrangian immersion $f:L\to\mathbb{C}^n$ is that $TL\otimes\mathbb{C}$ is trivial as a complex vector bundle.

    Reminder: Real vector bundle $\pi:E\to B$, $\pi$ continuous and surjective, $B$ connected.
    (1) $\forall b\in B$, there exists neighbourhood $U\subset B$ s.t.
    \begin{array}{ccc}
    \;\;E\supset
    \pi^{-1}(U) &\xrightarrow{\underset{(\text{home.})}{\approx}}& U\times\mathbb{R}^k\\
    &{}_{\pi}{\searrow}\;\;\;\;\;&\Big\downarrow\rlap{\scriptstyle\text{pr}_1}\;\;\;\;\;\;\;\;\;\\
    && U\;\;\;\;\;\;\;\;\;
    \end{array}
    This is called the local trivialization.

    (2) If $\varphi_U,\varphi_V$ are bundle charts as in (1), then
    \begin{array}{ccccc}
    (U\cap V)\times\mathbb{R}^k &\xrightarrow{\left.\varphi_U^{-1}\right|_{(U\cap V)\times\mathbb{R}^k}}& \pi^{-1}(U\cap V) &\xleftarrow{\left.\varphi_V\right|_{\pi^{-1}(U\cap V)}}& (U\cap V)\times\mathbb{R}^k\\
    &\searrow\;\;\;\;\;\;\;\;\;\;\;\;&\Bigg\downarrow\rlap{{}}&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\swarrow\\
    && U\cap V
    \end{array}
    The composition $\underset{(\text{id}|_{U\cap V}\,,\,\text{transition functions})}{\underbrace{\left.\varphi_V\right|_{\pi^{-1}(U\cap V)}\circ\left.\varphi_U^{-1}\right|_{(U\cap V)\times\mathbb{R}^k}}}:(U\cap V)\times\mathbb{R}^k\to(U\cap V)\times\mathbb{R}^k$
    here transition functions are linear isomorphisms depending on the base point represented as a family of matrices in $GL(k,\mathbb{R})$ (structure group).

    A complex bundle is to replace $\mathbb{R}^n$ with $\mathbb{C}^n$ above.
    A trivial vector bundle is that $U$ can be maximally extended to whole $B$ (global trivialization).

    $TL$ is a real vector bundle, to complexifiy it as $TL\otimes\mathbb{C}$, the new total space $E^\mathbb{C}$ is contructed so that it is locally trivialized as $\pi^{-1}(U)\to U\times(\mathbb{R}^k\otimes\mathbb{C})$. And those transition functions still remains the same matrices with original real entries, however they are now viewed as the matrices in the complexified structure group $GL(k,\mathbb{C})$, i.e. $GL(k,\mathbb{R})\subset GL(k,\mathbb{C})$. The fiber is complex vector space now.

    Pf. of lemma (sketch): Assume $f:L\to\mathbb{C}^n$ is a Lagrangian immersion. By Weinstein's tubular neighbourhood theorem: $f^*(NL)=T^*L$. The normal bundle is also $\mathfrak{i}f_*(T_pL)$ for $p\in L$.

    [attachment:5bef5a56a36f0]

    $\Rightarrow$ for each $p\in L$: $\underset{\cong T_{f(p)}\mathbb{C}^n=\mathbb{C}^n}{\underbrace{f_*(T_pL)\oplus\mathfrak{i}f_*(T_pL)}}=f_*(T_pL)\otimes\mathbb{C}=f_*(T_pL\otimes\mathbb{C})$. This shows the global trivialization. $\square$

    Reminder:
    $\bullet$ Pull back of bundle $f^*E=\{(b',e)\in B'\times E|f(b')=\pi(e)\}\xrightarrow{\text{pr}_1}B'$
    \[
    \begin{array}[c]{ccc}
    f^*E&\stackrel{{}}{\rightarrow}&E\\
    \downarrow\scriptstyle{{}}&&\downarrow\scriptstyle{\pi}\\
    B'&\underset{f}{\rightarrow}& B
    \end{array}
    \]
    $\bullet$ Complexification of vector space $V^\mathbb{C}:=V\otimes\mathbb{C}\cong V\oplus\mathfrak{i}V$
    $\mathfrak{i}V=JV$, $J=\left(
      \begin{array}{cc}
        0 & -\text{id} \\
        \text{id} & 0 \\
      \end{array}
    \right)$, $(a+ib)=av+bJv$ ($a,b\in\mathbb{R}$)
    e.g.

