# Connections, The Kähler Condition, and Curvature – IX. Holomorphic Sectional and Bisectional Curvature

Let ${M}$ be a complex manifold of complex dimension ${n}$ with a Kähler metric ${g_{\alpha\overline{\beta}}}$. Take a real tangent vector ${X}$. By the holomorphic sectional curvature in the direction of ${X}$, we mean the Riemannian sectional curvature for the plane spanned by ${X}$ and ${JX}$. We want to express this holomorphic sectional curvature in terms of local coordiantes. Let

$\displaystyle X = 2\,\text{Re}\left(\xi^\alpha {\partial\over{\partial z^\alpha}}\right).$

Then

$\displaystyle JX = 2\,\text{Re}\left(\sqrt{-1}\xi^\alpha{\partial\over{\partial z^\alpha}}\right).$

We have

$\displaystyle R(X, JX, X, JX) = R(\xi + \overline{\xi},\sqrt{-1}\xi - \sqrt{-1}\,\overline{\xi}, \xi + \overline{\xi}, \sqrt{-1}\xi - \sqrt{-1}\,\overline{\xi})$

$\displaystyle =-4R(\xi, \overline{\xi}, \xi, \overline{\xi}) = -4R_{\alpha\overline{\beta}\gamma\overline{\delta}}\xi^\alpha \overline{\xi^\beta} \xi^\gamma \overline{\xi^i}.$

Now, ${X}$ is perpendicular to ${JX}$ because

$\displaystyle g(X, JX) = 0.$

The length of ${X \wedge JX}$ is simply the square of the length of ${X}$, because ${X}$ and ${JX}$ have the same length. The square of the length of ${X}$ equals ${2g_{\alpha\overline{\beta}}}$. Hence, the holomorphic sectional curvature in the direction of ${2\,\text{Re}(\xi^\alpha (\partial/\partial z^\alpha))}$ is

$\displaystyle -{{R_{\alpha\overline{\beta}\gamma\overline{\delta}}\xi^\alpha \overline{\xi^\beta} \xi^\gamma \overline{\xi^\delta}}\over{(g_{\alpha\overline{\beta}} \xi^\alpha \overline{\xi^\beta})^2}}.$

Since in the case of Kähler manifold the curvature ${\Omega}$ of the complex metric connection of the tangent bundle agrees with the Riemannian curvature, we have

$\displaystyle R_{\alpha\overline{\beta}\gamma\overline{\delta}} \xi^\alpha \overline{\xi^\beta} \xi^\gamma \overline{\xi^\delta} = \langle \Omega(\xi, \overline{\xi})\xi, \xi\rangle.$

Thus the holomorphic sectional curvature of a complex submanifold is more than the corresponding holomorphic sectional curvature of the ambient Kähler manifold. Note that this statement is not true for Riemannian sectional curvatures and Riemannian manifolds, because the Riemannian sectional curvature of the unit sphere in the real Euclidean space is clearly greater than the corresponding Riemannian sectional curvature of the Euclidean space.

The decrease in holomorphic sectional curvature for complex submanifolds holds also for a more general kind of curvature, because it comes from the inequality involving ${\langle \Omega(\xi, \overline{\xi})\eta, \eta\rangle}$. So we want to see what curvature ${\langle \Omega(\xi, \overline{\xi})\xi, \xi\rangle}$ corresponds to. Let

$\displaystyle Y = 2\,\text{Re}\left(\eta^\alpha {\partial\over{\partial z^\alpha}}\right).$

We consider

$\displaystyle R(\xi, \overline{\xi}, \eta, \overline{\eta}) = {1\over{16}}R(X + iJ, X - iJX, Y + iJY, Y - iJY)$

$\displaystyle = -{1\over4}R(X, JX, Y, JY)$

$\displaystyle =-{1\over4}((R(X, Y, JX, JY) + R(X, JY, Y, JX)),$

where for the last equality the first Bianchi identity is used. One can easily check that a ${\mathbb{R}}$-bilinear form ${h(X, Y)}$ on ${T_M}$ satisfies

$\displaystyle h(JX, JY) = h(X, Y)$

if and only if

$\displaystyle h_{\alpha\beta} = h_{\overline{\alpha}\overline{\beta}} = 0$

when expressed in terms of complex basis of ${T_M^{1, 0}}$. Hence,

$\displaystyle R(\cdot,\cdot, JX, JY) = R(\cdot, \cdot, X, Y).$

Hence,

$\displaystyle R(\xi, \overline{\xi}, \eta, \overline{\eta}) = -{1\over4}(R(X, Y, X, Y) + R(X, JY, X, JY)).$

We call ${R(\xi, \overline{\xi}, \eta, \overline{\eta})}$ the holomorphic bisectional curvature in the direction of ${\xi}$ and ${\eta}$ (or in the drection of ${2\,\text{Re}\,\xi}$ and ${2\,\text{Re}\,\eta}$). After suitable normalization, it is equal to the sum of two Riemannian sectional curvatures, one for the plane spanned by ${X}$ and ${Y}$ and the other for the plane spanned by ${X}$ and ${JY}$. This is the reason for the name holomorphic bisectional curvature. The holomorphic bisectional curvature of a complex submanifold is no more than the corresponding holomorphic bisectional curvature of the ambient Kähler manifold.