Kaehler manifolds with constant sectional curvature

Question

I am reading "Structure on Manifolds" by Yano and Kon, and I don't understand the last step in the proof of this proposition that Kähler manifolds with constant sectional curvature is flat - Yano, Proposition 4.3? How they get this $(2n-1)cX=cX$? Is that Ricci tensor of both sides?

Answer 1

I think the computation we should take a contraction on both sides. On the left hand side this will give the Ricci curvature, while on the other many terms will vanish.

More precisely: fix a point $p\in M$ and an orthonormal basis $\{e_i\}$ for $T_pM$ with respect to the Kähler metric $g$ (that will be written with angled brackets). Using that the sectional curvature is constant equal to $c$ we find \begin{equation} \begin{split} Ric(X,Y)=&\sum_{a=1}^{2n}\left\langle e_i, R(e_i,X)Y\right\rangle=c\sum_a\left(\left\langle X,Y\right\rangle\left\langle e_i,e_i\right\rangle-\left\langle e_i,Y\right\rangle\left\langle e_i,X\right\rangle\right)=\\ =&c(2n-1)\left\langle X,Y\right\rangle. \end{split} \end{equation} In particular, $Ric(X,X)=c(2n-1)\lVert X\rVert^2$, for any tangent vector $X\in T_pM$. On the other hand, in Proposition $4.1$ shows that \begin{equation}\label{eqa} \tag{$\star$} R(X,Y)Y=R(JX,JY)Y. \end{equation} We can plug this in the definition of the Ricci curvature to obtain \begin{equation} \begin{split} c(2n-1)\lVert X\rVert^2=&Ric(X,X)=\sum_a\left\langle e_i, R(e_i,X)X\right\rangle=\\ \mbox{ (using \eqref{eqa}) }=&\sum_a\left\langle e_i, R(Je_i,JX)X\right\rangle=\\ \mbox{ (const. curv. condition) }=&c\sum_a\left\langle e_i,\left\langle JX,X\right\rangle Je_i\right\rangle-c\sum_a\left\langle e_i,\left\langle Je_i,X\right\rangle JX\right\rangle=\\ =&c\left\langle JX,X\right\rangle\sum_a\left\langle e_i,Je_i\right\rangle+c\sum_a\left\langle e_i,JX\right\rangle\left\langle e_i,JX\right\rangle. \end{split} \end{equation} The first term vanishes, as $\left\langle JX, X\right\rangle=\omega(X,X)$. Alternatively, notice that the sum in the first term is just the trace of the complex structure $J$. The second sum instead is $c\lVert JX\rVert^2=c\lVert X\rVert^2$, so that for every vector $X$ we get $c(2n-1)\lVert X\rVert^2=c \lVert X\rVert^2$.