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I'm struggling to prove that the sections for an affine open set $U$ of the cotangent sheaf $\Omega_X^1$ are isomorphic to the Kähler-differential $\Omega_{\mathcal{O}(X)\vert\mathbb{C}}$.

First I want to describe my case and my assumptions. I consider a complex analytic manifold $X$ with dimension $n$. Like in Claude Sabbahs book "Isomonodromic Deformations and Frobenius manifolds" chapter 0.2. He defines the tangent sheaf $\Theta_X$ to be the sheaf of derivations (chapter 0.9) and the sheaf of $1$-forms (cotangent sheaf) $\Omega_X^1$ to be $\mathcal{Hom}_{\mathcal{O}_X}(\Theta_X,\mathcal{O}_X)$.

Now I know that $\Theta_X\cong (\mathrm{Der}_\mathbb{C}(\mathcal{O}_X))^\sim$ (quasi-coherent sheaf). I also know that $\Omega_X^1(U)\cong \Omega_{\mathcal{O}_X(U)\vert\mathbb{C}}$ for a affine open subset $U$ (see https://stacks.math.columbia.edu/tag/01UM Lemma 29.32.5), but they use another definition of $\Omega_X^1$. So I think I should get the same result with my definition.

I showed (with Claude Sabbahs definitions) that $\Omega_X^1$ and $\Theta_X$ are locally free $\mathcal{O}_X$-modules of rank $n$. Also I showed that then $$\Omega_X^1{^\vee}\cong\Theta_X.$$ Now I wanted to show that for an affine open subset $U$ of $X$, $\Omega_X^1(U)$ fulfills the universal property of $\Omega_{\mathcal{O}_X(U)\vert\mathbb{C}}$, so there exists for a arbitrary $\mathcal{O}_X(U)$-module $M$ and a derivation $\alpha:\mathcal{O}_X(U)\to M$ a unique $A-$linear $\phi:\Omega_X^1(U)\to M$ such that $\phi\circ d=\alpha$ with $$d:\mathcal{O}_X(U)\to \Omega_X^1(U), \quad (f\mapsto (\theta\mapsto \theta(f)).$$ I thought its possible to show this via the following argument: $$\mathrm{Hom}_{\mathcal{O}_X(U)}(\Omega_X^1(U),M)=\Omega_X^1(U)^{\vee}\otimes M\cong\Theta_X(U)\otimes M=\mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U))\otimes M$$ Here the iso is given because I've shown that $\Omega_X^1$ is a locally free $\mathcal{O}_X(U)$-module. Finally I want to have $\mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U))\otimes M\cong \mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U),M)$ to finish my argument. But I don't know either if it'ss right or not, and if it's right how to prove it. I've constructed a map $$\mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U))\otimes M\to \mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U),M)\quad (\gamma\otimes m\mapsto (y\mapsto \gamma(y)\cdot m)),$$ but I don't know how to go on. Is there a way to prove that $$\mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U),M)\cong \mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U))\otimes M$$ or is this just the wrong 'ansatz'?

Martin Brandenburg
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nico_r
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Feb 25 '25 at 10:18
  • @nico_r: You write the following: $\Theta_X\cong (\mathrm{Der}_\mathbb{C}(\mathcal{O}_X))^\sim$. What do you mean by this? – hm2020 Feb 25 '25 at 10:25
  • sorry that was a bit unclear I meant that $\Theta_X$ is the quasi-coherent sheaf with $\Theta_X(U)\cong \mathrm{Der}_\mathbb{C}(\mathcal{O}_X(U)$ for every affine open subset $U$ of $X$. – nico_r Feb 25 '25 at 10:29
  • @nico-r: It seems to me your field is complex analysis and complex manifolds. In algebraic geometry we interpret the tangent sheaf in terms of derivations of the structure sheaf. If your manifold is projective it is algebraic and hence you may use this interpretation. If your manifold is not algebraic you should post this question to complex analysts/geometers - they may give you a good response. – hm2020 Feb 25 '25 at 11:24

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