Math 395: Geometric Group Theory

HNN Extensions

Posted on November 21, 2022 by Michaela Polley

Motivation and Definition

When we studied free products, we discussed the free product with amalgmation very briefly. To recall, given two groups $G = \langle X|R\rangle$ and $H= \langle Y|S\rangle$ , with subgroups $A \subseteq G$ and $B \subseteq H$ and an isomorphism $\phi: A\rightarrow B$ ; the free product of $G$ and $H$ , amalgamating the subgroups $A$ and $B$ by $\phi$ is the group

$\langle X,Y|R,S, a= \phi(a), a\in A \rangle$ .

Essentially, we are adding in a set of relators where every element of $A$ is equivalent to its image in $B$ . A Higman-Neumann-Neumann Extension (HNN Extension) of a group is similar, but both $A$ and $B$ are subgroups of the same group $G$ and we introduce a new, stable, generator to relate elements of $A$ to elements of $B$ . Without further ado, let us see the definition of an HNN Extension!

Definition. Let $G$ be a group with subgroups $A$ and $B$ such that there exists an isomorphism $\phi$ from $A$ to $B$ . Then the HNN Extension of $G$ with respect to $A,B,$ and $\phi$ is the group

$G^\ast = \langle G,t|t^{-1}at=\phi(a), a\in A\rangle$

Under this definition, we retain all of the generators and relators from $G$ and add in a single stable generator, $t$ . We then add in a set of relators whereby we conjugate elements of $A$ by $t$ to get their images under $\phi$ . While the definition officially requires relations for all elements of $A$ , in practice we only need to define relators for the generators of $A$ . Let $a= g_1g_2g_3\dots g_n \in A$ . Then,
$t^{-1}at=t^{-1}g_1g_2g_3\dots g_nt= t^{-1}g_1tt^{-1}g_2tt^{-1}g_3tt^{-1}\dots tt^{-1}g_nt=\phi(g_1)\phi(g_2)\phi(g_3)\dots\phi(g_n)=\phi(g_1g_2g_3\dots g_n)=\phi(a).$
Since $\phi$ is an isomorphism, we can get the relators we need for the other elements of $A$ using only the relators involving the generators.

Let us now see some examples! Let $G = F_2=\langle a,b|\rangle$ with $A = \langle a\rangle$ and $B= \langle b \rangle$ with $\phi: a\rightarrow b$ . Then, $G^\ast = \langle a,b,t|t^{-1}at=b\rangle.$ We will see a normal form for this group later, but for right now, we can think of words in this group as alternating between freely reduced elements of $F_2$ and either $t$ or $t^{-1}$ , where we can reduce the word by replacing $t^{-1}a^nt$ with $b^n$ and $tb^mt^{-1}$ with $a^m$ .

Additionally, the Baumslag-Solitar groups are all HNN extensions of $\mathbb{Z}$ with $BS(m,n) = \langle a,t|ta^mt^{-1}=a^n \rangle$ . While we do define it here as $ta^mt^{-1}$ instead of $t^{-1}a^mt$ , we can easily take $t^{-1}$ for $t$ and the presentations will be equivalent. For once they are an example and not a counter-example!

Our definition and example are of an HNN Extension with a single stable letter. We can also define a more general HNN extension for a family of subgroups $A_i$ and a family of subgroups $B_i$ with associated isomorphisms $\phi_i$ for each $A,B$ pair. We would then define $G^\ast = \langle G, t_1,\dots ,t_i | t^{-1}_ia_it_i=\phi_i(a_i), a_i\in A_i\rangle$ . However, for notational and simplification reasons we will focus the rest of this blogpost on the single stable letter case.

Britton’s Lemma

Our next goal is to define a normal form for $G^*$ . We will define something close to a normal form first before diving into an actual normal form. Let $g$ be any element of $G$ and let $\epsilon = \pm 1$ . Every element of $G^*$ can be written as a product $g_0t^{\epsilon_1}g_1t^{\epsilon_2}g_2\dots t^{\epsilon_n}g_n.$ We will allow $g_i$ to be equal to the identity in $G$ , so we can have higher powers of $t$ if we want to but we will write them as $t1t1t\dots t$ or $t^{-1}1t^{-1}1\dots t^{-1}$ . We then define a word to be reduced if and only if there is no subsequence of the form $t^{-1}g_it$ for $g_i \in A$ or $tg_jt^{-1}$ for $g_j\in B$ . Essentially, we just reduce the word using the relators we got from the extension.

J.L. Britton proved the following lemma:

Lemma. If the word $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ is reduced and $n \ge 1$ , then $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n \ne 1$ in $G^*$ .

The proof of this lemma is beyond the scope of this blogpost, but can be found in “The Word Problem for Groups” by John L. Britton, published in 1958 as Lemma 4 (The Principal Lemma).

This lemma tells us that if we can reduce a word in $G^*$ , we can easily tell if it is equivalent to the identity, assuming that the word problem is solvable in $G$ , so we can tell whether or not the resulting element of $G$ is equal to the identity or not. However, there may be two words which are both reduced in $G<em>^</em>*$ and are not equivalent, but are equal in the group. Thus, simply reducing words as described above is not enough to define a normal form. For example, consider again $G=F_2$ and let $x\in G^*$ be $a^2bt1tab$ . This is a reduced word in $G^\ast$ . We can, however, replace the second $t$ with $a^{-1}tb$ to get $a^2bta^{-1}tbab$ . This word is also reduced. Instead to define a normal form, we will need to add some restrictions on what the elements of $g$ look like.

