# Smoothness and Kähler differentials

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?

• Can you take the wedge product of a coherent sheaf which is not locally free? Dec 30, 2013 at 21:34
• 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). Dec 30, 2013 at 21:41
• Francesco, I think in general you need to take the reflexive hull there, that is, $\omega_X=(\Omega_X^p)^{**}$. Dec 31, 2013 at 10:47
• Daniel, you can take wedge products of any module. They're just not that nice. Dec 31, 2013 at 10:49
• Hey Patrick, welcome to mathoverflow! Dec 31, 2013 at 11:02

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.
• 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). Jan 2, 2014 at 17:02
• I would assume that the OP meant $p\leq \dim X$. Dec 31, 2013 at 10:48
• 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 $X$. Jan 2, 2014 at 17:01