## On finite covers of surfaces with boundary…

I have a new preprint on the arxiv, joint with Emily Stark. We provide the first known examples of one-ended hyperbolic groups which are not abstractly coHopfian. That means that there is a one ended hyperbolic group $$G$$ which contains a finite index subgroup $$G’ \leq G$$ that embeds $$G’ \rightarrow G$$ as an infinite index subgroup. I encourage you to look at the paper for details. The main example and proof can be drawn out on a single side of A4 — it’s a simple surface amalgam and we exploit the tremendous flexibility you have when you take a finite cover of a surface with boundary.

We use the following Lemma extensively. It’s from Walter Neumann’s 2001 paper Immersed and virtually embedded $$\pi_1$$-injective surfaces in 3-manifolds, although, as he says, it is apparently “well known”.

The utility of this Lemma is that it reassures you that if you can imagine your desired cover — such as the following I’ve drawn below — and it satisfies a basic necessary Euler characteristic computations, then the cover does in fact exist.

For our main example you can compute your desired covers by hand, but it is worth knowing what kind of covers of a surface with boundary you can take. This lemma tells you exactly how much control you have. And mathematical research, like all forms of insecurity, is really all about control.

The proof given above is brief, to say the least, so I think it is worth expanding on the details.

First, we remind ourselves how you might construct such a cover by hand. Take a surface with genus one and a single boundary component. From a group theoretic point of view this is just the free group generated by two element $$\mathbb{F}_2 = \langle x, y \rangle$$ bundled together with the conjugacy class of the commutator $$[x,y]$$. For me at least, finding finite index subgroups of the free group boils down to futzing around with graphs. Thus, we let $$X$$ be a bouquet of two circles and let $$\langle x,y \rangle = \pi_1X$$.

On the right I drew the surfaces with boundary and on the left I drew the corresponding graphs with the loop corresponding to the boundary. Once you have drawn the graphs out it’s easy to verify that you have the boundary components you want. The trick is knowing you can find the desired finite covers. The key insight is that $$\alpha$$-sheeted covers of a graph $$X$$ are in a correspondence with representations into the permutation group on $$n$$ elements: $$\pi_1 X \rightarrow \textrm{Sym}(\alpha)$$

We can see how this correspondence works in practice in the example I just drew:

Giving each of vertex a number we can see that the edges labelled by a given generator of our free group gives a permutation. In this example the generator $$x$$ gives the permutation $$(1,2)(3)$$ (see the red edges on the right), while the generator $$y$$ gives the permutation $$(1,2,3)$$ (see the blue edges on the left). Thus we have a homomorphism determined by mapping the generators to the corresponding permutation.

At this point we need to be careful because there are some left-right issues hidden here. When I multiply group elements $$xy$$ I am composing paths in the fundamental group. That means I concatenate the corresponding paths, starting with the $$x$$ path and then following it with the $$y$$ path. I’m reading the composition from left to right. In contrast, when I usually compose a pair of permutations $$\sigma_1 \sigma_2$$ I compute the composition by reading them from right to left. But in order to be able to interpret my homomorphism correctly I’m going to have to compose my permutations in reverse order, from left to right.

Now we can compute the image of the element corresponding to the boundary curve: $$xyx^{-1}y^{-1} \mapsto (1,2) \circ (1,2,3) \circ (1,2) \circ (3,2,1) = (1,2,3).$$ Tracing out how the boundary curve lifts is equivalent to computing this permutation element. This makes it clear that there is a single boundary component covering the previous with degree 3. (In this example it doesn’t matter in which direction we composed the permutations).

Conversely, choosing pair of permutions, say, $$(1,2)(3,4)\textrm{, and } (2,4,3) \in \textrm{Sym(4)}$$ to be the images of $$x$$ and $$y$$ we can construct a corresponding cover by taking 4 vertices and adding the appropriate labelled edges:

Now when we compute (remembering to compose our permutations from left to right) the image of our commutator element we get $$xyx^{-1}y^{-1} \mapsto (1,2)(3,4) \circ (2,4,3) \circ (1,2)(3,4) \circ (3,4,2) = (1,4)(2,3).$$ Thus the surface has two boundary components, each covering the boundary in the base surface with degree two.

