$\beta\mathbb{N}$ vs $\beta\mathbb{Z}$ - MathOverflow

$\beta\mathbb{N}$ vs $\beta\mathbb{Z}$

2012-09-29T04:10:38Z

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?

Answer by Andreas Blass for $\beta\mathbb{N}$ vs $\beta\mathbb{Z}$

2012-09-29T05:05:13Z

$\beta\mathbb Z$ 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.

Answer by unknown (google) for $\beta\mathbb{N}$ vs $\beta\mathbb{Z}$

2012-09-29T05:20:26Z

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.