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Let $X$ be a compact complex manifold, for arbitrary $\phi_1,\phi_2\in H^1(X,T_X)$, if the Lie bracket $[,]:H^1(X,T_X)\times H^1(X,T_X)\rightarrow H^2(X,T_X)$ always maps $\phi_1,\phi_2$ to zero, i.e.$[\phi_1,\phi_2]=0\in H^2(X,T_X)$, then $X$ admits an unobstructed deformation?

As we know, if we assume $H^2(X,T_X)=0$, then of course the Lie bracket gives a map $[,]:H^1(X,T_X)\times H^1(X,T_X)\rightarrow 0$, and for a Calabi-Yau manifold, by Tian-Todorov lemma and $\partial \bar\partial$-lemma, the Lie bracket also maps $H^1(X,T_X)\times H^1(X,T_X)$ to $0$, so, generally, if we assume that the Lie bracket always maps $H^1(X,T_X)\times H^1(X,T_X)$ to $0$, then can we say $X$ must have an unobstructed deformation?

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    $\begingroup$ No, this just tells you that there is no first order obstruction to deform $X$. If you can read french, I recommend this seminar lecture by Douady, in particular the counter-example at the end. $\endgroup$
    – abx
    Commented Jun 24, 2021 at 4:49
  • $\begingroup$ @abx, I think I have made a stupid mistake, for $\phi=\phi_1 t+\phi_2 t^2+\phi_3 t^3+...$, I wrongly think all the $\phi_i \in H^1(X,T_X)$, but actually, except for $i=1$, $\phi_i$ only in $\Gamma(X,A^{0,1}(T_X))$. $\endgroup$
    – Tom
    Commented Jun 24, 2021 at 8:09

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