Symplectic Geometry

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    1楼 10月21日 数学版主, 物理版主, 优秀回答者
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    This post is written on the lecture course ''Symplectic Geometry'' held by Prof. Dr. Thomas Vogel in LMU during WS18/19. The goal of the course is mainly an introduction to symplectic topology.
    The lecture note and excercises will be presented. Any repost and copy for business commercial purposes are forbidden. All rights are reserved.


  2. Phantom_Ghost

    2楼 10月22日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    [1] D. McDuff, D. Salamon: Introduction to symplectic topology (Oxford Math. Monographs)
    [2] Kai Cieliebak, Symplectic Geometry (lecture note), 2010.
    [3] A. Cannas da Silva, Lectures on symplectic geometry, Lecture Notes in Mathematics, 1764, Springer-Verlag, Berlin, 2001 and 2008.


    Symplectic Manifold

    Def. : A symplectic manifold $(M,\omega)$ is a $2n$-dimensional (smooth) manifold $M$ with a 2-form $\omega$ satisfying the following:
    i) $\omega$ is non-degenerate, i.e. $\forall v(\neq 0)\in T_x M$, $\exists w\in T_xM$ s.t. $\omega(v,w)\neq 0$;
    ii) $\omega$ is closed, i.e. $d\omega=0$.

    1) $M=\mathbb{R}^{2n}$, $\omega=dx_1\wedge dy_1+\cdots+dx_n\wedge dy_n=d(x_1dy_1+\cdots+x_ndy_n)$.

    2) $M=\Sigma$ an oriented surface, $\omega=``\text{area form''}$ is symplectic.

    3) $(M_1,\omega_1)$, $(M_2,\omega_2)$ symplectic $\Rightarrow$ $(M_1\times M_2,\text{pr}_1^*\omega_1+\text{pr}_2^*\omega_2)$ is symplectic.

    A very important exmaple: cotangent bundles
    Let $M$ be smooth, then $T^*M$ is also smooth and carries a tautological 1-form $\lambda$: let $v\in T_\mu(T^*M)$, $\mu\in T_x^*M$, $\lambda(v)=\mu(\text{pr}_* v)$ where
    \text{pr}&:T^*M\to M\;\;(\mu\mapsto x)\\
    \text{pr}_*&:T(T^*M)\to TM\;\;(T_\mu(T^*M)\to T_xM)

    Claim: $\omega=d\lambda$ is a symplectic form.
    This will be verified in a coordinate system. Let $(q_1,\dots,q_m)$ be a local coordinate in a neighborhood $U$ of $x$ in $M$. This induces a coordinate system $(p_1,\dots,p_m;q_1,\dots,q_m)$ on $\text{pr}^{-1}(U)\subset T^*M$:
    (p_1,\dots,p_m;q_1,\dots,q_m)\mapsto p_1dq_1+\cdots+p_mdq_m\in T_{(q_1,\dots,q_m)}^*M
    Express $\lambda$ in terms of these coordinates:
    v=a_1\frac{\partial}{\partial p_1}+\cdots+a_m\frac{\partial}{\partial p_m}+b_1\frac{\partial}{\partial q_1}+\cdots+b_m\frac{\partial}{\partial q_m}\in T_\mu(T^*M)\;,\;\mu=(p_1,\dots,p_m;q_1,\dots,q_m)
    Then $\lambda(v)=\mu(\text{pr}_*v)=\mu(b_1\frac{\partial}{\partial q_1}+\cdots+b_m\frac{\partial}{\partial q_m})=p_1b_1+\cdots+p_m b_m$
    $\Rightarrow$ $\lambda=p_1dq_1+\cdots+p_mdq_m$, $d\lambda=dp_1\wedge dq_1+\cdots+dp_m\wedge dq_m$ is non-degenerate. $\square$


    Linear theory —— symplectic vector spaces and their automorphisms

    Def.: Let $V$ be a real vector space; a symplectic form on $V$ is a non-degenerate skew 2-form $\omega\in\Lambda^2V^*$ (antisymmetric).

    1) Let $U$ be a $\mathbb{R}$-vector space, $V=U\oplus U^*$, on $V$ there is a tautological 2-form $\omega$: $\omega((u_1,u_1^*),(u_2,u_2^*))=u^*_2(u_1)-u_1^*(u_2)$, $u_{i\in\{1,2\}}\in U$.

