Let $X$ be a compact complex manifold, and let $\phi: \mathcal X \to B$ be a deformation of $X$, i.e. $\phi$ is a proper submersive morphism of complex manifolds, with central fiber $X_0 = X$. By the Ehresmann theorem we shrink $B$ such that $\mathcal X$ becomes a product
$$ \mathcal X = X \times B$$
in the category of $\mathcal C^\infty$-manifolds, and we can even assume that the fibers of the projection $\phi_1: \mathcal X \to X$ are complex submanifolds of $\mathcal X$, even if $\phi_1$ is not holomorphic [1, Prop 9.5]. So after choosing this trivialisation, we get $\mathcal C^{\infty}$-identifications $X \cong X_t$ for $t \in B$. So we may think of $\phi$ not as deformations of the manifold $X$, but rather as deformations of the complex structure on $X$. So for each $t \in B$ we get a complex structure $I_t: T_{X, \mathbb R} \to T_{X, \mathbb R}$ on the tangent bundle $T_{X, \mathbb R}$, which comes from $X_t$. From this we get an associated decomposition
$$ T_{X, \mathbb C} = T_t^{1,0} \oplus T_t^{0,1}. $$
Let's use this data to define a homomorphism
$$ \alpha_t: T_0^{0,1} \to T_0^{1,0}$$
as follows.
First note that if $t$ is nearby $0$, the projection $T_{X, \mathbb C} \to T_0^{0,1}$ induces an isomorphism if restricted to $T_t^{0,1}$
$$ f_t: T_t^{0,1} \xrightarrow{\cong} T_0^{0,1}.$$
Then $\alpha_t$ is defined as the composition of the inverse $f_t^{-1}$ with the inlcusion $T_t^{0,1} \to T_{X, \mathbb C}$, followed by the projection $T_{X, \mathbb C} \to T_0^{1,0}$. So
$$\alpha_t: T_0^{0,1} \xrightarrow{f_t^{-1}} T_t^{0,1} \subset T_{X,\mathbb C} \xrightarrow{\operatorname{pr}} T_0^{1,0}.$$
For $x \in X$ let $E_x$ denote the fiber of any vector bundle $E \to X$ over $x$.
Voisin then claims for all $x \in X$ the equality
$$ T_{t,x}^{0,1} = \{u – \alpha_t(u) \,|\, u \in T_{0,x}^{0,1}\}.$$
My question is if this is correct? I get a different sign:
Suppose $w \in T_{t,x}^{0,1}$, and write $w = u + u'$ for $u \in T_{0,x}^{0,1}$, $u' \in T_{0,x}^{1,0}$. Then by definition, $$\alpha_t(u) = (\operatorname{pr} \circ f_t^{-1})(u) = \operatorname{pr}(u + u') = u',$$ so I think it should be $u + \alpha_t(u)$ above.
Did I make a mistake?
[1] Claire Voisin, Hodge Theory and Complex Algebraic Geometry, I
Best Answer
I've come to the conclusion that Voisin deliberately includes the sign for $\alpha_t$ to make Proposition 9.7 work:
Here $\rho$ is the Kodaira-Spencer map $T_{B,0} \to H^1(X, T_X)$ obtained by the long exact sequence associated to $$0 \to T_X \to T_{\mathcal X}|_X \to \phi^* T_B|_X \to 0.$$