A Normal Form for HNN Extensions

As part of defining a normal form for $G^*$ we will choose a set of right coset representatives of both $A$ and $B$ and will allow $1$ to be the representative of both $A$ and $B$ . We also choose a normal form for $G$ We define a normal form for $G^*$ as follows:

Definition. A normal form is a word $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ for $n\ge 0$ such that the following hold:

1. $g_0$ is an arbitrary element of $G$ in normal form,
2. if $\epsilon_i=-1$ , then $g_i$ is a representative of a coset of $A$ in $G$ ,
3. if $\epsilon_i=1$ , then $g_i$ is a representative of a coset of $B$ in $G$ , and
4. there is no consecutive sequence $t^\epsilon 1 t^{-\epsilon}.$

we know that for all $a\in A$ , $t^{-1}at=\phi(a)$ , and since $\phi$ is an isomorphism we can define $\phi^{-1}$ and get $tbt^{-1}=\phi^{-1}(b)$ . We can rearrange these equalities to get $t^{-1}a=\phi(a)t^{-1}$ and $tb=\phi^{-1}(b)t.$ We will use these relationships as we put words into their normal form. Let us consider a couple of examples.

Example 1.
$G= F_2$ , $G^* = \langle a,b,t|t^{-1}at=b\rangle.$ Let $x = a^2bt1tab$ as before. We will choose the coset representatives of $A$ to be the freely reduced elements of $G$ that start with a power of $b$ and the coset representatives of $B$ to be the freely reduced elements of $G$ that start with a power of $a$ . Then, both 1 and $ab$ are representative of their cosets of $B$ , so $x$ is in normal form. However, the other word for $x$ that we discussed before is not in normal form. Recall that we had the word $a^2bta^{-1}tbab$ . We know that $bab$ is not a coset representative of $B$ , so we need to get rid of the second to last $b$ . We do this by using the relationship $tb=at$ , which after free reduction gives us back the sequence for $x$ .

Example 2.
Let $G,G^*$ , and the coset representatives be as above. Let $x = ab^2t^{-1}abab^2tbatba^2t^{-1}ab.$ This word is reduced, but not in a normal form as defined above. We will work right to left to put it into a normal form. $g_4=ab$ , and since $\epsilon_4=-1$ , we need $g_4$ to be a representative of a coset of $A$ . This is not currently the case. We use the fact that $t^{-1}a=\phi(a)t^{-1}=bt^{-1}$ to give us $ab^2t^{-1}abab^2tbatba^2bt^{-1}b.$ Now $g_4=b$ , which is in line with our definition. Next we consider $g_3 = ba^2b$ and apply the same relationship to give us $ab^2t^{-1}abab^2tbaata^2bt^{-1}b=ab^2t^{-1}abab^2tba^2ta^2bt^{-1}b.$ Applying the same idea to $g_2$ gets us $ab^2t^{-1}abab^2ata^2ta^2bt^{-1}b$ . Finally, $g_1$ should be a coset representative for $A$ , which means we need to use the fact that $t^{-1}a=bt^{-1}$ to give us $ab^2bt^{-1}bab^2ata^2ta^2bt^{-1}b=ab^3t^{-1}bab^2ata^2ta^2bt^{-1}b$ , which is finally in normal form.

Normal Form Theorem for HNN Extensions

Having defined a normal form for HNN Extensions, we now prove that each of these normal forms represents a unique element in the group. This is summarized in the Normal Form Theorem for HNN Extensions:

Theorem. Let $G^*= \langle G,t|t^{-1}at=\phi(a),a\in A\rangle$ be an HNN extension. Then,

(I) The group  is embedded in  by the map . If  in  where , then  is not reduced.

(II) Every element  of  has a unique representation as  where  is a normal form.

For the proof, we begin by showing that I implies II and vice versa so that I and II are equivalent. We do this by using Britton’s lemma and induction. We then show that statement II is actually true by defining a homomorphism from $G^*$ to the set of permutations of the normal forms of $G^*$ .

We begin by showing that I implies II. Let $G$ be embedded in $G^\ast$ and assume that if $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n = 1$ in $G<em>^</em>*$ and $n\ge 1$ , then $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ is not reduced. We wish to show that each element of $G^*$ has a unique normal form. Suppose that we have two normal forms for $w$ — $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ and $h_0t^{\delta_1}\dots t^{\delta_n}h_n$ . It is enough to show that these two words are equivalent as words and not just as elements in the group. Since the two words correspond to equivalent elements we know that \begin{center} $g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n = h_0t^{\delta_1}\dots t^{\delta_m}h_m$ . \end{center} If both $n$ and $m$ are zero, then we have $g_0=h_0$ , which are both elements of $G$ and since we write the first element of any word in $G^*$ in normal form, we know that $g_0$ and $h_0$ are the same word. Thus, suppose that $m > 0$ . By rearranging the statement above, we get

$1= g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_nh_m^{-1}t^{-\delta_m}\dots t^{-\delta_1}h_0^{-1}$ .