The take away from this discussion is that finding suitable covers corresponds to finding a suitable homomorphism $$\phi : \pi_1 X \rightarrow \textrm{Sym(\alpha)}$$ such that the image of the elements corresponding to boundary curves are permutaions with the desired decomposition into cycles. Our weapon of choice is the fact that any even permutation can be written as the commutator of an $$\alpha$$-cycle and an involution:

First we consider the case where $$\Sigma$$ has a single boundary component. So $$\Sigma$$ is a surface with genus $$g$$, Euler characteristic $$\chi(\Sigma) = 2 – 2g – |\partial \Sigma |$$ and a single boundary component, so $$|\partial \Sigma | = 1$$. This corresponds to the free group generated by $$2g$$ elements and the group element corresponding to the boundary, which is the product of commutators: $$( \langle x_1, y_1, \ldots x_g, y_g \rangle, [x_1,y_1]\cdots [x_g, y_g] ).$$

Suppose we wish to construct a cover of degree $$\alpha$$ with the boundary components of degrees $$\alpha_1, \ldots, \alpha_k$$. Then apply the above Theorem to the permutation $$\sigma = (1, \ldots, \alpha_1)(\alpha_1 +1, \ldots, \alpha_1 + \alpha_2) \cdots (\alpha_1 + \cdots + \alpha_{k-1} +1, \ldots, \alpha).$$ The theorem only applies if $$\sigma$$ is an even permutation, which we compute to be equivalent to $$\sum_i (\alpha_i -1) = \alpha – k$$ being even. As $$\chi(\Sigma) = 1-2g$$ this is equivalent to $$k$$ having the same parity as $$\alpha\chi(\Sigma)$$, the sufficient condition given in the statement of our theorem.

Thus there exists permutations $$\sigma_x, \sigma_y \in \textrm{Sym}(\alpha)$$ such that $$[\sigma_x, \sigma_y] = \sigma$$. The homomorphism $$\phi: \pi_1 \Sigma \rightarrow \textrm{Sym(\alpha)}$$ given by mapping $$x$$ to $$\sigma_x$$ and $$y$$ to $$\sigma_y$$ therefore corresponds to a cover with the desired boundary.

Now we consider the slightly trickier general case where we have multiple boundary components, which is to say $$|\partial \Sigma| = b$$. In which case the pair $$(\Sigma, \partial \Sigma)$$ corresponds to $$(\langle x_1, y_1, \ldots, x_g, y_g, t_1, \ldots t_{b-1} \rangle , \{t_1,\ldots, t_{b-1}, t_{b-1}\cdots t_{1}[x_1,y_1] \cdots[x_n,y_n] \} ).$$

Now suppose we desire that the $$i$$-th boundary component is covered with degrees $$\alpha_1^i, \ldots, \alpha_{k_i}^i$$, then let $$\sigma_i = (1, \ldots, \alpha_1^i)(\alpha_1^i +1, \ldots, \alpha_1^i + \alpha_2^i) \cdots (\alpha_1^i + \cdots + \alpha_{k_i-1}^i +1, \ldots, \alpha)$$ for $$1 \leq i \leq b$$. Now we wish to find $$\sigma_x, \sigma_y$$ such that $$[\sigma_x, \sigma_y] = \sigma_1 \cdots \sigma_b.$$ This requires that the product of the $$\sigma_i$$ is even. This means that the sum $$\sum_i \sum_j (\alpha_j^i – 1) = \sum_i (\alpha – k_i) =\alpha b – \sum_i k_i = \alpha (\chi(\Sigma) -2 + 2g) – \sum_i k_i$$ should be even, which is true precisely when the total number of prescribed boundary components $$\sum_i k_i$$ has the same parity as $$\alpha \chi(\Sigma)$$.

