A Constructive Proof that Integer-Valued Continuous Functions form a Free Abelian Group

Disclaimer: I am not a constructivist, nor do I entirely know what strictly speaking entails a constructive proof. The adjective “constructive” used in the title of this post is meant to be illustrative and not technical.

Let X be a compact Hausdorff space. The abelian group C(X,\mathbb{Z}) of continuous integer-valued functions on X is free (as a \mathbb{Z}-module). One proof of this is that the group of all integer-valued functions is free (prove using Zorn’s lemma, I suppose?) and that a subgroup of a free abelian group is also free (more generally, the submodule of a free module over a PID is free; this is sometimes proved during the classification of finitely generated modules over a PID, though not always).

When I was giving a talk last June, a senior mathematician found the provided proof unsatisfactory. He and a colleague of mine gave a more constructive proof for the case where the space is the Cantor set, basically illustrating the general case. I provide a sketch of the proof below.

First, we can obtain a quotient space by identifying the points within a connected component. This space, which we call Y, is totally disconnected, and since integer-valued continuous functions must be constant on connected components, we see that C(X,\mathbb{Z}) \cong C(Y,\mathbb{Z}). So it suffices to prove the theorem in the case where Y is totally disconnected. If Y contains finitely many points, then C(Y,\mathbb{Z}) is finitely generated and we are done (since the group has no torsion elements). Otherwise, as noted yesterday, Y is the inverse limit of finite spaces, and so C(Y,\mathbb{Z}) is the inductive limit of finitely generated free abelian groups. Of course, not every inductive limit of finitely generated free abelian groups is free. But in this case, since the connecting morphisms are induced from continuous maps of spaces, we know something about the connecting morphisms; characteristic functions (which help form bases) must map to characteristic functions.

Let C(Y,\mathbb{Z}) = \lim_{k\to\infty} \mathbb{Z}^{r(k)}. In terms of the original finite spaces, r(k) denotes the number of points in each finite space and characteristic functions of singletons form a basis for the group. Let e^k_i denote this basis for k=1,2,... and i=1,2,...,r(k). Let \sigma_k: \mathbb{Z}^{r(k)} \to \mathbb{Z}^{r(k+1)} denote the kth connecting morphism. Since characteristic functions map to characteristic functions, for each k and i, there exists a set S^k_i \subseteq \{1,2,\dotsc,r(k+1)\} such that \sigma_k(e^k_i) = \sum_{j\in S^k_i} e_j. Furthermore, these sets are disjoint (since otherwise \sigma_k(e_{i_1} + e_{i_2}) would not be a characteristic function) and the union of the sets is the entire set \{1,...,r(k+1)\} (if not, it can be arranged to do so). For the C*-algebraists, this is basically saying that the Bratteli diagram of every commutative AF-algebra is a rooted tree.

Now, here I get hand-wavy. The basis B is constructed so that for each i, k, B contains all but one element of S^k_i.

To see that B generates the group, it suffices to see that e^k_i is generated by B for all k and i. If e^k_i \not\in B, then by construction e^{k-1}_i\in B and all other elements of S^k_i are in B. So e^k_i can be written as the difference between e^{k-1}_i and the sum of the elements in S^k_i.

To prove independence, we first prove by induction on t that for every i,k,t, if the equation me^k_i = \sum_{i'=i+1}^{i+t}\sum_{k'} m_{i',j'}e^{k'}_{i'} holds where e^{k'}_{i'}\in B, then m = 0. From this, we can establish the general result by multiplying through by e^k_i in a general linear combination, where i is the smallest value of the index in a linear combination and k is chosen arbitrarily.

Unfortunately, this is a very visual concept and so my verbal description might not be so great. If you know about Bratteli diagrams, I suggest trying to draw a picture with rooted trees.

Correction (2016/12/12): The previous version of this proof was incorrect. In particular, I included too many elements in my basis and my proof of independence was also incorrect.

Correction (2017/01/28): A brief sketch of the proof of independence was added.

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About minimalrho
Unemployed guy with a PhD in math.

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