November 28, 2023
We show that a compact almost-Kähler four manifold \((M, g, \omega)\) with harmonic self-dual Weyl curvature and constant scalar curvature is Kähler if \(c_{1}\cdot\omega\geq0\). We also prove an integral curvature inequality for compact almost-Kähler four manifolds with harmonic self-dual Weyl curvature.
Let \((M, \omega)\) be a symplectic four-manifold. Then \((M, \omega)\) admits a compatible almost-complex structure \(J\), that is, \(\omega(X, Y)=\omega(JX, JY)\) and \(\omega(v, Jv)>0\) for any nonzero \(v\in TM\) [12]. If we define a metric by \(g(X, Y):=\omega(X, JY)\), then \((M, g, \omega, J)\) is called an almost-Kähler structure. When \(J\) is integrable, \(\omega\) is parallel and this defines a Kähler structure.
The curvature operator of an oriented four-dimensional riemannian manifold decomposes according to the decomposition of \(\Lambda^{2}=\Lambda^{+}\oplus \Lambda^{-}\) as follows.
\[\mathfrak{R}= \LARGE \begin{pmatrix} \begin{array}{c|c} \scriptscriptstyle{W_{+}\scriptstyle{+}\frac{s}{12}}I& \scriptscriptstyle{ric_{0}}\\ \hline\scriptscriptstyle{ric_{0}^{*}}& \scriptscriptstyle{W_{-}\scriptstyle{+}\frac{s}{12}}I\\ \end{array} \end{pmatrix}\]
If \(W_{-}=0\), then \(g\) is called to be self-dual. If \(ric_{0}=0\), then \(g\) is called to be Einstein. A compact almost-Kähler Einstein manifold with nonnegative scalar curvature is Kähler [14]. In a four-dimensional case, it was shown in [8] that a compact almost-Kähler four manifold with \(\delta W_{+}=0\) and nonnegative scalar curvature is Kähler with constant scalar curvature.
Using methods in proofs in [8], we prove two results. First, we show that a compact almost-Kähler four manifold with \(\delta W_{+}=0\) and constant scalar curvature is Kähler if \(c_{1}\cdot\omega\geq 0\). This shows in particular a compact almost-Kähler Einstein four manifold with \(c_{1}\cdot\omega\geq 0\) is Kähler-Einstein.
\(\mathbf{Keywords}\): almost-Kähler, four manifold, self-dual Weyl curvature, Einstein metric
\(\mathbf{MSC}\): 53C21, 53C24, 53C25.
Republic of Korea
Email address: kiysd5@gmail.com
This result is not new since there is a more general result in [4]. On the other hand, our hypothesis \(\delta W_{+}=0\) is weaker than an Einstein condition.
Secondly, we show that for a compact almost-Kähler four-manifold with \(\delta W_{+}=0\), we have \[\int_{M}\frac{s^{2}}{24}d\mu\geq \int_{M}|W_{+}|^{2}d\mu.\] The same result was shown in [10]. Our proof entirely depends on the formulas given in [8]. Using this inequality, it was shown that a compact almost-Kähler four manifold with \(J\)-invariant Ricci tensor is self-dual Kähler-Einstein if the scalar curvature is constant and \(b_{-}=0\) [5]. We include a proof in case of compact almost-Kähler Einstein four manifolds.
Dr\(\breve{a}\)ghici showed that a compact almost-Kähler four manifold \((M, g, \omega)\) with \(J\)-invariant Ricci tensor and constant scalar curvature is Kähler if \(c_{1}\cdot\omega\geq 0\) [4]. In this article, we consider \(\delta W_{+}=\) condition instead of \(J\)-invariant Ricci tensor. Let \((M, g, \omega)\) be a compact almost-Kähler four manifold. Then \(\omega\) is a self-dual harmonic 2-form with length \(||\omega||=\sqrt{2}\) and this gives the following formula. \[0=\Delta\omega=\nabla^{*}\nabla\omega-2W_{+}(\omega)+\frac{s}{3}\omega.\] LeBrun showed that this formula with \(\delta W_{+}=0\) implies the following integral equality [8].
Proposition 1. (LeBrun) Let \((M, g, \omega)\) be a compact almost-Kähler four-manifold with \(\delta W_{+}=0\). Then \[\int_{M}sW_{+}(\omega, \omega)d\mu=8\int_{M}(|W_{+}|^{2}-\frac{1}{2}|W_{+}(\omega)^{\perp}|^{2})d\mu.\]
The proof of Lemma 2 in [8] contains a stronger result that the statement in Lemma 2.
