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I thought in asking this question on Math StackExchange, but by my experience I don' t think anyone will notice me. Recently, I started studying deformation of complex manifolds in the sense of Kodaira and Spencer, but some things relating infinitesimal approximations are not clear for me.

Let $\varpi: \mathscr{M} \twoheadrightarrow D$ be a analytic family of compact complex manifolds such that $D \in \mathbb{C}^m$ is disk centered at the origin, $M = M_0$ is covered by coordinate charts $\{ U_i \}$ and the radius of $D$ is small enough to $\{ U_i \times D \}$ cover $\mathscr{M}$. Now, let $(\zeta_i^1 (z_i , t), …,\zeta_i^n(z_i^n, t), t^1, …,t^m)$ denote the coordinates of $U_i \times D$, then, in the literature, without any formal justificative, it's stated that $$\frac{\partial \zeta_i^j}{\overline{\partial z_i}^k} = \sum_{l} \varphi_{lk}^i (t) \frac{\partial \zeta_i^j}{\partial z_i^l}$$ holds. The informal justificative usually done is by saying that if $t$ is close enough to $0$, then $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ defines an isomorphism. So it is possible to construct a map $$\varphi(t) = pr^{(1, 0)}\circ (pr^{(0, 1)}|_{T_t^{(0, 1)}})^{-1}: T^{(0, 1)} \longrightarrow T^{(1, 0)} $$ and it satisfies $(1 + \varphi(t)) (v) \in T_t^{(0, 1)}$ for all $v \in T^{(0, 1)}$.

Clearly, $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ being an isomorphism is the same as the complex strutcture being the same for all $t$, so how is possible to make this construction of $\varphi$ more formal? Why the equality above holds?

Maybe the assumption "t close enough to 0" should be formalized in terms of nilpotent elements, something like a kind of prorepresentable functor, such that $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ being an isomorphism makes sense.

Thanks in advance.

EDIT As noted by Peter Dalakov, the assertion that the splitting of $T_{\mathbb{C}}$ does not change, is totally wrong. So the remaining question is: why $$\frac{\partial \zeta_i^j}{\overline{\partial z_i}^k} = \sum_{l} \varphi_{lk}^i (t) \frac{\partial \zeta_i^j}{\partial z_i^l}$$ holds?

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  • $\begingroup$ I gave some links to the literature for a somewhat similar question: mathoverflow.net/questions/99912/… $\endgroup$ – Peter Dalakov Jun 24 '14 at 17:06
  • $\begingroup$ @PeterDalakov Thanks, I have already seen this similar question and I have already studied the references. But they don't exactly solve my question. The condition for "t close enough to 0" stated as the isomorphism of the bundles still making no sense at all. As I said in my question , this is equivalent to the splitting of the complexified tangent bundle doesn't change. $\endgroup$ – user40276 Jun 24 '14 at 22:06
  • $\begingroup$ I am having hard time understanding what the question is. The projection being an isomorphism doesn't mean that the complex structure is "the same". But you seem to be unhappy with the very fact that it is an isom, so I am confused. I would suggest to 1) think about the case of curves 2) Look at the way we introduce charts in grassmannians, e.g., Voisin, vol.1, 10.1.1 Or maybe the question is about the dependence of $\phi$ on $t$: why is it holomorphic, etc? $\endgroup$ – Peter Dalakov Jun 25 '14 at 6:08
  • $\begingroup$ Notice $T^{1,0}$ is not the holomorphic tangent bundle of $M_0$, but rather the underlying smooth complex vector bundle. I.e., as sheaves of $C^\infty$-modules, $T_{M_0}\subset T^{1,0}\subset T_{\mathbb{C}}$ and $T^{1,0}=T\otimes_{\mathcal{O}}C^\infty$ $\endgroup$ – Peter Dalakov Jun 25 '14 at 8:10
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    $\begingroup$ OK, I won't comment more on this. The summands are isomorphic, but are different as subbundles of $T_{\mathbb{C}}$. Think of "rotation of axes" in $\mathbb{R}^2$. $\endgroup$ – Peter Dalakov Jun 25 '14 at 10:19
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I am not sure which book you are using . But from Kodaira, Morrow's book $\it complex$ $\it manifolds$ page 149-151, the construction of vector-value $(0,1)-$form $\varphi$ will tell you why the equality holds, also you will see why $t$ is required to be close enough to 0.

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