User khalid bou-rabee - MathOverflow

Answer by Khalid Bou-Rabee for Normal subgroups of finite index in free groups

2013-04-18T15:21:37Z

I am preparing a paper with Ian Biringer, Martin Kassabov, and Francesco Matucci, where we study the growth of the index of the intersection of all normal subgroups of index at most $n$ in a given group. We call this the study of intersection growth of the group. In your notation, for the free group of rank $s$, $F_s$, and every natural number $n$, the intersection growth function, $i_{F_s}(n)$, is defined to be the order of $H_{n,s}$. As general motivation for studying this growth, for a general group $\Gamma$, we show that the growth of $i_\Gamma(n)$ determines the dimension of the profinite completion of $\Gamma$.

This paper (which we may split into two) has some examples worked out: we have precise calculations for this growth for nilpotent groups and certain arithmetic groups. In the case of a rank $s$ free group, we found the lower bound $e^{n^{s-2/3}}$ (which we compute by finding the precise growth when one only intersects maximal subgroups).

Answer by Khalid Bou-Rabee for What is the length of the shortest law of $S_n$?

2013-04-10T13:17:07Z

As far as I know, the state of the art gives some improvements to those bounds, but they are not huge. For a better lower bound: given a nontrivial element $w \in F_2$ of word length $\ell$, using a result of Buskin (Economical separability in free groups, Sib. Math. J., 50 (2009), 603-608) there exists a subgroup, $H$, of index $\ell/2+2$ that does not contain $w$. By looking at the action of $F_2$ on $F_2/ H$ we get a representation of $F_2$ into $S_{\ell/2+2}$, that does not kill $w$. Therefore, in order for $w$ to be trivial in any representation of $F_2$ into $S_n$ we must have that $n \leq \ell /2 + 2$, or $2(n-2) \leq \ell$.

There are also better upper bounds known (see, for instance, Asymptotic growth and least common multiples in groups (me and Ben McReynolds), Bulletin of the LMS (2011)).

Your question is equivalent to quantifying residual finiteness of free groups (the non-normal case), for which the precise answer is still unknown (the best known bounds are from the papers above).

Answer by Khalid Bou-Rabee for Intersection of conjugates of subgroups in free groups

2013-04-05T19:18:56Z

I'm sorry I don't have a reference but here is a constructive proof of Fact 1 just using basic algebraic topology: Let $F$ be generated by $a_1$ and $a_2$ (the higher rank case is an easy generalization). Let $B$ be the figure eight with basepoint $b_0$. Let $X$ and $Y$ be covers corresponding to $A$ and $B$ with base points $x_0$ and $y_0$. Since $A$ and $B$ are finitely generated and of infinite-index, there exists infinite graphs that are isomorphic, with labelings, to the images of the lifts of all cyclically reduced words which start with $a_1$ (and similarly $a_1^{-1}$) in the universal cover. Call $I_X$ this infinite graph corresponding to $a_1$ (including the initial $a_1$ edge). Call $I_Y$ the infinite graph corresponding to $a_1^{-1}$ in $Y$ but do not include the edge $a_1^{-1}$ in $I_Y$ (so it does not include the initial $a_1^{-1}$ edge).

Take the graph $X$ and remove the infinite piece $I_X$ to obtain $X_0$. Take $Y$ and remove the infinite piece $I_Y$ to obtain $Y_0$. Glue $X_0$ to $Y_0$ in the obvious place. This gives a new graph $Z$ which is a cover of $B$. A path from $x_0$ to $y_0$ in $Z$ is an $f$ you want.

A zeta function using half of the primes

2013-02-08T03:48:01Z

It is well known that the zeta function satisfies the Euler product formula. See this wikipedia article.

Enumerate all primes by $p_1, p_2, \ldots $ in ascending order. Set $S$ to be the set of all $p_i$ where $i$ is odd. If, for $s > 1$, you define $$\zeta_0(s) = \prod_{p \in S} \frac{1}{1-p^{-s}},$$ is it true that $(\zeta_0(s))^2$ has a meromorphic continuation to the entire complex plane?

Using topology to characterize embedded Lie subgroups of Lie groups.

2010-02-27T18:27:36Z

Cartan's theorem states that any topologically closed subgroup of a Lie group is an embedded Lie subgroup.

This leads us to ask the following question:

Can we replace "topologically closed" with a different topological property and achieve the same result? For instance, is a semi-locally simply connected subgroup of a Lie group an embedded Lie subgroup? Is a locally connnected and semi-locally simply connected subgroup of a Lie group an embedded Lie subgroup?

Some observations: An arcwise connected subgroup of a Lie group is not always an embedded Lie subgroup. For instance, consider the following example taken from http://en.wikipedia.org/wiki/Lie_subgroup:

"...take G to be a torus of dimension ≥ 2, and let H be a one-parameter subgroup of irrational slope, i.e. one that winds around in G. Then there is a Lie group homomorphism φ : R → G with H as its image. The closure of H will be a sub-torus in G."