Since $m > 0$ , we know that the number of $t$ ‘s in the sequence is at least 1, and by statement I (our assumption), this word is not reduced. However, both of our original words were reduced so the only place where there can be cancellation is in the middle. From this we know that $\epsilon_n=\delta_m$ and $g_nh_m^{-1}$ must be an element of either $A$ or $B$ such that we can reduce. We have two cases to consider: $\epsilon_n=\delta_m=1$ and $\epsilon_n=\delta_m=-1$ . Let us consider the first case first. Since $\epsilon_n=\delta_m=1$ , we know that $g_n$ and $h_m$ are both coset representatives of $B$ and are thus not elements of $B$ . However, in order for the reduction to work, we must have that $g_nh_m^{-1}\in B$ . The only way this works is if $g_nh_m^{-1}=e$ . Hence, $g_n=h_m.$ A similar argument holds when $\epsilon_n=\delta_m=-1.$

Having now shown that $g_n=h_m$ and $\epsilon_n=\delta_m$ , we can multiply on the right to get

$g_0t^{\epsilon_1}\dots t^{\epsilon_{n-1}}g_{n-1} = h_0t^{\delta_1}\dots t^{\delta_{m-1}}h_{m-1}$ .

We can then follow the same process until we get to $m=0$ , in which case $n$ must also equal zero for both words to be in normal form. Thus, using an inductive proof we show that

$g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n = h_0t^{\delta_1}\dots t^{\delta_m}h_m$

not only as elements in the group but also as words, and thus, the normal form is, in fact, unique and statement II is satisfied.

Let us now assume statement II and show that statement I holds. It is clear that $G$ embeds in $G^*$ by $g\rightarrow g$ . We wish to show that if we have a word that represents the identity and we have at least one $t$ in the word, then the word is not reduced. This is true by Britton’s Lemma.

Having shown that the two statements are equivalent, let us show that the theorem is in fact true. We will prove statement II. Let $W$ be the set of all normal forms and $G^*$ and let $S(W)$ be the set of permutations of $W$ . We define a homomorphism $\Phi: G^ *\rightarrow S(W)$ . By showing that this is in fact a homomorphism, we are showing that each element of $G^*$ permutes the normal forms in exactly one way and each of the normal forms maps to exactly one other word, so they must represent a unique element.

The action of $\Phi$ is to multiply on the left and then convert to normal form from right to left. Since we are mapping to the set of permutations, we need to always show how $\Phi$ of some element acts on a generic normal form. In order to show that it is a homomorphism, we need to define $\Phi$ on the generators of $G^*$ (the generators of $G$ and $t$ ) and show that each of the relations under $\Phi$ still goes to 1. For any $g\in G$ , we define $\Phi(g)$ as follows. For $g_0t^{\epsilon_1}\dots t^{\ep_n}g_n\in W$ , $\Phi(g)(g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n) = gg_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ and then convert $gg_0$ to normal form. It is clear that $\Phi(gg^{-1})=\Phi(g)\Phi(g^{-1})=\Phi(g^{-1})\Phi(g)=\Phi(g^{-1}g)=\Phi(1)=1$ since we are simply multiplying on the left and can cancel out the $g$ and $g^{-1}$ before converting to normal form.

Let us now define $\Phi(t)$ . Essentially, we are still multiplying on the left and converting to normal form, but we need to carefully consider how we do it so that we end up with a normal form. If $\epsilon_1=-1$ and $g_0\in B$ , then once we multiply on the left we get something of the form $tbt^{-1}$ , which is equal to $\phi^{-1}(b)$ . This leaves us with

$\Phi(t)(g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n) = \phi^{-1}(g_0)g_1\dots t^{\epsilon_n}g_n$ .

Otherwise, we get

$\Phi(t)(g_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n) = tg_0t^{\epsilon_1}\dots t^{\epsilon_n}g_n$

which we can convert to normal form using the steps outlined in the examples to get

$\phi^{-1}(b)t\hat{g_0}t^{\epsilon_1}\dots t^{\epsilon_n}g_n$

where $g_0 = b\hat{g_0}$ and $\hat{g_0}$ is the coset representative of the coset of $B$ containing $g_0$ and $b\in B$ .

We must also check that $\Phi(t)$ actually permutes the normal forms of $G^*$ , which we do by showing that $\phi^{-1}(b)t\hat{g_0}t^{\epsilon_1}\dots t^{\epsilon_n}g_n$ is a normal form. We can see that this is true because we are starting with an element of $G$ , we know that $\hat{g_0}$ is a representative of a coset of $B$ and everything following was already a normal form. Finally, we need to show that $\Phi(tt^{-1})=\Phi(t^{-1}t)=\Phi(t)\Phi(t^{-1})=\Phi(t^{-1})\Phi(t)=1$ . We can verify this by considering all of the different cases (since there were two cases for $\Phi(t)$ ). I will not go into the details here, since this proof is already more than two pages long, but the cases are given in Lyndon and Schupp’s Combinatorial Group Theory if you are interested. Once all of the cases are considered and we have shown that the relators are still mapped to 1 under the homomorphism, we know that we have a homomorphism and then we get unique normal forms for each element.

