Now, for to be an exotic -sphere, must satisfy two properties: (1) is not diffeomorphic to , and (2) is homeomorphic to . This at first seems very counterintuitive- how can we possibly find a homeomorphism between smooth manifolds that is both nice enough to write down, but also not a diffeomorphism?

We know that any diffeomorphism of smooth manifolds naturally induces an isomorphism of tangent bundles , as gives us an isomorphism of tangent spaces at each point. Thus, we want but .

It is well known that for an -dimensional manifold, if there is a smooth map $f: M \to \mathbb R$ such that has exactly two critical points, (both of which are nondegenerate) then is homeomorphic to an -sphere. This is the criterion that we will use to check that our constructed manifold is homeomorphic to .

We now consider manifolds that are given as -sphere bundles over . These are classified by maps , in other words, by elements of . Note that this is isomorphic to , by the following correspondence: for each , we have the map , where quaternion multiplication is understood on the right. Let be the $3$-sphere bundle corresponding to a given pair . If we let be the standard generator for , it turns out that the Pontryagin class .

For be such that , define to be the total space of the bundle . Then, it turns out that satisfies our previous condition for being homeomorphic to . Furthermore, given the Pontryagin class, we find that , and therefore is not diffeomorphic to the standard -sphere, as desired.

While I only defined direct and inverse limits for totally ordered collections of objects, we can do the same for partially ordered collections as well, (where we say if there is a morphism ) by still defining the direct limit to be the universal object receiving morphisms from all of the objects in a way that is compatible with the given morphisms. For example, letting , and letting be the inclusion morphism if , and having no morphism from to otherwise, we have , as we would expect. In a similar manner, we can extend the definition of inverse limit to partially ordered systems as well (here, we define the ordering on the collection to be if there is a morphism ).

Let be some topological space, and let be a sheaf on . Given a point , we define the stalk of at , to be . If is a smooth manifold and is the sheaf of functions on , then is the ring of all locally convergent power series around . If is the sheaf of continuous real-valued functions on , then is the ring of all functions that are continuous in some open neighborhood of .

A morphism of presheaves on some space is simply a collection of morphisms for each open subset that commute with the restriction morphisms. Because all of the morphisms commute with the restriction morphisms, it is clear that this induces morphisms of stalks for each point $p$.

One may wonder whether if given an arbitrary presheaf, there is a canonical sheaf associated to it. The answer turns out to be yes. Recalling our discussion of presheaves in the last post, it is clear that we should be able to define the notion of a stalk for a presheaf as well, as neither of the two sheaf axioms are involved. In fact, if we consider all presheaves that have the same stalks at all points, it turns out that exactly one of them is a sheaf, which we will define as the sheaf associated to any of them. In other words, for sheaves, the set of stalks contains all of the original information of the sheaf, as we can use this to reconstruct it.

To define sheafification in a nice manner, we will first need to discuss morphisms. Fortunately, the category of sheaves is a full subcategory of the category of presheaves, meaning that in this case, if we have a morphism of two presheaves such that are actually sheaves, then is also a morphism of sheaves. Because of this, it will never be ambiguous what we mean by a morphism.

Given two presheaves on a topological space , a morphism between them is a collection of morphisms for every open subset , such that they commute with the restriction maps for .

Now, given a presheaf , we define , the sheafification of as the unique (up to unique isomorphism) sheaf, along with the sheafification morphism satisfying the following universal mapping property: for any sheaf and a morphism , there is a unique morphism such that $f = g \circ \psi$.

For example, suppose that is the constant sheaf that assigns to each open subset of . Then, is the sheaf that assigns to the open subset if is a disjoint union of exactly $n$ connected components.

Luckily, most of our immediate examples of presheaves are already sheaves. However, the presheaf quotient of two sheaves is not necessarily a sheaf, so when we refer to the cokernel of a map of sheaves, we will mean the sheafification of the presheaf cokernel, which is in fact the correct definition of the cokernel in the category of sheaves (you should convince yourself that this is true).

