# Is the center of the fundamental group of the double torus trivial?

I know that the fundamental group of the double torus is $\pi_1(M)=\langle a,b,c,d;a^{-1}b^{-1}abc^{-1}d^{-1}cd\rangle$.

How can I calculate its center subgroup $C$? Is $C$ trivial?

Let $p$ be the quotient map from $\pi_1(M)$ to $H_1(M)$, maybe it's easy to prove that $p(C)$ is $0$ in $H_1(M).$ That will also solve my problem.

Thanks, Yan

-
According to a theorem of Preissman, every nontrivial subgroup of the fundamental group of a compact negatively curved Riemannian manifold is isomorphic to $\mathbb{Z}$. This doesn't quite prove that the center of the fundamental group of a surface is trivial, but that would be my guess. – Paul Siegel Nov 6 '12 at 8:12
Like all of the surface groups (apart from the torus itself, where it is free abelian of rank 2), this one is torsion-free and word-hyperbolic. Its centre must be trivial, since word-hyperbolic groups do not contain non-cyclic free abelian subgroups. – Derek Holt Nov 6 '12 at 8:18
Two quick ways: (1) $G$ is an amalgamated product of two free groups (whose centers are trivial), so it itself has trivial center. (2) As Derek Holt remarked, all fundamental groups of hyperbolic surfaces have trivial centers. This can be seen "geometrically" by allowing $a$ and $b$ in $G$ to commute; then the isometries they represent in the hyperbolic plane have the same fixed points, and using discreteness, imply $a$ and $b$ are powers of the same word in $G$. This says centralizers are cyclic, and it is easy to go from there to trivial center (since $G$ itself is non-cyclic). – user641 Nov 6 '12 at 18:03

I want to expand on Derek Holt's comment, because it is interesting. I also want to add a little nugget of my own.

Both ideas generalise to the torus of genus $n$, $n>1$.

Firstly, note that your group has a single defining relation. There is a paper of G. Baumslag and Taylor which gives an algorithm to determine the centre of a group with a single defining relation (The centre of groups with one defining relator). That was the nugget of my own... (EDIT: actually, you do not need to go as far as this paper - it follows immediately from the fact that $G$ is generated by more than two elements and from Theorem 5.4 on p203 of Lyndon and Schupp's fine text Combinatorial Group Theory. This theorem says that for $M=\langle a, b, \ldots, c\rangle$ where $G=\langle X; R\rangle$ is a one-relator group, $a, b, \ldots, c\subsetneq X$, then if $g\not\in M$ then $M^g\cap M$ is cyclic (actually, it is trivial, but I won't go into that here...).)

I want to expand on Derek Holt's comment, but sticking with the fact that your group has a single defining relation. So, open Lyndon and Schupp's fine text Combinatorial Group Theory. In it, we discover two things. Firstly, because the relator is not a proper power (that is, not of the form $R^m$ for some $m>1$) our group is torsion free. Secondly, we learn of a very interesting theorem: B. B. Newmann's Spelling Theorem. (A spelling theorem is a theorem which tells us the form of a word which is (non-freely) equal to the identity in the group.)

In Lyndon and Schupp, the theorem is, if a remember correctly, slightly weaker than I need...the version I need is an addition by a chap called Gurevich, and can be found in a paper of Steve Pride and Jim Howie, A spelling theorem for staggered generalized 2-complexes, with applications. I'm sure we could do this a more simple way (we can - see the addendum), but I do quite like this theorem!

In modern language, the original version of B. B. Newmann's Spelling Theorem tells us that a one-relator group with torsion (that is, the relator is a proper power) is Hyperbolic in the sense of Gromov (see, for example, Bridson and Haelfinger's book - type in their names in google and you'll find the book. Alternatively, there are plenty of good introductions to hyperbolic groups - there is one of Jim Howie I quite like). The thing to note about Hyperbolic groups is that they do not contain a copy of $\mathbb{Z}\times\mathbb{Z}$ (indeed, they do not contain any subgroup of the form $\langle a, b; a^{-1}b^ma=b^n\rangle$, a so-called Baumslag-Solitar group. This is in Bridson and Haefinger).

For our uses, the Newmann-Gurevich Spelling Theorem tells us that the group you are looking at is hyperbolic. Here it is,

Theorem: (B. B. Newman, Gurevich)

Let $G=\langle X; R^m\rangle$ where $R$ is a freely and cyclically reduced word over $X^{\pm}$. If $W=_G 1$ then either $W\equiv 1$, $W$ is a cyclic shift of $R$ or $R^{-1}$, or some cyclic shift of $W$ contains two disjoint subwords $S^{m-1}S_0$ and $T^{m-1}T_0$ where $S$ and $T$ are cyclic shifts of $R$ or $R^{-1}$ and $S=S_0S_1$ $T=T_0T_1$ with $S_0$ and $T_0$ containing all the elements of $X$ which appear in $R$.

