Let $M$ be a Kaehler manifold with positive holomorphic sectional curvature. then the maximum of sectional curvatures at point $p$ is assumed at the holomorphic planes. I read this claim in Klingenberg's paper "On Compact Kaehlerian Manifolds with Positive Holomorphic Curvature" He refers this result to Berger "Pincement Riemannien et pincement holomorphe"


But I don't have access to Berger's paper nor do I read French, so I ask here to see if this can be proved very quickly or has been written in somewhere else.



Why the sectional curvatures assume maximum on holomorphic planes for positively curved Kaehler manifold?
There is a more enlightening proof of this statement than Berger's calculation and, in fact, it proves something a bit more general. First, a definition: Let $(M,g)$ be a Riemannian $n$-manifold with Riemann curvature tensor $R$ and let $E\subset T\_xM$ be a $p$-plane with orthonormal basis $e\_1,\ldots,e\_p$. Define
$$
\sigma(E) = \sum\_{i,j=1}^p R(e\_i,e\_j,e\_j,e\_i),
$$ 
which is easily seen not to depend on the choice of orthonormal basis.
The quantity $\sigma(E)$ is the *scalar curvature* of $E$. For $p=2$, this is twice the sectional curvature of the $2$-plane $E$, while $\sigma(T\_xM)$ is simply the scalar curvature of $g$ at $x$. In general, the scalar curvature of $E$ is $p(p{-}1)$ times the average of the sectional curvatures of the $2$-planes that lie in $E$.

**Fact:** If $(M,g,J)$ is a Kähler manifold, and $E\subset T\_xM$ is a $4$-plane that is complex (i.e., stable under $J$), then $\sigma(E)$ is $6$ times the average sectional curvature of the complex $2$-planes that lie in $E$. (See below for a proof.)


**Proposition:** If $(M,g,J)$ is a Kahler manifold and $E\subset T\_xM$ is a $4$-plane that is complex and satisfies $\sigma(E)>0$, then the only local maxima of the sectional curvature function on $\mathrm{Gr}\_2(E)$ occur at $2$-planes $L$ that are complex. 


*Remark:* If the holomorphic sectional curvatures of $M$ are positive, then, by the Fact, $\sigma(E)>0$ for all $4$-planes $E$ that are complex, and hence, by the Proposition, the maximum of the sectional curvature on $\mathrm{Gr}\_2(T\_xM)$ is attained only by complex $2$-planes (since each noncomplex tangent $2$-plane lies in some tangent $4$-plane that is complex). Meanwhile, one can have $\sigma(E)>0$ for all $4$-planes that are complex without having positive holomorphic sectional curvature; just consider a Kähler surface with positive scalar curvature that does not have positive holomorphic sectional curvature. Thus, this gives a stronger result than Klingenberg's claim.


