Suppose $(M,\omega)$ is a symplectic manifold, $J(M)$ is the space of all compatible complex structures. How can we show $J(T^4)$ is homotopic to the space of continuous maps $Map(T^4\rightarrow S^2)$?
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Let $J(\mathbb{R}^4)$ denote the set of almost complex structure on $\mathbb{R}^4$ which preserves the orientation on $\mathbb{R}^4$. Now we fixed this notation as a complex structure on any four dimensional manifold is section of the fiber bundle with fibers as $J(\mathbb{R}^4)$. Thus any section could be seen as a map $T^4 \rightarrow J(\mathbb{R}^4)$. Now it is a result in symplectic geometry that $J(\mathbb{R}^4)$ is homotopy equivalent to the space $S^2$. Thus we get the result.
The proof of the fact that $J(\mathbb{R}^4)$ is homotopy equivalent to the space $S^2$.
$J(\mathbb{R}^4) = GL^+(\mathbb{R}^4)/GL(2,\mathbb{C})$ which is homotopy equivalent to $SO(4)/U(2)$. Now consider an almost complex structure $J$ such that $J \in SO(4)$ then $Je_1 \in \lbrace e_1 \rbrace^{\perp} \cong \mathbb{R}^3$. Also $Je_1 \neq 0$ and $Je_1 \in S^2$. Now choose any vector $u \in span\lbrace e_1, Je_1 \rbrace^{\perp}$. Then $\lbrace e_1,Je_1,u,Ju \rbrace$ completely determine $J$ upto $U(2)$. Thus a vector $Je_1 \in S^2$ determines completely the almost complex structure upto $U(2)$.
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