Negative holomorphic sectional curvature

Question

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?

Answer 1

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.

Answer 2

Although I'm sure you know this, but as stated, the original question can be misleading. As you have shown, if $\omega$ is a Kähler metric with negative holomorphic sectional curvature, then $\omega$ has negative scalar curvature. If the metric is not Kähler, however, then the (Chern) holomorphic sectional curvature does not control the (Chern) scalar curvature.

If we write $R_{i \bar{j} k \bar{\ell}}$ for the components of the Chern connection in a local frame, then tracing the curvature with the metric yields four Ricci curvatures $\text{Ric}^{(1)} = g^{k \bar{\ell}} R_{i \bar{j} k \bar{\ell}}$, $\text{Ric}^{(2)} = g^{i \bar{j}} R_{i \bar{j} k \bar{\ell}}$, $\text{Ric}^{(3)}= g^{k \bar{j}} R_{i \bar{j} k \bar{\ell}}$, and $\text{Ric}^{(4)} = g^{i \bar{\ell}} R_{i \bar{j} k \bar{\ell}}$. Using the metric to trace again, we see that $\text{Ric}^{(1)}$ and $\text{Ric}^{(2)}$ trace to the same scalar curvature $s$, while $\text{Ric}^{(3)}$ and $\text{Ric}^{(4)}$ trace to $\hat{s}$. In general, $s \neq \hat{s}$, and the holomorphic sectional curvature only controls $s+\hat{s}$. This is essentially what your argument in the Kähler case shows if you don't use the Kähler symmetries :-) Of course, if the metric is balanced $d\omega^{n-1}=0$ and the manifold is compact, then $s = \hat{s}$.