Complex scalar curvature on Kahler manifold is a half of Riemannian scalar curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:43:46Z http://mathoverflow.net/feeds/question/95883 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95883/complex-scalar-curvature-on-kahler-manifold-is-a-half-of-riemannian-scalar-curvat Complex scalar curvature on Kahler manifold is a half of Riemannian scalar curvature Hee Kwon Lee 2012-05-03T15:55:10Z 2012-05-05T06:33:28Z <p>I want to prove the following statement : Complex scalar curvature $R$ on Kahler manifold is a half of Riemannian scalar curvature $s$</p> <p>This is an exercise. The problem is in [1, p. 63].</p> <p>[1] B. Chow et al, The Ricci flow : techniques and applications part I : geometric aspects, AMS</p> <p>The above problem is about the algebraic relation between scalar curvature and complex scalar curvature.</p> <p><strong>Notation :</strong> $Z_a \doteq \frac{1}{2} (\frac{\partial}{\partial x^a} -i \frac{\partial}{\partial y^a})$ and $\nabla_{Z_a} Z_b \doteq \Gamma_{ab}^c Z_c + \Gamma_{ab}^{\overline{c}} \overline{Z}_c$</p> <p>Assume that $J$ is a complex structure with $\nabla J =0$ and $J \frac{\partial }{\partial x_k} = \frac{\partial }{\partial y_k}$ and $J \frac{\partial }{\partial y_k} = - \frac{\partial }{\partial x_k}$, we have $g_{a\overline{b}} \doteq g(Z_a, \overline{Z}_b) =1/2 (g( \frac{\partial }{\partial x_a} , \frac{\partial }{\partial x_b}) + ig ( \frac{\partial }{\partial x_a} ,\frac{\partial }{\partial y_b}))$ where $g$ is $J$-invariant</p> <p>Let $Rm$ be a extension of Riemannian curvature tensor such that $Rm$ is complex linear.</p> <p>$Rm(Z_a,\overline{Z}_b)Z_c\doteq$</p> <p>$R_{a\overline{b} c}^k Z_k +R_{a\overline{b} c}^{\overline{k}} \overline{Z}_k$</p> <p>Let $R_{a\overline{b}} \doteq R_{c\overline{b}a}^c$ and $R\doteq R_{a\overline{b}} g^{a\overline{b}}$</p> <p><strong>Subquestion 1 :</strong> There exists a proper normal coordinates ${ x_a, y_a }_{a=1}^n$ on Kahler manifold such that ${ z_a = x_a + i y_a }_{a=1}^n$ is a local holomorphic coordinates. If there exists then the above problem is easy</p> <p><strong>Subquestion 2 :</strong> From $g_{a\overline{b}} g^{\overline{b}c} \doteq \delta_{a}^c$ we have $g^{\overline{b}c} =2 ( g^{-1} (dx_b,dx_c) - ig^{-1}(dy_b ,dx_c))$.</p> <p>For convenience $g^{fg} \doteq g^{-1}(df,dg)$</p> <p>In addition complex scalar curvature is $R=4(g^{x_bx_a} -i g^{y_b x_a})(g^{x_hx_\delta} -i g^{y_h x_\delta}) [ R( \frac{\partial }{\partial x_\delta}, \frac{\partial }{\partial x_b} , \frac{\partial }{\partial x_a}, \frac{\partial }{\partial x_h}) - R( \frac{\partial }{\partial y_\delta}, \frac{\partial }{\partial x_b}, \frac{\partial }{\partial y_a}, \frac{\partial }{\partial x_h}) +$</p> <p>$i ( -R( \frac{\partial }{\partial x_a}, \frac{\partial }{\partial x_h}, \frac{\partial }{\partial y_\delta}, \frac{\partial }{\partial x_b}) - R( \frac{\partial }{\partial x_\delta}, \frac{\partial }{\partial x_b}, \frac{\partial }{\partial y_a}, \frac{\partial }{\partial x_h}))]$</p> <p>On the other hand Riemannian scalar curvature $s$ is given by $s = Rc( \frac{\partial }{\partial x_b} , \frac{\partial }{\partial x_a} ) g^{x_a x_b} + Rc( \frac{\partial }{\partial x_b}, \frac{\partial }{\partial y_a} ) g^{y_a x_b}+ Rc( \frac{\partial }{\partial x_b}, \frac{\partial }{\partial y_a} ) g^{x_b y_a} + R( \frac{\partial }{\partial y_b}, \frac{\partial }{\partial y_a} )g^{y_b y_a }$</p> <p>To prove $R = 1/2 s$, I can not solve even though I tried to calculate.</p> <p>Is there a mistake in calculation ? Can we solve through this approach ? Or is there more simple method ?</p>