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Let $(M,g)$ be a $2n$-dimensional almost Hermitian manifold ($n\geq 2$) with a almost complex structure $\cal J$ (not necessary integrable). i.e., $${\cal J}^2=-I,\quad\qquad g({\cal J} X,{\cal J} Y)=g(X,Y).$$ Suppose that $\{X_i,{\cal J}X_i\}$ be any local orthonormal ${\cal J}$-frame and the following relations hold for $i\neq j$ $$K(X_i,X_j)-K(X_j,{\cal J}X_j)=\frac{1}{2n-2}\{\rho(X_i,X_i)-\rho({\cal J}X_j,{\cal J}X_j)\}\tag{1}$$ $$K(X_i,X_j)-K(X_i,{\cal J}X_j)=\frac{1}{2n-2}\{\rho(X_j,X_j)-\rho({\cal J}X_j,{\cal J}X_j)\}\tag{2}$$ where $\rho$ and $K$ are Ricci curvature tensor and sectional curvature respectively. Then

Can be deduce that $(M,\cal J,g)$ is of constant curvature?

This question comes from the study of conformally flat almost Hermitian manifolds.

Your suggestions will be appreciated.

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    $\begingroup$ The answer is 'no', at least when $2n=4$. In that case, your conditions only involve the Weyl curvature of the underlying metric $g$, so, in particular, if $g$ is conformally flat, then your conditions are satisfied, and there are many conformally flat metrics in dimension $4$ that do not have constant curvature. Then just choose any $g$-compatible almost complex struture $\mathcal{J}$, and you have a counterexample. Perhaps you meant to ask whether it can be deduced that $g$ is conformally flat? (This does not follow immediately since your conditions don't force the Weyl curvature to vanish.) $\endgroup$
    – Robert Bryant
    Commented Sep 6, 2017 at 14:17
  • $\begingroup$ Dear prof. Bryant, many thanks for your comment. May be the case dim $=4$ is a exception (such as Milnor exotic sphere! :) and I didn,t cheacked it. can be extend your conterexample to dim $=5$? That is surprising for me that the condition (2) is not useful for conformally flatness because the right hand side of (2) is $X_i$-free. $\endgroup$
    – C.F.G
    Commented Sep 6, 2017 at 14:42
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    $\begingroup$ I don't understand your comment or what you mean by 'not useful for conformal flatness'. The case $\dim = 5$ cannot happen, since we must be on an even dimensional manifold. The fact is that, in dimension $4$, your conditions only constrain the Weyl curvature of the underlying metric $g$, not the Ricci curvature, but they don't force the Weyl curvature to be zero; it's just required to lie in a certain $2$-dimensional subbundle that is defined by the $\mathrm{U}(2)$-structure given by $(g,\mathcal{J})$ on $M$. If you add that $g$ is Einstein, then, yes, you'll get constant curvature. $\endgroup$
    – Robert Bryant
    Commented Sep 6, 2017 at 14:51
  • $\begingroup$ Can you write your counterexample as a Answer? Please accept my apology for this request and thank you for sharing your wisdom with me. $\endgroup$
    – C.F.G
    Commented Sep 6, 2017 at 15:05

1 Answer 1

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The answer to the question as asked is 'No', at least when $2n=4$. In that case, the conditions (1) and (2) only involve the Weyl curvature of the underlying metric $g$, so, in particular, when $g$ is conformally flat, then the conditions are satisfied, and there are many conformally flat metrics in dimension $4$ that do not have constant curvature.

To construct an example of $(g,\mathcal{J})$ that does not have constant curvature just choose any conformally flat metric $g$ in dimension $4$ that does not have constant sectional curvature and any $g$-compatible almost complex structure $\mathcal{J}$.

Since the Ricci curvature is not constrained, there are probably examples that are not conformally flat, but they do, at least, have to be self-dual, i.e., $W_-$ must vanish (which is implied by (1) and (2)). Meanwhile, the conditions (1) and (2) imply that $W_+$ must take values in a certain rank $2$ subbundle of the self-dual Weyl curvatures that is defined using $\mathcal{J}$: If $\omega$ is the the (self-dual) $2$-form defined so that $\omega(v,w) = g(\mathcal{J}_xv,w)$ for all $v,w\in T_xM$, then the self-dual $2$-forms can be split as $\Lambda^2_+(T^*M) = \mathbb{R}\omega\oplus E$ where $E$ is a bundle of rank $2$. Then (1) and (2) are equivalent to the requirement that the Weyl curvature be self-dual and take values in the rank $2$ subbundle $\omega\circ E\subset S^2\bigl(\Lambda^2_+(T^*M)\bigr)$.

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