Lemma 1. (LeBrun) Let \((M, g, \omega)\) be an almost-Kähler four manifold. Then \[|W_{+}|^{2}-|W_{+}(\omega)^{\perp}|^{2}\geq\frac{3}{8}[W_{+}(\omega, \omega)]^{2},\] with equality at points where \(W_{+}(\omega)^{\perp}=0\).
Lemma 1 implies in particular that \[|W_{+}|^{2}-\frac{1}{2}|W_{+}(\omega)^{\perp}|^{2}\geq\frac{3}{8}[W_{+}(\omega, \omega)]^{2}.\] Using this in Proposition 1, we get the following result [8].
Proposition 2. (LeBrun) Let \((M, g, \omega)\) be a compact almost-Kähler four-manifold with \(\delta W_{+}=0\). Then \[\int_{M}W_{+}(\omega, \omega)(W_{+}(\omega, \omega)-\frac{s}{3})d\mu\leq 0.\]
Theorem 1. Let \((M, g, \omega)\) be a compact almost-Kähler four-manifold with \(\delta W_{+}=0\) and constant scalar curvature. If \(c_{1}\cdot\omega\geq0\), then the scalar curvature is nonnegative and \((M, g, \omega)\) is a Kähler surface with constant scalar curvature.
Proof. We note that \[s^{*}:=2R(\omega, \omega)=2\left(W_{+}(\omega, \omega)+\frac{s}{12}|\omega|^{2}\right).\] From this, we get \[W_{+}(\omega, \omega)=\frac{s^{*}}{2}-\frac{s}{6}.\] Using this in Proposition 2, we get \[\int_{M}(3s^{*}-s)(s^{*}-s)d\mu\leq 0.\] Suppose \(s\) is negative constant. Then from \[0\leq 4\pi c_{1}\cdot\omega=\int_{M}\frac{s+s^{*}}{2}d\mu,\] we have \[\int_{M}(3s^{*}-s)(s^{*}-s)d\mu=\int_{M}(3(s^{*})^{2}-4s^{*}s+s^{2})d\mu\] \[=\int_{M}(-4s(s^{*}+s)+5s^{2}+3(s^{*})^{2})d\mu\geq 0.\] Thus, we get \(s=0\), which is a contradiction. Thus, \(s\geq 0\). One of the main results in [8] is that a compact almost-Kähler four manifold with \(s^{*}\geq 0\) and \(\delta W_{+}=0\) is a Kähler surface with constant scalar curvature. Using \[0=\frac{1}{2}\Delta|\omega|^{2}+|\nabla\omega|^{2}-2W_{+}(\omega, \omega)+\frac{s}{3}|\omega|^{2} =|\nabla\omega|^{2}-2W_{+}(\omega, \omega)+\frac{s}{3}|\omega|^{2},\] we get \[s^{*}=2\left(W_{+}(\omega, \omega)+\frac{s}{12}|\omega|^{2}\right)=s+|\nabla\omega|^{2}.\] Thus, \(s^{*}-s\geq 0\), with equality on \(M\) if and only if \((M, g, \omega)\) is Kähler. Then if \(s^{*}\geq 0\), we have \[\int_{M}(3s^{*}-s)(s^{*}-s)d\mu=\int_{M}(s^{*}-s)^{2}+2s^{*}(s^{*}-s)d\mu\geq 0.\] Since \(\int_{M}(3s^{*}-s)(s^{*}-s)d\mu\leq 0\) by Proposition 2, we get \(s^{*}\equiv s\) if \(s^{*}\geq 0\). Since \(s^{*}\geq 0\) if \(s\geq 0\), we get a compact almost-Kähler four manifold with \(\delta W_{+}=0\) is Kähler if \(s\geq 0\). We note that \(s\) on a Kähler surface is constant if \(\delta W_{+}=0\). ◻
Since an Einstein metric has \(\delta W_{+}=0\) and constant scalar curvature, we get the following corollary.
Corollary 1. Let \((M, g, \omega)\) be a compact almost-Kähler Einstein four manifold. Suppose \(c_{1}\cdot\omega\geq 0\). Then \((M, g, \omega)\) is Kähler-Einstein.
This result can be also proved from Dr\(\breve{a}\)gichi’s result [4] since an almost-Kähler Einstein four manifold has \(J\)-invariant Ricci tensor and has constant scalar curvature.