    2) $\mathbb{R}^{2n}$ with basis $\{e_1,\dots,e_n;f_1,\dots,f_n\}$; $\omega(e_i,e_j)=\omega(f_i,f_j)=0$, $\omega(e_i,f_j)=-\omega(f_j,e_i)=\delta_{ij}$ is a non-degenerate 2-form.

    Def.: Let $(V,\omega)$ be a symplectic vector space and $U\subset V$ a subspace, $U^{\perp\omega}:=\{v\in V|\omega(v,u)=0,\;\forall u\in U\}$.

    $\bullet$ $U$ is isotropic iff. $U\subset U^{\perp\omega}$ $\Leftrightarrow$ $\omega|_U=0$
    $\bullet$ $U$ is coisotropic iff. $U^{\perp\omega}\subset U$
    $\bullet$ $U$ is symplectic iff. $U\cap U^{\perp\omega}=\{0\}$ $\Leftrightarrow$ $\omega|_U$ symplectic
    $\bullet$ $U$ is Lagrangian iff. $U=U^{\perp\omega}$

    lemma: $(V,\omega)$ symplectic and $\text{dim}V<+\infty$, then $V\to V^*$, $v\mapsto[w\mapsto\omega(v,w)]$ is an isomorphism.
    Proof: The map is injective, $V$ is finite dimentional $\Rightarrow$ the map is surjective $\square$

    lemma: $U\subset(V,\omega)$, then $\text{dim}V=\text{dim}U+\text{dim}U^{\perp\omega}$.
    Proof: $V\to U^*$, $v\mapsto[u\mapsto\omega(v,u)]$ is surjective by previous lemma. Then $\text{ker}(V\to U^*)=U^{\perp\omega}$. $\square$

    lemma: $U\subset(V,\omega)$, then the following hold
    (a) $(U^{\perp\omega})^{\perp\omega}=U$
    (b) $(U/(U\cap U^{\perp\omega}),\omega)$ is a symplectic vector space.
    Proof: exercise!

    Proposition: Let $(V,\omega)$ be a finite dimensional symplectic vector space. Then there is a basis $\{e_1,\dots,e_n;f_1,\dots,f_n\}$ of $V$ such that $\omega(e_i,e_j)=\omega(f_i,f_j)=0$, $\omega(e_i,f_j)=-\omega(f_j,e_i)=\delta_{ij}$ (such basis is called a symplectic basis).
    Proof: By induction, let $e_1(\neq 0)\in V$, by degeneracy there is $f'_1\in V$ such that $\omega(e_1,f'_1)\neq 0$, normalizing $\omega(e_1,f_1)=1$. Then $U_1=\text{span}_\mathbb{R}\{e_1,f_1\}$ is 2-dimensional and symplectic $\Rightarrow$ $U^{\perp\omega}_1$ is symplectic and has $\text{codim}=2$ in $V$.
    Complete the indution on $U_n$ and this is done.

    Corollary 1: $V$ is even dimensional.

    Corollary 2: Two fintie dimensional symplectic vector spaces are symplectic isomorphic iff. they has the same dimension.

    Corollary 3: $\omega\in\Lambda^2V^*$ ($\text{dim}V=2n$) is non-degenerate iff. $0\neq\omega^n\in\Lambda^{2n}V^*$.

    There are more statements:
    Let $U\subset(V,\omega)$ be a coisotropic subspace, then there exists a symplectic standard basis of $(V,\omega)$, i.e. $\{e_1,\dots,e_n;f_1,\dots,f_n\}$ such that $\{e_1,\dots,e_k;f_1,\dots,f_k\}\;\;(k\leq n)$ is a basis of $U$.

    $\bullet$ A symplectic manifold $(M,\omega)$ has a natural volume form $\omega^n$ and hence is orientable.

