In a previous post, we discussed the fundamental group of the harmonic archipelago in some detail.
One item that I skipped earlier is the fact that this group is uncountable. In this post, we’ll see why is uncountable. We’ll use a little bit of work from that original post.
Definition: The Baer-Specker group (or often just Specker group) is the infinite direct product . The elements of this abelian group are infinite sequences of integers and the operation is component-wise addition. Let denote the sequence which is zero everywhere except the n-th position where it is .
It is easy to see that the Specker group is torsion free and uncountable (since it projects onto which is the power set of ). It is less obvious that the Specker group is not isomorphic to a free abelian group (perhaps a future post!).
The Specker-group does have a natural free subgroup: the infinite direct sum freely generated by the basis . This subgroup consists of sequences where all but finitely many terms are . We’re going to make use of subgroups which are isomorphic to the direct sum .
Lemma 1: If is any homomorphism, then the quotient group is uncountable.
Proof. Notice that the group is countable since it is the countable union of countable free abelian groups . Thus has at most countable cardinality. If the quotient is countable, then is the countable disjoint union of the cosets of . But all of these cosets have the same cardinality as and must be countable. Since we have written the uncountable Specker group as the countable union of countable sets, we conclude the Specker group is countable, which is a contradiction.
Assuming the axiom of choice, if is a partition of a set with cardinality of the continuum (like the Specker group) and each is countable, then must have cardinality of the continuum. Thus, if we wanted to, we could strengthen the above lemma to say that has the cardinality of the continuum.
Now, we’re going to fix a certain homomorphism to show that naturally surjects onto one of these uncountable quotients.
Corollary 2: The homomorphism defined as on basis elements is an injection and is uncountable.
Proof. The second part of the corollary follows directly from Lemma 1. To see that is injective, suppose
Then for and . Thus for all .
It is interesting to notice that
Proposition 3: is torsion free.
Proof. Suppose for some element that for some . Then and thus . Thus showing that is torsion free.
Now we need to call upon some things we know about the fundamental group of the Hawaiian earring and the harmonic archipelago.
Fact I: The map , which collapses all of the circles of the Hawaiian earring except the n-th circle, induces a surjection which gives the winding number of a loop around the n-th circle. So if is the loop going once around in the clockwise direction, then , , , etc.
Together, these winding number homomorphisms uniquely induce a surjection to the Specker group, given by . Thus keeps track of all winding numbers around all circles. See the Hawaiian earring post for more details on .
Fact II: The natural inclusion of the Hawaiian earring induces a surjection where is (freely) generated by the classes for each . See the harmonic archipelago post for more details.
Theorem 4: is uncountable.
Proof. Recall the injective homomorphism from Corollary 2. Let be the canonical projection map.
We are looking to construct the following morphism of short exact sequences of groups. The top row is an exact sequence of non-abelian groups and the bottom row is an exact sequence of abelian groups
Recall that is the conjugate closure of the free subgroup of generated by the elements . If , then
Consequently, we have an induced homomorphism such that . Moreover, since all other maps in the right square are surjective, is also surjective. Since , it follows that . Since is injective, we get an induced map such that . Moreover, for each , we have where . It follows that and thus is surjective.
Ok…so we didn’t really need to do this homological algebra, but it personally helps me see what is going on here. By Corollary 2, is uncountable. Since surjects onto this uncountable group, must also be uncountable.
Remark 1: If you want to be a little more specific about the cardinality, you could notice this: is the quotient of , which has cardinality of the continuum (it projects onto the Specker group and is a subgroup of ). Additionally, surjects onto which also has the cardinality of the continuum, so must have the same cardinality as these groups.
Remark 2: All of the vertical maps , , and in the above diagram are surjective. The top row is non-abelian and the bottom row is abelian so it is tempting to conjecture that these three maps are abelianization maps, however, it seems that none of them are!