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Let $(M,\omega)$ is a symplectic manifold, and you can assume it is compact if necessary. Denote by $\mathcal J^k = \mathcal J^k(M, \omega)$ the set of all $\omega$-compatible almost complex structures of class $C^k$.

How can we see that the tangent space $T_J\mathcal J^k$ at $J$ consists of $C^k$-sections $Y$ of the bundle $End(TM,J,\omega)$ whose fiber at $x\in M$ is the space of linear maps $Y:T_xM\to T_xM$ such that $$JY+YJ=0,~~~~~~~~\omega(Yv,w)+\omega(v,Yw)=0$$ ? One intuitive way is to consider a "curve" $J_t=J+tY+o(t)$, and plug in this to $J_t^2=1$ and $J_t^*\omega(u,v)=\omega(u,v)$.

However, I prefer a rigorous way using explicit local coordinates of the space $\mathcal J^k$. Please help, thanks!

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  • $\begingroup$ See Section 2.2 of Wendl's notes (mathematik.hu-berlin.de/~wendl/pub/jhol_bookv33.pdf). Really your question is Exercise 4.4.6, but it reduces to Proposition 2.2.17 and Exercise 2.2.18. Hint: Use the map $Y\mapsto (1+\frac{1}{2}J_0Y)J_0(1+\frac{1}{2}J_0Y)^{-1}$ for any reference $J_0$. $\endgroup$ – Chris Gerig Nov 22 '16 at 3:09
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An almost complex structure gives you a reduction of the frame bundle $LM$ from $Sp(2n)$ to $U(n)$. Hence the space $\mathcal{J}$ of all almost complex structures compatible with the symplectic form is the space of sections of $LM \times_{Sp} Sp(2n) / U(n)$.

By general (non)sense, the tangent space of the section space $\Gamma(P \times_G F)$ at a section $\phi$ is $\Gamma(\phi^* vertical) = \Gamma(\phi^* (P \times_G TF))$. Now, in your case $P = LM$, $G = Sp(2n)$ and typical fibre $F = Sp(2n) / U(n)$. Unfolding the hidden identifications and using the explicit description of the tangent space to $Sp(2n)/U(n)$ gives you the identification $J^* (LM \times_{Sp} Sp(2n)/U(n)) = End(TM, J, \omega)$.

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