    [attachment:5bef5a5652113]

    $T_pS^2\oplus N_pS^2\cong T_pS^2\otimes\mathbb{C}$, $av+b\hat{n}_p\times v=(a+\mathfrak{i}b)v$, $\hat{n}_p\in N_pS^2$, $v\in T_pS^2$.

    Remark:
    (i) For all oriented closed manifolds of dimension 1,2,3, the bundle $TL\otimes\mathbb{C}$ is trivial. In higher dimensions, there are manifolds which do not admit Lagrangian immersions.
    (ii) The necessary condition for the existence of a Lagrangian immersion into $\mathbb{C}^n$ is also sufficient! The story for Lagrangian embeddings is far more complicated.

    \[\]

    Lagrangian embedding

    Theorem: Let $L$ be closed, oriented manifold and $f:L\to\mathbb{C}^n$ a symplectic embedding. Then $\chi(L)=0$ (null Euler characteristic).

    Proof: (1) Weinsten tubular neighbourhood thm.: $f(L)= L$ has a tubular neighbourhood sympletctomorphic to a tubular neighbourhood of the zero section on $T^*L$.

    [attachment:5bef5ec27b001]

    $T^*L\cong TL$ as bundle isomorphism $\Rightarrow$ $\chi(T^*L)=\chi(TL)$

    (2) The Euler charateristic of rank-$n$ oriented vector bundle ocer a closed manifold of dimension-$n$ can be computed as follows: Pick a section $s:L\to TL$ s.t. $s(L)\cap\underset{\text{zero section}}{\underbrace{L}}$ ($s(p)=(p,0)=p$) is transversal $\Rightarrow$ $s(L)\cap L$ is a submanifold of zero dimension, i.e. a collection of intersection points, it contains only finitely many points since $L$ is compact. These points are equipped with signs: '$+$' if $(T_pL,\text{orient.})\oplus s_*(T_pL,\text{orient.})=(T_p(TL),\text{(same) orient.})$, otherwise takes '$-$'.

    [attachment:5bef66515af61]

    Fact: sign-count gives the Euler characteristic of the bundle.

    (3) This sign-count actually computes the self-intersection number of the zero-section $0=[L]\in H_n(\mathbb{C^n})$ (i.e. the top homology class).
    $f_*([L]) f_*([L])=0$, $f_*([L])=0$ $\Rightarrow$ $\chi(TL)=\chi(L)=0$ $\square$

    Reminder: $\Lambda_n:=\{L\subset\mathbb{C}^n| L\;\text{is Lagrangian submanifold}\}\cong U(n)/O(n)$ is Lagrangian Grassmannian submanifold and
    \[\begin{array}{c} \pi_1(\Lambda_n)\xrightarrow{\sim}\pi_1(\underset{\subset\mathbb{C}}{\underbrace{S^1}})\cong\mathbb{Z}\\ [\gamma]\mapsto[(\text{det}(\gamma))^2]\end{array}\]

    Assume $f:L\to\mathbb{C}^n$ is Lagrangian immersion. This induced the Gauss map
    \[\begin{array}{c} \Gamma(f):L\rightarrow\Lambda_n\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;p\mapsto[f_*(T_pL)]\end{array}\]
    ($T_pL\to f_*(T_pL)=T_{f(p)}L\subset T_{f(p)}\mathbb{C}^n\cong\mathbb{C}^n$)

    Def.: The Maslov class of $f$ is the cohomology class
    \[
    \underset{\in H^1(L;\mathbb{Z})}{\underbrace{\mu(f)}}:\pi_1(L)\to\pi_1(\Lambda_n)\cong\pi_1(S^1)\cong\mathbb{Z}
    \]

    Example: $S^1\subset\mathbb{C}$ Lagrangian submanifold

    [attachment:5bef71f66a204]

    Maslov class: $(\mu(f))(\underset{\cong\pi_1(S^1)}{\underbrace{\pi_1(L)}})=2$ (winding number or degree of map is 2)

    Fact: Assume $f_0,f_1:L\to\mathbb{C}^n$ are Lagrangian immersion so that $\mu(f_0)\neq\mu(f_1)\in H^1(L;\mathbb{Z})$ $\Rightarrow$ $f_0$ is not homotopic to $f_1$ through Lagrangian immersion.