To prove the Fact and the Proposition, it clearly suffices to do it for a Kähler manifold of complex dimension $2$ with positive scalar curvature, in which case $E=T\_xM$ for some $x\in M$ that one can suppose fixed for the purposes of this discussion. This is a local argument, so one can choose an orthonormal coframing $\omega\_0,\ldots,\omega\_3$ such that the Kähler form is given by $\Omega=\omega\_0\wedge\omega\_1+\omega\_2\wedge\omega\_3$. Let $\phi\_{ij}=-\phi\_{ji}$ be the connection forms, so that $d\omega\_i=-\phi\_{ij}\wedge\omega\_j$ and, because of the Kähler condition, $\phi\_{02}+\phi\_{31}=\phi\_{03}+\phi\_{12}=0$. Then, by the Bianchi identities for a Kahler metric, one has the following expression for the curvature forms $\Phi\_{ij}=d\phi\_{ij}+\phi\_{ik}\wedge\phi\_{kj}\ $:
$$
\begin{pmatrix}
\Phi\_{01}+\Phi\_{23}\cr
\Phi\_{02}+\Phi\_{31}\cr
\Phi\_{03}+\Phi\_{12}\cr
\Phi\_{01}-\Phi\_{23}\cr
\Phi\_{02}-\Phi\_{31}\cr
\Phi\_{03}-\Phi\_{12}
\end{pmatrix} =
\begin{pmatrix}
3s&0&0&r\_1&r\_2&r\_3\cr
0&0&0&0&0&0\cr
0&0&0&0&0&0\cr
r\_1&0&0&s{+}w\_{11}&w\_{12}&w\_{13}\cr
r\_2&0&0&w\_{12}&s{+}w\_{22}&w\_{23}\cr
r\_3&0&0&w\_{13}&w\_{23}&s{+}w\_{33}
\end{pmatrix} 
\begin{pmatrix}
\omega\_0\wedge\omega\_1+\omega\_2\wedge\omega\_3\cr
\omega\_0\wedge\omega\_2+\omega\_3\wedge\omega\_1\cr
\omega\_0\wedge\omega\_3+\omega\_1\wedge\omega\_2\cr
\omega\_0\wedge\omega\_1-\omega\_2\wedge\omega\_3\cr
\omega\_0\wedge\omega\_2-\omega\_3\wedge\omega\_1\cr
\omega\_0\wedge\omega\_3-\omega\_1\wedge\omega\_2
\end{pmatrix}
$$
Here, $12s$ is the scalar curvature (which is positive by hypothesis) and $w\_{ij}=w\_{ji}$ satisfies $w\_{11}{+}w\_{22}{+}w\_{33}=0$. [The reader may recognize this as the classic presentation of the curvature of a Riemannian $4$-manifold due to Singer and Thorpe; in this case, it has been adapted to the case of a Kähler metric.]


Define functions $u\_i$ and $v\_i$ on the Grassmann bundle $\mathrm{Gr}^+\_2(M)$ of oriented tangent $2$-planes as follows: If $L$ is an oriented $2$-plane with oriented orthonormal basis $(X,Y)$, then 
$$
\begin{pmatrix}
u\_1(L)\cr
u\_2(L)\cr
u\_3(L)\cr
v\_1(L)\cr
v\_2(L)\cr
v\_3(L)
\end{pmatrix} = 
\begin{pmatrix}
\omega\_0\wedge\omega\_1+\omega\_2\wedge\omega\_3\cr
\omega\_0\wedge\omega\_2+\omega\_3\wedge\omega\_1\cr
\omega\_0\wedge\omega\_3+\omega\_1\wedge\omega\_2\cr
\omega\_0\wedge\omega\_1-\omega\_2\wedge\omega\_3\cr
\omega\_0\wedge\omega\_2-\omega\_3\wedge\omega\_1\cr
\omega\_0\wedge\omega\_3-\omega\_1\wedge\omega\_2
\end{pmatrix}
(X,Y).
$$
One then has ${u\_1}^2+{u\_2}^2+{u\_3}^2={v\_1}^2+{v\_2}^2+{v\_3}^2=1$, and the map $(u,v):\mathrm{Gr}^+\_2(M)\to S^2\times S^2$ carries each fiber of $\mathrm{Gr}^+\_2(M)\to M$ diffeomorphically onto $S^2\times S^2$. Note that a $2$-plane $L$ is complex if and only if $u\_2(L)=u\_3(L)=0$, i.e., if and only if $u\_1(L)=\pm1$.

Now, computing from the definitions, one finds
$$
\sigma(L) = 3s\ {u\_1}^2 + 2u\_1(r\_iv\_i) + s + w\_{ij}v\_iv\_j\ .
$$
(The proof of the Fact about the average of sectional curvatures of complex $2$-planes in $E$ follow immediately from this formula.) 

Now, $\sigma(L)$ does not depend on $u\_2$ or $u\_3$, and we can thus think of $\sigma$ restricted to a single fiber $\mathrm{Gr}^+\_2(T\_xM)$ of $\mathrm{Gr}^+\_2(M)\to M$ as a function on $[-1,1]\times S^2$ instead of $S^2\times S^2$. Because $s>0$, when one fixes a $v\in S^2$, the quadratic function of $u\_1$ above cannot attain a local maximum when $-1 < u\_1 <1 $. It follows that the local maxima of $\sigma$ on $\mathrm{Gr}^+\_2(T\_xM)$ must occur only at those $L$ for which $u\_1(L)=\pm1$, i.e., those $L$ that are complex. QED