    $\bullet$ If $M$ is closed, then $[\omega]\in H^2_\text{dR}(M;\mathbb{R})$ does not vanish. More generally, $[\omega]^k\in H^{2k}_\text{dR}(M;\mathbb{R}),\;\forall 0\leq k\leq n$ do not vanish.
    Proof: Assume $\omega^k=d\lambda$ with $\lambda\in\Omega^{2k-1}(M)$ ($2k$-cohomology vanishes!), then
    0\neq&\int_M\omega^n=\int_M\omega^k\wedge\omega^{n-k}=\int_M (d\lambda)\wedge\omega^{n-k}=\int_M d(\lambda\wedge\omega^{n-k})\\
    &\overset{\text{Stoke's Thm.}}{=}\int_{\underset{(M\text{ is closed})}{\partial M=\emptyset}}\lambda\wedge\omega^{n-k}=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\text{conflict!}\;\;\;\square


    Complex structure

    Def.: Let $V$ ve a real vector spaces, a complex structure on $V$ is a linear map $J:V\to V$ with $J^2=-\mathbb{I}$.

    If $(V,\omega)$ is symplectic, then a complex structure is compatible with $\omega$ if $g_J(v,\omega)=\omega(v,Jw)$ for all $v,w\in V$. This defines an Euclidean structure on $V$, i.e. $g_J(\cdot,\cdot)$ is symmetric and positive definite bilinear form.

    Lemma: $(V,\omega)$ symplectic, there is a continous map: $\{\text{Euclidean structure on}\;V\}\to\{\text{compatible complex structure on}\; (V,\omega)\}$, such that $g_J$ is mapped to $J$ (topology is constructed by the subspaces of vector space $V$).
    Proof: Let $g$ be Euclidean, there is a unique isomorphism $A:V\overset{\sim}{\longrightarrow}V$ s.t. $\omega(\cdot,\cdot)=g(A\cdot,\cdot)$.
    ($\omega(v,\cdot)$ defines a 1-form, $g(A v,\cdot)=\omega(v,\cdot)$ has a unique solution on $A v$)
    Because $g(\cdot,\cdot)$ symmetric and $\omega(\cdot,\cdot)$ is antisymmetric, $A$ is thus antisymmetric
    ($g(A v,w)=g(v,A^Tw)=g(A^Tw,v)$ $\forall v,w\in V$, $\omega(v,w)=-\omega(w,v)$ $\Rightarrow$ $A^T=-A$).
    Now $AA^T$ is a symmetric positive definite matrix $\Rightarrow$ $\exists!\;P$ s.t. $P$ is positive definite and $P^2=AA^T$. Moreover $P$ commutes with all matrices that commute with $AA^T$, e.g. $A$. Then there is a map: $g\mapsto J_g=P^{-1}A=\sqrt{AA^T}A$, where $J_g^2=P^{-1}A P^{-1}\overset{[P,A]=0}{=}-P^{-2}AA^T=-\mathbb{I}$. (This procedure is called polar decomposition)

    If $g=g_J$ for some $J$, then $J_{g_J}=J$.

    Claim: $A=J$, $J$ is compatible (with $\omega$).
    Proof: $\omega(Jv,Jw)=g_J(Jv,w)\overset{\text{symmetric}}{=}g(w,Jv)=\omega(w,J^2v)=-\omega(w,v)=\omega(v,w)$
    $J_{g_J}=\sqrt{JJ^T}J=J$, $J^T=-J$ by above. Hence $J_{g_J}=J$ is compatible.

    Namely, we have shown: $g\mapsto J_g=\sqrt{AA^T}A$, $J\mapsto\omega(\cdot,J\cdot)=g_J(\cdot,\cdot)$ $\square$

    Note:$\{\text{positive definite inner product on}\;V\}$ is a convex vector space (contractible), so $\{\text{compatible complex structure on}\;(V,\omega)\}$ is also contractible.

    Def.: A Hermitian structure on a real vector space $V$ is a pair $(\omega,J)$ where $\omega$ symplectic and $J$ compatible with $\omega$.

     $(v,w)=g_J(v,w)-i\omega(v,w)$ is a Hermitian inner product.
    e.g. $(Jv,w)=g_J(Jv,w)-\mathfrak{i}(J^2v,Jw)=\omega(Jv,Jw)+\mathfrak{i}\omega(v,Jw)=\omega(v,w)+\mathfrak{i}g_J(v,w)=\mathfrak{i}(v,w)$

    Note: If $J$ is compatible with $\omega$, then there is a symplectic basis of the form $\{e_1,Je_1,\dots,e_n,Je_n\}$ where $\{e_i\}_{i=1,\dots,n}$ orthonomal. Under this standard basis, $\omega$ has a standard form (denoted as $\omega_\text{st}$) and $J$ has also standard form (denoted as $J_\text{st}$):
        0 & -1 & & & \\
        1 & 0 & & & \\
         & & 0 & -1 & \\
         & & 1 & 0 & \\
         & & & & \ddots \\
    We now consider this standard structure.

