# Why is the fundamental group of a compact Riemann surface not free ?

Consider a compact Riemann surface $X$ of genus $g$.
It is well-known that its fundamental group $\pi_1(X)$ is the free group on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by the single relator $[a_1,b_1]\cdot \ldots\cdot [a_g,b_g]$.
(This has of course nothing to do with the complex structure of $X$, but may be computed by considering the underlying topological manifold as a cell complex.)
This group is trivial for $g=0$ and free abelian on two generators for $g=1$.

For $g\geq 2$, however, I had always taken for granted that it is not free but I have just realized that I cannot prove that.

So, although I guess the answer is no, I'll ask my official question in an open way : Is $\pi_1(X)$ free for $g\geq 2$ ?

Edit: Users have now brilliantly solved the problem in multiple ways.
Non-freeness is definitely established, with 12 proofs plus sketches of proofs in the comments!
It is clearly impossible to select in a reasonable way an answer for "acceptance" among all these great answers .
Since the software forces me to make only one choice, I have chosen Daniel's answer because of its merit, but also because it acknowledges Vitali's contribution: Vitali was the first to sketch a solution (in the comments) .

• it's not free because it has nontrivial cohomology in dimension 2 which free groups don't. Apr 6, 2012 at 15:52
• @ Mark Sapir. I think your remark is very condescending and useless , since it has zero mathematical content. If this question is beneath your level, I'll be more than compensated by the friendly and informative comments and answers I am receiving right now. Apr 6, 2012 at 18:31
• I miss the days when I could idly peruse MathOverflow and learn a lot of interesting math from great answers to "low level" questions such as this one. Even though I was able to find an answer to this specific question, I still managed to learn quite a bit from all the other nice answers---about things I doubt I would have come across otherwise. It was this sort of experience that attracted me to MO in the first place, and I lament the fact that it is now rarer to come by. Anyway, thank you, Georges, for asking this question. Apr 6, 2012 at 23:31
• Speaking as someone who did manage to get some medium-good papers into the Annals, I can honestly say that I did not know the answer to this question, and learned something from the various responses given. Apr 7, 2012 at 0:44
• Dear Mark: I am sure there are many excellent graduate students at Chicago in algebra, algebraic geometry, algebraic topology, analysis, differential geometry, logic, number theory, PDEs, probability, and representation theory who do not know what a quasi-isometry is. Moreover, you must know this is the case, since you wouldn't use quasi-isometries in a colloquium talk without explaining or defining them. In any case, I'd be happy to continue this discussion over e-mail, but I don't think it would contribute anything for me to say anything further here. Best regards, Tom Apr 7, 2012 at 5:09

As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment. Let $F_n$ be the free group on $n$ letters; $K(F_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$. On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible. But then $$H^2(\pi_1(X), \mathbb{Z})=H^2_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero. In particular, $\pi_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.

• You don't need to appeal to uniformization to conclude that compact Riemann surfaces are aspherical; once you know that you can construct the orientable surface of genus $g$ as a topological space by identifying sides of a $4g$-gon, it suffices to construct a tiling of the hyperbolic plane that provides the right identifications. Apr 7, 2012 at 15:36
• Sure, that's essentially one proof of the weak (non-conformal) uniformization. And indeed, constructing such a tiling is pretty easy, e.g. by using pairs of pants. Apr 7, 2012 at 17:51
• Just a tiny remark, that your proof works for non-orientable surfaces as well, if only you drop the claim that $H^2(X,\mathbb Z)=\mathbb{Z}$ and change it to $H^2(X,\mathbb Z)=\mathbb{Z}$ or $\mathbb Z_2$, that is, non-zero. Apr 10, 2020 at 20:03

You can use covering spaces instead of cohomology.

(Edit: This proof does not use a presentation of the group, either -- just the weaker fact that its abelianization is free of rank $2g$.)

As Daniel pointed out in his comment, if the group were free it would have to be free on $2g$ generators. The genus $g$ surface has a $2$-sheeted covering space which is a genus $2g-1$ surface. Every index $2$ subgroup of a free group on $r$ generators is free on $2r-1$ generators, because it is the fundamental group of a $2$-sheeted covering space of a wedge of $r$ circles. So $2(2g-1)=2(2g)-1$, a contradiction.

