Are the sheaves of $n$-forms well-defined for singular varieties?

Solution 1:

Question: "Are the sheaves of $n$-forms $\Omega_X^n$ always well-defined on $X$? I suspect it, because, in the sense of geometry, given a (topological) manifold that doesn't admit a smooth structure, then its (co)tangent bundles are not well-defined (think about the (co)tangent space of a singular point of an orbifold)."

Answer: If $X/S$ is any scheme (or any algebraic variety in the sense of Hartshorne, CH.I) you may define $\Omega:=\Omega^1_{X/S}$ (Hartshorne, CH.II.8) and since $\Omega$ is a quasi coherent sheaf you may define its $n$'th exterior product $\wedge^n \Omega:=\Omega^n$.

Example: If $X$ is a scheme (or variety) of finite type over a field $k$ and if $X$ is non-regular this sheaf will be a coherent sheaf but it will not be locally trivial.

Example: If $A:=k[x,y]/(f(x,y))$ where $f=x^2-y^3$,let $df:=x^2dx-3y^2dy$. It follows $\Omega:=\Omega^1_{A/k} \cong A\{dx,dy\}/(df)$. This follows from a formula proved in Matsumuras book "Commutative ring theory". Outside of the singular point $(x,y)$ this is locally free of rank one. The rank of $\Omega$ at $(x,y)$ is 2.

Note: The fiber of the cotangent sheaf $\Omega$ at a $k$-rational point $x$ is the cotangent space $\mathfrak{m}_x/\mathfrak{m}_x^2$, and $dim(X) \leq dim_k(\mathfrak{m}_x/\mathfrak{m}_x^2)$. It follows $dim(X)= dim_k(\mathfrak{m}_x/\mathfrak{m}_x^2)$ iff $x$ is a non-singular point. You find this explained in Hartshorne, Chapter I.5 and II.8.

In general if you have a morphism $f: X\rightarrow Y$ of schemes (over $S$) you get an exact sequence

$$f^*\Omega^1_{Y/S} \rightarrow \Omega^1_{X/S} \rightarrow \Omega^1_{X/Y} \rightarrow 0,$$

and the relative cotangent sheaf $\Omega^1_{X/Y}$ is used in the study of the "ramification" of $f$.

Example: If $k$ is an algebraically closed field of char $0$ and if $X,Y$ are of finite type over $k$ and irreducble, it follows $X,Y$ are non-singular iff $\Omega^1_{X/k}, \Omega^1_{Y/k}$ are locally trivial of rank $dim(X), dim(Y)$. In this case $\Omega^1_{X/Y}$ is locally trivial iff $f$ is smooth of relative dimension $d:=dim(X)-dim(Y)$ (Hartshorne, Prop.III.10.4). The notion "smoothness" has to do with the fibers (HH Thm.III.10.2). It is smooth iff it is flat and the fibers are equidimensional and regular.

Many questions have been asked on this site on on derivations, differentials and polynomials. You may find other answers by searching. If you want to understand this topic you should start with polynomials and polynomial rings: How do you calculate the module of Kahler differentials $\Omega^1_{A/k}$ and module of derivations $Der_k(A)$ when $A:=k[x_1,..,x_n]$ is a polynomial ring?

Note: The module of derivations $Der_k(A)$ is the dual $Der_k(A) \cong Hom_A(\Omega^1_{A/k},A)$ and when $A$ is non-regular you lose information. Hence if you want to study "ramification" you must study $\Omega^1_{A/k}$.

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