Functoriality of Tensor Algebras

The three constructions we’ve just shown — the tensor, symmetric tensor, and exterior algebras — were all asserted to be the “free” constructions. This makes them functors from the category of vector spaces over to appropriate categories of -algebras, and that means that they behave very nicely as we transform vector spaces, and we can even describe exactly how nicely with explicit algebra homomorphisms. I’ll work through this for the exterior algebra, since that’s the one I’m most interested in, but the others are very similar.

Okay, we want the exterior algebra to be the “free” graded-commutative algebra on the vector space . That’s a tip-off that we’re thinking should be the left adjoint of the “forgetful” functor which sends a graded-commutative algebra to its underlying vector space (Todd makes a correction to which forgetful functor we’re using below). We’ll define this adjunction by finding a collection of universal arrows, which (along with the forgetful functor ) is one of the many ways we listed to specify an adjunction.

So let’s run down the checklist. We’ve got the forgetful functor which we’re going to make the right-adjoint. Now for each vector space we need a graded-commutative algebra — clearly the one we’ll pick is — and a universal arrow . The underlying vector space of the exterior algebra is the direct sum of all the spaces of antisymmetric tensors on .

Yesterday we wrote this without the , since we often just omit forgetful functors, but today we want to remember that we’re using it. But we know that , so the obvious map to use is the one that sends a vector to itself, now considered as an antisymmetric tensor with a single tensorand.

But is this a universal arrow? That is, if is another graded-commutative algebra, and is another linear map, then is there a unique homomorphism of graded-commutative algebras so that ? Well, tells us where in we have to send any antisymmetric tensor with one tensorand. Any other element in is the sum of a bunch of terms, each of which is the wedge of a bunch of elements of . So in order for to be a homomorphism of graded-commutative algebras, it has to act by simply changing each element of in our expression for into the corresponding element of , and then wedging and summing these together as before. Just write out the exterior algebra element all the way down in terms of vectors, and transform each vector in the expression. This will give us the only possible such homomorphism . And this establishes that is the object-function of a functor which is left-adjoint to .

So how does work on morphisms? It’s right in the proof above! If we have a linear map , we need to find some homomorphism . But we can compose with the linear map , which gives us . The universality property we just proved shows that we have a unique homomorphism . And, specifically, it is defined on an element by writing down in terms of vectors in and applying to each vector in the expression to get a sum of wedges of elements of , which will be an element of the algebra .

Of course, as stated above, we get similar constructions for the commutative algebra and the tensor algebra .

Since, given a linear map the induced homomorphisms , , and preserve the respective gradings, they can be broken into one linear map for each degree. And if is invertible, so must be its image under each functor. These give exactly the tensor, symmetric, and antisymmetric representations of the group , if we consider how these functors act on invertible morphisms . Functoriality is certainly a useful property.

Direct sums of Abelian groups

Let’s go back to direct products and free products of groups and consider them just in the context of abelian groups.

The direct product of abelian groups and works as we expect it to, because the elements coming from and already commute inside . The free product of and as groups gives as before, but now this is not an abelian group. Let’s consider the property that defined free products a little more closely. Here’s the diagram.

We want to read it slightly differently now. The new condition is that for any abelian group and homomorphisms and there is a unique homomorphism of abelian groups from the free product to making the diagram commute. We know that there’s a unique homomorphism from already, but we need to “abelianize” this group. How do we do that?

We just move to the quotient of by its commutator subgroup of course! Recall that any homomorphism to an abelian group factors uniquely through this quotient: . So now is an abelian group with a unique homomorphism to making the diagram commute, it works for a free product in the context of abelian groups. This sort of thing feels odd at first, but you get used to it: when you change the context of a property (here from all groups to abelian groups) the implications change too.

Okay, so is like the free product , but we’ve thrown in relations making everything commute. We started with abelian groups and , so all we’ve really added is that elements coming from the two different groups commute with each other. And that gives us back (wait for it..) the direct product! When we restrict our attention to abelian groups, direct products and free products are the same thing. Since this is such a nice thing to happen and because we change all our notation when we look at abelian groups anyhow, we call this group the “direct sum” of the abelian groups and , and write it .

Now I didn’t really talk about this much before in the context of groups, but I’m going to need it shortly. We can take the direct sum of more than two groups at a time. I’ll leave it to you to verify that the groups and are isomorphic (use the universal property), so we can more or less unambiguously talk about the direct sum of any finite collection of groups. Infinite collections (which we’ll need soon) are a bit weirder.