• I was going to post that :( Apr 6, 2012 at 17:25
• But this only covers the odd genus case, right? Apr 6, 2012 at 18:15
• It should be pointed out that this answer, with little modification, could also be used to show that $\pi_1$ of any compact Kahler manifold is never free (unless it's trivial); cf. Mohan Ramachandran's comment. Apr 6, 2012 at 23:35
• The key observation is that in this case $b_1(X)$ is even, say equal to $2g$. So if $\pi_1(X)$ were free, it would be free of rank $2g$, and would therefore contain a subgroup $H$ of index 2, which is itself free of rank $|G:H|(\text{rank }\pi_1(X)-1)+1=4g-1$. But now consider the 2:1 covering $\tilde{X}\to X$ corresponding to $H$: it has $b_1(\tilde{X})=4g-1$, which is odd. This is a contradiction because $\tilde{X}$ must be compact Kahler as well. Apr 7, 2012 at 1:03
• @Elizabeth: The fundamental group of a torus is a free abelian group, not a free group. Apr 7, 2012 at 21:17

There are many good answers already, but maybe you'll be interested in a counting argument.

Let $Q_8$ denote the quaternion group. Consider the set

$$\mbox{Hom}(\pi_1(X),Q_8)$$

where $X$ is the closed surface of genus $g$. Since we have a presentation for $\pi_1(X)$, we note that a homomorphism to $Q_8$ is given by a $2g$-tuple of elements in $Q_8$ satisfying some group word. In particular, the cardinality of the set is finite. Indeed, if the fundamental group were free, the cardinality would be a power of $8$. However, combinatorics (or gentle representation theory) gives the cardinality of the set exactly:

$$|\mbox{Hom}(\pi_1(X),Q_8)| = 2^{6g-1}+2^{4g-1}$$

which (for $g>0$) is never a power of $2$, still less a power of $8$.

---Edit---

The "gentle representation theory" to which I refer can be found here: http://arxiv.org/abs/1102.4353

We show in section 4 that the cardinality of $\mbox{Hom}(\pi_1(X),G)$ can be written explicitly in terms of the dimensions of the irreducible representations of $G$.

• "Combinatorics" could mean anything. Could you elaborate? Same with "gentle rep. theory"... Apr 7, 2012 at 3:55
• @Igor Riven I provided a reference which gives the representation-theoretic argument Apr 7, 2012 at 4:17
• This is a beautiful answer. One highlight is that you don't need the fact that free groups are Hopfian, as Andy and I did in our answers. Moreover if you start by showing that $\pi_1(X)$ would have to be of rank $2g$, it suffices to exhibit a single finite group $G$ with a single tuple of elements not satisfying the group word, since then $|\text{Hom}(\pi_1(X),G)|<|G|^{2g}$. I think this is the most elementary argument possible. The precise count you give in your paper is very satisfying, however. Apr 7, 2012 at 4:45
• Very interesting. It is amazing that after so many good answers there is still room for another good one: thank you John! Apr 7, 2012 at 6:30
• Awesome "algebraist's answer"!
– JSE
Apr 11, 2012 at 12:29

I have a feeling that almost every field of pure mathematics has its "own" way to see that surface groups $\pi_g$ (of genus $g\ge 2$) are not free. The two arguments below are, of course, more complicated than the ones which are based on group theory, geometry, topology or homological algebra, but the idea is to see connection to other fields of mathematics. Here are two examples:

1. Algebraic geometry (combination of abelian and nonabelian Hodge theory): If $\pi_g\cong F_r$, free group of rank $r$, then, by looking at the 1-st Betti numbers and using the usual Hodge theory we see that $r$ has to be even. On the other hand, by Narasimhan-Seshadri theorem, the moduli space ${\mathcal M}_g$ of semistable rank 2 holomorphic bundles (with fixed determinant) on genus $g$ surface is analytically isomorphic to $Hom(\pi_g, SU(2))/SU(2)=Hom(F_r, SU(2))/SU(2)$, which would have odd (real) dimension $3r-3$. Contradiction, since (being a complex-projective variety) ${\mathcal M}_g$ is even-dimensional. (This is, of course, an argument similar to, but more complicated, than Mohan's.)

