# Is there a reason for defining the differential forms before the vector fields ?

Hi, my question is the following :

In EGA IV chapter 16, given $X$ a scheme over $S$, Grothendieck defines first $\Omega^1_{X/S}$, the $O_{X}$-module of the 1-differentials. He then defines the tangent sheaf : $T_{X/S}:= Hom_{O_X} (\Omega^1_{X/S}, O_X)$, which is equal to $Der(O_X, O_X)$. Why, one does not do the opposite and first define $T_{X/S}:=Der(O_X, O_X)$ and then $\Omega^1_{X/S}:=Hom_{O_X} (T_{X/S}, O_X)$ ?

I suspect there are several reasons for this :

1) This new object, whose definition seems to be easier, is maybe less handy to work with.

2) In some important cases, it gives the wrong object

3) Some other philosophical reason

I would like to have the opinion on this question of mathematicians who know more than me geometry and differential forms.

Thanks.

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You will have to wait for an algebraic geometer to get a precise answer (perhaps you already have one below), but there is good motivation for taking differential forms as more fundamental in differential topology. Specifically, differential forms are easily organized into a cohomology theory while vector fields are not. I suspect, but cannot verify from experience, that one's interest in differential forms on a scheme also stems from their relation to cohomology. – Paul Siegel Jun 3 '10 at 20:20

Unless $X/S$ is smooth, or something close to it, $\Omega_{X/S}^1$ may fail to equal $Hom(T_{X/S},O_X)$. The precise condition should be that $\Omega_{X/S}^1$ is reflexive. So it does appear that differentials are more fundamental.

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I wonder if there are useful conditions for $\Omega^1$ to be reflexive... – Mariano Suárez-Alvarez Jun 3 '10 at 19:36
Yes, but, in the case $X/S$ not smooth, is $\Omega^1_{X/S}$ useful ? – user2330 Jun 3 '10 at 20:07
I think it's useful, but one has to be more careful with it. – Donu Arapura Jun 3 '10 at 20:23
If $X/S$ is not smooth, it is common to consider the cotangent complex rather than $\Omega^1.$ This is a derived version of Kahler differentials that agrees with $\Omega^1$ if $X/S$ is smooth. You can then define the tangent complex to be the dual of the cotangent complex. On a ring theoretic level, this is related to what's known as Andre-Quillen (co)homology. – Mike Skirvin Jun 3 '10 at 20:40

Dear Nicojo, since you mention philosophical reasons let me remark that, since a scheme $(X,\mathcal O_ X)$ is a locally ringed space, the most primitive concepts should be as close as possible to the data, the (generalized) functions encapsulated in the sheaf $\mathcal O_X$ .

At a point $x\in X$ , what could be more natural to consider as the COtangent Zariski space $\mathcal M/\mathcal M^2$ ? It just consists of the functions vanishing at $x$ modulo those vanishing at higher order. And the dimension $d$ of this space will already tell you if the (locally noetherian) scheme $X$ is regular or not at $x$, according as $d=dim\mathcal O_{X,x}$ or $d> dim\mathcal O_{X,x}$

In the relative case $X/S$ the sheaf $\Omega_{X/S}$ will give you a lot of information. Just its nullity at a point already tells you (under a mild finiteness condition) everything about nonramification:

$\Omega_{X/S,x}=0 \iff$ $f:X\to S$ is unramified at $x$

And this is only the beginning: by taking the top exterior product of $\Omega_{X/k}$ you get (in the smooth projective case over an algebraically closed field $k$) the canonical sheaf $\omega_{X/k}$ which is a key concept for Serre duality. This canonical sheaf also plays a fundamental role in the classification of curves, surfaces and higher dimensional varieties, arguably the very heart of classical algebraic geometry . For example to $X$ you associate its canonical ring $R$, a graded ring whose degree $m$ component is $R_m=\Gamma(X,\omega^m)$. The Kodaira dimension of $X$ is $\kappa(X)=trdeg_k (R)-1$ and general varieties are defined as those with $\kappa(x)=dim(X)$. They are supposed to be generic in some sense and have been intensively studied, in particular by the Japanese school (Kodaira, Iitaka, Kawamata, Ueno, Mori, ... )

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Note that the universal property of Kaehler differentials is just

$Hom(\Omega_{X/S},M)=Der_S(\mathcal{O}_X,M)$

so in a certain sense, vector fields do come first.

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