Let \((M, \omega)\) be a symplectic four manifold. If \(c_{1}\cdot\omega>0\), then by [15], \(b_{+}=1\). If a symplectic four manifold has \(b_{+}=1\) and \(c_{1}\cdot\omega>0\), then \(M\) is diffeomorphic to a rational or ruled surface [11]. Conversely, there is a following result [9].
Proposition 3. (LeBrun) Let \(M\) be a smooth four-manifold which is diffeomorphic to a rational or ruled surface. Suppose \(2\chi+3\tau\geq0\). Then \(c_{1}\cdot\omega>0\) for any symplectic form \(\omega\) on \(M\).
This gives the following Corollary, which was pointed out in [2].
Corollary 2. Let \(M\) be a smooth four manifold which is diffeomrophic to a rational or ruled surface. Then an almost-Kähler Einstien metric \((M, g, \omega)\) is Kähler-Einstein with positive scalar curvature.
Proof. From \[2\chi+3\tau=\frac{1}{4\pi^{2}}\int_{M}\left(\frac{s^{2}}{24}+2|W_{+}|^{2}-\frac{|ric_{0}|^{2}}{2}\right)d\mu,\] we get \(2\chi+3\tau\geq 0\) for an Einstein metric. By Corollary 1, \((M, g, \omega)\) is Kähler-Einstein since \(c_{1}\cdot\omega>0\) by Proposition 3. For a Kähler surface, we have \[4\pi c_{1}\cdot[\omega]=\int_{M}s d\mu.\] From this, we get \(s>0\). ◻
We prove Proposition 1 in case of compact almost-Kähler Einstein four manifolds using the curvature formula of Blair connection given in [7]. Let \((M, g, \omega)\) be an almost-Kähler four manifold. We consider \[\tilde{\nabla}_{X}Y:=\nabla_{X}Y-\frac{1}{2}J(\nabla_{X}J)Y,\] where \(\nabla\) is Levi-Civita connection. Then \(\tilde{\nabla}\) induces the connection on the anti-canonical line bundle. Then the curvature of \(\tilde{\nabla}\) on the anti-canonical line bundle of an almost-Kähler Einstien four manifold is given by \[iF_{\tilde{\nabla}}=W_{+}(\omega)^{\perp}+\frac{s+s^{*}}{8}\omega+\frac{s-s^{*}}{8}\hat{\omega}.\] The anti-self-dual 2-form \(\hat{\omega}\) is defined on \(U\) and \(|\hat{\omega}|=\sqrt{2}\), where \(U=\{p\in M|s^{*}(p)\neq s(p)\}\).
Corollary 3. Let \((M, g, \omega)\) be a compact almost-Kähler Einstein four-manifold. Then \[\int_{M}\left(\frac{ss^{*}}{8}-\frac{s^{2}}{24}\right)d\mu=\int_{M}(2|W_{+}|^{2}-|W_{+}(\omega)^{\perp}|^{2})d\mu.\]
Proof. For a compact almost-Kähler Einstein four-manifold, we have \[c_{1}^{2}=\frac{1}{4\pi^{2}}\int_{M}(|iF_{\nabla}^{+}|^{2}-|iF_{\nabla}^{-}|^{2})d\mu\] \[=\frac{1}{4\pi^{2}}\int_{M}\left(|W_{+}(\omega)^{\perp}|^{2}+\left(\frac{s+s^{*}}{8}\right)^{2}|\omega|^{2} -\left(\frac{s-s^{*}}{8}\right)^{2}|\hat{\omega}|^{2} \right)d\mu.\]
Using \[c_{1}^{2}=2\chi+3\tau=\frac{1}{4\pi^{2}}\int_{M}\left(\frac{s^{2}}{24}+2|W_{+}|^{2}\right)d\mu,\] the result follows. ◻
Using Seiberg-Witten theory, LeBrun showed the following result [6].
Theorem 2. (LeBrun) Let \((M, g)\) be a compact, Einstein four manifold. Assume \(M\) admits an almost-complex structure \(J\) for which the Seiberg-Witten invariant is nonzero. Suppose \(M\) is not finitely covered by \(T^{4}\). Then with respect to the orientation by \(J\), \(\chi\geq 3\tau\). Moreover, \(\chi=3\tau\) if and only if up to rescaling, the universal cover of \((M, g)\) is complex-hyperbolic 2-space with the standard metric.
The following result was also shown in [10].