2. Geometric analysis: Suppose that $\pi_g\cong F_r$. Realize this isomorphism $\rho$ by a harmonic map $h$ from $S_g$ (genus $g$ compact Riemann surface) to a metric graph $\Gamma_r$ (the rose with $r$ leaves and unit edges). Preimages of generic points in $\Gamma_r$ under $h$ will be compact 1-dimensional submanifolds. By the maximum principle (look at the lifted harmonic map from the universal cover of $S_g$ to the tree), components of these submanifolds cannot bound disks in $S_g$, hence, they are not nul-homotopic. Hence, $\rho$ cannot be injective.

So, are there proofs (even difficult ones) which make essential use of other fields of mathematics, e.g.:

a. Number theory (algebraic or analytic number theory, or arithmetic algebraic geometry)?

b. Measure theory? (This might be difficult since $\pi_g$ and $F_r$ are "measure-equivalent".)

c. Probability? (An argument using random walk on graphs maybe?)

d. Dynamical systems/ergodic theory?

e. Functional analysis? (Maybe infinite-dimensional unitary representations of $\pi_g$ and $F_r$?)

f. Logic?

g. Commutative algebra?

Update: Here is a proof by the commutative algebra. Consider the affine schemes $S=Hom(\pi_g, GL(2))$ and $S'=Hom(F_r, GL(2))\cong GL(2)^r$. The latter, being an open subscheme of the affine space, has the same dimension of Zariski tangent space at every point. Consider $S$, let $R$ be its coordinate ring and $m_1, m_2\subset R$ be the ideals corresponding to the points $\rho_1$ (the trivial representation) and $\rho_2$, the representation which sends all but one standard generators to $1$ and the remaining generator to any noncentral matrix in $GL(2)$. Then (by a reasonably simple computation) $d_1=dim(R/m_1)=8g$ while $d_2=dim(R/m_2)=8g-2$. Hence, $d_1>d_2$ and dimensions of Zariski tangent spaces to $S$ at $\rho_1, \rho_2$ are different. In particular, the schemes $S, S'$ (equivalently, their rings) cannot be isomorphic and $\pi_g$ cannot be isomorphic to $F_r$.

• Dear Misha, thanks for this very original contribution and your quite creative questions: I would indeed love a proof from commutative algebra! Apr 7, 2012 at 6:24
• I put one in for universal algebra :) Apr 7, 2012 at 14:06
• @Georges Elencwajg: Dear Georges, actually, there should be a commutative algebra proof and you can probably even write one for MO! The idea is that the scheme $Hom(F_r, GL(2))\cong GL(2)^r$ is smooth, while the scheme $Hom(\pi_g, GL(2))$ has quadratic singularity at the trivial representation (actually, at every reducible representation). You should be able to detect this by looking at the coordinate ring of $Hom(\pi_g, GL(2))$. Apr 7, 2012 at 15:21
• Along the lines of 1., I computed Hom$(\pi_1 M^g, U)/U$ (up to homotopy) and Tyler Lawson computed Hom$(F_r, U)/U$ (up to homotopy) and they're not homotopy equivalent (for any $g, r>0$). The former is the infinite symmetric product of $M^g$ and the latter is $(S^1)^r$ (which is homotopy equivalent to the infinite symmetric product of a wedge of $r$ circles). Here U = colim U(n) is the infinite unitary group. Apr 7, 2012 at 18:22
• @Dan: the only way I know how to prove that infinite symmetric products aren't homotopy equivalent is to appeal to Dold-Thom: the homotopy groups of the infinite symmetric product of some nice space $X$ are just its singular homology groups, so this is a roundabout way of reproducing the homological argument (once you accept that Eilenberg-Maclane spaces are determined up to homotopy by their fundamental groups). Did you have another argument in mind? Apr 9, 2012 at 6:40

Since the abelianization of $\pi_1(\Sigma_g)$ is $\mathbb{Z}^{2g}$, if this group were free it would be free of rank $2g$. Since free groups are Hopfian, any $2g$ generators for $F_{2g}$ are free generators. The universal property of free groups then gives

$$\text{Hom}(\pi_1(\Sigma_g),G)\simeq \text{Hom}(F_{2g},G)\simeq G^{2g}$$ for any group $G$, i.e. every collection of $2g$ elements $a_1,b_1,\ldots,a_g,b_g$ in any group $G$ would have to satisfy $[a_1,b_1]\cdots[a_g,b_g]=1$. This would imply that every torsion-free group is abelian, since for any $x$ and $y$ in $G$ setting $a_i\mapsto x$ and $b_i\mapsto y$ gives $[x,y]^g=1$.