First, we will define the category of open sets associated to a topological space . The objects of , as you may have guessed, are simply the open sets of . If are open subsets, then we say that

Let be any category. A presheaf of objects of is simply a contravariant functor . That is, to each open subset , we assign an object . Additionally, if , we have a morphism , known as a restriction morphism. We will often write .

However, in order for this to be useful geometrically, we’d like to be able to study such an object locally. That is, if is some open subset of , and is an open cover of , then we would like to be able to determine from . It turns out that if we impose two simple axioms (or alternatively, the single axiom known as the sheaf axiom, although this one is less intuitive), we can do just that.

The first is known as the identity axiom. Let be an open cover of . For any such that for each , we have .

The second is known as the gluing axiom. Again, let be an open cover of , and suppose that we have such that . Then, there is some such that . Note that by the previous axiom, we know that such an must be unique.

Let’s look at some examples. Let be a smooth manifold, and let be the ring of functions . It is not hard to see that is a presheaf. Additionally, both axioms are satisfied, as a function is determined by the values it takes at each point (you should check these details if you’ve never done so before – the gluing axiom in this case is known as the pasting lemma). Actually, replacing with for any nonnegative integer also gives us a sheaf, by exactly the the same arguments.

Let be a topological space, and let . Suppose that has a zero object; that is, an object which is both an initial and a final object (an initial object is an object with exactly one morphism to any other object, and a final object is an object with exactly one morphism from any other object). Then, define the skyscraper sheaf for an object by setting when and if .

Note that sometimes, the condition that must be the terminal object of is added to the definition of a sheaf. In all of the cases anyone cares about, this always happens to be true anyways, so we need not concern ourselves with it.

Now, let be the category of abelian groups and let be some abelian group. Let be a topological space, and let be the presheaf obtained by setting for each nonempty open , and . In most cases, will contain two disjoint nonempty open sets, and because of this, will not be a sheaf, as the gluing axiom is not satisfied. However, we can construct a sheaf by letting if is connected, and .

There is, in fact, a way to do this for any presheaf, which I will talk about in the next post.

A category consists of a class of objects , as well as a set of morphisms between each pair of objects . Given we have a composite morphism $g \circ f \in \hom(X, Z)$, such that composition is associative. Additionally, for each object , we have an identity morphism , such that and for any object and morphisms . We will often get lazy and write when we mean .

Given two categories , a (covariant) functor is an assignment of an object for each object , and a function for each pair of objects that commutes with composition (ie. ). We also require that for each .

A contravariant functor is defined similarly, with the only difference being that we now have maps for each pair of objects . In other words, it reverses the direction that the arrows go.

Given two or more objects in a category, we sometimes want to form a new object from them. For example, in the category of groups, we might want to form the group that is the direct sum of several groups. The two most natural ways of doing this, that are well-defined (although not guaranteed to exist) for any category, are the product and coproduct.

Let . We define their product to be an object , along with projection morphisms such that for any object along with morphisms , there is a unique morphism such that . In the category of sets or topological spaces, this is the usual notion of product. For groups, rings, modules, or vector spaces, it is the direct product that you are probably familiar with.

After seeing the above definition, it is natural to wonder about the dual notion, where we have an object equipped with morphisms to it from and , rather than the other way around, which we shall refer to as the coproduct. We find (conveniently) that this is also a useful notion, and is a familiar construction in the categories that we are used to working with. More formally, the coproduct is an object together with inclusion morphisms , such that for any object with morphisms , there is a unique morphism such that . In the category of sets or topological spaces, this is simply the disjoint union. For groups, modules, and vector spaces, it is the direct sum (note that this is why the infinite direct sum differs from the infinite direct product, and furthermore, why they are so named). In the category of rings, this gives us the tensor product, which is a bit less intuitive than the other examples, so you should think about why this is true.

