2
$\begingroup$

Reading the proof of the Hales-Jewett theorem the author defines $W_L$ as the set of finite words over some alphabet $L$, $W_{L_v}$ as the set of variable-words over $L$, i.e. finite words over $L \cup \{v\}$ where $v \notin L$ and $v$ occurs at least once, $S = W_L \cup W_{L_v}$ and the substitution map from $S$ into $W_L$ obtained by replacing every occurrence of $v$ in the word by some $\lambda \in L$

Consider $(S, ^\frown)$ as a discrete topological space with the concatenation operation and let $\left(\beta S, ^\frown \right)$ be the space of ultrafilters over $S$ (the Stone–Čech compactification of $S$) with the operation given by $$\mathcal{U} ^\frown \mathcal{V} = \left\{ A \subseteq S | \left\{x \in S | \left\{y \in S | x ^\frown y \in A \right\} \in \mathcal{V} \right\} \in \mathcal{U} \right\}$$

Let $S_{L}^{*} = \left\{\mathcal{U} \in S^{*}: W_{L} \in \mathcal{U}\right\}$ and $S_{L}^{*} = \left\{\mathcal{U} \in S^{*}: W_{L v} \in \mathcal{U}\right\}$

at the beginning of the proof the author asserts:

Each letter $\lambda \in L$ determines the substitution map $x \mapsto x[\lambda]$ from $W_{L v} \cup W_{L}$ into $W_{L},$ which is clearly the identity on $W_{L}$ and which extends to a continuous homomorphism $\mathcal{U} \mapsto \mathcal{U}[\lambda]$ from $S_{L v}^{*} \cup S_{L}^{*}$ into $S_{L}^{*}$, which is the identity on $S_{L}^{*}$

I'm assuming that the extension is given by the universal property of the Stone–Čech compactification of $S$ since you can suppose that the substitution map is continuous and has domain in $S$ and codomain in $\beta S$. I can easily verify that the extension is a continuous function that maps $S_{L v}^{*} \cup S_{L}^{*}$ into $S_{L}^{*}$ and that it is the identity on $S_{L}^{*}$ using the fact that the extension given by the Stone–Čech compactification is

$$ \mathcal{U}[\lambda] = \left\{ B \subseteq S | \exists A \in \mathcal{U} \left( A[\lambda] \subseteq B \right) \right\}$$

I've been trying to prove that the extension is a homomorphism using the above equality but it gets very cumbersome and it is not clear how to prove it, I know that one inclusion would prove the equality (since they are ultrafilters) but i can't prove either of the two so I think that is not the way to prove it, a hint would help me a lot

$\endgroup$
5
  • 3
    $\begingroup$ Do you know about ultrafilter quantifiers? You write $\mathcal{U}x\ \varphi(x)$ to mean that $\{ x: \varphi(x) \} \in \mathcal{U}$. Then, treat $\mathcal{U}x$ as a "quantifier" (like $\forall$ and $\exists$). The advantage is that $\mathcal{U}^\frown \mathcal{V}$ has a much simpler definition as $\{ A \subseteq S : \mathcal{U}x\ \mathcal{V} y\ (x^\frown y \in A) \}$. $\endgroup$ – Jordan Mitchell Barrett Feb 11 at 8:41
  • 2
    $\begingroup$ Then use the fact that ultrafilter quantifiers commute with all the logical connectives $\land$, $\lor$, $\to$, $\neg$. This makes proofs much easier - you don't have to deal with sets of sets of sets... $\endgroup$ – Jordan Mitchell Barrett Feb 11 at 8:42
  • 3
    $\begingroup$ Also, this feels like the kind of thing you should get "for free", in the following sense. I conjecture that for any semigroup homomorphism $f\colon S \to T$, its extension $\beta f\colon \beta S \to \beta T$ is a continuous homomorphism of topological semigroups. See if you can prove that, or look at Hindman and Strauss' book - I'm sure they prove it. $\endgroup$ – Jordan Mitchell Barrett Feb 11 at 9:13
  • 1
    $\begingroup$ As @JordanMitchellBarrett says, the $\beta$-semigroup is functorial. I think this is easier to see using the description of the product as limits. There are two ways to do the product based on whether you want left multiplication to be continuous or right multiplication. I'm not sure which convention corresponds to yours. I'll do once choice. For each $s\in S$, we have a continuous map $\lambda_s\colon S\to S$ by left translation $x\mapsto sx$. (Part 1) $\endgroup$ – Benjamin Steinberg Feb 11 at 19:03
  • 1
    $\begingroup$ (ctd) This extends by the universal property to $\lambda_s\colon \beta S\to \beta S$. Then for each $\sigma\in \beta S$ we have $\rho_{\sigma}\colon S\to \beta S$ by $\rho_{\sigma}(s) = \lambda_s(\sigma)$. This extends to a mapping $\rho_{\sigma}\colon \beta S\to \beta S$ and we can now define the product $\tau\cdot \sigma=\rho_{\sigma}(\tau)$. This should be easy to check is functorial. $\endgroup$ – Benjamin Steinberg Feb 11 at 19:04
3
$\begingroup$

