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Assume $(M,J)$ is a smooth manifold equipped with an almost complex structure. Let $p$ be a point of $M$, and define the linear map $\Theta_p \colon T_pM \to T_pM$ so that $$\Theta_p = -\frac{1}{2}\mathrm{id}_{T_pM} + \frac{\sqrt{3}}{2}J_p.$$ One has that $\Theta_p^3 = \mathrm{id}_{T_pM}$, so $\Theta_p$ is a linear isomorphism of $T_pM$. The goal is to show there exists a neighbourhood of $p$, $U(p)$, and a map $\theta_p \colon U(p) \to U(p)$ such that $\theta_{p*} = \Theta_p$.

I have tried to solve this exercise finding a candidate for $\theta$: locally around $p$ one can assume to have a metric and then a connection, so we can construct an exponential map $\exp_p \colon T_pM \to U(p)$, where $U(p)$ is some neighbourhood of $p$. If $t$ is a real parameter in a sufficiently small open interval containing $0$, then $\exp_p tX_p = \gamma(t,p)$, where $\gamma$ is a curve on $M$ such that $\gamma(0,p) = p$ and $\gamma'(0,p) = X_p$. Hereafter, $Y_p$ is a tangent vector at $p$ and $\alpha(t,p)$ its integral geodesic. Defining $\theta_p := \exp_p \circ \thinspace \Theta_p \circ\exp_p^{-1}\colon U(p) \to U(p)$, I get \begin{align} \theta_{p*}(Y_p) & = \frac{\mathrm{d}}{\mathrm{dt}}\left(\exp_p \circ \thinspace \Theta_p \circ \exp_p^{-1}(\alpha(t,p)) \right)\Bigr\lvert_{t=0} \\ & = \frac{\mathrm{d}}{\mathrm{dt}}\left(\exp_p \circ \thinspace \Theta_p(tY_p) \right)\Bigr\lvert_{t=0} \\ & = \frac{\mathrm{d}}{\mathrm{dt}}\left(\exp_p\left(t\left( -\frac{1}{2}Y_p+\frac{\sqrt{3}}{2}J_pY_p\right)\right)\right)\Biggr\lvert_{t=0} \\ & = -\frac{1}{2}Y_p+\frac{\sqrt{3}}{2}J_pY_p = \Theta_p(Y_p). \end{align} What I am not convinced of is that I am fixing a curve $\alpha$ which is the geodesic associated to some vector $Y_p$. I do not know if this can be an arbitrary curve, and then I am not quite sure I solved the problem. Probably the point is the definition of the exponential map. Other solutions are also welcome, of course.

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