    Def.: The sympletic group is $Sp(2n):=\{\psi\in GL(\mathbb{R}^{2n})|\psi^*\omega_\text{st}=\omega_\text{st}\}$, $(\psi^*\omega)(v,w)=\omega(\psi(v),\psi(w))=\omega(v,w)$.

    Warning: There is a second class of group which are called sympletic.
    On $\mathbb{H}$, there is a norm $N(h)=h\bar{h}$, where
    $Sp(n)$ is the preserving $N(h)$ (compact) subgroup of $\text{Aut}(\mathbb{H})$.

    Remark: $Sp(2n)=\{\psi\in GL(2n,\mathbb{R})|\psi^T J_\text{st}\psi=J_\text{st}\}$
    Proof: $\omega_\text{st}(v,w)=-g_\text{st}(v,J_\text{st}w)$
    On the other hand, $\omega_\text{st}(v,w)=-g_\text{st}(v,J_\text{st}w)$ $\forall v,w\in\mathbb{R}^{2n}$
    $\Rightarrow$ $\psi^TJ_\text{st}\psi=J_\text{st}$

    Lemma: Properties of $Sp(2n)$
    i) it is (closed) Lie subgroup pf $GL(2n,\mathbb{R})$ which is closed under transpose.
    ii) $A\in Sp(2n)$ $\Rightarrow$ $\text{det}(A)=1$
    iii) If $\lambda$ is an eigenvalue of $\psi\in Sp(2n)$, then so are $\lambda^{-1},\bar{\lambda},\bar{\lambda}^{-1}$
    iv) If $\lambda,\mu$ are eigenvalue and $\lambda\mu\neq 1$ $\Rightarrow$ corresponding eigenspaces are $\omega$-orthogonal.
    i) closed under transpose: $\psi\in Sp(2n)$ $\Rightarrow$ $\psi^T J\psi=J$, then $J=\psi^{-1}J(\psi^T)^{-1}$
    $((\psi^{-1})^T)^T J(\psi^{-1})^T=J$ $\Rightarrow$ $\psi^T,(\psi^{-1})^T\in Sp(2n)$
    ii) $\psi^*\omega=\omega$ $\Rightarrow$ $\psi$ preserves the volume form $\omega\wedge\cdots\wedge\omega$
    namely $\psi^*(\omega\wedge\cdots\wedge\omega)=\omega\wedge\cdots\wedge\omega=\text{det}(\psi)\omega\wedge\cdots\wedge\omega$ $\Rightarrow \text{det}(\psi)=1$
    iii) Let $v$ be an eigenvector for $\lambda$, $Jv=(\psi^TJ\psi)(v)=(\psi^TJ)(\lambda v)=\lambda\psi^T(Jv)$
    $\Rightarrow$ $\psi^T(Jv)=\lambda^{-1}Jv$, i.e. $\psi\,,\,\psi^T$ have the same eigenvalue. (similarly for $\bar{\lambda},\bar{\lambda}^{-1}$)
    iv) $\omega(\psi(v_\lambda),\psi(w_\mu))=\lambda\mu\omega(v_\lambda,w_\mu)$, on the other hand $\omega(\psi(v_\lambda),\psi(w_\mu))=\omega(v_\lambda,w_\mu)$ $\Rightarrow$ If $\lambda\mu\neq 1$, then $\omega(v_\lambda,w_\mu)=0$. $\square$

    Example: $Sp(2)=SL(2,\mathbb{R})$ (non-compact)

    Identifying $(\mathbb{R}^{2n},J)$ and $(\mathbb{C}^n,\mathfrak{i})$
    A complex matrix $Z=X-\mathfrak{i}Y$ ($n\times n$) acts on $v=x+\mathfrak{i}y$ in the way $Zv=(Xx-Yy)+\mathfrak{i}(Yx+Xy)$. If we write it under $\mathbb{C}^n\cong\mathbb{R}^n\oplus\mathfrak{i}\mathbb{R}$, then $Z=\left(
        X & -Y \\
        Y & X \\
    $. This identifies $GL(n,\mathbb{C})$ with a subgroup of $GL(2n,\mathbb{R})$.