    Let $f:L\to\mathbb{C}^n$ be a Lagrangian immersion, then define $||\mu(f)||=\text{non-negative generator of }\text{im}(\mu(f))$, i.e. $(\mu(f))(\pi_1(L))\subseteq\mathbb{Z}$.

    Theorem(Viterbo): If $\mathbb{T}^n\xrightarrow{f}\mathbb{C}^n$ is a Lagrangian embedding, then $2\leq ||\mu(f)||\leq n+1$.

    \[\]

    Classical (Hamilton) mechanics system

    ...

  2. 2018-11-13 05:40:52
    Phantom_Ghost 更新于 Symplectic Geometry

    Exercise Sheet 3

    [attachment:5be9f3d36f785]

    [attachment:5be9f3d268166]

  3. 2018-11-12 04:42:32
    Phantom_Ghost 更新于 Conformal Field Theory

    Exercise sheet 3

    [attachment:5be894329c88f]

  4. 2018-11-12 02:39:57
    Phantom_Ghost 更新于 Conformal Field Theory

    Simplest continuum (field) theory

    Euclidean field (for statistical model)
    Action: $S[\phi(x)]=\int d^nx\left(\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\frac{1}{2}m^2\phi^2\right)$

    Correlation function: $\langle\phi(x_1)\cdots\phi(x_N) \rangle=\frac{1}{Z}\int\mathcal{D}[\phi(x)]\phi(x_1)\cdots\phi(x_N)e^{-S[\phi]}$, $Z=\int\mathcal{D}[\phi] e^{-S[\phi]}$.

    $\phi(x)=\sum_n c_n\phi_n(x)$, $(-\Delta+m^2)\phi_n(x)=\lambda_n\phi_n(x)$; $\int\mathcal{D}[\phi]=\int\prod_n dc_n$.

    $Z[J]=e^{-W[J]}:=\int\mathcal{D}[\phi]e^{-S[\phi]-\int d^nx J(x)\phi(x)}$, $\langle\phi(x_1)\cdots\phi(x_N) \rangle=\left.\frac{\delta^{N}W[J]}{\delta J(x_1)\cdots\delta J(x_M)}\right|_{J=0}$.

    Integrate out $\phi$:
    \begin{align*}
    e^{-W[J]}&=\int\mathcal{D}[\phi]e^{-\int d^nx ( \frac{1}{2}\phi(-\Delta+m^2)\phi+J\phi )}\\
    &\propto\frac{1}{\text{det}(\Delta+m^2)}e^{\int dx^n\int d^ny\;J(x)(-\Delta+m^2)^{-1}J(y)}\\
    &=\frac{1}{\text{det}(\Delta+m^2)}e^{\int dx^n\int d^ny\;J(x)G(x,y)J(y)}
    \end{align*}
    $(-\Delta+m^2)G(x,y)=\delta^{(n)}(x-y)$. Then the two-point correlation function yields the Green function $\langle\phi(x)\phi(y)\rangle=\left.\frac{\delta^2W[J]}{\delta J(x)\delta J(y)}\right|_{J=0}=G(x,y)$.
    \[
    G(x,y)=
    \left\{
    \begin{array}{rl}
    \frac{e^{-m|x-y|}}{|x-y|^{n-2}}\;\;\;,\;\;\;|x-y|\to\infty\;(n\neq 2)\\ \frac{1}{|x-y|^{n-2}}\;\;\;,\;\;\;|x-y|\to\infty\;(n\neq 2)\\
    \ln|x-y|\;\;\;,\;\;\;|x-y|\to 0\;(n=2)\\
    e^{-m|x-y|}\;\;\;,\;\;\;|x-y|\to\infty\;(n=2)
    \end{array}
    \right.
    \]
    Note:
    \begin{align*}
    \frac{\delta^2W[J]}{\delta J(x)\delta J(y)}&=-\frac{\delta}{\delta J(x)}\frac{\delta\ln Z[J]}{\delta J(y)}=\frac{\delta}{\delta J(x)}\left(\frac{1}{Z[J]}\frac{\delta Z[J]}{\delta J(y)}\right)\\
    &\xrightarrow{J\to 0}\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y)e^{-S[\phi]}-\left(\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)e^{-S[\phi]}\right)\left(\frac{1}{Z}\int\mathcal{D}[\phi]\phi(y)e^{-S[\phi]}\right)\\
    &=\langle\phi(x)\phi(y)\rangle-\underset{=0}{\underbrace{\langle\phi(x)\rangle}}\underset{=0}{\underbrace{\langle\phi(y)\rangle}}=\langle\phi(x)\phi(y)\rangle_\text{connected}\end{align*}