For us $G=\langle a, b, c, d; [a, b][c, d]\rangle$, and so we can only think about the $S_0$ and $T_0$. However, this is enough. If $W=_G1$ then $W$ contains $a^{-1}b^{-1}abc^{-1}d^{-1}$, or something similar. That is, it contains "most of" the relator. So, we can replace the big slice of the relator with the smaller slice, to get a new word $W_1=_G1$. As $W_1=_G1$ we can apply the theorem again and apply this trick to get a new word $W_2$ with $|W_1|>|W_2|$, and repeat until we get a word which is either the trivial word or the relator $R=[a, b][c, d]$. This algorithm is called Dehn's algorithm, and implies that our group us hyperbolic (in fact, it is equivalent - look up one of the hyperbolic references for this).

Thus, $G$ is torsion-free and hyperbolic. As it is hyperbolic it cannot contain $\mathbb{Z}\times\mathbb{Z}$. As it is torsion-free and does not contain $\mathbb{Z}\times \mathbb{Z}$, it cannot have a centre.

Now, I wrote this all down for two reasons. Firstly, it was good procrastination. Secondly, the groups you are looking at were the "original" hyperbolic groups. You see, sometime in around 1910-1920 Max Dehn came up with a version of B.B. Newman's spelling theorem which worked for these groups, and he applied this algorithm which bears his name (B.B. Newman's spelling theorem is basically a generalisation of Dehn's observation). It wasn't until Gromov came along in the 1980s that people finally understood what was going on with Dehn's algorithm and these surface groups, and the field of Hyperbolic Groups was introduced. The study of hyperbolic groups has turned out to be a rather fruitful field and has led to many important results, most recently the proof of the Virtual Haken Conjecture and Thurston's Virtually Fibering Conjecture.

Addendum: When I was using the strong version of Newman's Spelling Theorem I had a feeling that it was too much; that we didn't have to leave the pages of Lyndon and Schupp and delve into the world of research papers...and I was right! The groups we are looking at (so long as the genus $>1$) are $C^{\prime}(1/6)$ small cancellation groups. The standard reference for such groups is...Lyndon and Schupp! Basically, between Dehn and Gromov there was a class of groups called "Small Cancellation" groups which also exploited Dehn's algorithm (today they are the standard examples of Hyperbolic groups). (Okay, so that isn't quite right - there are lots of different flavours of small cancellation groups, and only the metric small cancellation ones are always hyperbolic - the $C^{\prime}(1/\lambda)$ ones, $\lambda \geq 6$.)

Small cancellation theory talks about things called pieces. Take all the relators in your presentation and their inverses, and try and lie one relator on top of another (the ends don't have to be "in line"). Anywhere where the two relators agree is called a piece. (In small cancellation theory we think of relators as "tiles" and the pieces are where the tiles can connect together.)

If every piece has length less than $1/6$ of the relator it is a part of then the presentation satisfies the $C^{\prime}(1/6)$ small cancellation condition.

You should realise that with your groups every piece has length $1$, so your group satisfies $C^{\prime}(1/7)$ (in general, $C^{\prime}(1/(4g-1))$).

Thus, $G$ is hyperbolic and we're done!

-

Using marked groups, the fundamental group $G$ of the double torus is a non abelian limit group (you can find a morphism from $G$ to $\mathbb{Z} \ast \mathbb{Z}$ showing that $[a,c] \neq 1$) and so $G$ has the same universal theory than a nonabelian free group. We deduce that $\forall x,y \left( x=1 \vee y=1 \vee [x,y]\neq 1 \right) \in \text{Th}_{\forall}(G)$, that is $Z(G)$ is trivial.

Also, you can say that because $G$ is a limit group, it is commutative transitive. So either $Z(G)$ is a trivial or $G$ is abelian.

Without using marked groups, notice that $G$ is a pinched word group $\mathbb{F}_2 \underset{\langle w \rangle}{\ast} \mathbb{F}_2$ where $w=[x,y]$. Here, it is shown that a pinched word group is fully residually free (proposition 2.9). So, if $g \in Z(G)$ and $x,y \in G$ such that $[x,y] \neq 1$, there exists a morphism $\varphi$ from $G$ to a free group such that $\varphi(g),\varphi(x),\varphi(y),\varphi([x,y])\neq 1$, supposing that $g \neq 1$. But $[\varphi(g),\varphi(x)]=[\varphi(g),\varphi(y)]=1$ so $\varphi([x,y])=[\varphi(x),\varphi(y)]=1$ (a free group is clearly commutative transitive). This is a contradiction, so $Z(G)$ is trivial.

-
$G$ a limit group $\Leftrightarrow$ $G$ fully residually free... – user1729 Apr 26 '13 at 9:45