# The Unapologetic Mathematician

## More on tensor products and direct sums

We’ve defined the tensor product and the direct sum of two abelian groups. It turns out they interact very nicely.

One thing we need is another fact about the tensor product of abelian groups. If we take three abelian groups $A$, $B$, and $C$, we can form the tensor product $A\otimes B$, and then use that to make $(A\otimes B)\otimes C$. On the other hand, we could have started with $B\otimes C$ and then built $A\otimes(B\otimes C)$. If we look at the construction we used to show that tensor products actually exist we see that these two groups are not the same. However, they are isomorphic.

To see this, let’s make a bilinear function from $(A\otimes B)\times C$ to $A\otimes(B\otimes C)$. By our construction, any element of $A\otimes B$ can be represented as a sum $\sum\limits_i a_i\otimes b_i$, so linearity says we just need to consider elements of the form $a\otimes b$. Define $f(a\otimes b,c)=a\otimes(b\otimes c)$. This induces a unique linear function given by $\bar{f}((a\otimes b)\otimes c)=a\otimes(b\otimes c)$ and extending to sums of such elements. Similarly we get a linear function $\bar{f}^{-1}(a\otimes(b\otimes c))=(a\otimes b)\otimes c)$, so we have an isomorphism of abelian groups. We can thus (somewhat) unambiguously talk about “the” tensor product $A\otimes B\otimes C$.

Now let’s take a collection of abelian groups $A_i$ with $i$ running over an index set $\mathcal{I}$, and let $B$ be any other abelian group. We want to consider the tensor product
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$

Since the direct sum is a direct product of groups, it comes with projections $\pi_k:\bigoplus_i A_i\rightarrow A_k$. Since the free product is in general a subgroup of the direct sum (a proper subgroup when the index set is infinite), we also have injections $\iota_k:A_k\rightarrow\bigoplus_i A_i$ coming from the free product. We can use these to build homomorphisms
$\iota_i\otimes1_B:A_i\otimes B\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$
applying $\iota_i$ to $A_i$ and the identity to $B$. By the universal property of direct sums (the one it gets from free products of groups) this gives us a homomorphism
$\alpha:\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$

On the other hand, for each $k$ we have a bilinear function sending $(a,b)$ in $\left(\bigoplus_i A_i\right)\times B$ to $\pi_k(a)\otimes b$ in $A_k\otimes B$. By the universal properties of tensor products this gives a linear function $\left(\bigoplus_i A_i\right)\otimes B\rightarrow A_k\otimes B$. The universal property of direct sums (the one it gets from direct products of groups) gives us a linear function
$\beta:\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B\rightarrow\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)$

Now there’s a lot of juggling of functions and injections and projections here that I really don’t think is very illuminating. The upshot is that $\alpha$ and $\beta$ are inverses of each other, giving us an isomorphism of the two abelian groups. There’s nothing really special about the left side of the tensor product either. A similar result holds if the direct sum is the right tensorand. We can even put them together to get the really nice isomorphism:

$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes\left(\bigoplus\limits_{j\in\mathcal{J}}B_j\right)\cong\bigoplus\limits_{(i,j)\in\mathcal{I}\times\mathcal{J}}(A_i\otimes B_j)$

Neat!

April 17, 2007

## 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

## A few more groups

I want to throw out a few more examples of groups before I move deeper into the theory.

First up: Abelian groups. These are more a general class of groups than an example like permutation groups were. They are distinguished by the fact that the composition is “commutative” — it doesn’t matter what order the group elements come in. The composition $ab$ is the same as the composition $ba$.

All the groups I’ve mentioned so far, except for permutations and rotations, are Abelian. It’s common when dealing with an abelian group to write the composition as “+”, the identity as “0”, and the inverse of $a$ as $-a$. Let’s recap the axioms for an Abelian group in this notation.

1. For any $a$, $b$, and $c$: $(a+b)+c=a+(b+c)$
2. There is an element 0 so that for any $a$: $a+0=a=0+a$
3. For every $a$ there is an element $-a$ so that: $a+(-a)=0=(-a)+a$
4. For any $a$ and $b$: $a+b=b+a$

Abelian groups are really fantastically important. Many later algebraic structures start with an Abelian group and add structure to it, just as a group starts with a set and adds structure to it. We’ll see many examples of these later.

The other thing I want to mention is a free group. As the name might imply, this is a group with absolutely no restrictions other than the group axioms. We start by picking some basic pieces, sometimes called “generators” or “letters”, and then just start writing out whatever “words” the rules of group theory allow.

Let’s start with the free group on one letter: $F_1$. We definitely have the identity element — written “1” — and we throw in our single letter $a$. We can compose this with itself however many times we like by just writing letters next to each other: $aa$, $aaa$, $aaaa$, and so on. We also need an inverse, $a^{-1}$. We can use $a$ and $a^{-1}$ to build up long words like $aaa^{-1}aaa^{-1}aaaa^{-1}a^{-1}aaaaa^{-1}aaaa$. But notice that whenever an $a$ and an $a^{-1}$ sit next to each other they cancel. That collapses this long word down to $aaaaaaaaa$. We see that in $F_1$ all words look like $a^n$, where a positive $n$ means a string of $n$ $a$s, a negative $n$ means a string of $|n|$ $a^{-1}$s, and $n=0$ for the identity. We compose just by adding the exponents.

The free group on two letters, $F_2$ gets a lot more complicated. We again start with the identity and throw in letters $a$ and $b$. Now we can build up all sorts of words like $aba^{-1}aa^{-1}abba^{-1}b^{-1}aaabb^{-1}baab$. But now we can’t do anything to pull $a$ and $b$ past each other. Letters only cancel their inverses when they’re right next to each other, so this word only collapses to $abbba^{-1}b^{-1}aaabaab$. That’s the best we can do. Free groups on more generators are pretty much the same, but with more basic symbols.

Composition of words $w_1$ and $w_2$ just writes them one after another, cancelling whatever possible in the middle. For example, in $F_3$ let’s say $w_1=abcbac$ and $w_2=c^{-1}a^{-1}bc^{-1}ab$. We write them together (in order!) as $abcbacc^{-1}a^{-1}bc^{-1}ab$ and cancel inverses where we can to get $w_1w_2=abcbbc^{-1}ab$.

Free groups seem hideously complicated at first, but they aren’t so bad once you get used to them. They’re also tremendously useful, as we’ll soon see. They’re the primordial groups, with absolutely nothing extra beyond the bare minimum of what’s needed to make a group.

Some points to ponder

• What is the inverse of a word in a free group?
• What should the free Abelian group on $n$ letters look like?

February 6, 2007