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COURSE NOTES FOR THE SPRING 2010 TUTORIAL IN COMBINATORIAL AND GEOMETRIC GROUP THEORY THOMAS KOBERDA Contents 1. Free groups and free products 1 2. Basics of hyperbolic geometry and background from diﬀerential geometry 3 3. Quasi-isometries 8 4. Geometrization of surfaces (INCOMPLETE) 11 5. The geometry of H2 /P SL2 (Z) 11 6. HNN extensions and embedding theorems for ﬁnitely generated groups 13 7. Unsolvability of certain decision problems in group theory 15 These course notes will follow the material covered in lecture. They will be updated regularly as the course progresses. The reader is assumed to have a basic knowledge of group theory, topology and geometry, equivalent to that of an intermediate-to-advanced undergrad- uate. The material here is supplemented by the extensive problems. 1. Free groups and free products Let G be a group. If we forget the operation, we obtain a set. This is called the “forgetful functor” in the language of category theory. We wish to construct an left adjoint to this functor, called the “free group functor”. In less nonsensical terminology, we are given a group G and a set X, and we want a group F (X) such that for each map of sets f :X→G we obtain a unique group homomorphism F (f ) : F (X) → G which extends f . Intuition tells us that F (X) should just be the set of words in X and the formal inverses of elements of X under concatenation, and that F (f ) should be deﬁned in the unique way that makes this map a homomorphism. Let w be a formal word in X and its formal inverses. We say that w is reduced if there is no occurrence of xx−1 or x−1 x as a subword of w. 1 2 T. KOBERDA Theorem 1.1 (Free groups exist). The free group on a set X, denoted F (X) consists of the set of reduced words in X and its formal inverses. The identity (1) is the empty word. The group operation is given by formal concatenation. The reason that this is a theorem is that we need to do some work to prove that our intuitive deﬁnition really does give rise to a group law. The existence of the identity is be deﬁnition and its properties are immediate. The existence of inverses is also obvious, since we take a word in X, write it backwards, and replace each letter with its inverse. The diﬃculty is in verifying that the concatenation operation is associative. If we can embed the set of reduced words in X into a group of automor- phisms of some object, then the associativity of the composition of permu- tations does all the work for us. Proof that free groups exist. We will make F (X) act on a certain graph. Let the vertices of a graph T (X) be the reduced words in X. We draw a single edge between two vertices if the two words diﬀer by multiplication by an element of X on the right. We label the edge by the element by which we multiply. Note that this is not quite well deﬁned: to traverse an edge in one direction we multiply by an element x ∈ X and in the other direction we multiply by x−1 . The reader should verify that T (X) is a tree. To do this, use the notion of length of words: for w ∈ F (X), (w) is the smallest n such that w is a reduced product of n elements of X. For w ∈ F (X), we deﬁne a symmetry of T (X) which we denote by fw . The action of fw is to multiply on the left by w and reduce. The reader should verify that w acts without a ﬁxed point if w = 1 and preserves the edge relation and hence is an automorphism of T (X), and that thus we embed F (X) as a group of automorphisms of T (X). In group theory, we are interested in three fundamental problems: (1) Given G = S | R and w ∈ F (S), can we algorithmically determine if w = 1 in G (word problem)? (2) Given g, h ∈ G, can we algorithmically determine if g and h are conjugate (conjugacy problem)? In the same automorphism orbit (Aut(G)-orbit problem)? (3) Given two groups G, H together with ﬁnite presentations, can we algorithmically determine whether G and H are isomorphic (iso- morphism problem)? For the free group we can easily answer all of these questions (we will return to the automorphism orbit problem later). A word in a free group (it must be presented as a group with generators and no relations) is nontrivial if and only if it is reduced and nontrivial. Two words are conjugate if and only if they are cyclically reduced and equal. Two free groups are isomorphic if and only if they have bases of the same cardinality, equivalently if their abelianizations are isomorphic. In general, these problems are not solvable, even for ﬁnitely presented groups. Now that free groups are on sound footing, we may speak of presentations for groups. A presentation is an expression of