Peter Li and Jiaping Wang defined holomorphic bisectional curvature in their paper as follows:
Assume that $M^m$ is a Kahler manifold of complex dimension $m$. Let $ \{e_1, \cdots , e_m\} $ be a unitary frame for the $(1, 0)$-part of the complexified tangent space, $T_x^{1,0}M$. The holomorphic bisectional curvature is denoted by $$R_{\alpha\bar\alpha\beta\bar\beta} = g(R(e_\alpha,e_{\bar \alpha})e_{\bar \beta},e_\beta)$$ for $\alpha, \beta = 1, \cdots , m$.
Another common definition is:
Let $\sigma_1$ and $\sigma_2$ are two planes in $T_xM$ each invariant under $J$. Let $X$ and $Y$ be unit vectors in $\sigma_1$ and $\sigma_2$, respectively. Then the holomorphic bisectional curvature of $\sigma_1$ and $\sigma_2$ is defined by: $${\rm bisec}(\sigma_1, \sigma_2) = g(R(X,JX)JY,Y).$$
My questions is: Why do the two definitions differ by a factor of $2~$? I tried for a long time, but my calculation result was never $2$.
Here is my calculation:
Since $\{e_1, \cdots , e_m\}\subset T_x^{1,0}M$ is a unitary frame, write $e_i=\frac12(X-iJX)$ and $e_j=\frac12(Y-iJY)$, where $X$, $Y\in T_pM$ are unit, then by anti-commutative law $R(X,Y)Z=-R(Y,X)Z$, and $g(R(X,Y)Z, W) = −g(R(X,Y)W, Z)$, we have \begin{equation*} \begin{aligned} 16g(R(e_i,e_{\bar i})e_{\bar j},e_j) & = g(R(X-iJX,X+iJX)(Y+iJY),Y-iJY) \\ & =g(R(X,iJX)Y,-iJY)+g(R(X,iJX)iJY,Y) \\ & \quad +g(R(-iJX,X)Y,-iJY)+g(R(-iJX,X)iJY,Y) \\ & =4g(R(X,iJX)iJY,Y) \\ & =-4g(R(X,JX)JY,Y). \end{aligned} \end{equation*} The result is $\frac{16}{-4}=-4$, not $2$. Any help would be appreciated!
The norm of $X-iJX$ is $\|X-iJX\|^2 = \|X\|^2 + \|JX\|^2 = 2\|X\|^2$. Hence, if $X$ is a unit real vector field, the associated unit complex vector field is $\frac{1}{\sqrt{2}}(X-iJX)$, not $\frac{1}{2}(X-iJX)$. With this proper renormalisation, you should end up with the same tensors.