Conclusions

HNN extensions are important because they generate groups with interesting properties (think Baumslag-Solitar groups) and also because they are an extension of amalgamated free products. Given the Normal Form Theorem and Britton’s Lemma, we see that $G^*$ will have solvable word problem is everything acts “nicely” together — $G$ has solvable word problem, $A$ and $B$ have solvable word problem, and $\phi$ and $\phi^{-1}$ are easily defined. HNN extensions are cool and if you want more information, look at Combinatorial Group Theory by Lyndon and Schupp!

References

Baumslag–Solitar group. (2021). In Wikipedia. \url{https://en.wikipedia.org/w/index.php?title=Baumslag%E2%80%93Solitar_group&oldid=1026392287}

Britton, J. L. (1958). The Word Problem for Groups. Proceedings of the London Mathematical Society, s3-8(4), 493–506. \url{https://doi.org/10.1112/plms/s3-8.4.493}

Britton, J. L. (1963). The Word Problem. Annals of Mathematics, 77(1), 16–32. \url{https://doi.org/10.2307/1970200}

Lyman, R. (2019, April 18). Answer to “Understanding HNN extensions: Intuition, examples, exercises.” Mathematics Stack Exchange. \url{https://math.stackexchange.com/a/3192668}

Polak, J. (2018, October 8). Britton’s lemma and a non-Hopfian fp group – Aleph Zero Categorical. \url{https://blog.jpolak.org/?p=2072}

Roger C. Lyndon and Paul E. Schupp. (2001). Higman-Neumann-Neumann Extensions and Free Products with Amalgmation. In Combinatorial Group Theory (pp. 178–186). Springer.

user1729. (2021, March 28). Answer to “Baumslag-Solitar Group.” Mathematics Stack Exchange. \url{https://math.stackexchange.com/a/4080463}

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Automaticity of Hyperbolic Groups

Posted on November 21, 2022 by Sam Hiken

This term has been lots of fun! It has been great meeting seven more people who would make great theoretical computer scientists if they really put their hearts into it!

ggt_final_project-1 Download

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Residual Finitude and Hyperbolicity!

Posted on November 21, 2022 by John Eichelberger

It’s been a very long but very fun term. Cheers!

Final-Project Download

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Braid Groups Blog Post!

Posted on November 21, 2022 by Osip Surdutovich

GGT_Final_Project-2 Download

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Fuchsian Groups

Posted on November 21, 2022 by Akash Ganguly

It’s pronounced fook-see-an not fyooshun 🙁

Thanks for an awesome class!

Fuchsian_Groups Download

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What groups are Amenable, and what are not?

Posted on November 19, 2022 by Shuhang Xue

What to continue to learn about amenable groups after my presentation? Hope you will find this reading enjoyable! Finally, it was a good class with you all. I wish everyone a merry holiday season. QED

TEMPLATE-copy Download

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Proper(T)

Posted on November 18, 2022 by Horace Fusco

I hope you enjoy reading about Property $(T)$ and expanding graphs! Thank you everyone for making Geometric Group theory a wonderful class!!

(Edit: Sorry for the super late update, I meant to do this at the beginning of break but apparently I forgot… anyway there was a mistake in one of the propositions in the original post where I supposed that a group was finite in a way that rendered the proposition redundant. It’s fixed below. Hope everyone’s break has been good!)

PropT Download

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9th Friday: CAT(0) Cube Complexes!

Posted on November 16, 2022 by Osip Surdutovich

Last time we dug into random groups — MurphyKate’s favorite kind of group (until she heard my presentation on braids). We also covered the Gromov density model and small cancellation. Some of the key points picked up can be summed up in the following four theorems.
Theorem Small cancellation implies hyperbolicity

Theorem If the Gromov density $d<\frac{1}{2}$ implies small cancellation

Theorem If the Gromov density $d<\frac{1}{12}$ implies hyperbolicity

Theorem If the Gromov density $d<\frac{1}{n}$ implies hyperbolicity.

Now that we have recalled all that fun stuff, let’s get on to business! It’s time for CAT(0) Cube Complexes (CCCs).
In essence, what we seek to achieve is a generalization of hyperbolicity. We have seen it in triangles and quadrilaterals, but how can we talk about hyperbolicity in $n$ -dimensions. What we seek to do is avoid positive curvature. In the Euclidean plane this means to avoid circles. Today, e will get closer to this idea and find what we need to avoid in $n$ -dimensions. But first, a whole lot of definitions are needed…

Definition A group is cubulated if it acts geometrically by isometries on a CCC.

This definition utterly meaningless — since we don’t know what a CCC is. The overall notion is that we want a group on which we can tile our cubes and cube complexes, which respects isometries, and behaves hyperbolically in $n$ -dimensions. Furthermore, we can fix that by actually defining what a CCC is, and for that matter what a cube complex is.

Definition A cube complex is a space made by gluing together cubes along faces isometrically.

Now this, we can actually decipher. It is likely easiest to do so with a few drawn out examples. Here are some cube complexes:

And here is a NON cube complex:

A few notes we should make are that cube complexes can be infinite and furthermore can tile an $n$ -dimensional space!

Ok, great… but what was that CAT(0) part about. Well, we call a cube complex a CAT(0) cube complex if the cube complex has no corners unless they are “filled in” for dimension of $n\geq 3$ . What we seek to do here isto avoid positive curvature. As with $\delta$ -slim triangles, we know that we are good with negative and flat curvature. However, when curvature turns positive and that is when hyperbolicity breaks. Thus it is useful to think of the CAT(0) condition as a requirement for a cube complex to be hyperbolic.
We have encountered CCCs before! In fact, every tree is a CCC. On the other hand, any graph which is not a tree is not a CCC. We won’t learn too many more details about cube complexes, but if you would like to learn more, this source should direct you to some wonderful ideas on low dimensional topology.