This is patently absurd. For explicit counterexamples, send $a_i$ and $b_i$ to reflections across two parallel lines, or reflections across two lines meeting at an irrational angle, or to the matrices $\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 1 & 0\\ 1 & 1\end{bmatrix}$.

• This can be simplified slightly, since setting $a_i=b_i = 1$ for $i>1$ will show that ANY group is abelian, and $S_3$ is a counterexample. Apr 6, 2012 at 22:00
• Dear Tom, This is essentially the second proof I gave in my answer! best, Andy. Apr 6, 2012 at 22:47
• @Igor: great point! @Andy: sorry, I didn't see that part of your answer (maybe it wasn't there when I opened the page?) I think I will leave mine only because it answers the implicit question "but how do you know that a free group satisfies no relation?" with "well, for a given relation it suffices to show that some group doesn't satisfy that law, and this can be done using some concrete group arising in nature". Here the benefit of using a concrete group (rather than one given by generators and relations) is that there's no worry about checking whether an element is equal to the identity. Apr 6, 2012 at 22:52
• Dear Tom, Since $[a_1,b_1]\cdots [a_g,b_g]$ is a reduced word in the generators for the free group on the $a_i$ and the $b_i$, it certainly is not the identity! best, Andy. Apr 6, 2012 at 23:02
• Apparently Tom doesn't think that a free group is a 'concrete group arising in nature'. De gustibus non est disputandum.
– HJRW
Apr 7, 2012 at 16:48

Original proof. How about a surface group is 1-ended since its universal cover is the hyperbolic plane and the non-abelian free group has infinitely many ends since the universal cover of the wedge of circles is a tree.

Second proof. Since Misha has suggested a proof from each area here is a proof from universal algebra. Let V be the variety of extensions of elementary abelian 2-groups by elementary abelian 2-groups. The word problem for the relatively free group on X in this variety is well-known. A word is trivial if and ony if it is trivial in $(\mathbb Z/2)^X$ and the loop it labels in the Cayley graph of $(\mathbb Z/2)^X$ with respect to $X$ traverses each geometric edge an even number of times.

If the surface group were free it would have to be free on $2g$-generators because of the abelianization. So if we factor by the verbal subgroup associated to V we would get a free group in this variety on 2-generators. But the group in V on 2g-generators with the surface defining relation is a proper quotient of the relatively free group because the product of commutators in question uses each edge of the Cayley graph of $(\mathbb Z/2)^{2g}$ exactly once. Since these are finite groups, a proper quotient is not free.

• very nice argument Apr 6, 2012 at 19:49
• Using geometric group theory as in the original proof, one can show that $\pi_1(\Sigma_g)$ is not isomorphic to a non-trivial free product at all (as opposed to just a free product of $\mathbb{Z}$'s). Sep 14, 2017 at 13:34
• @MichaelAlbenese, that is true since a free product has two or more ends. I would expect though the second proof can be adapted because the free product will get turned into the varietal coproduct. Sep 14, 2017 at 16:31

FINAL EDIT : This edit cleans up the first proof (and simplifies it -- there are no longer any references to the free nilpotent group) and adds some remarks to the second proof following the discussion in the comments.

PROOF 1.

Here's a low-tech way to see that a surface group is not free (though cohomology is secretly lurking in the background). Let $G_g = \langle a_1,b_1,\ldots,a_g,b_g\ |\ [a_1,b_1]\cdots [a_g,b_g]=1 \rangle$ be the surface group. Form the group

$$\tilde{G}_g = \langle a_1,b_1,\ldots,a_g,b_g,t\ |\ [a_1,b_1]\cdots [a_g,b_g]=t, [a_i,t]=1, [b_i,t]=1\ \text{for all 1 \leq i \leq g} \rangle$$

The subgroup of $\tilde{G}_g$ generated by $t$ is contained in the center and the quotient is $G_g$. Below I will show that this subgroup is infinite cyclic. We thus have a central extension

$$1 \longrightarrow \mathbb{Z} \longrightarrow \tilde{G}_g \longrightarrow G_g \longrightarrow 1.$$

If $G_g$ were free, then this would split as a direct product. However, since $t$ becomes zero when we abelianize $\tilde{G}$, there is no splitting homomorphism $\tilde{G}_g \rightarrow \mathbb{Z}$. Thus $G_g$ cannot be free.

