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Questions in category: 辛流形上的 Hodge 结构 (Hodge Structure on Symplectic Manifolds).

## [Dong Yan|1996]Hodge structure on symplectic manifolds[Section 1]

Posted by haifeng on 2013-01-06 15:03:02 last update 2013-01-22 16:15:30 | Answers (0) | 收藏

## 1. $\Omega^*(M)$ 上的算子

$*:\ \Omega^k(M)\rightarrow\Omega^{2m-k}(M),\quad\forall\ 0\leqslant k\leqslant 2m,$

$\beta\wedge *\alpha=\wedge^k(G)(\beta,\alpha)v_M,\quad\forall\ \alpha,\beta\in\Omega^k(M),$

$\omega=dp_1\wedge dq_1+\cdots+dp_m\wedge dq_m,$

$G=\frac{\partial}{\partial q_1}\wedge\frac{\partial}{\partial p_1}+\cdots+\frac{\partial}{\partial q_m}\wedge\frac{\partial}{\partial p_m}.$

$\wedge^k(G)=k!\sum_{i_1 <\cdots < i_k}\frac{\partial}{\partial q_{i_1}}\wedge\frac{\partial}{\partial p_{i_1}}\wedge\cdots\wedge\frac{\partial}{\partial q_{i_k}}\wedge\frac{\partial}{\partial p_{i_k}}$

$d^*=d\circ i(G)-i(G)\circ d=[d,i(G)].\tag{1}$

$\begin{array}{rcl} L:\ \Omega^k(M)&\rightarrow&\Omega^{k+2}(M)\\ \alpha&\mapsto& L(\alpha):=\omega\wedge\alpha. \end{array}$

$[L,d]=L\circ d-d\circ L$. 对任意 $\alpha\in\Omega^k(M)$, 有
$\begin{split} &L\circ d\alpha-d\circ L(\alpha)\\ =&\omega\wedge d\alpha-d(\omega\wedge\alpha)\\ =&\omega\wedge d\alpha-[d\omega \wedge\alpha+(-1)^2\omega\wedge d\alpha]\\ =&-d\omega\wedge\alpha\\ =&0. \end{split}$

Lemma 1.1. 若 $\alpha\in\Omega^k(M)$ ($0\leqslant k\leqslant 2m$), 则

$[L,i(G)]\alpha=\omega\wedge i(G)\alpha-i(G)(\omega\wedge\alpha)=(k-m)\alpha.$

Lemma 1.2. $[L,d^*]=-d$.
Pf. 设 $\alpha\in\Omega^k(M)$. 根据恒等式 (1) 和 Lemma 1.1, 我们有
$\begin{split} [L,d^*]\alpha &=Ld^*\alpha-d^*(\omega\wedge\alpha)\\ &=L\bigl(d\circ i(G)\alpha-i(G)(d\alpha)\bigr)-\bigl(d\circ i(G)(\omega\wedge\alpha)-i(G)d(\omega\wedge\alpha)\bigr)\\ &=\omega\wedge(d\circ i(G)\alpha)-\omega\wedge\bigl(i(G)(d\alpha)\bigr)-d\circ i(G)(\omega\wedge\alpha)+i(G)d(\omega\wedge\alpha)\\ &=d\bigl(\omega\wedge i(G)\alpha\bigr)-\omega\wedge\bigl(i(G)(d\alpha)\bigr)-d\bigl(i(G)(\omega\wedge\alpha)\bigr)+i(G)(\omega\wedge d\alpha)\\ &=d\bigl(\omega\wedge i(G)\alpha-i(G)(\omega\wedge\alpha)\bigr)-\bigl(\omega\wedge i(G)(d\alpha)-i(G)(\omega\wedge d\alpha)\bigr)\\ &=d(k-m)\alpha-(k+1-m)d\alpha\\ &=-d\alpha. \end{split}$

$L^*:\ \Omega^k(M)\rightarrow\Omega^{k-2}(M)$

Corollary 1.3. $[L^*,d^*]=0$, $[L^*,d]=d^*$.

Pf. 若将 $[L^*,d^*]$ 和 $[L^*,d]$ 限制在 $\Omega^{k-2}(M)$ 上, 则我们有
$[L^*,d^*]=(-1)^{k+1}*[L,d]*=0.$

$\begin{split} [L^*,d]&=L^*\circ d-d\circ L^*\\ &=(-1)*L*d-d(-1)*L*\\ &=*(-L*d*+*d*L)*\\ &=(-1)^{2m-k}*\bigl((-1)^k *d*L-(-1)^k L*d*\bigr)*\\ &=(-1)^{2m-k}*[d^*, L]*\\ &=(-1)^k *d*\\ &=d^* \end{split}$