Existence of metric-compatible almost complex structure on $S^2$ I'm trying to understand this answer, since I'm stuck on the same proof as OP.

$2)$ A two-dimensional surface admits an almost complex structure if and only if it is orientable. If $X$ is an oriented surface, then choosing a Riemannian metric on $X$, we obtain the corresponding Hodge dual $\ast : T^*X \to T^*X$ which satisfies $\ast^2 = -\operatorname{id}_{TX}$. The Riemannian metric also gives rise to an isomorphism $\Phi_g : TX \to T^*X$ which allows us to construct the map $J : TX \to TX$ given by $J = \Phi_g^{-1}\circ\ast\circ\Phi_g$. As $J$ satisfies $J^2 = -\operatorname{id}_{TX}$, it is an almost complex structure on $X$.

As I have never worked with the Hodge operator before, could some one clarify to me how one would check why (if?) the thus obtained almost complex structure $J$ is metric-compatible?
 A: First note that $g$ induces a metric $g^*$ on $T^*X$ given by $g^*(\alpha, \beta) = g(\Phi_g^{-1}(\alpha),\Phi_g^{-1}(\beta))$; compare with this answer. It is with respect to this metric, and a choice of orientation form $\omega$, that we obtain the Hodge dual $\ast : T^*X \to T^*X$ which is defined by $\alpha\wedge\ast\beta = g^*(\alpha, \beta)\omega$. Note that
$$g^*(\ast\alpha,\ast\beta)\omega = \ast\alpha\wedge\ast\ast\beta = \ast\alpha\wedge -\beta = \beta\wedge\ast\alpha = g^*(\beta, \alpha)\omega = g^*(\alpha, \beta)\omega$$
so $g^*(\ast\alpha, \ast\beta) = g^*(\alpha, \beta)$, i.e. $\ast$ is an isometry of $g^*$. More generally, on a $2n$-dimensional oriented Riemannian manifold $(M, g)$, the Hodge dual $\ast : \bigwedge^nT^*M \to \bigwedge^nT^*M$ is an isometry of the metric on $\bigwedge^nT^*M$ induced by $g$.
Now note that
\begin{align*}
g(J(u), J(v)) &= g(\Phi_g^{-1}(\ast(\Phi_g(u))), \Phi_g^{-1}(\ast(\Phi_g(v))))\\
&= g^*(\ast(\Phi_g(u)), \ast(\Phi_g(v))\\
&= g^*(\Phi_g(u), \Phi_g(v))\\
&= g(\Phi_g^{-1}(\Phi_g(u)), \Phi_g^{-1}(\Phi_g(v)))\\
&= g(u, v)
\end{align*}
so $J$ is compatible with $g$.
