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Sheafification

This is a blog post on sheafification. I am broadly going to be following Ravi Vakil’s notes on the topic.

Sheafification is the process of taking a presheaf and giving the sheaf that best approximates it, with an analogous universal property. In a previous blog post, we’ve discussed examples of pre-sheaves that are not sheaves. A classic example of sheafification is the sheafification of the presheaf of holomorphic functions admitting a square root on \Bbb{C} with the classical topology.

Let \mathcal{F} be a presheaf. Then the morphism of presheafs \mathcal{F}\to\mathcal{F}^{sh} is a sheafification of \mathcal{F} if \mathcal{F}^{sh} is a sheaf, and for any presheaf morphism \mathcal{F}\to \mathcal{G}, where \mathcal{G} is a sheaf, there exists a unique morphism \mathcal{F}^{sh}\to \mathcal{G} such that the required diagram commutes. What this means is that \mathcal{F}^{sh} is the “smallest” or “simplest” sheaf containing the presheaf \mathcal{F}.

Because of the uniqueness of the maps, it is easy to see that the sheafification is unique upto unique isomorphism. This is just another way of saying that all sheafifications are isomorphic, and that there is only one (one each side) isomorphism between each pair of sheafifications. Also, sheafification is a functor. This is because if we have a map of presheaves \phi:\mathcal{F}\to \mathcal{G}, then this extends to a unique map \phi':\mathcal{F}^{sh}\to\mathcal{G}^{sh}. How does this happen? Let g:\mathcal{G}\to\mathcal{G}^{sh}. Then g\circ\phi:\mathcal{F}\to\mathcal{G}^{sh} is a map from \mathcal{F} to a sheaf. Hence, there exists a unique map from \mathcal{F}^{sh}\to\mathcal{G}^{sh}, as per the definition of sheafification. Hence, sheafification is a covariant functor from the category of presheaves to the category of sheaves.

We now show that any presheaf of sets (groups, rings, etc) has a sheafification. If the presheaf under consideration is \mathcal{F}, then define for any open set U, define \mathcal{F}^{sh} to be the set of all compatible germs of \mathcal{F} over U. What exactly are we doing? Are we just taking the union of all possible germs of that presheaf? How does that make it a sheaf? This is because to each open set, we have now assigned a unique open set. These open sets can easily be glued, and uniquely too, to form the union of all germs at each point of \mathcal{F}. Moreover, the law of the composition of restrictions holds too. But why is this not true for every presheaf, and just the presheaves of sets? Are germs not defined for a presheaf in general?

A natural map of presheaves sh: \mathcal{F}\to\mathcal{F}^{sh} can be defined in the following way: for any open set U, map a section s\in \mathcal{F}(U) to the set of all germs at all points of U, which in other words is just \mathcal{F}^{sh}(U). We can see that all the restriction maps to smaller sets hold. Moreover, sh satisfies the universal property of sheafification. This is because sh can be extended to a unique map between \mathcal{F}^{sh} and \mathcal{F}^{sh}: the unique map is namely the identity map.

We now check that the sheafification of a constant presheaf is the corresponding constant sheaf. We recall that the constant sheaf assigns a set S to each open set. Hence, each germ at each point is also precisely an element of S, which implies that the sheaf too is just the set of all elements of S: in others words, just S. The stalk at each point is also just S, which implies that this is a constant sheaf.

What is the overall picture that we get here? Why is considering the set of all germs the “best” way of making a sheaf out of a pre-sheaf? I don’t know the exact answer to this question. However, it seems that through the process of sheafification, to each open set, we’re assigning a set that can be easily and uniquely glued. It is possible that algebraic geometers were looking for a way to glue the information encoded in a presheaf easily, and it is that pursuit which led to this seemingly arbitrary method.

Toric Varieties: An Introduction

This is a blog post on toric varieties. We will be broadly following Christopher Eur’s Senior Thesis for the exposition.

