Hausdorff Derived Series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:11:31Z http://mathoverflow.net/feeds/question/4636 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4636/hausdorff-derived-series Hausdorff Derived Series QPeng 2009-11-08T15:03:54Z 2009-12-14T06:08:06Z <p>There is a short section in the book <a href="http://books.google.com/books?id=3%5FBPupMDRr8C&amp;printsec=frontcover&amp;source=gbs%5Fv2%5Fsummary%5Fr&amp;cad=0#v=onepage&amp;q=&amp;f=false" rel="nofollow">Locally Compact Groups</a> by Markus Stroppel (Chapter B7) on the notion of a "Hausdorff Solvable Group", which he defines as a topological group with a descending chain of closed normal subgroups, but it ends rather abruptly. I've tried searching around for the Hausdorff Dervied Series but I can not find a reference other than in this book. Perhaps this goes by a different name in other texts, but assuming not:</p> <p>If we equip the automorphism group of a field extension with the compact-open topology (by assumption of material present in the same book, chapter C9), then this shares separation properties of the field (9.2). What is the significance of an automorphism group being Hausdorff Solvable, or being solvable but not Hausdorff Solvable? Can we construct some examples?</p> http://mathoverflow.net/questions/4636/hausdorff-derived-series/8848#8848 Answer by Greg Kuperberg for Hausdorff Derived Series Greg Kuperberg 2009-12-14T06:08:06Z 2009-12-14T06:08:06Z <p>The point of this section of Stroppel's book is to show, ultimately, that nothing new happens. Stroppel shows that each term in the Hausdorff derived series is nothing other than the closure of the same term in the usual derived series. A topological group is Hausdorff-solvable if and only if it is solvable, and the solvable height equals the Hausdorff-solvable height. In a sense, you can't construct interesting examples. :-) </p> <p>One thing that you can do is make an example of a topological group whose commutator subgroup isn't closed. I cheated with Google to find this, but here goes anyway. There exists a finite group $G_n$ which is 2-step nilpotent and such that the commutator subgroup requires a product of $n$ commutators. Namely, take a central extension of a $2n$-dimensional vector space $V$ over an odd finite field by its exterior square $\Lambda^2 V$, such that the commutator of $a,b \in V$ is $a \wedge b \in \Lambda^2 V$. The point is that you need $k$ commutators to reach a tensor in $\Lambda^2 V$ of rank $k$. Now let $G$ be the product of all $G_n$ in the product topology. The algebraic commutator subgroup of $G$ isn't closed, because it does not include elements in the closed commutator subgroup whose commutator length in $G_n$ is unbounded as $n \to \infty$. Amazingly, this group $G$ is even compact.</p> <p>The implication for a Galois algebraic field extension, say, is as follows. The Galois group $G$ of such a field extension is a topological group, in fact a profinite group. You might have wondered if the algebraic field extension is "solvable" in the group-theoretic sense, but without leading to solvability by radicals. Happily, it doesn't happen, because what you should do is replace the solvable series of $G$ by the closed solvable series.</p>