We now define direct and inverse limits. To motivate this, let be the category of rings, and let . If , then we have an inclusion morphism . As becomes larger and larger, the ring becomes “closer” to , so we’d like to be able to say that in some sense. It turns out that once we define the direct limit in the correct manner, we do in fact have .

Suppose that we have some collection of objects , for , with maps . Then, we define the direct limit , if it exists, to be the universal object , along with maps , such that . By universal, we mean that for any object and morphisms , such that , there is a unique morphism such that for all .

Letting , and the inclusion , we find that is in fact the direct limit of the system .

There is another kind of limit, which, as you may have guessed, is simply the dual notion of the direct limit, and is known as the inverse limit. Suppose that we have a collection of objects and morphisms between them . The inverse limit is the universal object , with morphisms such that . Unfortunately, there is not as simple of an example here. Let , and let be the natural quotient map. Then, the inverse limit , the -adic integers.

To conclude, inverse limits and products are both examples of what is called a limit in the more general case, which is a universal object with maps to every object in some diagram. Similarly, direct limits and coproducts are both examples of colimits, a universal object receiving morphisms from every object in some diagram.

Let be commutative rings, and let . Suppose that we have a unital ring homomorphism . Then, since prime ideals pull back to prime ideals under unital ring homomorphisms, defines a map by , for prime ideals . Certainly, we want to allow such maps as morphisms .

Since and are topological spaces, we can also look at the set of all continuous maps from to . One natural question to ask then is the following: does every continuous map arise from a ring homomorphism ?

The answer, not unsurprisingly, is no. Only a small subset of all topological spaces arise as the prime spectrum of a ring, so we should expect an arbitrary continuous map to respect this structure, especially as th is an algebraic condition. Consider , and the map given by , for any nonzero prime ideal of . This is clearly Assume for the sake of contradiction that this comes from some ring homomorphism . We must have , so . But, , so as well, since is a field. Thus, must be the zero map. But, then , which is a contradiction. Therefore, this map does not arise from a ring homomorphism.

The reason we choose schemes, and in particular, affine schemes, as an object of study is that we can convert geometric statements to algebraic statements and vice versa. Thus, we want our morphisms of spectra to be related to the rings themselves, and so we will only consider morphisms which arise from ring homomorphisms. Given this definition of , it is clear that the contravariant function from the category of rings to the category of topological spaces is an equivalence of categories (while I won’t give the formal definition of this here, you should have some idea of what it should mean intuitively).

Let’s look at some examples. Let be any ring and let , where is some ideal. Consider the quotient map . What exactly is the map induced by this? Recall that we have a natural bijection between the (prime) ideals of and the (prime) ideals of that contain . Furthermore, if is an ideal of , then , so . This is also the natural bijection mentioned before, giving us a natural identification of with , the set of all prime ideals of containing . We can then think of as an inclusion , which is a closed immersion, since is closed.

You may have seen from classical algebraic geometry that any affine variety, that is the vanishing set of some ideal of polynomials in , as a subset of , where is some algebraically closed field, can be identified with the set of maximal ideals of . From the previous article and the above discussion, we know that this is just the set of closed points of . Furthermore, the map of spectra induced by any -homomrphism between finitely-generated -algebras takes maximal ideals to maximal ideals, so in this case, we can literally ignore the non-closed points.

On first inspection, it seems like using this construction is nicer- all of the points in our topological space are now closed, and we still have just as much information as before (we can certainly recover the original ring, and the non-closed points simply correspond to the irreducible closed subsets of our space). However, using just the maximal ideals seriously limits us when we want to move to more exotic rings. For a general homomorphism of rings, maximal ideals do not necessarily pull back to maximal ideals. As a counter-example, consider the inclusion . Here, pulls back to , which is not maximal in . The actual issue here is related to the fact that the all of the residue fields of a finitely-generated -algebra obtained by modding out by a maximal idea are isomorphic to , whereas in general it depends on the maximal ideal chosen (I won’t go into this in detail here, but Qiaochu discusses this in his post “Max-Spec is not a functor”).

