The Unapologetic Mathematician

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 $\mathbb{F}$ to appropriate categories of $\mathbb{F}$-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 $\Lambda(V)$ to be the “free” graded-commutative algebra on the vector space $V$. That’s a tip-off that we’re thinking $\Lambda$ should be the left adjoint of the “forgetful” functor $U$ 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 $U$) 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 $U$ which we’re going to make the right-adjoint. Now for each vector space $V$ we need a graded-commutative algebra — clearly the one we’ll pick is $\Lambda(V)$ — and a universal arrow $\eta_V:V\rightarrow U(\Lambda(V))$. The underlying vector space of the exterior algebra is the direct sum of all the spaces of antisymmetric tensors on $V$.

$\displaystyle U(\Lambda(V))=\bigoplus\limits_{n=0}^\infty A^n(V)$

Yesterday we wrote this without the $U$, since we often just omit forgetful functors, but today we want to remember that we’re using it. But we know that $A^1(V)=V$, so the obvious map $\eta_V$ to use is the one that sends a vector $v$ to itself, now considered as an antisymmetric tensor with a single tensorand.

But is this a universal arrow? That is, if $A$ is another graded-commutative algebra, and $\phi:V\rightarrow U(A)$ is another linear map, then is there a unique homomorphism of graded-commutative algebras $\bar{\phi}:\Lambda(V)\rightarrow A$ so that $\phi=U(\bar{\phi})$? Well, $\phi$ tells us where in $A$ we have to send any antisymmetric tensor with one tensorand. Any other element $\upsilon$ in $\Lambda(V)$ is the sum of a bunch of terms, each of which is the wedge of a bunch of elements of $V$. So in order for $\bar{\phi}$ to be a homomorphism of graded-commutative algebras, it has to act by simply changing each element of $V$ in our expression for $\upsilon$ into the corresponding element of $A$, 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 $\bar{\phi}$. And this establishes that $\Lambda(V)$ is the object-function of a functor which is left-adjoint to $U$.

So how does $\Lambda$ work on morphisms? It’s right in the proof above! If we have a linear map $f:V\rightarrow W$, we need to find some homomorphism $\Lambda(f):\Lambda(V)\rightarrow\Lambda(W)$. But we can compose $f$ with the linear map $\eta_W$, which gives us $\eta_W\circ f:V\rightarrow U(\Lambda(W))$. The universality property we just proved shows that we have a unique homomorphism $\Lambda(f)=\overline{\eta_W\circ f}:\Lambda(V)\rightarrow\Lambda(W)$. And, specifically, it is defined on an element $\upsilon\in\Lambda(V)$ by writing down $\upsilon$ in terms of vectors in $V$ and applying $f$ to each vector in the expression to get a sum of wedges of elements of $W$, which will be an element of the algebra $\Lambda(W)$.

Of course, as stated above, we get similar constructions for the commutative algebra $S(V)$ and the tensor algebra $T(V)$.

Since, given a linear map $f$ the induced homomorphisms $\Lambda(f)$, $S(f)$, and $T(f)$ preserve the respective gradings, they can be broken into one linear map for each degree. And if $f$ is invertible, so must be its image under each functor. These give exactly the tensor, symmetric, and antisymmetric representations of the group $\mathrm{GL}(V)$, if we consider how these functors act on invertible morphisms $f:V\rightarrow V$. Functoriality is certainly a useful property.

October 28, 2009

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 $G\times H$ of abelian groups $G$ and $H$ works as we expect it to, because the elements coming from $G$ and $H$ already commute inside $G\times H$. The free product of $G$ and $H$ as groups gives $G*H$ 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 $X$ and homomorphisms $f_G:G\to X$ and $f_H:H\to X$ there is a unique homomorphism of abelian groups from the free product to $X$ making the diagram commute. We know that there’s a unique homomorphism from $G*H$ already, but we need to “abelianize” this group. How do we do that?