    Lemma: $Sp(2n)\cap GL(n,\mathbb{C})=Sp(2n)\cap\underset{\subset GL(2n,\mathbb{R})}{\underbrace{O(2n,\mathbb{R})}}$
    \psi\in Sp(2n)\;&:\;\;\;\Psi^TJ\psi=J\\
    \psi\in O(2n)\;&:\;\;\;\psi^T\psi=\mathbb{I}\\
    \psi\in GL(n,\mathbb{C})\;&:\;\;\;\psi J=J\psi
    \right\} \text{any two imply the third}
    $\Rightarrow$ All three interior coincide.
    If $\psi\in O(2n)\cap GL(n,\mathbb{C})$, then $\psi=\left(
        X & -Y \\
        Y & X \\
    \right)\in GL(n,\mathbb{C})$ where $X,Y\in O(n)$.
        X^T & Y^T \\
        -Y^T & X^T \\
        X & -Y \\
        Y & X \\
        X^TX+Y^TY & -X^TY+Y^TX \\
       -Y^TX+X^TY & X^TX+Y^TY \\
        \mathbb{I} & 0\\
        0 & \mathbb{I} \\
    $X+\mathfrak{i}Y$ is then uinitary. $\square$

    Prop.: $\psi\in Sp(2n)$, $\psi$ has a unique decomposition $\psi=PQ$, where $P$ is symplectic and posotive definite while $Q$ is unitary.
    Proof: We have seen $\psi=(\sqrt{\psi^T\psi})(\sqrt{\psi^T\psi})^{-1}\psi$

    • $\sqrt{\psi^T\psi}$ is positive definite, need to show it is also symplectic.
    • $(\sqrt{\psi^T\psi})^{-1}\psi$ is orthogonal and symplectic $\Leftrightarrow$ unitary

    Fact: $(\psi^T\psi)^s=P^s$ is symplectic if $\psi$ is symplectic for all $s\geq 0$.
    Proof the fact: $v\in\text{ker}(\lambda\mathbb{I}-P)$, $w\in\text{ker}(\mu\mathbb{I}-P)$
    If $\lambda\mu=1$, fine ; $\lambda\mu\neq 1$, also fine (i.e. $\omega(v,w)=0$ by previous lemma above).

    Note: $Sp(2n)\times[0,1]\to Sp(2n)$, $(\psi,s)\mapsto P^sQ$ $(\psi=PQ)$
    This map establishes $U(n)$ as a deformation restriction of $Sp(2n)$. This allows to establish $\Pi_1(Sp(2n))=\Pi_1(U(n))=\mathbb{Z}$.

    We now want to apply the correspondence: $\{\text{inner product on}\;V\}\to\{\text{compatible complex structure on}\;(V,\omega)\}$ to symplectic manifolds.

    Def.: An almost complex structure on a smooth manifold $M$ is a smooth map $J:TM\to TM$ s.t. $J|_{T_xM}$ is a complex structure on $T_xM$ and $J^2=-\text{id}$.

    Compare this def. with the following:

    Def.: A complex structure on a smooth manifold $M$ is a smooth atlas $(U_i,\psi_i)$ s.t. coordinatestransition $\psi_i\circ\psi_j^{-1}:\psi_j(I_i\cap U_j)\mapsto\psi_i(U_i\cap U_j)\subset\mathbb{C}\cong\mathbb{R}^{2n}$ is holomorphic.

    Note: A complex structure induces an almost complex structure
    \mathbb{C}^n&\stackrel{{}}{ \underset{\text{holomorphic}}{\xrightarrow{{\;\;\;\;\;\;\;}} }
    However, not every almost complex structure arise in this way (e.g, Nijenhuis tensor).

    Theorem: A symplectic manifold $(M,\omega)$ admits an almost complex structure which is compatible, i.e. $\omega(\cdot, J\cdot)$ defines a Riemannian metric.

    Def.: $(M,\omega,J)$ is Kähler manifold if $\omega$ is symplectic and $J$ is compatible complex structure.

    Remark: If $N\subset (M,\omega,J)$ satisfies $J TN=TN$ for a compatible almost complex structure $J$, then $N$ is symplectic. ($\omega(v,Jv)>0$ if $v\neq 0$)

    Example: Assume $M$ is Kähler and $f:M\to E$ is a holomorphic section of a holomorphic bundle s.t. $f^{-1}(0)$ is Kähler and symplectic submanifold.
    e.g. $M=\mathbb{C}\text{P}^3$, $X=\{[z_0,z_1,z_2,z_3]\in\mathbb{C}\text{P}^3|z_0^4+z_1^4+z_2^4+z_3^4=0\}\subset M$ (Fermat quartic, a K3-surface).