While I was tempted to write this in the more general context of noncommutative rings, it turns out that prime spectra of noncommutative rings are pretty tricky, as the definition of a prime ideal of a noncommutative ring is a subset such that for all , if for all , then either or . This is certainly much harder to work with than the definition of a prime ideal of a commutative ring (although it can easily be seen to be equivalent in this case).

Let be a commutative ring. Let be the set of all prime ideals of . We will now give a topology, known as the Zariski topology. For any subset , define to be the set of all prime ideals such that . Note that if is the ideal generated by , then we have , so we need not consider all subsets of .

Now, we claim that declaring such sets to be closed gives us a topology on . It is not hard to see that arbitrary intersections of sets of the form can be written in this form, as can finite unions. Additionally, and , so this does in fact define a topology. We will refer to the topological space as . This seems rather strange, and indeed it should, as it turns out that this space is not even Hausdorff (although it is still compact- it is not hard to show this, and is a good exercise if you haven’t done this before). The closed points of are precisely the maximal ideals of . It is clear that for any maximal ideal , . Furthermore, since every ideal is contained in a maximal ideal, if is prime and not maximal, and is in some closed set , then for some maximal ideal , and since , we have , and therefore . Thus, not only are the maximal ideals the closed points, but every closed subset of contains at least one maximal ideal. (It should also be obvious that the closure of is simply .

To see what’s actually going on, let’s look at some examples. First, let , our usual starting block for thinking about rings. We have . Each ideal is a closed point of , while the closure of the ideal is all of , and is known as the generic point for this reason.

Next, we look at the ring . We have . We can think of this as the complex plane, together with a generic point , whose closure is again the whole space. However, the topology here is different from the standard topology on . Giving the subspace topology (by removing the generic point), we see that the only nonempty open sets are exactly those with finite complement. However, this is at least comparable to the standard topology on , as it is easily seen to be coarser (have fewer open sets) than in the usual topology.

Both of these cases were pretty straightforward, so we will now look at one last example that is a bit more complicated. The first gives us a look into the natural question to ask: how much more complicated do things get when we work with for . It turns out that looking at gives us enough insight to be able to correctly guess what happens in all higher dimensions. Recall that there are three types of prime ideals in : the maximal ideals, the principal ideals generated by irreducible polynomials, and the zero ideal. In general, for any algebraically closed field , the maximal ideals of can all be expressed in the form , where .

Here, the maximal ideals are the ideals of the form , which are in a natural bijection with the points of (and in general, it is straightforward to see that the maximal ideals of have a natural bijection with the points of for any algebraically closed field ). So can be thought of as . Again, is simply the generic point- it’s closure is the whole space (this will always be true whenever is a prime ideal, for example in any reduced ring). But what exactly are the principal prime ideals here? Well, as we said before, each of these can be expressed in the form , where is an irreducible monic polynomial. This suggests that they may be related to the curve given by the zero set of the polynomial. Suppose that . This occurs if and only if under the quotient map , which is equivalent to saying that . Thus, if and only if , so the closure of the singleton set is just along with all of the points on which vanishes.

So, we conclude that . While the formal definition of a curve is more complicated in general, a curve in is simply the vanishing set of some nonzero polynomial in . A curve is said to be irreducible if we cannot write it as the union of two distinct curves, neither of which is all of . It turns out that a curve defined by a polynomial is irreducible if and only if the polynomial is irreducible (which is good, since this agrees with our intuition). We see that the three types of prime ideals became three types of points- the maximal ideals became the “zero-dimensional” closed points, the non-zero principal prime ideals became “one-dimensional” curves, and the zero ideal became the “two-dimensional” generic point.

I guess we can no longer use “contains chlorophyll” as the distinction between plants and animals.