We just move to the quotient of $G*H$ by its commutator subgroup of course! Recall that any homomorphism $f:G\to A$ to an abelian group $A$ factors uniquely through this quotient: $G\to G/[G,G]\to A$. So now $(G*H)/[(G*H),(G*H)]$ is an abelian group with a unique homomorphism to $X$ 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 $(G*H)/[(G*H),(G*H)]$ is like the free product $G*H$, but we’ve thrown in relations making everything commute. We started with abelian groups $G$ and $H$, 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 $G$ and $H$, and write it $G\oplus H$.

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 $(G_1\oplus G_2)\oplus G_3$ and $G_1\oplus(G_2\oplus G_3)$ 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 $S$ and for each of its elements $s$ an abelian group $G_s$. We can define the infinite direct sum $\bigoplus\limits_{s\in S}G_s$ as the collection of all “$S$-tuples” $(g_s)$ where $g_s\in G_s$ for all $s\in S$, and where all but a finite number of the $g_s$ are the zero element in their respective groups. This satisfies something like the free product’s universal property — each $G_s$ has a homomorphism $\iota_s:G_s\rightarrow\bigoplus\limits_{s\in S}G_s$, and so on — but with an infinite number of groups on the top of the diagram: one for every element of $S$.

The direct product $\prod\limits_{s\in S}G_s$, 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 $G_s$ comes with a homomorphism $\pi_s:\prod\limits_{s\in S}G_s\to G_s$, and so on. We can realize this property with the collection of all $S$-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.

April 12, 2007

Tensor products of abelian groups

Often enough we’re going to see the following situation. There are three abelian groups — $A$, $B$, and $C$ — and a function $f:A\times B\rightarrow C$ 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 $f$ is a homomorphism from $A\times B$ to $C$. That would mean the following equation held:
$f(a+a',b+b')=f(a,b)+f(a',b')$

Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is
$f(a,b+b')=f(a,b)+f(a,b')$
$f(a+a',b)=f(a,b)+f(a',b)$
$f(a+a',b+b')=f(a,b)+f(a,b')+f(a',b)+f(a',b')$
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 $A$ and $B$ is an abelian group $T$ and a bilinear function $t:A\times B\rightarrow T$ so that for every other bilinear function $f:A\times B\rightarrow C$ there is a unique linear function $\bar{f}:T\rightarrow C$ so that $f(a,b)=\bar{f}(t(a,b))$. 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 $T$ has a presentation by generators and relations as an abelian group. For generators take all elements of $A\times B$, and for relations take all the elements $(a+a',b)-(a,b)-(a',b)$ and $(a,b+b')-(a,b)-(a,b')$ for all $a$ and $a'$ in $A$ and $b$ and $b'$ in $B$.

By the properties of such presentations, any function of the generators — of $A\times B$ — defines a linear function on $T$ if and only if it satisfies the relations. That is, if we apply a function $f$ to each relation and get ${}0$ every time, then $f$ defines a unique function $\bar{f}$ on the presented group. So what does that look like here?
$f(a+a',b)-f(a,b)-f(a',b)=0$
$f(a,b+b')-f(a,b)-f(a,b')=0$
So a bilinear function $f$ gives rise to a linear function $\bar{f}$, just as we want.

Usually we’ll write the tensor product of $A$ and $B$ as $A\otimes B$, and the required bilinear function as $(a,b)\mapsto a\otimes b$.

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 $R$ equipped with a linear map $m:R\otimes R\rightarrow R$ for multiplication.

April 6, 2007

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 $G*G$ we must use two different — but isomorphic — copies of $G$. For example, in $S_3*S_3$ we might take the first copy of $S_3$ to be permutations of $\{1,2,3\}$, and the second to be permutations of $\{A, B, C\}$.