    We can see that $G(x,y)$ and $\langle\sigma_x\sigma_y\rangle^*$ for statistical lattice model at critical point coincide iff. $m=0$.

    \[\]

    Ginzburg-Landau theory

    $S[\phi]=\frac{1}{2}\int d^nx[(\partial\phi)^2+m^2\phi^2+V(\phi)]$, $m^2=\frac{T-T_c}{T_c}$, $V(\phi)=\lambda\phi^4$. $\phi$ is order parameter.

    [attachment:5be88aaf02239]

    \begin{align*}
    \langle\phi(x)\phi(y)\rangle&=\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y) e^{-\int d^nx[\frac{1}{2}(\partial\phi)^2+\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4]}\\
    &\approx\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y)\left(1+\frac{\lambda}{4!}\int d^nx\phi^4\right)e^{-\int d^nx[\frac{1}{2}(\partial\phi)^2+\frac{1}{2}m^2\phi^2]}+O(\lambda^2)
    \end{align*}
    [attachment:5be88ea57a8ca]

    \begin{align*}
    G(x,y)&=G(|x-y|)=\int\frac{d^nk}{(2\pi)^n}e^{ik\cdot(x-y)}\Delta(k)\\
    \Delta(k)&=\underset{=\frac{1}{k^2+m^2}}{\underbrace{\Delta_0(k)}}-\frac{\lambda}{2}\Delta_0(k)\underset{=\Sigma(0)}{\underbrace{\left(\int d^nq\frac{1}{q^2+m^2}\right)}}\Delta_0(k)+O(k)
    \end{align*}
    self-energy: $\Sigma(0)\overset{\text{dim. regularization}}{=}\frac{(m^2)^{\frac{n-2}{2}}\Gamma(\frac{2-n}{2})}{(\sqrt{4\pi})^n}$

    Classical approximation
     extremizes the potential in Lagrangian to find the critical point (only classical motion invovled)

    [attachment:5be88a955a1ee]

    critical exponent: $\chi=\left.\frac{\partial M}{\partial B}\right|_{B=0}=\frac{1}{(T-T_c)^\gamma}$ $(T\geq T_c)$.
    \begin{align*}
    \chi&=\int d^nx C(x)\propto\xi^2\;,\;\gamma\approx 1.25\\
    C(x-y)&=\langle\sigma_x\sigma_y\rangle^*=\frac{e^{-|x-y|/\xi}}{|x-y|^{n-2}}
    \end{align*}
    $\lambda=0$: $C(x-y)=G(x,y)$ $\Rightarrow$ $\chi\propto\frac{1}{m^2}\propto\frac{1}{T-T_c}$ $\Rightarrow$ $\gamma=1$.

  5. 2018-11-12 02:28:15
    Phantom_Ghost 更新于 Conformal Field Theory

    Exercise sheet 2

    [attachment:5be874a291905]

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  6. 2018-11-06 04:58:11
    Phantom_Ghost 更新于 Symplectic Geometry

    Generating functions for Lagrangian immersion

    Linear algebra version
    ...
    \[\]

    Construction of Lagrangian immersion by reduction

    ...
    \[\]

  7. 2018-11-06 04:18:24
    Phantom_Ghost 更新于 Symplectic Geometry

    Exercise Sheet 2

    [attachment:5be88fb109c4b]

  8. 2018-11-05 03:30:04

    @DTSIo 如果接下来想了解 de Rham 定理, 不妨直接进入层论, 了解一下层的上同调, 层的分解, 以及抽象的 de Rham 定理.

    可否推荐下哪本书从这个角度讲的?

