# Maurer-Cartan integrability condition of an almost complex structure

Suppose $$(X,I)$$ is a complex manifold, and let $$I_t: T_{X, \mathbb R} \to T_{X, \mathbb R}$$ be another almost complex structure on $$X$$. Then both $$I$$ and $$I_t$$ determine decompositions of the complexified tangent bundle $$T^{1,0} \oplus T^{0,1} = T_{X, \mathbb C} = T^{1,0}_t \oplus T^{0,1}_t.$$ Suppose that $$I_t$$ is "a small deformation of $$I$$", in the sense that the projection $$T_{X, \mathbb C} \to T^{0,1}$$ maps $$T^{0,1}_t$$ isomorphically to $$T^{0,1}$$. Then over a point $$x$$ we can write a tangent vector $$v \in (T^{0,1}_t)_x$$ uniquely in the form $$u + \phi_t(u)$$ for $$u \in T^{0,1}$$ and $$\phi_t(u) \in T^{1,0}$$. This defines a homomorphism of bundles $$\phi_t: T^{0,1} \to T^{1,0}.$$ See this question for a further discussion about the definition of $$\phi_t$$ (there called $$\pm \alpha_t$$, but the sign does not matter for now).

We may also think of $$\phi_t$$ as a section $$\phi_t \in \mathcal A^{0,1}(T^{1,0})$$, i.e. a differential form of type $$(0,1)$$ with values in the holomorphic tangent bundle $$T_X = T^{1,0}$$.

Huybrechts  then has the following lemme:

Lemma 6.1.2 The integrability condition $$[T^{0,1}_t, T^{0,1}_t] \subset [T^{0,1}_t]$$ is equivalent to the Maurer-Cartan equation $$\bar \partial \phi_t + [\phi_t, \phi_t] = 0.$$

However, other sources claim the Maurer-Cartan equation to involve a factor $$\frac 1 2$$: $$\bar \partial \phi_t + \frac{1}{2} [\phi_t, \phi_t] = 0,$$ For example Voisin [2, Prop 2.4] and Wikipedia (admittetly the last one has a different context).

Question Is this a mistake by Huybrechts or are there different conventions at play here?

I followed the proof by Huybrechts and tried to pin down the missing factor $$2$$. Omit the $$t$$ for now and write in local coordinates $$\phi_t = \phi = \sum_{i,j} \phi_{ij} d\bar{z_i} \otimes \frac{\partial}{\partial z_j}$$ Then Huybrechts claims

$$\tag{*}\sum_{j,l} \bigg[ \phi_{ij} \frac{\partial}{\partial z_j}, \phi_{kl} \frac{\partial}{\partial z_l} \bigg] = [\phi,\phi] \left(\frac{\partial}{\partial \bar z_i}, \frac{\partial}{\partial \bar z_k} \right)$$ and I think there is a missing factor $$2$$ on the left hand side. So let's try to evaluate the right hand side. First the bracket $$[\phi,\phi]$$ is defined by $$[\phi, \phi] = \sum_{i,j,k,l} d\bar z_i \wedge d\bar z_k \otimes \left[\phi_{ij} \frac{\partial}{\partial z_j}, \phi_{kl} \frac{\partial}{\partial z_l} \right]$$ Now if we plug in $$\left(\frac{\partial}{\partial \bar z_i}, \frac{\partial}{\partial \bar z_k}\right)$$, there are three types of terms to consider. Following [3, XIX, Prop 1.5 / Exercise 3] we get

\begin{align*}\tag{**} (d\bar z_i \wedge d\bar z_k)\left(\frac{\partial}{\partial \bar z_i}, \frac{\partial}{\partial \bar z_k}\right) & = \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, \\ (d\bar z_k \wedge d\bar z_i)\left(\frac{\partial}{\partial \bar z_i}, \frac{\partial}{\partial \bar z_k}\right) & = \begin{vmatrix}0 & 1 \\ 1 & 0 \end {vmatrix} = -1, \\ (d\bar z_n \wedge d\bar z_m)\left(\frac{\partial}{\partial \bar z_i}, \frac{\partial}{\partial \bar z_k}\right) & = \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} = 0 \end{align*} for all $$(n,m) \neq (i,k)$$ and $$(n,m) \neq (k,i)$$. So I think the term $$\left[ \phi_{ij} \frac{\partial}{\partial \bar z_i}, \phi_{kl} \frac{\partial}{\partial \bar z_k}\right]$$ should appear twice in ($$*$$).

My guess is that Huybrechts either forgot the second term of ($$**$$), or sneaks in a factor $$\frac{1}{2}$$ in the evaluation of a wedge product of forms against tangent vectors.

 Huybrechts, Complex Geometry, An Introduction

 Claire Voisin, Sur la stabilité des sous-variétés lagrangiennes des variétés symplectiques holomorphes

 Serge Lang, Algebra

• that $1/2$ in the MC equation is definitely standard and the only way how to make it disappear is to divide $\phi$ by $2$ Jun 10, 2021 at 7:51
• I did the computation and have the same issue. I think it all boils down to whether the wedge product of two one-forms $\eta$ and $\nu$ is defined to be $\eta\otimes\nu - \nu\otimes\eta$ or $\frac{1}{2}(\eta\otimes\nu - \nu\otimes\eta)$. The former is consistent with your computation, while the latter is consistent with Huybrechts'. I'm not sure if Huybrechts is using the second convention or if he just made a mistake. Jul 1, 2021 at 14:32