Theorem 3. Let \((M, g, \omega)\) be a compact almost-Kähler four-manifold with \(\delta W_{+}=0\). Then \(\int_{M}\frac{s^{2}}{24}d\mu\geq \int_{M}|W_{+}|^{2}d\mu\), with equality if and only if \((M, g, \omega)\) is Kähler with constant scalar curvature.
Proof. By Proposition 1, we have \[\int_{M}|W_{+}|^{2}d\mu=\int_{M}\left((|W_{+}(\omega)^{\perp}|^{2}-|W_{+}|^{2})+\frac{ss^{*}}{8}d\mu-\frac{s^{2}}{24}\right)d\mu.\] Using Lemma 1, we get \[|W_{+}|^{2}-|W_{+}(\omega)^{\perp}|^{2}\geq \frac{3}{8}|W_{+}(\omega, \omega)|^{2}.\] Then we get \[\int_{M}\frac{s^{2}}{24}d\mu-\int_{M}|W_{+}|^{2}d\mu \geq\int_{M}\left(\frac{s^{2}}{24}+\frac{3}{8}|W_{+}(\omega, \omega)|^{2}-\frac{ss^{*}}{8}+\frac{s^{2}}{24}\right)d\mu\] \[=\int_{M}\left(\frac{s^{2}}{12}+\frac{3}{8}\left(\frac{s^{*}}{2}-\frac{s}{6}\right)^{2}-\frac{ss^{*}}{8}\right)d\mu =\int_{M}\frac{9(s^{*}-s)^{2}}{96}d\mu\geq 0.\] Moreover, \(\int_{M}\frac{s^{2}}{24}d\mu=\int_{M}|W_{+}|^{2}d\mu\) if and only if \(s^{*}\equiv s\), which implies \((M, g, \omega)\) is Kähler. ◻
Using this, we get the following result in case of almost-Kähler Einstein four manifolds without using Seiberg-Witten theory.
Corollary 4. Let \((M, g, \omega)\) be a compact almost-Kähler Einstein four-manifold. Then we have \(\chi\geq 3\tau\), with equality if and only if \((M, g, \omega)\) is self-dual Kähler-Einstein.
Proof. From the following formulas, \[\chi=\frac{1}{8\pi^{2}}\int_{M}\left(\frac{s^{2}}{24}+|W_{+}|^{2}+|W_{-}|^{2}-\frac{|ric_{0}|^{2}}{2}\right)d\mu,\] \[\tau=\frac{1}{12\pi^{2}}\int_{M}(|W_{+}|^{2}-|W_{-}|^{2})d\mu,\] we have \[\chi-3\tau=\frac{1}{8\pi^{2}}\int_{M}\left(\frac{s^{2}}{24}-|W_{+}|^{2}+3|W_{-}|^{2}-\frac{|ric_{0}|^{2}}{2}\right)d\mu.\] For an Einstein metric, we have \(\delta W_{+}=0\), and therefore by Theorem 3, we get \[\chi-3\tau\geq0,\] with equality if and only if \((M, g, \omega)\) is Kähler-Einstein and self-dual. ◻
Corollary 5. Let \((M, g, \omega)\) be a compact almost-Kähler Einstein four manifold. Suppose \(b_{-}=0\). Then \((M, g, \omega)\) is self-dual Kähler-Einstein and \(b_{1}=0\).
Proof. By Corollary 4, we have \(\chi\geq 3\tau\). Since \(\chi=2-b_{1}+b_{+}-b_{3}\) and \(\tau=b_{+}\), we have \[\chi-3\tau=2-b_{1}-b_{3}-2b_{+}\geq 0.\] Since \(b_{+}\geq 1\), we get \(b_{1}=b_{3}=0\) and \(b_{+}=1\). Moreover, \(\chi=3\tau\). The result follows from Corollary 4. ◻
Corollary 6. (Satoh [13]) Let \((M, g, \omega)\) be a compact almost-Kähler four manifold with \(\delta W_{+}=0\). Suppose \(\int_{M}s^{2}d\mu=32\pi^{2}(2\chi+3\tau)\). Then \((M, g, \omega)\) is Kähler-Einstein.
Proof. From \(\int_{M}s^{2}d\mu=32\pi^{2}(2\chi+3\tau)\), we get \[\int_{M}\left(2\left(\frac{s^{2}}{24}-|W_{+}|^{2}\right)+\frac{|ric_{0}|^{2}}{2}\right)d\mu=0.\] Since \(\int_{M}\frac{s^{2}}{24}d\mu\geq\int_{M}|W_{+}|^{2}d\mu\), we get \(\int_{M}\frac{s^{2}}{24}d\mu=\int_{M}|W_{+}|^{2}d\mu\) and \(ric_{0}=0\). Then by Theorem 3, \((M, g, \omega)\) is Kähler-Einstein. ◻
Corollary 7. Let \((M, g, \omega)\) be a compact simply connected almost-Kähler Einstein four manifold with degenerate spectrum of \(W_{+}\). Then \((M, g, \omega)\) is Kähler-Einstein.