I guess it makes sense- I often end up solving the problems I get stuck on for a while when I’m just letting my mind wander.

Let be the set of all compactly supported functions on , and let , where is topologized as follows: a sequence converges to if (1) there is a compact set containing all of the supports of the , and (2) for each , converges uniformly to .  Elements of are referred to as distributions.

Given , we can define by .  It is clear that if for all , then as elements of .  In this way, we can think of as containing , and therefore of distributions as a generalization of Lebesgue-integrable functions.

The linear functional given by clearly satisfies properties (1) and (2), so it is an element of .  Note that we also have , by our previous definition of the delta function.  Thus, while we cannot really think of the delta function as a function, it gives a perfectly well-defined distribution.

You may have seen mention of before (possibly in the context of a physics textbook), and wondered what this meant.  After all, even when we pretend that the delta function is a function, it is not even continuous, and certainly not differentiable.  However, given any distribution, we can make a perfectly well-defined definition of its derivative.  Let , and define the distribution by .  Here, the obvious definition for is .  Then, we observe that , using integration by parts.  Since we know that whenever , we can simply define the derivative of any distribution to be the linear functional .  In this way, we can now talk about the derivative of the dirac delta function as the linear operator which takes to .  Physicists may say that this is the “function” such that , agreeing with our definition.  Continuing in this way, we can define the ‘th derivative of the distribution as the linear functional .

We can generalize all of the above constructions to , or any open subset .  In this case, if , and , we define .

You have probably come across a partial differential equation of the form , where is a linear differential operator.  One strategy for solving such an equation is to first find a solution to , known as a fundamental solution of the operator , and then use it to obtain an actual solution (the Green’s function method).  While we can do this while still thinking of the delta function as a function (say, as the derivative of the unit step function), this requires a good deal of hand-waving.  Fortunately, when thinking of this as an equation whose solution is some distribution, we can still be completely rigorous.  Because of this, distribution theory allows us to make certain methods, such as this, of solving PDEs mathematically rigorous.

A division ring is defined to be a ring such that every nonzero element of has an inverse, that is, for each there is a such that $ab = ba = 1$.  Note that a commutative division ring is simply a field.

Exercise: Show that every finite division ring is a field.

This is actually a well-known result, but still a good problem.

Let be a ring, and let .  We call $I$ a left ideal if , right ideal if , and an ideal if .  If is commutative, then these three conditions are equivalent.  In this case, we also know that is a field iff every ideal of is trivial, that is the only nonzero ideal of is all of .  However, we have to be careful when formulating the noncommutative analogue of this statement.  While it is clear that if is a division ring, then every ideal of is trivial (a ring in which this is true is known as a simple ring), the converse does not hold (why?).  It turns out that the necessary and sufficient condition for to be a division ring is that every left ideal of is trivial (equivalently, every right ideal of is trivial).  You should convince yourself of this.

This alone motivates the study of simple rings, which by necessity requires understanding noncommutative rings to be able to say anything at all (as the only commutative simple rings are fields).

As in the commutative case, we call an element idempotent if .  If is idempotent, then we again have , as in the commutative case, so is idempotent as well.  Another familiar relation that still holds is .  It is clear from this that .  Applying this relation a second time yields the following decomposition: .

We can also write this in the form .  It is clear that the left and right hand sides are isomorphic as additive groups.  Recalling how matrix multiplication is defined, we see that the ring multiplication of both sides agrees as well.  This is known as the Peirce decomposition.

More generally, if are a complete orthogonal set of idempotents, that is if and , then we have , where is a matrix ring with .  Note  that and , so this is in fact a generalization of our previous result.  Additionally, if is commutative, then if , so becomes the diagonal matrix whose nonzero entries are , trivializing the matrix algebra, as each is now a ring and is simply the product ring .  Thus, the Peirce decomposition is only interesting in the case where is not commutative.