  9. 2018-11-03 23:29:29
    Phantom_Ghost 更新于 Symplectic Geometry

    Normal forms for symplectic manifolds/Neighbourhoods of submanifolds

    Bassic description:
    (1) Cartan's formula: $\underset{\text{Lie derivative}}{\underbrace{L_X}\alpha}=di_X\alpha+i_Xd\alpha$

    (2) If $\phi_t$ is a family of diffeomorphisms of $M\to M$, $\phi_0=\text{id}$ (isotopy) , then
    \begin{align*}
    \left.\frac{d}{dt}\right|_{t=t_0}\phi^*\alpha_t=\phi^*_{t_0}(\dot{\alpha}_t+L_{X_{t_0}}\alpha_{t_0})
    \end{align*}
    where $\alpha_t$ is a family of forms, $X_{t_0}(\phi_{t_0}(x))=\left.\frac{d}{dt}\right|_{t=t_0}\phi_t(x)$ (i.e. $\frac{d\phi_t}{dt}=X_t\circ\phi_t$).

    (3) Same tool to find smooth families of solutions of $d\alpha_t=\beta_t$ for a smooth family of exact forms $\beta_t$.
    There are two theories relating to this problem that worth mentioning:

    • Hodge theory: for closed (Riemannian) manifolds $\Omega^{k-1}$ is a harmonic $k-1$-form s.t. it is exact and $\delta\Omega^{k-1}\neq 0$ ($\delta$ is the adjoint of $d$, also denoted as $d^*$).
    • There exists an explicit formula (see Bott-Tu): $H^\bullet_\text{dR}(M;\mathbb{R})\cong \check{H}{}^\bullet(M;\mathbb{R})$ ---$\check{\text{C}}\text{ech}$ cohomology of $M$ for (constant sheaf) $\mathbb{R}$.

    The problem we want to solve: Given a family of $k$-forms $\omega_t$, $t\in[0,1]$, is there a family of diffeomorphisms (isotpy) $\phi_t$ s.t. $\phi^*_t\omega_t=\omega_0$ (for zero forms, i.e. functions, it yields simply $\phi_t^* f=f\circ\phi_t$) ?
    $\phi_t$ will be constructed from vector fields (non-autonomous solution of an ODE system).

    Differentiating $\phi^*\omega_t=\omega_0$ and get $0=\phi^*_t(L_{X_t}\omega_t+\dot{\omega}_t)$. So we now look for vector fields solving
    \[
    \boxed{i_{X_t}d\omega_t+di_{X_t}\omega_t=-\dot{\omega}_t}\;\;\;\;\;(*)
    \]
    Theorem(Moser): Let $\Omega_{t\in[0,1]}$ be a family of volume forms on the connected closed manifold $M$ so that $\int_M\Omega_t=\text{const.}$. Then there is a family of diffeomorphisms (isotopy) $\phi_t:M\to M$ s.t. $\phi^*_t\Omega_t=\Omega_0$ for all $t\in[0,1]$.

    Corollary: Any two volume forms $\Omega_0,\Omega_1$ on $M$ as above with the same volume ($\int_M\Omega_0=\int_M\Omega_1$) are diffeomorphic: $\Omega_t=t\Omega_1+(1-t)\Omega_0$, $\forall t\in[0,1]$.

    Proof of Moser's thm.: We look for $X_t$ solving $(*)$. $\Omega_t$ has maximal degree $\Rightarrow$ $d\Omega_t=0$. $(*)$ becomes $di_{X_t}\Omega_t=-\dot{\Omega}_t$ ($\dot{\Omega}_t$ is exact since $\int_M\dot{\Omega}_t=\frac{d}{dt}\underset{=\text{const.}}{\underbrace{\int_M\Omega_t}}=0$) $\Rightarrow$ there exists a smooth family of $(\underset{=\text{dim}\,M}{\underbrace{n}}-1)$-forms $\alpha_t$ s.t. $d\alpha_t=-\dot{\Omega}_t=di_{X_t}\Omega_t$. To find $X_t$, only need to solve $\alpha_t=i_{X_t}\Omega_t$.
    (The map $\begin{array}{c}
    T_xM\to\Lambda^{n-1}T_x^*M\\
    X\mapsto\Omega(X,\dots)
    \end{array}$ is obviously injective, since $\text{dim}\,T_xM=n=\text{dim}\,\Lambda^{n-1}T_x^*M$ it is also surjective.)
    Since $M$ is closed, $X_t$ defines a family of diffeomorphisms (isotopy) $\phi_t$ s.t. $\phi^*_t\Omega_t=\Omega_0$. $\square$

    ....

  10. 2018-10-25 04:10:05
    Phantom_Ghost 更新于 Symplectic Geometry

    Exercise Sheet 1

    [attachment:5bd77d705f752]

    [attachment:5bd77d745e036]

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