Proof. By Derdzi\(\acute{n}\)ski’s theorem [3], there exists a Kähler metric \(h\) which is conformally equivalent to \(g\) or \(W_{+}=0\). Suppose \(W_{+}\neq 0\). Since \(L^{2}\)-norm of \(W_{+}\) is conformally invariant, by Theorem 3, we have \[\int_{M}\frac{s^{2}_{g}}{24}d\mu_{g}\geq \int_{M}|W_{+, g}|^{2}d\mu_{g}=\int_{M}|W_{+, h}|^{2}d\mu_{h}=\int_{M}\frac{s^{2}_{h}}{24}d\mu_{h}\] On the other hand, we have \[2\chi+3\tau=\frac{1}{4\pi^{2}}\int_{M}\left(\frac{s^{2}_{g}}{24}+2|W_{+, g}|^{2}-\frac{|ric_{0, g}|^{2}}{2}\right)d\mu_{g}\] \[=\frac{1}{4\pi^{2}}\int_{M}\left(\frac{s^{2}_{h}}{24}+2|W_{+, h}|^{2}-\frac{|ric_{0, h}|^{2}}{2}\right)d\mu_{h}.\] Since \(g\) is Einstein, we have \[\frac{1}{4\pi^{2}}\int_{M}\left(\frac{s^{2}_{g}}{24}+2|W_{+, g}|^{2}\right)d\mu_{g} \leq \frac{1}{4\pi^{2}}\int\left(\frac{s^{2}_{h}}{24}+2|W_{+, h}|^{2}\right)d\mu_{h}.\] Then we get \[\int_{M}\frac{s^{2}_{g}}{24}d\mu_{g}= \int_{M}|W_{+, g}|^{2}d\mu_{g}=\int_{M}|W_{+, h}|^{2}d\mu_{h}=\int_{M}\frac{s^{2}_{h}}{24}d\mu_{h}\] From Theorem 3, we get \((M, g, \omega)\) is Kähler-Einstein.
Suppose \(W_{+}=0\). From \(W_{+}(\omega, \omega)=\frac{s^{*}}{2}-\frac{s}{6}\), we get \(s^{*}=\frac{s}{3}\) for an anti-self-dual almost-Kähler metric. Moreover, there is a point where \(s^{*}=s\) unless \(5\chi+6\tau=0\) [1]. Suppose \(5\chi+6\tau\neq 0\). Then, \(s=0\) since \(s\) is constant. Then by [14], \((M, g, \omega)\) is Kähler-Einstein. Suppose \(5\chi+6\tau=0\). Since \(2\chi+3\tau\geq 0\) for an Einstien manifolds, we get \(\chi\leq 0\). Then, from the following formula, \[\chi=\frac{1}{8\pi^{2}}\int_{M}\left(\frac{s^{2}}{24}+|W_{+}|^{2}+|W_{-}|^{2}-\frac{|ric_{0}|^{2}}{2}\right)d\mu,\] we get \(s=0\). ◻
Let \(M\) be a smooth oriented four manifold which is diffeomorphic to a rational or ruled surface. Suppose \(M\) admits an almost-Kähler structure \((g, \omega)\) and \(2\chi+3\tau\geq0\). Then LeBrun showed the following inequality, \[\int_{M}|W_{+}|^{2}d\mu\geq\frac{4\pi^{2}}{3}(2\chi+3\tau)(M).\] For more general results, we refer to [9]. If \(M\) admits an almost-Kähler Einstein metric, then from this inequality, we get \(\int_{M}\frac{s^{2}}{24}d\mu\leq \int_{M}|W_{+}|^{2}d\mu\). On the other hand, from Theorem 3, we have \(\int_{M}\frac{s^{2}}{24}d\mu\geq \int_{M}|W_{+}|^{2}d\mu\). Thus, we get the equality and therefore, \((M, g, \omega)\) is Kähler-Einstein positive scalar curvature. This gives another proof for Corollary 2.
\(\mathbf{Acknowledgments}\): The author is thankful to Prof. Claude LeBrun for helpful comments.