# 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?

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 $$g$$.

• Thank you very much. That was very clear. Jul 26, 2022 at 8:20