    Fact: Kähler manifolds have even odd-order Betti number, i.e. $b_{2k+1}\in 2\mathbb{Z},\;\forall k\in\mathbb{Z}_{\geq 0}$.

    Example: Thurston, 1976.

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    3楼 10月25日 数学版主, 物理版主, 优秀回答者
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    Exercise Sheet 1



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    4楼 11月3日 数学版主, 物理版主, 优秀回答者
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    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
    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
    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 $\begin{array}{c}
    \end{array}$ 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. $\phi^*_t\Omega_t=\Omega_0$. $\square$


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    Exercise Sheet 2


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    6楼 11月6日 数学版主, 物理版主, 优秀回答者
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    Generating functions for Lagrangian immersion

    Linear algebra version

    Construction of Lagrangian immersion by reduction


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    Exercise Sheet 3



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    8楼 11月13日 数学版主, 物理版主, 优秀回答者
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    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 $F(p)=(H(p),f(p))$
    & & \mathbb{R}\times M\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\
    &\underset{\underset{\;}{\;}}{\stackrel{F\;\;\;\;\;}{\nearrow}} & \;\Bigg\downarrow\scriptstyle{\text{pr}_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\\
    L &\underset{\;\;\;f\;\;\;\;}{\xrightarrow{\;\;\;\;\;\;\;}} & M\;\;\;\;\;\;\;\;\;\;\;\;\;\;
    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 $\text{ker}(dz-\alpha)=(\pi\circ F)_*(T_pL)$.


    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 $X=\pi_*\hat{X}$.
    $\Rightarrow$ $f(p)(\pi_*\hat{X})=\lambda_\text{st.}(\hat{X})$
    $(dH)|_p(Y)=(f^*\lambda_\text{st.})|_p(Y)=\lambda_\text{st.}(f_* Y)=f(p)(\pi_*f_*Y)$, $Y\in T_pL$.

    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.
    \pi^{-1}(U) &\xrightarrow{\underset{(\text{home.})}{\approx}}& U\times\mathbb{R}^k\\
    && U\;\;\;\;\;\;\;\;\;
    This is called the local trivialization.

    (2) If $\varphi_U,\varphi_V$ are bundle charts as in (1), then
    (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\\
    && U\cap V
    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 $p\in L$.


    $\Rightarrow$ 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. $\square$

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


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

    (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 $T^*L$.


    $T^*L\cong TL$ 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 '$-$'.


    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


    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


  9. Phantom_Ghost

    9楼 11月13日 数学版主, 物理版主, 优秀回答者
    8周前Phantom_Ghost 重新编辑

    Exercise Sheet 4


  10. Phantom_Ghost

    10楼 11月13日 数学版主, 物理版主, 优秀回答者



    Theorem(Poincaré-Birkhoff): If $h: A\to A$ is an area preserving homeomorphism satisfying the twist (boundary) condition, then $h$ has at least two fixed points.

    (i) $(r,\varphi)$ polar coordinate on $A$ and $h(r,\varphi)=(r,\varphi+\alpha)$ is area preserving and its lifts to universal cover are translations $\Rightarrow$ $h$ does not satisfy the twist condition and if $\alpha\neq 2\pi\mathbb{Z}$ then there are no fixed points.

    (ii) $h(r,\varphi)=(\rho(r),\varphi+r-\frac{a+b}{2})$ where $\rho(r)$ is a function.


    This satisfies the twist condition and has no fixed points (for $a,b\notin 2\pi\mathbb{Z}$), $h$ is not area preserving.

    (iii) Let $H$ be a Hamilton function on $A$, whose level sets (the trajectories of Hamiltonian flow) are like in the figure bellow:


    assume: along $\partial A$, $\nabla H\neq 0$. Hamiltonian flow satisfies the twist condition, it is area preserving and for short time the Hamiltonian flow has two fixed points (i.e. the critical points of $H$).


  11. 8周前


    11楼 11月17日 数学版主, 物理版主, 优秀回答者

    Exercise Sheet 5