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 $H$ up at the top and homomorphisms from $H$ into each of $G_1$ and $G_2$, and we insist that the square commutes. That is, we send an element of $H$ into $G_1$ and on into $G_1*_HG_2$ or we can send it into $G_2$ and on into $G_1*_HG_2$, and the result will be the same either way. In many cases $H$ will be a common subgroup of $G_1$ and $G_2$, and the homomorphisms from $H$ are just the inclusions.

So how do we read this diagram? We start with three groups and two homomorphisms — $G_1\leftarrow H\rightarrow G_2$ — and we look for groups that “complete the square”. The amalgamated free product $G_1*_HG_2$ is the “universal” such group — given any other group $X$ that completes the square there is a unique homomorphism from $G_1*_HG_2$ to $X$.

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 $G_1*G_2$ to $G_1*_HG_2$ by the universal property of $G_1*G_2$. Then we have two ways of sending $H$ into $G_1*G_2$: send $H$ to $G_1$ sitting inside $G_1*G_2$ or send it to $G_2$ inside $G_1*G_2$. Let’s call the first way $f_1:H\rightarrow G_1*G_2$ and the second way $f_2:H\rightarrow G_1*G_2$. Now we want to add the relation $f_1(h)=f_2(h)$ for each element of $H$. That is, $f_1(h)f_2(h)^{-1}$ should be the identity. It isn’t the identity in $G_1*G_2$, but we can make it the identity by taking the smallest normal subgroup $N$ of $G_1*G_2$ containing all these elements and moving to the quotient $(G_1*G_2)/N$. This is the group we’re looking for.

We’ve shown that $(G_1*G_2)/N$ does complete the square. Now if $X$ is any other group that completes the square it has homomorphisms into it from each of $G_1$ and $G_2$, so there’s a unique homomorphism $f:G_1*G_2\rightarrow X$. But now $f$ sends every element of $N$ to the identity in $X$ because we assumed that $X$ makes the outer square in the diagram commute. Since $N$ is in the kernel of $f$ we get a well-defined homomorphism from $(G_1*G_2)/N$ to $X$, 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.

March 3, 2007

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 $A_1$, $A_2$, $B_1$, and $B_2$, and four homomorphisms:

• $f_{1,1}:A_1\rightarrow B_1$
• $f_{1,2}:A_1\rightarrow B_2$
• $f_{2,1}:A_2\rightarrow B_1$
• $f_{2,2}:A_2\rightarrow B_2$

Use these to construct two homomorphisms from $A_1*A_2$ to $B_1\times B_2$, and show that they’re the same.

March 1, 2007

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 $G$ and $H$, the free product is a group $G*H$ with homomorphisms $\iota_G:G\rightarrow G*H$ and $\iota_H:H\rightarrow G*H$ so that given any other group $X$ with homomorphisms $f_G:G\rightarrow X$ and $f_H:H\rightarrow X$ there is a unique homomorphism $f_G*f_H:G*H\rightarrow X$.

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

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

So how do those two copies interact? Let’s put $F_2$ in for $X$ and let $f_G$ send everything in $G$ to some power of the generator $a$ — effectively defining a homomorphism from $G$ to $F_1\cong\mathbb{Z}$ — while $f_H$ sends everything in $H$ to some power of the generator $b$. If there’s any relation between the copies of $G$ and $H$ sitting inside $G*H$, it won’t be respected by the function to $F_2$ required by the universal property. So there can’t be any such relation.

$G*H$ is actually very much like $F_2$, only instead of alternating powers of $a$ and $b$, we have alternating members of $G$ and $H$. An arbitrary element looks something like $g_1h_1g_2h_2...g_kh_k$. Of course it could start with an element of $H$ or end with an element of $G$. 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 $G$ and the next sequence starts with an element of $G$, we compose those elements in $G$ so the whole sequence is still alternating.

