A brief post today:
I was talking about an algebraic topology problem from Hatcher's book (available freely on his website) with two of my colleagues. In short, we were finding the fundamental group of some terrible space, and we thought that there might be a really slick almost entirely algebraic way to do a problem. We had a group $G$ and the exact sequence $0 \to \mathbb{Z} \to G \to \mathbb{Z} \to 0$, in short, and we wondered what we could say about $G$. Before I go on, I mention that we had been working on things all day, and we were a bit worn. So the calibre of our techniques had gone down.
In particular, we could initially think of only two examples of such a $G$, and we could show one of them didn't work. Of the five of us there, two of us thought that there might be a whole family of nonabelian groups that we were missing, but we couldn't think of any. And if none of us could think of any, could there be any? At the time, we decided no, more or less. So $G \approx Z \times Z$, which is what we wanted in the sense that we sort of knew this was the correct answer. As is often the case, it is very easy to rationalize poor work if the answer that results is the correct one.
We later made our work much better (in fact, we can now show that our group in question is abelian, or calculate is in a more geometric way). But this question remained - what counterexamples are there? There are infinitely many nonabelian groups satisfying that exact sequence! But I'll leave this question for a bit -
Find a nonabelian group (or family of groups) that satisfy $0 \to \mathbb{Z} \to G \to \mathbb{Z} \to 0$
The second quick problem of this post. It's found in Ahlfors - Find a closed form for $\displaystyle \sum_{n \in \mathbb{Z}} \frac{1}{z^3 - n^3}$.
When I first had to do this, I failed miserably. I had all these divergent series about, and it didn't go so well. I try to factor $z^3 - n^3 = (z - n)(z - \omega n)(z - \omega^2 n)$, use partial fractions, and go. And... it's not so fruitful. You get three terms, each of which diverge (if taken independently from each other) for a given $z $. And you can do really possibly-witty things, like find functions that have the same poles and try to match the poles, and such. But the divergence makes things hard to deal with. But if you do $-(n^3 - z^3) = -[(n-z)(n-\omega z)(n - \omega^2 z)]$, everything works out very nicely. That's the thing with complex numbers - the 'natural factorization' may not always be unique.
Info on how to comment
To make a comment, please send an email using the button below. Your email address won't be shared (unless you include it in the body of your comment). If you don't want your real name to be used next to your comment, please specify the name you would like to use. If you want your name to link to a particular url, include that as well.
bold, italics, and plain text are allowed in comments. A reasonable subset of markdown is supported, including lists, links, and fenced code blocks. In addition, math can be formatted using
$(inline math)$or$$(your display equation)$$.Please use plaintext email when commenting. See Plaintext Email and Comments on this site for more. Note also that comments are expected to be open, considerate, and respectful.