It remains to show that the subgroup generated by $t$ is infinite cyclic. Let $H$ be the $3$-dimensional Heisenberg group, ie the group of upper-triangular $3 \times 3$ integer matrices with $1$'s on the diagonal. As is well-known, $H$ has a presentation $$H = \langle x,y,z\ |\ [x,y]=z, [x,z]=1, [y,z]=1 \rangle.$$ Examining the presentations, there is a homomorphism $\psi : \tilde{G}_g \rightarrow H$ with

$$\psi(a_1) = x \quad \text{and} \quad \psi(b_1) = y \quad \text{and} \quad \psi(t) = z$$

and

$$\psi(a_i) = \psi(b_i) = 1 \quad \quad (2 \leq i \leq g)$$

Since $z$ generates an infinite cyclic subgroup of $H$ (as a matrix, $z$ is the matrix with $1$'s on the diagonal and at position $(1,3)$ and $0$'s elsewhere), it follows that $t$ generates an infinite cyclic subgroup of $\tilde{G}_g$.

PROOF 2.

It is known that free groups are Hopfian, i.e. that all surjections from a free group to itself are isomorphisms. A simple-minded cancellation-based proof (using Nielsen reduction) can be found in Proposition 2.7 of Lyndon and Schupp's book "Combinatorial group theory". Alternatively, Malcev proved that all residually finite groups are Hopfian (this can also be found in Lyndon and Schupp), and there are many proofs that free groups are residually finite; see the answers to the question Why are free groups residually finite?

This implies that if $F$ is a free group on $n$ generators and $S$ is a generating set for $F$ which has $n$ elements, then $S$ is a free generating set. But this implies the result -- letting $G_g$ be the surface group as above, by abelianizing we see that if $G_g$ were a free group, then it would be free on $2g$ generators. But $a_1,b_1,\ldots,a_g,b_g$ is a generating set of size $2g$ which is not free since it satisfies a relation. Thus $G_g$ is not free.

• I guess you are interpreting $H_2(\pi_1(X), \mathbb{Z})$ as the kernel of a maximal stem extension, which is the sense in which cohomology is lurking in the background? Apr 6, 2012 at 17:26
• I think your new proof is about as elementary as one could hope for, and satisfies my desire for a proof that a mathematician at the beginning of the 20th century would like. Apr 6, 2012 at 18:36
• The proof a free group is residually finite is 3 lines. Write the word as a finite line automaton and complete to a permutation automaton. So one doesn't even need anything. Malcev's result that fg implies Hopfian for residually finite groups is also elementary and short. Apr 6, 2012 at 20:39
• Stallings proof is not really geometric. It is language only. The proof in content is the same as the accepted answer in your link which is elementary. It is a consequence of the fact that any partial permutation of a finite set can be completed to a permutation. Apr 6, 2012 at 21:11
• With no disrespect to Stallings who's is one of my math heroes. Apr 6, 2012 at 21:21

Here is a Stallings-style argument. Suppose that $S$ is a closed connected surface. Let $G = \pi_1(S)$. Suppose $T$ is a graph and $v \in T$ a vertex. Let $F = \pi_1(T,v)$ and suppose that $\phi \colon G \to F$ is any homomorphism.

Theorem: If $\phi$ is injective then $S$ is the two-sphere.

Here's the proof. Let $e$ be any edge of $T$ and let $p \in e$ be the midpoint. The given generating set $\{a_i,b_i\}$ for $G$ gives a one-skeleton for $S$, with one vertex; call the vertex $u$. Let $Q$ be the single two-cell remaining in $S$.

We may now define a map $f$ from the one-skeleton of $S$ to $T$ by mapping $u$ to $v$ and by mapping the edges of $Q$ to paths in $T$ as instructed by $\phi$. Note that the relation in $G$ is killed by $\phi$ -- so, thinking of the image as a word $w$, we find $w$ is a completely reducible word. The reduction of $w$ tells us how to extend $f$ to the two-cell $Q$. (Draw $Q$. Subdivide and label the edges of $Q$ by their images. The first reduction of $w$ gives a triangle cutting off a pair of these new, smaller edges. Etc.)

