Where is the factor of 2 missing in my calculation?

89 Views Asked by At

I am trying to follow a computation done in Chern's Complex Manifolds without Potential Theory. I find an extra factor of 2 in my computation which differs from the book and the thing I knew.

Let $\theta=C^m/\Gamma$ where $C^m$ is m-dimensional complex plane and $\Gamma$ is a rank 2m lattice. The goal is to see existence of restricted type Kahler structure on $\theta$ under some condition of $\Gamma$. Assume $\Gamma=\langle\pi_1,\dots,\pi_{2m}\rangle$

Restricted type Kahler structure is demanding that Kahler form $\hat{H}\in H^2(\theta,Z)$. Let $\tau_{ij}$ be the torus generated by $C^2/\langle\pi_i,\pi_j\rangle\subset\theta$. Set $\pi_i=(\pi^1_i,\dots,\pi^m_i)$. Then $\tau_{ij}$ is a set of basis for $H^2(\theta,Z)$. Consider Poincare duality $H^2(\theta,Z)\times H_2(\theta,Z)\to Z$ by integration. So to test a closed form's integrality, it suffices to test by Poincare duality.

Assume there restricted Kahler structure $H=\sum_{ij}h_{ij}dz_i\otimes d\bar{z}_j$. Then corresponding Kahler form is given by $\hat{H}=\frac{i}{2}\sum_{ij}h_{ij}dz_i\wedge d\bar{z}_j$. Thus $\int_{\tau_{ij}}\hat{H}\in Z$ for all $i,j$ iff $\hat{H}$ is restricted Kahler structure. WLOG, we can integrate $H$ over $\theta$ to obtain invariant restricted Kahler structure as it is a group. Thus we can assume $h_{ij}$ are all constants.

Then I got, $\frac{i}{2}\int_{\tau_{ij}}\sum_{lm}h_{lm}dz_l\wedge d\bar{z}_m=\frac{i}{2}\sum_{lm}h_{lm}\int_{\tau_{ij}}dz_l\wedge d\bar{z}_m$

$=\frac{i}{2}\sum_{lm}h_{lm}(\pi^i_l\bar{\pi}_m^j-\pi_m^i\bar{\pi}_l^j)\in Z$.

However, the book does not have $\frac{1}{2}$ in computation above.(i.e. $i\sum_{lm}h_{lm}(\pi^i_l\bar{\pi}_m^j-\pi_m^i\bar{\pi}_l^j)\in Z$)

$\textbf{Q:}$ Where did I miss a factor of $2$?

Ref. Chern, Complex Manifolds Without Potential Theory, Sec 7, equation 7.30 pg 65