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Just started learning the Stone-Cech compactification of discrete groups this week. My motivation comes from a question on $\beta\mathbb{Z}$. Surprisingly, I realized there are muchhhh more literature devoted to $\beta\mathbb{N}$ than to $\beta\mathbb{Z}$. I wonder why is that? After all, algebraically $\mathbb{N}$ is a semigroup while $\mathbb{Z}$ is a group, and as discrete topological spaces they are homeomorphic. From your experience, how far $\beta\mathbb{N}$ and $\beta\mathbb{Z}$ are different in behaviour? Also, is $\beta\mathbb{N}$ ( or ($\beta\mathbb{N}\setminus\mathbb{N}$) easier to deal with?

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Not an answer, but: have you looked at the book of Hindman and Strauss for some of the striking properties of $\beta {\mathbb N}$? I think David Gunderson might have a copy. – Yemon Choi Sep 29 '12 at 5:22
Hi Yemon. Yes! The library has a 2012 copy of it! In fact, it's open in front of me right now, working on chapter 3! Seems like I need a quite good deal of this book! Thanks for mentioning it! – Alvin Sep 29 '12 at 5:32

$\beta\mathbb Z$ consists of just two copies of $\beta\mathbb N$, one at "positive infinity" and one at "negative infinity". It's generally easier to think about just one copy rather than both, so people tend to write more about $\beta\mathbb N$. The advantage you mentioned for $\mathbb Z$ over $\mathbb N$, namely that the former is a group while the latter is only a semigroup, doesn't carry over to the Stone-Cech compactifications, both of which are (under the natural extensions of the addition operation) only semigroups.

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Are there any complications when it comes to the semigroup structure? that is, when we let the "negative infinity" bits of $\beta Z$ act on the "positive infinity" bits of $\beta Z$, does anything tricky/interesting occur? – Yemon Choi Sep 29 '12 at 5:19
Which of the group axioms fail for $\beta\mathbb Z$? – John Pardon Sep 29 '12 at 5:24
@unknown (google): $\beta{\mathbb N}$ isn't cancellative, so $\beta{\mathbb Z}$ certainly isn't. – Yemon Choi Sep 29 '12 at 6:01
@unknown: Associativity is fine for $\beta\mathbb Z$ but cancellation and therefore existence of inverses fail badly. There are lots of idempotent elements in $\beta\mathbb N and therefore in $\beta\mathbb Z$, whereas in a group the identity element is the only idempotent. – Andreas Blass Sep 29 '12 at 16:33
@Yemon: One thing that may not be obvious about the interaction of positive and negative points at infinity is that, when you add one of each, the sum is infinite in the same direction as the second summand. (That's assuming you use the "usual" convention, whereby the sum is a continuous function of the first argument. Some people prefer the opposite convention, in which case "second" should be "first" in int the unparenthesized part of this comment.) – Andreas Blass Sep 29 '12 at 16:37

Since $\mathbb N$ and $\mathbb Z$ are homeomorphic, so are $\beta\mathbb N$ and $\beta\mathbb Z$, though of course the semigroup structure will be different.

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I'm sorry, I didn't see (and still don't see) where the OP said that $\beta\mathbb N$ and $\beta\mathbb Z$ are homeomorphic. – John Pardon Sep 29 '12 at 5:37
I stand corrected - I was reading too quickly. I still think this is not a particularly useful observation in the present context, but perhaps I am mistaken and it is useful to the OP – Yemon Choi Sep 29 '12 at 5:59
[deleted earlier erroneous/unfair comment] – Yemon Choi Sep 29 '12 at 5:59
@unknow: It's not what I am looking for. I appreciate your comment though. Thanks! As you also mentioned, the extension of the homeomorphism between $'matbb{N}$ and $\mathbb{Z}$, shows that for the topologist these two spaces are the same. I am wondering how differently can these two objects $\beta\mathbb{N}$ and $\beta\mathbb{Z}$ behave? I consider the Stone-Cech compactification of a discrete semigroup as a special well-studied case of a more general notion namely compact Hausdorff right topological semigroups. So to me it's not only a topological object but also algebraic. – Alvin Sep 29 '12 at 7:28

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