Let’s say we have an infinite set and for each of its elements an abelian group . We can define the infinite direct sum as the collection of all “-tuples” where for all , and where all but a finite number of the are the zero element in their respective groups. This satisfies something like the free product’s universal property — each has a homomorphism , and so on — but with an infinite number of groups on the top of the diagram: one for every element of .

The direct product , on the other hand, satisfies something like the product condition but with an infinite number of groups down on the bottom of the diagram. Each of the comes with a homomorphism , and so on. We can realize this property with the collection of all -tuples, whether there are a finite number of nonzero entries or not.

What’s really interesting here is that for finite collections of groups the free product comes with an epimorphism onto the direct product. Now for infinite collections of abelian groups, the free product (direct sum) comes with a monomorphism into the direct product. The free product was much bigger before, but now it’s much smaller. When all these weird little effects begin to confuse me, I find it’s best just to plug my ears and go back to the universal properties. They will never steer you wrong.

Tensor products of abelian groups

Often enough we’re going to see the following situation. There are three abelian groups — , , and — and a function that is linear in each variable. What does this mean?

Well, “linear” just means “preserves addition” as in a homomorphism, but I don’t mean that is a homomorphism from to . That would mean the following equation held:

Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is



we call such a function “bilinear”.

The tensor product is a construction we use to replace bilinear functions by linear functions. It is defined by yet another universal property: a tensor product of and is an abelian group and a bilinear function so that for every other bilinear function there is a unique linear function so that . Like all objects defined by universal properties, the tensor product is automatically unique up to isomorphism if it exists.

So here’s how to construct one. I claim that has a presentation by generators and relations as an abelian group. For generators take all elements of , and for relations take all the elements and for all and in and and in .

By the properties of such presentations, any function of the generators — of — defines a linear function on if and only if it satisfies the relations. That is, if we apply a function to each relation and get every time, then defines a unique function on the presented group. So what does that look like here?


So a bilinear function gives rise to a linear function , just as we want.

Usually we’ll write the tensor product of and as , and the required bilinear function as .

Now, why am I throwing this out now? Because we’ve already seen one example. It’s a bit trivial now, but catching it while it’s small will help see the relation to other things later. The distributive law for a ring says that multiplication is a bilinear function of the underlying abelian group! That is, we can view a ring as an abelian group equipped with a linear map for multiplication.

Amalgamated free products

I was thinking that one thing might have been unclear in my discussion of free products. The two groups in the product must be completely disjoint — there are no elements in common. Particularly, if we take the free product we must use two different — but isomorphic — copies of . For example, in we might take the first copy of to be permutations of , and the second to be permutations of .

So what if the groups aren’t disjoint? Say they share a subgroup. There’s a tool we can use to handle this and a few other situations: the amalgamated free product. Here’s the picture:

This diagram looks a lot like the one for free products, but it has the group up at the top and homomorphisms from into each of and , and we insist that the square commutes. That is, we send an element of into and on into or we can send it into and on into , and the result will be the same either way. In many cases will be a common subgroup of and , and the homomorphisms from are just the inclusions.

So how do we read this diagram? We start with three groups and two homomorphisms — — and we look for groups that “complete the square”. The amalgamated free product is the “universal” such group — given any other group that completes the square there is a unique homomorphism from to .

Just like for free products we’ve given a property, but we haven’t shown that such a thing actually exists. First of all there should be a homomorphism from to by the universal property of . Then we have two ways of sending into : send to sitting inside or send it to inside . Let’s call the first way and the second way . Now we want to add the relation for each element of . That is, should be the identity. It isn’t the identity in , but we can make it the identity by taking the smallest normal subgroup of containing all these elements and moving to the quotient . This is the group we’re looking for.

We’ve shown that does complete the square. Now if is any other group that completes the square it has homomorphisms into it from each of and , so there’s a unique homomorphism . But now sends every element of to the identity in because we assumed that makes the outer square in the diagram commute. Since is in the kernel of we get a well-defined homomorphism from to , which is the one we need.

Amalgamated free products will become very important somewhere down the road, especially because they (and related concepts) figure very prominently in my own work.

A difficult exercise

While proctoring the make-up exam for my class, I thought of an exercise related to my group theory posts on direct and free products that should cause even the more experienced mathematicians in the audience a bit of difficulty.

Consider four groups , , , and , and four homomorphisms:

• 
• 
• 
• 

Use these to construct two homomorphisms from to , and show that they’re the same.

Free products of groups

On Tuesday I talked about how to put a group structure on the Cartesian product of two groups, and showed that it satisfied a universal property. For free products I’m just going to start from the property. First, the picture.