A toric variety is an irreducible variety with a torus as an open dense subset. What does a dense subset of a variety look like? For instance, in \Bbb{R} consider the set of integers. Or any infinite set of points for that matter. The closure of that set, under the Zariski Topology, is clearly the whole real line. Hence, a dense set under the Zariski topology looks nothing like a dense set under the standard topology.

An affine algebraic group V is a variety with a group structure. The group operation is given by \phi:V\times V\to V, which is interpreted as a morphism of varieties (remember that the cartesian product of two varieties is a variety). The set of algebraic maps of two algebraic groups V,W, denoted as \text{Hom}(V,W) is the set of group homomorphisms between V and W which are also morphisms between varieties. Are there variety morphisms which are not group homomorphisms? Yes. Consider the morphism f:\Bbb{R}\to \Bbb{R} defined as x\to x+1.

The most important example for us is (\Bbb{C}^*)^n\simeq \Bbb{C}^n-V(x_1x_2\dots x_n). This is the same as removing all the hyperplanes x_i=0 from \Bbb{C}^n. Again, this is the same as V(1-x_1x_2\dots x_n y)\subset \Bbb{C}^{n+1}, which is the same as embedding a variety in a higher dimensional space. The coordinate ring of (\Bbb{C}^*)^n looks like \Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}]\simeq \Bbb{C}[\Bbb{Z}^n]. Why does the coordinate ring look like this? This is because in the ring \Bbb{C}[x_1,x_2,\dots,x_n,y]/(1-x_1x_2\dots x_ny), all the x_i's become invertible (in general, all of the n+1 variables become invertible. However, y can be expressed in terms of the x_i‘s).

A torus is an affine variety isomorphic to (\Bbb{C^*})^n for some n, whose group structure is inherited from that of (\Bbb{C^*})^n through a group isomorphism.

Example: Let V(x^2-y)\subset \Bbb{C}^2, and consider V_{xy}=V\cap (\Bbb{C}^*)^n. We will now establish an isomorphism between \Bbb{C}^* and V_{xy}. Consider the map t\to (t,t^2) from \Bbb{C}^* to V_{xy}. This map is bijective. How? If t is non-zero, then so is each coordinate of (t,t^2). Also, each point in X_{xy} looks like (t,t^2), where t is a non-zero number, and each such point has been mapped to by t\in\Bbb{C}^*. Hence, we have a bijection. How does V_{xy} inherit the group structure of \Bbb{C}^*? By the following relation: (a,a^2).(b,b^2)=(ab,(ab)^2). Remember that V_{xy} had no natural group structure before. Now it has one.

A map \phi:(\Bbb{C}^*)^n\to (\Bbb{C}^*)^m is algebraic if and only if the map \phi^*: \Bbb{C}[y_1^{\pm},y_2^{\pm},\dots,y_m^{\pm}]\to \Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}] is given by y_i\to x^{\alpha_i} for \alpha_i\in \Bbb{Z}^n. In other words, the maps correspond bijectively to lattice maps \Bbb{Z}^m\to \Bbb{Z}^n. What does this mean? The condition that the variety morphism also be a group homomorphism was surely expected to place certain restrictions on the the nature of the nature of the morphism. The way that this condition places restrictions is that a unit can only map to a unit. And the only units in \Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}] are monomials times a constant. Why’s that? Why isn’t an expression of the form x_1+x_2, for instance, a unit? Because \Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}] is not a field! It is just a polynomial ring in which the variables happen to be invertible. Polynomials in those variables need not be! This is not the same as the rational field corresponding to the polynomial ring \Bbb{C}[x_1,x_2,\dots,x_n]. Returning to the proof, the constant is found to be 1, and one side of the theorem is proved. The converse is trivial.

A character of a Torus T is an element \chi\in\text{Hom}(T,\Bbb{C}^*). An analogy that immediately comes to mind is that of a functional on an n-dimensional vector space. Characters are important in studying toric varieties.