Next we must make a subsequent definition for a surface on a CCC. This is that of a hyperplane. First let’s draw one on a cube:

2 hyperplanes drawn on a cube complex. Namely, H and H’ along with their neighborhoods.

In general, if we have a CCC then we have a hyperplane. Furthermore, if the hyperplanes are “nice" then the CCC is “special”. For the figure above the CCC is special! OK, what… But of course there is an explanation. In essence, let’s imagine a subgroup. If this subgroup has certain properties then suppose they are passed to the group itself. This is the identical concept to what we are dealing with: the hyperplane is the subgroup and the CCC is the group to which the nice properties are passed onto.

Puzzle A puzzle for the reader is to show what a self -intersecting hyperplane would like on a CC and what would a self-parallel hyperplane look like?!

We covered similar material when we talked about virtually free groups.
This begs the question, when is a CCC virtually “special”?

Suppose some group $G$ acts on a set $X$ and I can find subspaces permuted by $G$ which behave like hyperplanes. Then we can produce a CCC that my group acts on. We demonstrate

this claim with an example by picking some set $X$ with an accompanying cube complex. Notice, however, that this is NOT a CCC because we have a 3D cube which has non-filled in

corners — that’s not good.

Hyperplanes (pink) bisecting the space $X$ in blue and a vertex in each complimentary section is connected to neighboring ones creating a cube complex which tiles the space.

What is happening here is that we see something that looks like the corner of a cube in 2D, in this case thee squares glued together around one vertex, then we have a spot that would be a spot of positive curvature. So we must either forbid it, or fill it in with a solid 3D cube so that we have flat curvature. How do we get rid of this? We introduce orientations. For this we show another example, where we suppose 3 hyperplanes intersect to form a triangle and we put a vertex in each region. We state that we can only connect two vertices if there is a change of 1 orientation between them, where orientation is given by the arrows by the hyperplanes in the figure.

Once we have done this, our initial example can be redrawn as follows, which satisfies a CCC.

An example of orientations where the initial (black) lines are drawn as a 2D complex with positive curvature. But the red lines fix our issues by following the steps defined above and making in a 3D cube with flat curvature.

We end this short lecture with an important theorem which is somewhat of an open question with regards to precision.

Theorem When the density $d<\frac{1}{5}$ then CCCs are embedded trees.

An example of a loop of a hyperplane (green) and two cells tiling the area as defined below.

The proof, which we shall omit is done by contradiction. The general idea is provided in the figure below. In essence, we want to show that we cannot have an embedded tree on the small scale and thus we suppose we have a loop. We draw a space of two-cells which are outlined by a loop of a hyperplane. It is then possible to show that each block depicted on the loop would have to have a length of less than $\frac{l}{2}$ where $l$ is the isoperimetric measurement of the hyperplane for each section. The contradiction is that the box where the the hyperplane intersects itself must contribute more to the length than $\frac{l}{2}$ .

If we generalize the theorem stated on small cancellation we can use the above result to bump up the $d>\frac{1}{12}$ to $d<\frac{1}{5}$ .

Finally, it can be shown that for a group with Property T, we can never get an action on a CCC. This opens up the idea that Property T and cubulation are not allowed at the same time.

This concludes our lectures in the Geometric Group Theory seminar. It may be extremely difficult about what went down in class today. The fact is, this is state of the art mathematics, and that is itself a matter of great intrigue. What we have learned about groups acting on sets during week 1, can now be applied to a group acting on a CCC in an n-dimensional space, such as an n-dimensional sphere (whatever that means). We can now think about hyperbolic groups in $n$ -dimensions as well as the curvature that goes along with them. Although we have only went a small way into the depths, it is beautiful that we can know such a structure exists on a space we cannot even imagine.

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Eighth Monday: Geometric Actions and a Schwarz Lemma

Posted on November 2, 2022 by John Eichelberger

On this fine Halloween, we discussed what it means to act geometrically, vindicating our work on hyperbolicity with a discussion of the Schwarz Lemma. First, recall the definition from last week:

Definition. A group G is hyperbolic if it acts geometrically on a $\delta$ -hyperbolic space.

We’ve established what it means for a space to be $\delta$ -hyperbolic, but what is a geometric action?

Definition. A group G acts on a metric space X geometrically if the following are true:

1) G acts by isometries;

2) G acts properly discontinuously;

3) G acts cocompactly.

This is all well and good, but what do these conditions mean? Fortunately, we have the tools to answer these questions.

Definition. A group G acts by isometries on a metric space X if given a metric $d_X$ on $X$ , for all $x, y \in X$ and $g \in G$ it’s the case that

$d_X (x, y) = d_X (g \cdot x, g \cdot y).$

We see that this isn’t too unfamiliar a notion; we’re basically just asserting that an action of $G$ corresponds to an isometry on the space.

