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Let $X \subset \mathbb{C}^n$ be an analytic set which means that it is the zero locus of holomorphic functions $f_1,f_2,\dotsc,f_n$ on $\mathbb{C}^n$ and suppose that there is a singular locus denoted $\mathrm{Sing}$. Let $\mathcal{O}_X(U)$ be the holomorphic functions on an open set $U$ of $X$ and denote by $\mathcal{O}_X$ the sheaf of the holomorhic function on $X$. We can define the sheaf of Kähler differentials as follow: consider for every open set $U$ in $X$ the $\mathcal{O}_X(U)$-module generated by the set of symbols $\{df : f \in \mathcal{O}_X(U)\}$ which satisfies the relations \begin{gather*} d(fg)=gdf+fdg \\ d(af+bg)=ad(f)+bd(g). \end{gather*} This is the sheaf of Kähler differentials, denoted $\Omega_X^1$. When $\Omega_X^1$ is restricted to $X \setminus \mathrm{Sing}$ it is free of rank $n$ where $n$ is the dimension of X.

Question: Is the vector bundle generated by this sheaf on $X \backslash \mathrm{Sing}$ isomorphic to the cotangent bundle on $X \setminus \mathrm{Sing}$? If not, what is the sheaf we should define on $X$ to get the cotangent bundle on $X \backslash \mathrm{Sing}$?

LSpice
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singularity
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    Yes it is. You can see this directly by choosing a system of local holomorphic coordinates $(z_i)$ on a neighborhood $U$ of a smooth point $x\in X$. Then, you can see that $\Omega_X^1|_U$ is generated by the $dz_i$'s, with no relations – Henri Mar 16 '22 at 14:56
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    No, $\Omega^1_{X\smallsetminus Sing}$ is not (locally) free of rank $n$. See the discussion in this MO question. For instance, $d(e^z)$ is not proportional to $dz$. – abx Mar 16 '22 at 14:57
  • thanks abx, i find in a lot of papers the notion of kahler differential and they link it with the cotangent bundle, maybe they don't use the same definition ? – singularity Mar 16 '22 at 15:02
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    But they are the same for smooth algebraic varieties, right? – Francesco Polizzi Mar 16 '22 at 15:06
  • yes i think it coincides for algebraic varieties but i need kahler differential for complex manifold and i don't find anything – singularity Mar 16 '22 at 15:09
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    For $X$ a complex manifold, you'll find the definition of $\Omega^1_X$ in any introductory book, e.g. Wells (Differential Analysis on Complex Manifolds). Why do you need this algebraic definition of Kähler differentials? – abx Mar 16 '22 at 15:30

1 Answers1

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Question: does the vector bundle generated by this sheaf on X∖Sing is isomorphic to the cotangent bundle on X∖Sing is ? If not what is the sheaf we should define on X to get the cotangent bundle on X∖Sing ?

To answer both questions: Kähler differentials do indeed coincide with the cotangent bundle of X∖Sing X, as long as we take the correct definition of Kähler differentials.

The correct definition does not amount to some explicit formula, but rather defines Kähler differentials as the left adjoint to the square-zero-extension functor from modules to algebras.

An important subtlety here is to choose the correct notion of algebras:

  • In algebraic geometry we take ordinary algebras, which give rise to ordinary derivations and ordinary Kähler differentials. These recover the cotangent bundle of a smooth algebraic variety.

  • In differential geometry we take C^∞-rings, which give rise to C^∞-derivations and C^∞-Kähler differentials. These recover the smooth cotangent bundle of a smooth manifold.

  • In complex geometry, we take entire functional calculus algebras (EFC-algebras), which give rise to EFC-derivations and EFC-Kähler differentials. These recover the holomorphic cotangent bundle of a holomorphic manifold.

For a reference that treats Kähler differentials in this manner, see, for example, Pridham's A differential graded model for derived analytic geometry, Definition 3.16.

Dmitri Pavlov
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