This example is an arc-wise connected (but not locally connected) subgroup of a Lie group that is not an embedded Lie subgroup. The issue is that in the definition of an embedded Lie subgroup you require that the subgroup be nice with respect to the subset topology, in order for the Lie subgroup to be an embedded submanifold. See the section on embedded submanifolds in

http://en.wikipedia.org/wiki/Submanifold

So whatever topological constraint we use to replace "closed" it has to be stronger than arcwise-connectedness.

Generation in a group versus generation in its abelianization.

2010-03-01T20:20:36Z

Background

I have been spending a lot of time in my research with subsets of groups that are very close to being generating sets. To make this precise:

Let $G$ be a group. If a subset $S$ of $G$ projects onto a generating set of $G/[G,G]$, we say that $S$ weakly generates $G$. The following fact (see page 350 in this book for a proof) shows that weak generation in nilpotent groups is a strong condition.

Fact. Let $G$ be a nilpotent group. If $S$ weakly generates $G$, then $S$ generates $G$.

In light of this result, we ask the following question:

Does there exist a finitely presented but non-nilpotent group $G$ such that every weakly generating subset of $G$ generates $G$?

If we drop the condition "finitely presented" then the first Grigorchuk group suffices. I'd be pretty surprised if no finitely presented examples exist.

Surface groups and free groups

In response to Matt's question below: For the free group $F(a,b)$, the set ${a[[a,b],a],b }$ weakly generates but doesn't generate (you can show this directly using uniqueness of freely reduced word form in a free group). You can use this to show that any non-abelian closed surface group has subsets that weakly generate but don't generate. For instance, in the genus two case, suppose $G$ has the standard presentation with generators $a,b,c,d$ and relation $[a,b][c,d]$. Consider the set $S = ${$a,b[[b,c],b],c,d$}. If this set generates G, then it generates the image $G/N$, where $N$ is the normal subgroup generated by $a$ and $d$. This image is a free group generated by the images of $b$ and $c$. The set S projects to a set which does not generate.

Geometrically interpreting the answer to a vector calculus question involving tangent line segments to ellipses.

2010-02-28T16:11:49Z

Let E be an ellipse centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question:

Can you give a geometric reason for why the length is a+b?

This was originally asked to me by Frank Jones a few years ago.

Comment by Khalid Bou-Rabee

2013-04-16T11:37:28Z

For a concrete example consider the discrete Heisenberg group in the 3-dimensional case: en.wikipedia.org/wiki/…

Comment by Khalid Bou-Rabee

2013-04-11T01:58:08Z

Oh, I see. Forgive my misunderstanding.

Comment by Khalid Bou-Rabee

2013-04-10T20:58:42Z

Oh, you wanted $K$ to be connected?

Comment by Khalid Bou-Rabee

2013-04-10T20:57:30Z

Did you want proper containment? If so, let $K$ be the pro-p completion of $\mathbb{Z}$. This is the $p$-adic integers--a complete, totally disconnected, and Hausdorff topological group. It contains the closed subgroup $p \mathbb{Z}_p$ which is totally disconnected and of the same cardinality of $\mathbb{Z}_p$.

Comment by Khalid Bou-Rabee

2013-02-08T15:30:56Z

Thank you. I wrote this specific question because it is appearing in my work. I agree that the more general question is also interesting.

Comment by Khalid Bou-Rabee

2013-02-08T02:58:33Z

You probably know this, but if you relax f.p. to f.g. then I believe the first Grigorchuk group is an example (right?).

Comment by Khalid Bou-Rabee

2013-01-28T05:03:27Z

I believe you can modify the example so that the direct products become semidirect products.

Comment by Khalid Bou-Rabee

2010-03-02T00:35:34Z

Thank you for the informative post!

Comment by Khalid Bou-Rabee

2010-03-01T22:08:47Z

Oh I see! Thank you.

Comment by Khalid Bou-Rabee

2010-02-28T23:18:10Z

That is a beautiful answer. Thank you!

Comment by Khalid Bou-Rabee

2010-02-28T19:13:38Z

I added "embedded" in front of Lie subgroup. The reason why I ask this is because I am curious of whether there are other versions of Cartan's theorem.

Comment by Khalid Bou-Rabee

2010-02-28T19:11:12Z

Thank you, I should clarify the question.

Comment by Khalid Bou-Rabee

2010-02-28T16:02:25Z

Thank you for your post. I am asking for another form of Cartan's theorem. Let H be a subgroup of a Lie group G. In Cartan's theorem the topological property, closed, refers to H with the subset topology from G. I am not asking for you to invent a new topology (I ruled out irrationally sloped lines on a flat torus, which are analytic subgroups but not Lie subgroups). Recall that stating that H is Lie subgroup of G means more than just that H is a Lie group that is also a subgroup of G. Please see the wikipedia article en.wikipedia.org/wiki/Lie_subgroup for a complete definition.

Comment by Khalid Bou-Rabee

2010-02-28T01:17:55Z

You are correct. Sorry for the confusion.

Comment by Khalid Bou-Rabee

2010-02-28T00:23:20Z

I added a few observations to the question. On your post: I don't believe this answers the question. Please let me know if I am misunderstanding you.