= \omega (u,v) = -\omega (v,u) = -g_1(A_1v,u) = -g_1(u,A_1v).$$ Thus $A_1^*=-A_1$. So $A_1^*A_1=A_1A_1^*=-A_1^2$ is positively definite and self-adjoint $g_1$. Let $Q_1:=\sqrt{-A_1^2}$ be a positive square root of $-A_1^2$. Put $$J_1=A_1Q_1^{-1}.\eqno(1.2.2)$$ $J_1$ is an almost-complex structure. Indeed $$J_1^2=A_1Q_1^{-1}A_1Q_1^{-1}=A_1^2(Q_1^{-1})^2=A_1^2(-A_1^{-2})=-\id .$$ Let us check that on $U_1$ one has $J_1=J$. In fact, on $U_1$ we have $g=g_1$ and so $$\omega (Ju,Jv) = \omega (u,v) = g(A_1u,v) = \omega (A_1u,Jv),$$ and thus $A_1=J$ on $U_1$. So $Q_1=\sqrt{-A_1^2}=\id $. From here we have $J_1=A_1Q_1^{-1}=A_1=J$ on $U_1$.

Remark. Note that at this point of the proof we construct a correspondence $P:g\to J$, which maps a Riemannian metric $g$ to an $\omega $-calibrated a.-c. structure $J$ with $g(u,v)=\omega (u,Jv)
$. This correspondence is obviously a continuous map from the space of metrics to the space of structures. Moreover, note that if $g$ appears as $g=g_J(u,v)=\omega (u,Jv)$ for some $\omega $-calibrated $J$, then $P(g_J)=J$.

Let us now check that $\omega $ is $J_1$-calibrated. First: $$\omega (J_1u,J_1v)=g_1(A_1J_1u,J_1v)=g_1(A_1^2Q_1^{-1}u,J_1v)=
g_1(Q_1u,-J_1v)=$$ $$=-g_1(Q_1u,J_1v)=-g_1(u,Q_1J_1v)=-g_1(u,A_1v)=-g_1(A_1v,u)=-\omega (v,u)=$$ $$=\omega (u,v).$$ Second: $$\omega (u,J_1u)=g_1(A_1u,J_1u)=g_1(A_1u,A_1Q_1^{-1}u)=$$ $$=g_1(u,u)>0$$ for nonzero $u$.

This remark leads to the following

Corollary 1.2.2.

*Let $(X,\omega )$ be as above. Then:*

1\. There exists a $\omega $-calibrating almost-complex structure on $X$.

2\. The space of 