Given that our parity condition is satisfied, we define our homormorphism $$\pi_1 \Sigma \rightarrow \textrm{Sym}(\alpha)$$ as follows:

\begin{align} x_1 \mapsto & \sigma_x \\ y_1 \mapsto & \sigma_y \\ x_2 \mapsto & 1 \\ \vdots \\ y_g \mapsto & 1 \\ t_1 \mapsto & \sigma_1^{-1} \\ t_2 \mapsto & \sigma_2^{-1} \\ \vdots \\ t_{b-1} \mapsto & \sigma_{b-1}^{-1} \end{align}

Then it only remains to verify that $$t_{b-1}\cdots t_{1} [x_1,x_2] \cdots [x_g, y_g] \mapsto \sigma_{b-1}^{-1} \cdots \sigma_1^{-1} [\sigma_x, \sigma_y] = \sigma_{b-1}^{-1} \cdots \sigma_1^{-1} \sigma_1 \cdots \sigma_{b-1} \sigma_b = \sigma_b$$ and conclude this gives us our desired cover.

QED.

(I’d like to thank Emily for informing me about Neumann’s Lemma, and Nir for various discussions related to this.)

## Gromov, cheese, pretending to quit mathematics, and French.

In December last year, the Notices of the AMS ran a collection of reminiscences in memory of Marcel Burger (1927-2016), the late French differential geometer. He was also a former director of the Institut des Haute Etudes Scientific and, according to the Wikipedia, played a major role in getting Gromov positions in Paris and at the IHES in the 80s. Gromov contributed to the article, listing Berger’s mathematical achievements, before sharing a more personal anecdote:

Within my own field Gromov has had a profound influence. His essay Hyperbolic Groups led to the term “combinatorial group theory” being more or less abandoned and replaced with “geometric group theory”. As a graduate student I found the monograph frequently cited as the origin for an astounding range of ideas. At some point I had trouble finding a copy of the paper online and for a brief moment wondered if the paper itself were just an urban myth or elaborate hoax.

Gromov’s foray into group theory is just one episode in a long career. His first major breakthrough was in partial differential equations; the “h-principal” which, according to Larry Guth, was analogous to observing that you don’t need to give an explicit description of how to put a wool sweater into a box in order to know that you can actually put it into the box. It is actually a little hard, at least for myself, to get a full grasp on Gromov’s other contributions as they span unfamiliar fields of mathematics, but I recommend this nice What is… article, written by the late Marcel Berger, describing Gromov’s contribution to the understanding of isosystolic inequalities.

There is the perception about great mathematicians that, while they are no doubt very clever, somehow they have lost a little of the common sense that the rest of us possess. I’m naturally inclined to discount such thinking; I am far happier believing that we are all fool enough to take absence of common sense, on certain occasions. However, it is hard to dismiss the  idea entirely given the following admission by Gromov, in his personal autobiographical recollections he wrote on receipt of the Abel prize in 2009:

The passage speaks for itself, but I wish to emphasize that Gromov’s discovery of the correct pronunciation of French verb endings came after ten years of living in Paris. You don’t have to have extensive experience learning foreign languages to appreciate how remarkable an oversight this is. It certainly puts his remarks in other interviews about dedicating one’s life to mathematical pursuits in a rather strong light.

If you read the entire autobiographical essay you will find it rather short on biography. The best biographical details I have found came from this La Monde article written on his being awarded the Abel prize. Even then the details seems to be coming second hand. The most interesting parts concern his leaving the Soviet Union:

I wanted to leave the Soviet Union from the age of 14. […] I could not stand the country. The political pressure there was very unpleasant, and it did not come only from the top. […] The professors had to teach in such a way as to show respect for the regime. We felt the pressure of always having to express our submission to the system. One could not do that without deforming one’s personality and each mathematician that I knew ended up, at a certain age, developing a neurosis accompanied  by severe disorders. In my opinion, they had become sick. I did not want to reach that point.

Gromov, according to Georges Ripka, via La Monde (apologies for my translation)

As the article goes on to explain, Gromov decided his best chance to escape was to hide his mathematical talent. He quit math, quit his university, and burned all his academic bridges. He stopped producing mathematics. Or at least writing it. He joined some meteorological institute and did research on paper pulp. Eventually, he was granted permission to emigrate to Israel, but on landing in Rome in 1974 he set off for the States instead, where Jim Simons secured him a position at Stony Brook.

(As a side note, this is Jim Simons of Renaissance Technology and Simons foundations fame. Simons, aside from his mathematical contributions, is probably one of the most important mathematicians alive in terms of funding, supporting, and propagandizing for mathematics. He is considered influential enough for the New Yorker to profile. Alongside Gromov, he is one of the names that every mathematicians should know.)