It follows that $f$ induces the homomorphism $\phi$. (It does the correct thing to the generators and to the relator.) Now consider $C = f^{-1}(p)$. By construction, $C$ is a collection of circles in $S$. If any component of $C$ is non-trivial in $G$, then $\phi$ is not injective, a contradiction. From the Jordon curve theorem deduce that all components of $C$ bound disks. Thus we may homotope $f$ so that $p$ is not in the image. Thus we may homotope $f$ so that the interior of $e$ is not in the image. Thus we may reduce the number of edges in $T$. We now induct downwards. In the base case, where $T = v$, we find that $f$ is the constant map, so $G$ is the trivial group, so $S$ is the two-sphere.

Riffing off of Ben Steinberg's answer, which amounts to the statement that a tree is not quasi-isometric to the hyperbolic plane, here's a proof which doesn't require knowing anything about ends.

Tile $H^2$ by regular $2g$-gons which are fundamental domains for the action of $\pi_1 X$. The boundaries of these $2g$-gons form a Cayley graph $\Gamma$ of $\pi_1 X$. If $\pi_1 X$ were a free group then $\Gamma$ would be quasi-isometric to a tree, and so there would exist constants $C \ge 0$ and $s \in (0,1)$ such that every closed edge path $\gamma$ in $\Gamma$ can be written as a concatenation $\gamma = \gamma_1 * \gamma_2$ so that $Length(\gamma_1), Length(\gamma_2) \ge s Length(\gamma)$, and the initial and terminal endpoints of $\gamma_1$ have distance $\le C$ in $\Gamma$. But for sufficiently large $r > 0$, choosing $\Gamma$ to be a closed edge path that stays within a uniform distance of the radius $r$ circle in $H^2$, we get a contradiction.

• Isn't the dual graph to the picture of $2g$-gons the Cayley graph? Apr 7, 2012 at 2:03
• Both the boundary graph and the dual graph are Cayley graphs. Apr 7, 2012 at 2:15
• In addition, you can arrange matters so that boundary graph and the dual graph are isomorphic as graphs! (I think, if one is careful, the induced map on labels can be an isomorphism of groups. It isn't an automorphism of the surface group because the basepoint has to move.) Apr 7, 2012 at 13:27
• I don't know whether this is true in general. It's true in genus 2 where there are just 4 ways to glue the octagon: $a b \bar a \bar b c d \bar c \bar d$; $a b c d \bar a \bar b \bar c \bar d$; $a b c d \bar a \bar d \bar b \bar c$; $a b \bar a c \bar b d \bar c \bar d$. These are all self-dual. But I don't know that this holds in higher genus. Apr 7, 2012 at 16:58
• I meant, I don't know if your parenthetical comment is true. Apr 7, 2012 at 16:59

There are already many excellent answers, and I'm adding another more out of solidarity than anything else. Unlike the other answers, this one is perversely complicated, although perhaps it addresses the question raised in Misha's answer about the existence of an arithmetic proof. I might also add my own question that I have wondered about:

Question: There are many known obstructions for group to be Kahler. Which of these extend to tame fundamental groups?

Lemma: Let $X$ be a smooth projective variety defined over an algebraically closed field. Let $\ell$ be a prime different from the characteristic, then the pro-$\ell$ part of the etale fundamental group $\pi_1^{et}(X)$ is not the pro-$\ell$ completion of a free group.

Sketch. (This is just a translation of the argument indicated in Mohan's comment to the original question.) Suppose it was, then after passing to a suitable etale cover, we get a new variety $Y$ with $$b_1(Y) := \dim H^1(Y_{et},\mathbb{Z}_\ell)= \dim Hom(\pi^{et}_1(Y),\mathbb{Z}_\ell)$$ odd. This is impossible, because by the hard Lefschetz theorem (Deligne) $H^1$ has a nondegenerate symplectic structure.

