Revision of Bundle a family of objects into a single 'universal' object from Mon, 27/04/2009 - 17:29

This article is incomplete. More examples needed.

Quick description

There are many situations in mathematics, especially in algebraic contexts, where a question about some infinite family of objects is equivalent to a question about a single 'universal' object of the same type. Often, it is easier to consider the question just for the universal object.

Prerequisites

Example 1

This section could use additional contributions. The Hasse principle seems a good example here, but I don't know enough number theory to do it justice. This example needs to be finished off.

Let $f$ be a polynomial, and suppose we wish to determine if there are any integers $x$ for which $f(x)=0$ .

Certainly, if $f(x) = 0$ then $f(x) \equiv 0 \mod n$ , for any positive integer $n$ . In other words, the equation $f(x)=0$ has solutions in the ring $\mathbb{Z}/n \mathbb{Z}$ . Consider the other direction: if we have solutions in $\mathbb{Z}/n \mathbb{Z}$ , when do we have solutions in $\mathbb{Z}$ ? Ideally, we want to incorporate all the 'modulo $n$ ' cases into a single case. The way to do this is by forming an inverse limit, as explained below.

We have a family of rings $\mathbb{Z}/n \mathbb{Z}$ , but this family has some structure to it: given positive integers $m$ and $n$ , there is a natural map from $\mathbb{Z}/mn \mathbb{Z}$ to $\mathbb{Z}/n \mathbb{Z}$ given by 'reduction modulo $n$ '. Now consider sequences $(x_n)_{n \in \mathbb{N}}$ , where $x_n$ is an element of $\mathbb{Z}/n \mathbb{Z}$ . Define addition and multiplication of sequences pointwise, and note that this forms a commutative ring, with identity given by the 'all 1' sequence. We say such a sequence is compatible if, for any positive integers $m$ and $n$ , we have $x_{mn} \equiv x_n$ modulo $n$ . Observe that the compatible sequences form a subring. This is the inverse limit of the rings $\mathbb{Z}/n \mathbb{Z}$ , and is written $\hat{\mathbb{Z}}$ .

It is clear that listing all the reductions modulo $n$ of a given integer will define a compatible sequence, so $\hat{\mathbb{Z}}$ contains $\mathbb{Z}$ as a subring. Hence if want to rule out solutions in $\mathbb{Z}$ , it suffices to rule out solutions in $\hat{\mathbb{Z}}$ .

We could instead have only looked at solutions in the rings $\mathbb{Z}/p^n \mathbb{Z}$ , where $p$ is some prime. This time, the inverse limit is $\mathbb{Z}_p$ , known as the $p$ -adic integers, which is the ring of integers of the field $\mathbb{Q}_p$ of $p$ -adic numbers. Despite the appearance of the prime $p$ here, this field actually has characteristic zero, and contains a copy of the rational numbers. It turns out that $\hat{\mathbb{Z}}$ is isomorphic as a ring to the (unrestricted) Cartesian product of the rings $\mathbb{Z}_p$ , as $p$ ranges over all primes. So finding solutions in $\hat{\mathbb{Z}}$ is equivalent to finding them separately in $\mathbb{Z}_p$ for each prime $p$ .

Example 2

(This is an example of a popular trick in group theory, 'try it for the free group'. This could perhaps be spun off to its own article.)

Let $G$ be a group generated by $d$ elements, and let $H$ be a subgroup of index $n$ . How many elements could it take to generate $H$ ?

At first sight, this seems an unreasonable question, as for $d>1$ the family of $d$ -generated groups is rather diverse, and apparently has little structure in common. For instance, every finite simple group is generated by $2$ elements. Moreover, $G$ could be infinite; in this case, how do we even know that $H$ is finitely generated? We don't want to be considering such a wide range of different examples.

However, we have a group that is in some sense universal for $d$ -generator groups: this is the free group on $d$ generators. Let's call our set of generators $X = \{ x_1, \dots, x_d \}$ ; we also need the symbols $X^{-1} = \{ x^{-1}_1, \dots, x^{-1}_d \}$ . Then the free group $F(X)$ on $X$ consists of all equivalence classes of strings of symbols in $X \cup X^{-1}$ , with the rule that we can 'cancel out' $x_i$ and $x^{-1}_i$ if we see them next to each other. Multiplication is given by concatenating the strings.

Now $F(X)$ is universal in the following sense: given any function from the set $X$ to a group $G$ , this function extends to a homomorphism from $F(X)$ to $G$ . In particular, if the function takes $X$ to a generating set for $G$ , we obtain a surjective homomorphism $\phi$ from $F(X)$ to $G$ . With a little more thought, it is clear how to reduce our original question to the special case of the free group: if $H$ is a subgroup of $G$ of index $n$ , then its preimage $\phi^{-1}(H)$ must have index $n$ in $F(X)$ . Now if $Y$ is a generating set for $\phi^{-1}(H)$ , then $\phi(Y)$ generates $H$ , and evidently $|\phi(Y)| \le |Y|$ . So we can rephrase our question as follows:

Let $G$ be a *free* group generated by $d$ elements, and let $H$ be a subgroup of index $n$ . How many elements could it take to generate $H$ ?

The answer turns out to be surprisingly simple (although the proof is not obvious, and involves tricks beyond the scope of this article). The number of generators needed for $H$ in this case is determined entirely by $n$ and $d$ , and is always precisely $n(d-1) +1$ . This is known as the Schreier index formula. As an immediate consequence, we get the following answer to our original question:

Let $G$ be a group generated by $d$ elements, and let $H$ be a subgroup of index $n$ . Then the number of elements needed to generate $H$ is at most $n(d-1) +1$ . The free groups show that this bound cannot be improved.