I am working on:
Prove that the Lie bracket of two symplectic vector fields is a Hamiltonian vector field. More precisely, show that for all $X, Y ∈ \scr X(M, \omega)$ we have
$i_{[X,Y ]}\omega = −dH$ where $H = \omega(X, Y )$
My attempt:
My definition of Hamiltonian and symplectic vector fields are:
A vector field $X \in \scr {X}(M)$, is Hamiltonian if $i_X\omega$ is exact, that is, there exists a smooth map $H : M \to \Bbb R$ such that $i_X\omega = −dH$
A vector field $X \in \scr X(M)$, is symplectic if $i_X\omega$ is closed as a one-form, i.e $d i_X\omega=0$
To prove the statement I will use the following formula
$i_{[X,Y ]}\omega = d(i_Xi_Y \omega) + i_Xdi_Y (\omega) − i_Y di_X(\omega) − i_X(i_Y (d\omega))$
so: since $\omega$ is symplectic, $d\omega$ is closed, $d\omega=0$ and by definition of symplectic vector field $di_Y (\omega)=0$= $di_X (\omega)$
$i_{[X,Y ]}\omega = d(i_Xi_Y \omega)+0+0+0= d(i_Xi_Y \omega)= d i_X\omega(Y,\cdot)=d \omega (X,Y)$
So my problem is that I am missing a minus sign. What am I doing wrong? I suspect the problem could be in $i_Xi_Y \omega= i_X\omega(Y,\cdot)= i_X\omega(X,Y)$. But the insertion operator always inserts into the first slot of the form so how could it be wrong? First Y is inserted into the first slot and then to insert X into the first slot Y is pushed to the right following the definition inLee's book:
You did the interior multiplication in the wrong order.
Since $\omega$ is a 2-form, we have that $i_Y\omega$ is a 1-form characterized by: $$ (i_Y\omega)(V)=\omega(Y,V). $$ Similarly, since $i_Y\omega$ is a 1-form, $i_X(i_Y\omega)$ is a 0-form (i.e., a function) characterized by: $$ i_X(i_Y\omega)=(i_Y\omega)(X). $$ But now, putting these together, we see $$ i_X(i_Y\omega)=(i_Y\omega)(X)=\omega(Y,X)=-\omega(X,Y). $$ In general: if $\theta$ is an $n$-form, then $i_{V_k}\cdots i_{V_1}\theta$ is an $(n-k)$-form characterized by $$ (i_{V_k}\cdots i_{V_1}\theta)(V_{k+1},\ldots,V_n)=\theta(V_1,\ldots,V_n). $$