Definition. A group $G$ acts properly discontinuously on a space $X$ if for any ball of radius $r$ around a point $x \in X$ , then

$|\{g \in G \big| gB \cap B \neq \emptyset\}| < \infty$

Definition. A group $G$ acts cocompactly if the quotient $X/G$ is compact. This is a topological notion, and for the purposes of our class, it will suffice to show that every orbit $G_x$ forms a net in $X$ , although this is technically not the same.

All of three conditions of what it means for an action to be geometric seem to enforce some sort of “nice behavior.” But what can we pull out from these assumptions? As it happens, quite a bit.

Theorem. (Schwarz’ Lemma.) For some finitely generated group $G$ and a metric space $X$ , suppose $G \curvearrowright X$ geometrically. Then for all $x \in X$ , the function

$f: G \to X$

$f: g \mapsto g \cdot x$

is a quasi-isometry.

Proof. By assumption of the action of $G$ being geometric we know that $f(G)$ forms a net in $X$ , and by definition of being a net there is some $R$ such that $N_R (G \cdot x) = X$ . Define the set $S$ as

$S := \{g \in G \big| gN_{2R+1} (x) \cap N_{2R+1} (x) \neq \emptyset \}.$

We will seek to show that $\langle S \rangle = G$ . As a consequence of the action of $G$ being geometric we know the action must be properly discontinuous, and so $S$ must be finite. Consider an arbitrary $g \in G$ and $x \in X$ , and let $\gamma$ be a geodesic connecting $x$ to $g \cdot x$ . We’ll parameterize $\gamma$ into some finite number $L$ many segments $i$ , with $\gamma(i) = x_i$ . Since $f(G)$ forms a net, there’s some $R$ such that for all $i$ , $\gamma(i) \in N_R (g_i \cdot x)$ .

Significantly, we can pull from this relationship that $d(g_i \cdot x, g_{i + 1} \cdot x) \leq 2R + 1$ , because $g_i \cdot x$ and $g_{i + 1} \cdot x$ can both differ from $x$ by at most $R$ , plus 1 to represent going through $x$ . Consider the picture:

We can take this relationship and multiply by $g_{i+1}^{-1}$ for

$d(g_{i + 1}^{-1} g_i x, x) \leq 2R + 1,$

which implies that $g_{i + 1}^{-1} g_i \in S$ . Fortuitously, we may apply this process over the entirety of $\gamma$ , since $g_k = (g_k g_{k - 1}^{-1})(g_{k - 1} g_{k - 2}^{-1}) \ldots (g_1 e)$ :

$d \big( (g_{L + 1} g_L^{-1})(g_L g_{L - 1}) \ldots (g_2 g_1^{-1})(g_1 e_1^{-1}) \cdot x \big) = d(g \cdot x) \leq 2R + 1.$

Since our choice of $g$ was arbitrary, this shows that all $g$ are in $S$ , and since elements of $S$ are defined as elements of $G$ , it must be the case that $\langle S \rangle = G$ .

This is good! We are now ready to put a bound on subsets of our geodesic. The strategy we’re going to employ won’t necessarily work for non-hyperbolic spaces, but here we’ll be able to turn this statement about subpaths into a conclusion about least distances between arbitrary points.

Let $g, h \in G$ be arbitrary . Note that by multiplying by $g^{-1}$ we have

$d(gx, hx) = d(x, g^{-1}hx).$

Great news! Since $g$ and $h$ are arbitrary, $g^{-1}h$ spans all of $G$ , so it’s equivalent to consider bounds on $d(x, g \cdot x)$ for some new, arbitrary $x \in G$ . Let $\gamma$ be a geodesic from $x$ to $g \cdot x$ . Then

$|\gamma| + 1 = d_x(x, g \cdot x) + 1 \geq d_S(e, g),$

since $G$ acts by isometries. We see from the above that it suffices for the lower bound of our purported quasi-isometry to let $K = \epsilon = 1$ . We’ll now seek an upper bound. Suppose $d_S(e, g) = k$ , and thus $g = s_1 s_2 \ldots s_k$ where $s_i$ are letters on $G$ . As per the graphic, we can subdivide the geodesic from $x$ to $g_x$ into $k$ -many pieces, each of which must have length less than or equal to $2(R + 1)$ .

The total distance of the path is $d_S (e, g) \cdot (4R + 2)$ . Since $G$ acts isometrically, this means that the inequality $d_X (x, gx) \leq d_S (e, g) \cdot (4R + 2)$ implies that the upper bound is $d_X(x, gx)$ times $4R + 2$ , which becomes our choice of $K$ .

The total distance of the path is $d_S (e, g) \cdot (4R + 2)$ . Since $G$ acts isometrically, this means that the inequality $d_X (x, gx) \leq d_S (e, g) \cdot (4R + 2)$ implies that the upper bound is $d_X(x, gx)$ times $4R + 2$ , which becomes our choice of $K$ .

This gives us our upper bound as $(4R + 2, 0)$ . Selecting the highest values of $K$ and $\epsilon$ from our lower and upper bounds, we conclude that $f: G \to X$ is a $(4R + 2, 1)$ quasi-isometric map. ⬛

This result is one of the big reasons we spent a week pushing through definitions for isometry and hyperbolicity. Having made it to the peak of this conceptual mountain, we may gaze on a valley full of beautiful corollaries. On this voyage through some implications, we will use $\simeq$ to indicate quasi-isometries.

