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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.

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the paper of Berger is here: numdam.org/numdam-bin/fitem?id=ASNSP_1960_3_14_2_151_0 –  YangMills Dec 24 '12 at 19:01
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the proof that you want is on pages 154-155 there (even if you can't read French, I think you'll have no problem following the proof). Note that some inequalities on page 155 are wrong, and are corrected in the erratum (linked above). –  YangMills Dec 24 '12 at 19:09
    
@yangMills, That's fantastic! Merry Christmas –  Ralph Dec 24 '12 at 22:23

1 Answer 1

up vote 11 down vote accepted

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

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wow!........... –  Gil Kalai Dec 25 '12 at 15:25
    
This is really a holiday gift! –  Ralph Dec 26 '12 at 22:24
    
@Robert, I am confused by "$\sigma(E)$ is 6 times the average sectional curvature of the complex 2-planes that lie in $E$". Suppose $e_1, Je_1, e_2, Je_2$ be a o.n. basis of $E$. Then the $\sigma(E)$ involves all sectional curvatures including the space $L:= span(e_1, e_2)$, not necessary just holomorphic ones $L_1:=span(e_1, Je_1)$ and $L_2:=span(e_2, Je_2)$. Then how to get the average only over complex 2-planes? –  Ralph Dec 26 '12 at 22:45
    
@unknown (google): The complex $2$-planes (oriented positively) that lie in the $4$-plane $E$ are mapped by $(u,v)$ to the subset $\lbrace(1,0,0)\rbrace\times S^2$, and on this set, which can be identified with the $2$-sphere in the $v$-variables, we have $\sigma=4s+2(r_iv_i)+w_{ij}v_iv_j$. (Remember that $\sigma(L)$ is twice the sectional curvature of $L$.) Since $w_{11}+w_{22}+w_{33} =0$, averaging this $\sigma$ over the unit $2$-sphere in $v$-space results in the value $4s$, so it follows that the average scalar curvature of the complex $2$-planes is $2s$. However, $\sigma(E)=12s$. QED –  Robert Bryant Dec 27 '12 at 1:19
    
@unknown (google): Oops! I should have written 'the average sectional curvature of the complex $2$-planes [in $E$] is $2s$'. Sorry about the slip. –  Robert Bryant Dec 27 '12 at 1:39

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