Differential Forms and the Canonical Bundle

We’re going to need to start out the day with a bit of algebra, because we’re going to talk about differential forms. Once we have forms, we’ll make a sheaf out of them, and then we’ll use this sheaf to construct other things.

We start out by needing the notion of a derivation. Given a ring $R$, an $R$-algebra (that is, a ring which is also an $R$-module) $S$ and an $S$-module $M$, we define an $R$-derivation to be a function $D:S\to M$ such that $D(fg)=fD(g)+D(f)g$ for all $f,g\in S$, $D(f+g)=D(f)+D(g)$, and $D(f)=0$ for all $f\in R$. (We say that an element $f\in S$ is in $R$ if it is of the form $f\cdot 1_S$ for $f\in R$ using the module structure.)

So really, what this is is just a function that acts like taking a derivative. The product rule is there, we think of $R$ as the set of constants, so they have derivative zero, it’s additive, etc. There is, in fact, a universal such derivation. Specifically, there’s an $S$-module called $\Omega_{S/R}$ and a derivation $S\to \Omega_{S/R}$ such that any other derivation is given by composing with a module homomorphism $\Omega_{S/R}\to M$. Even better, we can describe it.

We take the collection of symbols $d(f)$ for $f\in S$ and take formal finite sums of the form $\sum_{k=1}^n g_i d(f_i)$ where $g_i,f_i\in S$. These are the elements, but we do identify some of them. Specifically, we identify $d(g+f)$ with $d(g)+d(f)$, $d(fg)$ with $fd(g)+gd(f)$ and $d(f)$ with zero for $f\in R$. We define the universal derivation to be $d_{S/R}:S\to \Omega_{S/R}$ which takes $f\mapsto d(f)$. We call this module the module of Kähler differentials

To get a bit of a sense of this, we’ll note that if $R=k$ and $S=k[x_1,\ldots,x_n]$, so we’re just looking at affine space, then $\Omega_{k[x_1,\ldots,x_n]/k}$ is $\sum_{i=1}^n f_i dx_i$ where the $f_i$ are polynomials. For those who have done a bit of differential geometry, this should be looking familiar: it is an algebraic analogue of 1-forms.

Now, Kähler differentials have an extremely nice property: they commute with localization. That is, if $S$ is an $R$-algebra and $U\subset S$ is multiplicatively closed we have $\Omega_{U^{-1}S/R}=U^{-1}\Omega_{S/R}$. And by the way, I do apologize for the notation, using $S$ for an $R$-algebra is standard, as far as I know, and in the localization post I used it for a multiplicatively closed set.

So now we must come up with a sheaf theoretic version of this, because varieties are only locally equivalent to rings. Let $X$ be a topological space and $\mathscr{R},\mathscr{S}$ sheaves of rings, with $\mathscr{R}\to\mathscr{S}$ a homomorphism of sheaves of rings. This makes $\mathscr{S}$ into an $\mathscr{R}$-modules. We now define $\mathrm{pre}-\Omega_{\mathscr{S}/\mathscr{R}}$ to be the presheaf taking $U$ to $\Omega_{\mathscr{S}(U)/\mathscr{R}(U)}$. The restriction maps are given by taking $\mathscr{S}(U)\to\mathscr{S}(V)\to\mathrm{pre}-\Omega(V)$, which is then an $\mathscr{R}(U)$-derivation, and so factors through $\mathrm{pre}-\Omega(U)$, giving is the desired restriction maps. Now, once we’ve done all of this, we sheafifiy, and now we have a sheaf which we will call $\Omega_{\mathscr{S}/\mathscr{R}}$, the sheaf of relative differentials.

So now, we can specify to varieties. Let $f:X\to Y$ be a morphism of varieties. By definition we have a map $\mathscr{O}_Y\to f_*\mathscr{O}_X$. But we want the sheaf to be on $X$. So recall that when we talked about morphisms of sheaves, we briefly mentioned the inverse image sheaf. Using this, we get a morphism $f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$, which makes $\mathscr{O}_X$ into a $f^{-1}\mathscr{O}_Y$-module. We define $\Omega_{X/Y}$ to be $\Omega_{\mathscr{O}_X/f^{-1}\mathscr{O}_Y}$, and call it the relative cotangent sheaf.

As a special case, we define $\Omega_X=\Omega_{X/\{pt\}}$ to be the cotangent sheaf of $X$, and as an immediate consequence of the above definition, we can see that if $X\to Y$ is a morphism of affine varieties, then $\Omega_{X/Y}$ is the sheaf associated to $\Omega_{k[X]/k[Y]}$, and this fact implies that $\Omega_{X/Y}$ is always a coherent sheaf.

