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Let $X$ be a complex variety. It is well-known that $X$ is smooth if and only if the sheaf of Kähler differentials $\Omega_X^1$ is locally free (Hartshorne p. 177).

Question: What happens for forms of higher degree? I.e. define $\Omega_X^p := \bigwedge^p \Omega_X^1$ (no reflexive hull or something). For what values of $p$ and under what additional assumptions on $X$ does $\Omega_X^p$ locally free imply $X$ smooth?

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  • $\begingroup$ Can you take the wedge product of a coherent sheaf which is not locally free? $\endgroup$ Dec 30 '13 at 21:34
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    $\begingroup$ When $p=\dim X$ you can easily find examples of varieties such that $K_X=\Omega^p_X$ is free but $X$ is not smooth. Typical examples are Gorenstein varieties (for instance hypersurfaces in $\mathbb{P}^n$, or projective surfaces with only rational double points). $\endgroup$ Dec 30 '13 at 21:41
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    $\begingroup$ Francesco, I think in general you need to take the reflexive hull there, that is, $\omega_X=(\Omega_X^p)^{**}$. $\endgroup$ Dec 31 '13 at 10:47
  • $\begingroup$ Daniel, you can take wedge products of any module. They're just not that nice. $\endgroup$ Dec 31 '13 at 10:49
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    $\begingroup$ Hey Patrick, welcome to mathoverflow! $\endgroup$ Dec 31 '13 at 11:02
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I think that if you do not take reflexive hulls, then all $p\leq \dim X$ should work. Since you said "variety" I assume you mean "reduced". In that case, $\mathscr F$ being locally free is equivalent to $\dim \mathscr F_x\otimes \kappa(x)$ being constant. Since tensor operations commute, it seems to me that a coherent sheaf is locally free if and only if any non-zero exterior power of it is locally free, since you can compute the above value for any exterior power and if that is constant, then the original had to be constant.

If you allow reflexive hulls, then obviously there are varieties with $\omega_X$ being a line bundle. The next question could be whether having all reflexive powers locally free would imply smoothness. Of course, this would follow from the Lipman-Zariski conjecture, so this may be known or easy.

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  • $\begingroup$ Nice answer. It seems to me now that even any $p \leq$ embedding dimension of $X$ should work (which is of course the same thing a posteriori). $\endgroup$
    – pgraf
    Jan 2 '14 at 17:02
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Generically the rank of the coherent sheaf Omega^1_X is the dimension of X (because X is a variety, hence reduced, hence generically smooth over C). So the fibres of Omega^1_X in closed points are vector spaces of dimension at least dim(X) (by upper semi-continuity). The fibres of Omega^2_X are the second wedge powers of the fibres of Omega^1_X (because taking wedge powers commutes with taking fibres). Thus we see that the same thing works for Omega^2_X. (This even works if Omega^2_X is zero; in that case either X is a point or X is a smooth curve.)

But it doesn't work for Omega^3. Why? Well, because a nodal curve will have Omega^3 equal to zero, but a nodal curve is not smooth.

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    $\begingroup$ I would assume that the OP meant $p\leq \dim X$. $\endgroup$ Dec 31 '13 at 10:48
  • $\begingroup$ If $p >$ the embedding dimension of $X$, then obviously $\Omega_X^p = 0$ is free, so we need to assume at least that $p \leq$ embedding dimension of $X$. $\endgroup$
    – pgraf
    Jan 2 '14 at 17:01

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