MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too?

share|cite|improve this question
This appears to me to be a straightforward consequence of the definitions of scalar and holomorphic sectional curvature and of a metric compatible with a complex structure. – Deane Yang Oct 13 '10 at 20:50
On second thought I was careless. I do not see how to prove this and am not sure it is true. You can define something called holomorphic scalar curvature and that is negative. – Deane Yang Oct 14 '10 at 0:51
For the Kahler case you can see this on pages 177-178 of "Complex differential geometry" by Fangyang Zheng. For the non-Kahler case... do you like long calculations with curvature tensors in local coordinates? – Gunnar Þór Magnússon Oct 14 '10 at 5:18
Oh! Hey brother! – Gunnar Þór Magnússon Oct 14 '10 at 5:21
Hey little brother, thank you very much! I go and check ! – diverietti Oct 15 '10 at 16:03
up vote 7 down vote accepted

Here is the answer.

Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=2\sum_{j,k=1}^nc_{jjkk}. $$ So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ It is classically known that $$ \int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n, $$ and $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n. $$ Then, we get $$ \begin{aligned} \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) & =\sum_{j,k=1}^nc_{jjkk}\left(\delta_{jk}\frac 2{n(n+1)}+(1-\delta_{jk})\frac 2{n(n+1)}\right) \\ & = \frac 2{n(n+1)}\sum_{j,k=1}^nc_{jjkk}=\frac 1{n(n+1)}s(x_0), \end{aligned} $$ where we have used the Kähler identity $c_{jklm}=c_{jmlk}$.

Thus, if $\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v)$ is negative, so is its average and we are done.

share|cite|improve this answer
Dear diverietti - The volume containing the article you request can be viewed in 4 libraries near you, the closest being the one at Institut Henri Poincaré. If Liliane is as efficient as what she used to be, she should be able to have the paper brought to you in a few days. – ACL Dec 1 '10 at 7:08
Thanks a lot ACL... It's nice to hear you calling me "Diverietti"... :) – diverietti Dec 1 '10 at 22:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.