This is the same diagram as before, but now it’s upside-down. All the arrows run the other way. So we read this property as follows: Given two groups and , the free product is a group with homomorphisms and so that given any other group with homomorphisms and there is a unique homomorphism .

The exact same argument from Tuesday shows that if there is any such group , it is uniquely determined up to isomorphism. What we need is to have an example of a group satisfying this property.

First of all, has to be an monomorphism. Just put itself for , the identity on for , and the trivial homomorphism (sending everything to the identity) for . Now anything in the kernel of is automatically in the kernel of the composition of with the coproduct map. But this composite is the identity homomorphism on , which has trivial kernel, and so must . This means that a copy of must sit inside . Similarly a copy of must sit inside .

So how do those two copies interact? Let’s put in for and let send everything in to some power of the generator — effectively defining a homomorphism from to — while sends everything in to some power of the generator . If there’s any relation between the copies of and sitting inside , it won’t be respected by the function to required by the universal property. So there can’t be any such relation.

is actually very much like , only instead of alternating powers of and , we have alternating members of and . An arbitrary element looks something like . Of course it could start with an element of or end with an element of . The important thing is that the entries from the two groups alternate. We compose by just sticking sequences together like for a free group. If one sequence ends with an element of and the next sequence starts with an element of , we compose those elements in so the whole sequence is still alternating.

Does sit inside here? Of course! It’s just sequences with only an element of in them. The same goes for . And given any group and homomorphisms and , we can send to . That’s our . As I said above, any other group that has this property is isomorphic to , so we’re done.

If we compare the free product with the direct product , we see that the main difference is that elements of and don’t commute inside , but they do inside . We can check that . In fact, take a presentation of with generators and relations , and one of with generators and relations , and with and sharing no elements. Then has generators and relations , while has generators and relations .

Since we’ve only added some relations to the presentation to get from to , the latter group is a quotient of the former. There should be some epimorphism from to . I’ll leave it to you to show that some such epimorphism does exist in two ways: once by the universal property of and once by the universal property of .

Direct Products of Groups

There are two sorts of products on groups that I’d like to discuss. Today I’ll talk about direct products.

The direct product says that we can take two groups, form the Cartesian product of their sets, and put the structure of a group on that. Given groups and we form the group as the set of pairs with in and in . We compose them term-by-term: . It can be verified that this gives us a group.

There’s a very interesting property about this group. It comes equipped with two homomorphisms, and , the “projections” of onto and , respectively. As one might expect, , and similarly for . Even better, let’s consider any other group with homomorphisms and . There is a unique homomorphism — defined by — so that and . Here’s the picture.

The vertical arrow from to is , and I assert that that’s the only homomorphism from to so that both paths from to are the same, as are both paths from to . When we draw a diagram like this with groups on the points and homomorphisms for arrows, we say that the diagram “commutes” if any two paths joining the same point give the same homomorphism between those two groups.

To restate it again, has homomorphisms to and , and any other group with a pair of homomorphisms to and has a unique homomorphism from to so that the above diagram commutes. This uniqueness means that has this property is unique up to isomorphism.

Let’s say two groups and have this product property. That is, each has given homomorphisms to and , and given any other group with a pair of homomorphisms there is a unique homomorphism to and one to that make the diagrams commute (with or in the place of ). Then from the diagram with in place of we get a unique homomorphism . On the other hand, from the diagram with in place of , we get a unique homomorphism . Putting these two together we get homomorphisms and .

Now if we think of the diagram for with itself in place of , we see that there’s a unique homomorphism from to itself making the diagram commute. We just made one called , but the identity homomorphism on also works, so they must be the same! Similarly, must be the identity on , so and are inverses of each other, and and are isomorphic!

So let’s look back at this whole thing again. I take two groups and , and I want a new group that has homomorphisms to and and so any other such group with two homomorphisms has a unique homomorphism to . Any two groups satisfying this property are isomorphic, so if we can find any group satisfying this property we know that any other one will be essentially the same. The group structure we define on the Cartesian product of the sets and satisfies just such a property, so we call it the direct product of the two groups.

This method of defining things is called a “universal property”. The argument I gave to show that the product is essentially unique works for any such definition, so things defined to satisfy universal properties are unique (up to isomorphism) if they actually exist at all. This is a viewpoint on group theory that often gets left out of basic treatments of the subject, but one that I feel gets right to the heart of why the theory behaves the way it does. We’ll definitely be seeing more of it.