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 [1] 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.

[1] Huybrechts, Complex Geometry, An Introduction

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

[3] Serge Lang, Algebra

  • $\begingroup$ that $1/2$ in the MC equation is definitely standard and the only way how to make it disappear is to divide $\phi$ by $2$ $\endgroup$
    – user8268
    Commented Jun 10, 2021 at 7:51
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 1, 2021 at 14:32


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