As I mentioned in comments, I will prove a more general fact: that for any for any semigroup homomorphism $f\colon S→T$, its extension $βf\colon βS→βT$ is a continuous homomorphism of topological semigroups. Here, I use the definitions

$$f(\mathcal{U}) = \{ A \subseteq T: f^{-1}(A) \in \mathcal{U} \}$$

and, using ultrafilter quantifiers $\mathcal{U}x\ \varphi(x) \iff \{ x : \varphi(x) \} \in \mathcal{U}$,

$$\mathcal{U} \oplus \mathcal{V} = \{ A \subseteq S: \mathcal{U}x\ \mathcal{V}y\ (x+y \in A) \}$$

First, as a lemma, we observe that $f(\mathcal{U})x\ \varphi(x)$ and $\mathcal{U}x\ \varphi\big( f(x) \big)$ are equivalent, as

$\begin{align} f(\mathcal{U})x\ \varphi(x) &\iff \{ x: \varphi(x) \} \in f(\mathcal{U}) \\ &\iff f^{-1}\{ x: \varphi(x) \} \in \mathcal{U} \\ &\iff \big\{ x: \varphi\big(f(x) \big) \big\} \in \mathcal{U} \\ &\iff \mathcal{U}x\ \varphi\big( f(x) \big) \end{align}$

You previously noted that $\beta f$ is already continuous - this is essentially by definition. That $\beta f$ is a semigroup homomorphism now follows from a bit of algebra:

$\begin{align} f(\mathcal{U} \oplus \mathcal{V}) &= \big\{ A \subseteq T: f^{-1}(A) \in \mathcal{U} \oplus \mathcal{V} \big\} \\ &= \big\{ A \subseteq T: \mathcal{U}x\ \mathcal{V}y\ \big(x+y \in f^{-1}(A) \big) \big\} \\ &= \big\{ A \subseteq T: \mathcal{U}x\ \mathcal{V}y\ \big(f(x+y) \in A \big) \big\} \\ &= \big\{ A \subseteq T: \mathcal{U}x\ \mathcal{V}y\ \big( f(x) + f(y) \in A \big) \big\} \\ &= \big\{ A \subseteq T: f(\mathcal{U})x\ f(\mathcal{V})y\ \big( x + y \in A \big) \big\} && \text{by lemma} \\ &= f(\mathcal{U}) \oplus f(\mathcal{V}) \end{align}$


Now, that the substitution map $\mathcal{U} \mapsto \mathcal{U}[\lambda]$ is a homomorphism on ultrafilters follows from the substitution map $v \mapsto v[\lambda]$ on words being a homomorphism. Hopefully you can convince yourself of that...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.