So now, the cotangent sheaf gives us a new way to test for a point being nonsingular. If $X$ has dimension $r$, and $P\in X$, then $X$ is nonsingular at $P$ if and only if $\Omega_{X,P}$ is isomorphic to $\mathscr{O}_{X,P}^{\oplus r}$. This tells us that $X$ is nonsingular if and only if $\Omega_X$ is a locally free sheaf of rank $r$!

So this gives us a cotangent bundle! Now, we’re going to change notation slightly, and call this bundle (and the sheaf associated to it) $\Omega_X^1$. We can get the sheaf of $k$-forms by looking at $\Omega_X^k=\bigwedge^k\Omega_X^1$, where this is the exterior algebra construction. So in that case, we see that if $X$ has dimension $n$, then $\Omega^{n+1}_X$ is zero, so we only get $n$ of these sheaves. Aside from $\Omega_X^1$, there is another special one among these: $\Omega_X^n$. This one always turns out to be a line bundle, and we denote it by $\omega_X$, because it is so important. In fact, we call it the canonical bundle, and it will be showing up in the future regularly.

Before we stop for the day, we’re going to use $\Omega^1_X$ to define one more object that is very important. We define $\mathscr{H}om(\Omega_X^1,\mathscr{O}_X)={\Omega_X^1}^\vee$, the dual, to be the tangent sheaf $\mathscr{T}_X$. If $\Omega_X^1$ is a vector bundle, so is $\mathscr{T}_X$, and it is called the tangent bundle. We’ve now brought over a lot of valuable objects from differential geometry, and they’ll be quite important in our further study of varieties.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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12 Responses to Differential Forms and the Canonical Bundle

1. Emmit Sisk says:

Hi,
(Omega_{X})^1 is confusing me, perhaps it is so many things at the same time.

Is (Omega_{X})^1 ismorphic to the structure sheaf O_{X}. What map maps one to the other? Thank you.

2. No, $\Omega_X^1$ is not isomorphic to the structure sheaf, except in the case of elliptic curves. The structure sheaf is the trivial line bundle, whereas in the smooth variety case, $\Omega_X^1$ is the cotangent bundle, which is rarely trivial, and of rank equal to the dimension of your base.

3. Emmit sisk says:

I thank you for the quick response.
I think I am understanding what you have just said. In the case of elliptic
curves, how is the correspondence? How do we establish an isomorphism? I
hope I am not boring anybody with my unrigorous trivialities:)
Regards

4. Well, first off, you’ll need to be on a smooth curve so that $\Omega_X^1$ is a line bundle for there to be any hope. So then, it’s just a matter of determining which curves, if any, can have trivial cotangent bundle. For any curve, this bundle has degree $2g-2$, and so the genus must be 1. Now, among degree zero line bundles, there’s a unique one with a global section, so we just need to show that there’s a global section of the cotangent bundle, which is just a 1-form. But any genus 1 curve is a quotient of $\mathbb{C}$ by a lattice of rank 2. The global 1-form $dz$ then descends to the quotient, and so $\Omega^1_X$ has a global section, and is trivial. This is the proof over the complexes, but the result still works in general.

5. Emmit says:

Thi helps Charles, thank you. Simple question such as this one expose my lack of understanding. I think I need to review the the whole ‘section’, no pun intended. Thanks again

6. freddy says:

is there a relation between the cotangent bundle in differential geometry and in algebraic geometry, and is there a relation between differential forms in differential geometry and the stuff here?

7. In a word: yes. If you fix your base field to be $\mathbb{C}$, and only work on complete smooth varieties, then the global algebraic differential forms are the same thing as the global holomorphic differential forms. This is generally true, so in particular, the bundle associated to the sheaf $\Omega_X^1$ is dual to the holomorphic tangent bundle, etc. I recommend that you check out this wikipedia page for a bit more on the GAGA philosophy (and theorems, though I’m not necessarily using anything as high powered as Serre’s GAGA paper) http://en.wikipedia.org/wiki/Algebraic_geometry_and_analytic_geometry

8. Brian says:

Hi!

I’m confused about the tangent bundle and the tangent sheaf.

Let $X$ be a variety over $k$. On one hand, we have the definition of the tangent bundle $\hom(Spec k[\epsilon]/(\epsilon^2), X)$. On the other hand, we have the tangent sheaf, which is $\hom_{O_x}(\Omega_{X/k}, O_x)$.

Could you please explain the relationship between the two?

Thanks so much!
Brian

• Well, you’ll want to focus on the case where $X$ is smooth so that it’s actually a bundle. Then, well, use the bundle-sheaf correspondence. Take an open subset of $X$ and show that a section of the tangent sheaf over $U$ is the same as a map from the open set to the geometric tangent bundle which is a section of the projection.