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Jun 24, 2021 at 8:09 comment added Tom @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 $\Gamma(X,A^{0,1}(T_X))$.
Jun 24, 2021 at 4:49 comment added abx 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.
Jun 24, 2021 at 2:15 history asked Tom CC BY-SA 4.0