Does $G$ sit inside here? Of course! It’s just sequences with only an element of $G$ in them. The same goes for $H$. And given any group $X$ and homomorphisms $f_G$ and $f_H$, we can send $g_1h_1g_2h_2...g_k$ to $f_G(g_1)f_H(h_1)f_G(g_2)f_H(h_2)...f_G(g_k)$. That’s our $f_G*f_H$. As I said above, any other group that has this property is isomorphic to $G*H$, so we’re done.

If we compare the free product $G*H$ with the direct product $G\times H$, we see that the main difference is that elements of $G$ and $H$ don’t commute inside $G*H$, but they do inside $G\times H$. We can check that $(g,e_H)(e_G,h)=(ge_G,e_Hh)=(g,h)=(e_Gg,he_H)=(e_G,h)(g,e_H)$. In fact, take a presentation of $G$ with generators $X$ and relations $R$, and one of $H$ with generators $Y$ and relations $S$, and with $X$ and $Y$ sharing no elements. Then $G*H$ has generators $X\cup Y$ and relations $R\cup S$, while $G\times H$ has generators $X\cup Y$ and relations $R\cup S\cup\{xyx^{-1}y^{-1} (x\in X, y\in Y)\}$.

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

March 1, 2007

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 $G$ and $H$ we form the group $G\times H$ as the set of pairs $(g,h)$ with $g$ in $G$ and $h$ in $H$. We compose them term-by-term: $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$. 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, $\pi_G$ and $\pi_H$, the “projections” of $G\times H$ onto $G$ and $H$, respectively. As one might expect, $\pi_G(g,h)=g$, and similarly for $\pi_H$. Even better, let’s consider any other group $X$ with homomorphisms $f_G:X\rightarrow G$ and $f_H:X\rightarrow H$. There is a unique homomorphism $f_G\times f_H:X\rightarrow G\times H$ — defined by $f_G\times f_H(x)=(f_G(x),f_H(x))$ — so that $\pi_G(f_G\times f_H(x))=f_G(x)$ and $\pi_H(f_G\times f_H(x))=f_H(x)$. Here’s the picture.

The vertical arrow from $X$ to $G\times H$ is $f_G\times f_H$, and I assert that that’s the only homomorphism from $X$ to $G\times H$ so that both paths from $X$ to $G$ are the same, as are both paths from $X$ to $H$. 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, $G\times H$ has homomorphisms to $G$ and $H$, and any other group $X$ with a pair of homomorphisms to $G$ and $H$ has a unique homomorphism from $X$ to $G\times H$ so that the above diagram commutes. This uniqueness means that has this property is unique up to isomorphism.

Let’s say two groups $P_1$ and $P_2$ have this product property. That is, each has given homomorphisms to $G$ and $H$, and given any other group with a pair of homomorphisms there is a unique homomorphism to $P_1$ and one to $P_2$ that make the diagrams commute (with $P_1$ or $P_2$ in the place of $G\times H$). Then from the $P_1$ diagram with $P_2$ in place of $X$ we get a unique homomorphism $f_1:P_2\rightarrow P_1$. On the other hand, from the $P_2$ diagram with $P_1$ in place of $X$, we get a unique homomorphism $f_2:P_1\rightarrow P_2$. Putting these two together we get homomorphisms $f_1f_2:P_2\rightarrow P_2$ and $f_2f_1:P_1\rightarrow P_1$.

Now if we think of the diagram for $P_1$ with $P_1$ itself in place of $X$, we see that there’s a unique homomorphism from $P_1$ to itself making the diagram commute. We just made one called $f_2f_1$, but the identity homomorphism on $P_1$ also works, so they must be the same! Similarly, $f_1f_2$ must be the identity on $P_2$, so $f_1$ and $f_2$ are inverses of each other, and $P_1$ and $P_2$ are isomorphic!

So let’s look back at this whole thing again. I take two groups $G$ and $H$, and I want a new group $G\times H$ that has homomorphisms to $G$ and $H$ and so any other such group with two homomorphisms has a unique homomorphism to $G\times H$. 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 $G$ and $H$ 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.

February 27, 2007