• Thanks a lot, Donu. I am really amazed (but happy) to see Deligne invoked in an answer to this question! Apr 15, 2012 at 20:09

I just wanted to point out that Vitali's comment, and Daniel Litt's elaboration, can be explained without any hyperbolic geometry. As they point out, once we know that an orientable surface $M^g$ of genus $g>1$ has contractible universal cover, the desired result follow by computing $H_2 (M^g)$ (or $H^2 (M^g)$). So, here are 3 other proofs that the universal cover $X$ of $M^g$ is contractible:

1. $X$ is a 2-dimensional manifold, and it is non-compact because the fiber of the universal covering is the fundamental group, which is infinite (it has infinite abelianization). Any non-compact n-manifold has $H^i (M) = 0$ for $i>n-1$ (this is Proposition 3.29 in Hatcher). So in our case, $H_i (X) = 0$ for $i>1$ and since $X$ is simply connected, $H_1 (X)=0$ also. By the Hurewicz Theorem, all the homotopy groups of $X$ must be trivial as well. Since $M^g$ is a CW complex, so is its universal cover, so Whitehead's theorem says $X$ is contractible.

2. Hatcher Example B.14 proves this by describing $M^g$ in terms of a graph of groups in which the maps on the edges are injective.

3. Topologically, the only non-compact simply connected surface is $R^2$ by the classification of (non-compact) surfaces.

• This is quite nice! Jul 12, 2013 at 23:12
• Thanks, Daniel. Sometimes I use MO as a convenient place to record arguments I like; this would be one example. Jul 13, 2013 at 0:28

By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on $H_1(S, \mathbb{Z}))$ is symplectic (preserves the symplectic form, and the image under the Torelli map is $Sp(2g, \mathbb{Z})).$ On the other hand, the action of the outer automorphism group of $F_{2g}$ on its abelianization is the whole $SL(2g, \mathbb{Z})$ of which $Sp(2g, \mathbb{Z})$ is a proper subgroup. So, since $Out(\pi_1(S_{g})) \neq Out(F_{2g})$ the groups themselves are not isomorphic.

• Two small corrections. First, the action of $Out(F_{2g})$ on $\mathbb{Z}^{2g}$ lands in $GL(2g,\mathbb{Z})$. Second, the outer automorphism group of the surface group is the extended mapping class group (you have to allow orientation-reversing homeomorphisms). The action of this on $\mathbb{Z}^{2g}$ lands in a group containing the symplectic group as an index $2$ subgroup. Apr 9, 2012 at 1:44
• @Andy: Yes, I sit corrected! Apr 9, 2012 at 1:46
• The proof I know of the Baer-Dehn-Epstein-Nielsen theorem requires machinery from coarse geometry, such as Gromov boundaries and quasi-isometries. Is there a more direct proof of BDEN? Apr 9, 2012 at 9:20
• Since all the names you mention (with the exception of Epstein, I guess) preceded Gromov, I assume the answer is "yes"... Apr 9, 2012 at 14:05
• @Sam Nead : The proof you describe is actually essentially the original one due to Dehn! In fact, lots of Gromov's ideas in geometric group theory have precursors in work of Dehn. There are other proof of the theorem by now, though. See Section 8.3 of Farb-Margalit's book "A primer on mapping class groups" for a bibliography of alternate proofs. Apr 9, 2012 at 14:25
• Free groups are residually finite.
• Finitely generated and residually finite implies Hopfian.

So $$F\{a_1,b_1,...,a_g,b_g\}$$ is Hopfian so it is not isomorphic to the quotient group by any nontrivial normal subgroup, hence

$$F\{a_1,b_1,...,a_g,b_g\}\not\cong \pi_1(X)=\dfrac{F\{a_1,b_1,...,a_g,b_g\}}{\langle[a_1,b_1]\cdot...\cdot[a_g,b_g]\rangle}\quad (1)$$

But taking the epimorphism $$\pi_1(X)\to\pi_1(X)_{ab}\cong \mathbb{Z}^{2g}$$ we see that $$\pi_1(X)$$ has least set of generators of size $$2g$$ and from $$(1)$$ it isn't isomorphic to the only free group in $$2g$$ generators.

• This proof (which I like a lot!) is in fact already contained in Andy Putman's answer.
– HJRW
Feb 7, 2019 at 13:00
• Generally speaking, one shouldn't use less-than signs < and greater-than signs > as delimiters. I edited to replace << and >> with $\langle$ \langle and $\rangle$ \rangle, but I'm not sure if that was what you meant by the doubled-up signs. Feb 7, 2019 at 19:12