Corollary. For a group $G$ and spaces $X$ and $Y$ , if $G \curvearrowright X, Y$ geometrically, then $X \simeq Y$ .

Corollary. If groups $G, H$ act on a space $X$ geometrically, then $G \simeq H$ .

These results follows from the fact that, as proven previously, quasi-isometry is an equivalence relation, and thus transitive.

Corollary. If $H \leq G$ has finite index, then $H \simeq G$ .

Proof. We know $G \curvearrowright \Gamma_{G, S}$ geometrically, as does $H$ . Since the index of $H$ is finite with respect to $G$ , $H_x$ is a net.

Corollary. If $n \geq 2$ , then $F_2 \simeq F_n$ . This is unsettling!

Next we stated another theorem that helps tie together the definitions we’ve been considering.

Theorem. For a group $G$ and a space $X$ , the following are equivalent:

1) G is hyperbolic;

2) $G \curvearrowright X$ geometrically implies that $X$ is $\delta$ -hyperbolic;

3) $\Gamma_{G, S}$ is $\delta$ -hyperbolic.

The following are examples of hyperbolic groups:

every finite group
free groups
virtually free groups
virtually hyperbolic groups
subgroups of the isometry group of hyperbolic space Isom( $\mathbb{H}^n$ )

The following are some non-examples of hyperbolic groups:

$\mathbb{Z}^2$
most Baumslag-Solitar groups

There are far fewer non-examples. As previously discussed, that’s because even though it took us a lot of machinery to see what we were looking at, hyperbolic groups are quite common, for fairly rigorous notions of “common.” As discussed on previous blog posts, “random” groups are almost always hyperbolic!

Last, we tied things up with some discussion of Dehn presentations.

Definition. A Dehn presentation for a finitely presented group $G = \langle G \big| R \rangle$ is a presentation such that

1) for all $r \in R$ , there exist finitely many ways to write $r = uv$ such that $|u| > |v|$ ;

2) if $w$ is a word on $S$ representing the identity, then either $w$ isn’t reduced, or $w$ contains $u$ as a subword.

Theorem. If $G$ is hyperbolic, then $G$ admits a Dehn presentation.

This is quite a claim; we didn’t finish a proof. But to get the juices flowing, we did consider some thought-provoking examples, and set up a lemma.

Lemma. For a space $X$ , if $X$ is $\delta$ -hyperbolic and $\gamma$ is a path in such that every subpath of length $\leq 8 \delta$ is geodesic, then $\delta$ has no loops.

Much like how our algorithm in the proof of the Schwarz lemma doesn’t always yield the shortest expression of a word, instead yielding a path which satisfies $\delta$ -slimness, the Dehn presentation isn’t the only or necessarily the most efficient way to present a given group. However, what these results suggest is that there’s a deep relationship between hyperbolicity — that is, with having upper bounds on the distance between points on paths — and the ability to bound the number of letters needed to express an arbitrary word on generators and relators.

Posted in Uncategorized | 3 Comments

$\delta$ -Hyperbolicity is Quasi-Isometric Invariant

Posted on October 30, 2022 by Shuhang Xue

On Wednesday, when we introduced delta-hyperbolicity, we used the usually Cayley graph $\Gamma$ of $\mathbb{Z}_3*\mathbb{Z}_4$ (triangles glued to rectangles) as an example. Specifically, triangles in $\Gamma$ are 2-slim so that $\Gamma$ is 2-hyperbolic. On the other hand, we note that $\delta$ -hyperbolicity is a generalization of trees because all triangles in trees are 0-slim. Indeed, by using cosets of $\mathbb{Z}_3$ and $\mathbb{Z}_4$ as vertices, $\mathbb{Z}_3*\mathbb{Z}_4$ also acts on the Bass-Serre Tree, which we call $T$ .

By observing that $T$ and $\Gamma$ are quasi-isometric (the idea is to inject $T$ onto $\Gamma$ so that a net can be formed and $R$ , in this case, is 2), we ask if we can deduce that $\Gamma$ is $\delta$ -hyperbolic from the fact that $T$ and $\Gamma$ are quasi-isometric? In other words, is $\delta$ -hyperbolicity a quasi-isometric invariant?

The title and the class on Friday spoil the answer, so it is a firm YES! In this post, I will walk you through proving the following Theorem.

Theorem. If $X$ is $\delta$ -hyperbolic, and $X$ and $Y$ are quasi-isometric through $f:X\to Y$ . Then, there exists $\delta'$ such that $Y$ is $\delta'$ -hyperbolic.

To prove the Theorem, we first need to define quasi-geodesics.

Definition. A quasi-geodesic is the image of an interval $[0, L]$ ( $L$ can also be $\infty$ ) under a quasi-isometric map.

Note that the above definition is generalized from the fact that a geodesic is the image of the interval $[0, L]$ under an isometric map. For the reasoning, we defined geodesics as paths with the shortest length, and we know that the shortest distance on $\mathbb{R}$ with the Euclidean metric is realized by intervals. Thus, since isometry preserves geodesics (a good exercise for you! Hint: use the inverse isometry), we see that a geodesic is the image of the interval $[0, L]$ under the isometric map.

While geodesics are rigid and friendly, quasi-geodesics are, in general sneaky! Since quasi-isometric maps can stretch and translate spaces, quasi-geodesics can wind around the space for a while before it hits the final destination. I provide two examples.

