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This is on p. 37 of The Ricci Flow and the Sphere Theorem by Simon Brendle. The author is going to show that a gradient Ricci soliton on $S^2$ has constant scalar curvature.

Here is the setting: Let $(S^2,g)$ be a gradient Ricci soliton, i.e., there exists $\rho\in\mathbb{R}$ and $f:S^2\to\mathbb{R}$ such that $D^2f=(\rho-\operatorname{scal}_g/2)g$. Let $\xi$ be the gradient vector field of $f$. The author then says:

We assume that $J$ is an almost complex structure on $S^2$ which is compatible with the metric $g$.

I have several questions around this sentence.

Firstly, what does it mean to be compatible? Does it mean that $g(-,J-)$ should be a symplectic form on $g$? In particular, it should be closed as a $2$-form?

Secondly, if my understanding is correct, then why does such an almost complex structure exist?

Thirdly, I am confused because this is used in the next proposition, which says that $J\xi$ is a Killing vector field. The proof the author gave goes thus: $$\begin{aligned}\mathcal{L}_{J\xi}g(X,Y)&=-g(D_X\xi,JY)-g(D_Y\xi,JX)\\&=-D^2f(X,JY)-D^2f(Y,JX)\\&=0.\end{aligned}$$ Here $\mathcal{L}$ denotes the Lie derivative, and $D$ denotes the Levi-Civita connection. However, what I get is $$\mathcal{L}_{J\xi}g(X,Y)=g(D_XJ\xi,Y)+g(D_YJ\xi,X).$$ So how do I get the above formula? Should something similar to $D_XJ=0$ hold?

Yuxiao Xie
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1 Answers1

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$1)$ An almost complex structure $J$ is compatible with a Riemannian metric $g$ if $g(Jv, Jw) = g(v, w)$ for all $v$ and $w$. If that is the case, then $\omega(v, w) = g(Jv, w)$ is a two-form. If this form is closed, then $\omega$ is a symplectic form and $g$ is called a Kähler metric. In real dimension two, $\omega$ is automatically closed.

$2)$ 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 $X$. Moreover, it is compatible with $g$, see this question.

$3)$ For any almost complex structure $J$, we have $DJ = 0$ if and only if $\omega$ is closed and $N_J = 0$ where $N_J$ denotes the Nijenhuis tensor, see this question. That is, $DJ = 0$ if and only if $J$ is integrable and $g$ is Kähler. Every almost complex structure on a surface is integrable, see this question, and every compatible metric is Kähler, so $DJ = 0$ in this case. Therefore,

$$D_X(J\xi) = (D_XJ)(\xi) + J(D_X\xi) = J(D_X\xi)$$

so

\begin{align*} \mathcal{L}_{J\xi}g(X,Y) &= g(D_XJ\xi,Y) + g(D_YJ\xi,X)\\ &= g(JD_X\xi, Y) + g(JD_Y\xi, X)\\ &=g(J^2D_X\xi, JY) + g(J^2D_Y\xi, JX)\\ &= -g(D_X\xi, JY) - g(D_Y\xi, JX) \end{align*}

where we have used compatability of the almost complex structure $J$ with the Riemannian metric $g$ in the third equality.

Michael Albanese
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