Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There is a short section in the book Locally Compact Groups 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:

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?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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. :-)

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.

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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