Example (1) Logarithmic Spiral. Let $g:[1,\infty) \to \mathbb{E}^2$ be defined by $x\to (x\cos \ln x, x \sin \ln x)$ . Then, the image of $g$ , for example, restricted to $[1,,10]$ is a quasi-geodesic (we will check this claim in homework!).

Note that in this example, our quasi-geodesics are very quasi! They take too many unnecessary detours. Indeed, let’s compare quasi-geodesics and geodesics with endpoints on the positive $x$ -axis so that geodesics are just straight lines. There is no common constant $C$ such that all quasi-geodesics are within $C-$ neighborhood of geodesics since the spiral’s radius is strictly increasing.

(2) I give another example of a three-dimensional Swirl. Let $g:[0,\infty) \to \mathbb{E}^3$ be defined by $x\to (x\cos x, x \sin x, x)$ .

Then, we can also check that $g$ is a quasi-isometry. Similarly, this curve always takes too many details rather than directly traveling in a straight line. Also, we cannot find a common constant $C$ such that all quasi-geodesics following $g$ are in $C$ -neighborhood of geodesics with the same endpoints.

After those two examples, we understand that quasi-geodesics are generally hard to control. However, as a recurring theme that quasi-ness works well in hyperbolic geometry, we state the following lemma needed to prove the Theorem. Essentially, it says that we can find a constant $C$ in $\delta-$ hyperbolic spaces in the context of the previous two examples.

Lemma (Morse Lemma). If $\gamma$ is a quasi-geodesics in $\delta$ -hyperbolic space $X$ , then there exists a constant $C$ , depending only on $\delta,K,\epsilon$ such that there exists a geodesic $\gamma'$ so that $\gamma \subset N_C(\gamma')$ and $\gamma' \subset N_C(\gamma).$ In particular, if $\gamma$ has endpoints, we may assume $\gamma'$ has the same endpoints.

We do not give proof of this lemma. But the rough idea is to use the lemma proved last time about the “closeness” of projection points and the “thinness” of rectangles to restrict quasi-geodesics close to geodesics by triangle inequalities.

Next, we are ready to prove the Theorem!

Proof of the Theorem. Let $a,b,c$ form an arbitrary geodesic triangle $T$ in $Y$ . To show that $Y$ is $\deta'-$ hyperbolic, it suffices to show that this arbitrary geodesic triangle is $\delta'$ -slim.

For the first step, we note that by the existence of quasi-isometric inverse $g$ to $f$ , we name $x=g(a), y=g(b), z=g(c)$ such that $d(f(x), a), d(f(y),b), d(f(z),c)\leq R$ . Then, we claim that quasi-geodesics are quasi-isometrically mapped to quasi-geodesics. To prove the claim, note that quasi-isometries are transitive. Therefore, we compose the quasi-isometry from the quasi-geodesic with the given quasi-isometry to obtain a quasi-geodesic in the codomain.

With the above claim, we may choose an appropriate $g$ so that the geodesic triangle $T$ is mapped to a quasi-geometric triangle $S$ in X (isometry is trivially quasi-isometric). Then, we connect $x,y,z$ in $X$ with geodesics into a triangle $T'$ and use Morse Lemma to conclude that there exists constant $C$ so that $T'$ and $S$ are close.

We choose an arbitrary point $r$ on $T$ . Note that $g(r)$ is on $S$ by construction. Therefore, we may find a point $m$ on $T'$ so that $d(g(r),m)<C$ . Using that $T'$ is $\delta$ -thin, we know that there exists $n$ on some other side of the triangle so that $d(m,n)<\delta$ . Finally, there exists a point $w$ back to $T'$ so that $d(n,w)<C$ . All together, we note the following:

$d(g(r),w)\leq d(g(r),m)+ d(m,n) + d(n,w)<2C+\delta.$

Now, we have that $f(w)$ is at most $R$ distance away from a point a point $s$ on $T$ at a side other than the one $r$ sits on. Thus, we derive $d(r,s)<2C+\delta+R$ . We complete the proof by letting $\delta'=2C+\delta+R$ .

To close out the discussion, I give two following corollaries.

Corollary 1. If space $X$ is $\delta$ -hyperbolic and $Y$ is not $\delta$ -hyperbolic, then, there exists no quasi-isometry (with the net condition) between them.

From the above Corollary, we immediately know that there exists no quasi-isometry between the hyperbolic plane and $\mathbb{E}^2$ .

Corollary 1. If $X$ is quasi-isometric to a tree, then it is $\delta$ -hyperbolic.

Note that this corollary implies precisely that all Cayley graphs of free products are $\delta$ -hyperbolic.

Posted in Uncategorized | Tagged Quasi-geodesics, quasi-isometry | 5 Comments

HNN Extensions

Motivation and Definition

Britton’s Lemma

A Normal Form for HNN Extensions

Normal Form Theorem for HNN Extensions

Conclusions

References

Automaticity of Hyperbolic Groups

Residual Finitude and Hyperbolicity!

Braid Groups Blog Post!

Fuchsian Groups

What groups are Amenable, and what are not?

Proper(T)

9th Friday: CAT(0) Cube Complexes!

Eighth Monday: Geometric Actions and a Schwarz Lemma

$\delta$ -Hyperbolicity is Quasi-Isometric Invariant

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