# When are the cotangent and tangent sheaves isomorphic?

Let $$X$$ be an $$S$$-scheme. Under what conditions, if any, is the cotangent sheaf $$\Omega_{X/S}$$ isomorphic to the tangent sheaf $$\Theta_{X/S}$$ as $$\mathcal{O}_X$$- modules? For example, given a nonsingular variety $$V$$ over a field $$k$$, what conditions could $$V$$ satisfy to ensure that such an isomorphism would exist?

@PraphullaKoushik Here are my thoughts: If $$X$$ is a smooth variety of dimension $$n$$ over a field $$k$$, then we know that the $$\Omega_{X/k}$$ is locally free of rank $$n$$, and so is $$\Theta_{X/k}$$. At each point, we could find various isomorphisms between the stalks of $$\Omega_{X/k}$$ and $$\Theta_{X/k}$$ (this just follows because they are free modules over the respective local rings). However, this does not admit globally, i.e. there does not neccesarily exist a global section admitting the desired isomorphism. If we were to remove a particular subset of points from the variety, then we might be able to find such an isomorphism. However, I was curious if there were particular characterizations of the variety for which this is known!

• This'd be a nondegenerate bilinear form on the (co)tangent bundle, so a section of T*(X/S) ⊗ T*(X/S). Not sure if there's more you can say at this level of generality.
– skd
Commented Apr 7, 2020 at 17:13
• If you take arbitrary smooth varieties (over a field, say complex numbers), as you said, any point has a neighborhood where what you say is true, since $\Omega$ and $T$ both will be free, not just locally free. So, it is difficult to answer such a question. So, let us further assume that $X$ is projective. Then, the only varieties satisfying your hypothesis that comes to mind are abelian varieties. Commented Apr 7, 2020 at 19:36
• If $X$ is a smooth surface, then we always have an isomorphism $T_X \cong \Omega_X \otimes \omega_X$, coming from the perfect pairing $\Omega_X\times \Omega_X\to k$. So, $T_X\cong \Omega_X$ if and only if $\omega_X$ is trivial, i.e., for abelian and K3 surfaces. Commented Apr 7, 2020 at 20:22
• On the other hand, for Calabi-Yau threefolds $T_X=\Omega^2_X$, and so a necessary condition for this type of variety is $h^{11}=h^{21}$, ie the Hodge diamond is symmetric. This is quite rare.
– ssx
Commented Apr 7, 2020 at 23:02
• Two remarks: firstly (this hasn't been said yet), it's always true on an abelian variety (more generally, group scheme over $k$), because both are trivial. Secondly, if $T_X \cong \Omega_X$ for $X$ smooth, then taking determinants we conclude that $\omega_X$ is $2$-torsion. In even dimension an argument like @DevlinMallory's (but on $\Omega^{n/2}_X$) shows that $\omega_X$ is in fact trivial. Commented Apr 8, 2020 at 0:40