the form G = S | R . The set COURSE NOTES 3 S is called the set of generators and R ⊂ F (S) is the set of relations. These specify a group G by taking F (S) modulo the smallest normal subgroup containing R. G is ﬁnitely generated if S is ﬁnite. It is ﬁnitely related if R is ﬁnite. It is ﬁnitely presented if both S and R are ﬁnite. Given two groups A and B we want to make sense of their free product. The universal property should be that if G is any group and φ, ψ are any two homomorphisms A, B → G respectively, then there should exist a unique A ∗ B and φ ∗ ψ extending φ and ψ. This group is called the free product. Its existence is witnessed by taking A = SA | RA and B = SB | RB and A ∗ B = SA , SB | RA , RB . Uniqueness is, as usual, a consequence of the universal property. Every element w ∈ A ∗ B can be written uniquely (verify this!) as a product of the form w = a1 b1 · · · an bn , where ai ∈ A, bi ∈ B, and each ai , bi (except for possibly a1 and bn ) is nontrivial. There is a diﬀerent useful characterization of free products: Lemma 1.2 (Ping-pong lemma). Let G act on a set X and let A, B < G be two subgroups which together generate G. Let X1 , X2 be nonempty subsets of X which are disjoint. Suppose that g(X2 ) ⊂ X1 for all g ∈ A and g(X1 ) ⊂ X2 for all g ∈ B, and suppose that |A| ≥ 3, |B| ≥ 2. Then G ∼ A ∗ B. = Proof. Assume w = a1 b1 · · · ak ∈ A ∗ B, each letter nontrivial in A or B. Observing w(X2 ) ⊂ X1 shows w is nontrivial. If w = b1 a2 · · · bk , conjugate by an element of A and apply the previous observation. If w = a1 b1 · · · bk , choose a ∈ A with a = a−1 to use the previous observation. The remaining 1 case is analogous. Corollary 1.3. The matrices 1 2 1 0 , 0 1 2 1 generate a free group. Proof. Use the usual action of SL2 (Z) on R2 , and use as X1 and X2 the regions above and below the line y = x. 2. Basics of hyperbolic geometry and background from differential geometry We proved in class that the stabilizer of H2 in Aut(C) is P SL2 (R). Theorem 2.1 (Classiﬁcation of elements of P SL2 (R)). Every element A ∈ P SL2 (R) is conjugate to a diagonal matrix, a unipotent matrix which is upper triangular, or a rotation. This can be determined by looking at the absolute value of the trace of A and determining whether it is larger, equal to, or less than 2, respectively. Proof. Since we are consider the absolute value of the trace, we will have no problems lifting A to SL2 (R). The characteristic polynomial pA is given by x2 − tr(A)x + 1. The discriminant of pA is (tr(A)2 − 4). In the ﬁrst case, this number is positive so that A is diagonalizable, which was the ﬁrst claim. In the second case, the discriminant if 0 so that pA has a unique root which 4 T. KOBERDA is 1 (up to a sign) so that A is conjugate to a unipotent upper triangular matrix by the Jordan normal form. In the last case, pA has two conjugate non-real roots. The reader should verify that then A has a ﬁxed point in H2 , which up to conjugacy is i (or the origin in the disk model of hyperbolic space). The Schwarz lemma gives our conclusion. We want to describe simply connected Riemann surfaces together with intrinsic metrics which are invariant under their groups of holomorphic au- tomorphisms. The Riemann Mapping Theorem says that any simply con- nected Riemann surface is biholomorphic to either the Riemann sphere C, the complex plane C, or upper half space H2 . The fact that C is not con- tractible implies that it is not biholomorphic to either of the other two. Liouville’s Theorem implies that any holomorphic map C → H2 is constant. Let α be a holomorphic automorphism of C. Let r1 , . . . , rn be its roots and p1 , . . . , pm its poles (it is a nontrivial fact that n = m). Comparing α to (z − ri ) q(z) = (z − pj ) implies by Liouville’s Theorem that α and q diﬀer by a nonzero constant multiple. It is only possible for α to by a bijection if both the numerator and denominator have degree one, and one veriﬁes that each α ∈ Aut(C) is a fractional linear transformation. If α ∈ Aut(C), then the removable singularity theorem imlies that α extends to an automorphism of C which ﬁxes ∞ and is therefore an aﬃne transformation. If α ∈ Aut(H2 ), we use the fact that H2 is biholomorphic to the unit disk, so that the Schwarz reﬂection principle shows that we can holomorphically extend α to C by 1/α(z). It follows that P SL2 (R) is precisely the holomorphic automorphism group of H2 . Let M be a Riemannian manifold. This means that we have a smoothly varying family of positive-deﬁnite inner products gx : Tx M × Tx M → R. Abstractly, these inner products are suppose to give a notion of length to each tangent vector at x ∈ M . The positive-deﬁnite condition says that for each x and v ∈ Tx M , gx (v, v) ≥ 0 with equality holding if and only if v = 0. If f : M → M is smooth map, we say that f preserves g if g = f ∗ g. Precisely what this means is that gx (v, v) = gf (x) (Df (v), Df (v)). In classical diﬀerential geometry, if U ⊂ R2 is an open set with coordinates x, y, then a metric is speciﬁed by a 2-tensor E(x, y) dx2 + 2F (x, y) dxdy + G(x, y) dy 2 (this notation goes back to Gauss). In more modern terminology, this expression should be thought of as a smooth section of Sym2 (T ∗ M ), the second symmetric power of the cotangent bundle, but this level of ab- straction is unnecessary for our purposes. Fixing a point (x, y) ∈ R2 , we can reconstruct the Riemannian metric from the 2-tensor above by putting the terms into a matrix: E(x, y) F (x, y) g(x, y) = . F (x, y) G(x, y) The inner product of two vectors (a, b) and (c, d) at (x, y) is given by ap- plying the linear map g(x, y) to the matrix (c, d) and then taking the usual inner product of the resulting vector with (a, b). One should verify that the COURSE NOTES 5 resulting real number is unchanged by switching the role of (a, b) and (c, d), and this follows from the easy observation that A is self-adjoint. Let g1 and g2 be Riemannian metrics on M . We say that g1 and g2 are conformal, or are in the same conformality class, if there is an everywhere positive function M → R which satisﬁes g1 = f (x)g2 . It is obvious that there can exist metrics which are not in the same conformality class when the dimension of M ≥ 2, since it is not true that all positive-deﬁnite inner products are multiples of each other. We call the metrics on U which are in the conformality class of the Euclidean metric conformal. If we identify U ⊂ C, the metric becomes expressible as the symmetric 2- tensor ϕ(z) dz 2 = E(x, y)(dx2 +dy 2 ). The reader should verify that there are no conformal metrics on C or C which are invariant under all holomorphic automorphisms of the respective Riemann surfaces. Theorem 2.2. The metric 2|dz| 1 − |z|2 is the unique (up to scale) conformal metric on the unit disk in C which is invariant under all of its holomorphic automorphisms. This metric is called the hyperbolic metric. Proof. Let ϕ(0) > 0 be any candidate. There is a holomorphic automor- phism of the disk which sends a particular z to zero, and one such is given by ζ −z ζ→ . 1 − zζ If v is a tangent vector at z, we see that v must be sent to v , 1 − |z|2 which can be seen by computing the derivative. Since the metric at z is the pullback of the metric at zero, we must have ϕ(0) ϕ(z) = . 1 − |z|2 Corollary 2.3. The hyperbolic metric on H2 is given by |dz| , (z) or in Euclidean coordinates by |dz| . y Proof. We use the fact that H2 is biholomorphic to the unit disk via the map z−i z→ , z+i and hence the hyperbolic metric on H2 must be the pullback of the holo- morphic metric on the unit disk via this biholomorphism. 6 T. KOBERDA We want to develop the appropriate notion of lines in H2 , which are called geodesics. They are locally distance minimizing. Theorem 2.4. Let b > a > 0. The hyperbolic distance between bi and ai is ln(b/a). Proof. Let ai and bi be connected by a path (x(t), y(t)) parametrized over the unit interval. We have a sequence of inequalities: b 1 dy (dx/dt)2 + (dy/dt)2 ln(b/a) = ≤ dt. a y 0 y The last expression is the hyperbolic length of (x(t), y(t)), and the second to last is the result if x(t) = 0 for all t. The reader should ﬁll in the details necessary to establish that the vertical segment connecting ai and bi is the unique such shortest path. Therefore, straight vertical lines in H2 are geodesics. Since P SL2 (R) acts by isometries, we see that semicircles which intersect the real line perpen- dicularly are all geodesics in the hyperbolic metric. If x, y ∈ H2 are any two points, then they lie on either a straight vertical line or on such a semicircle. Therefore any two points in H2 can be connected by a unique geodesic. It is easy at this point to see that a hyperbolic element of P SL2 (R) stabilizes a unique geodesic in H2 . We are now in a position to give another interesting application of the ping-pong lemma. Theorem 2.5. Let A and B be two hyperbolic elements in P SL2 (R) which ﬁx diﬀerent geodesics in H2 . Then there exists an N > 0 such that AN , B N is free. Proof. Consider the four endpoints of the two geodesics. Cut out four dis- joint half-planes about each of the endpoints. By considering the dynamics of a hyperbolic element we see that one of the endpoints of each geodesic is a repelling ﬁxed point and the other is an attracting ﬁxed point for pos- itive powers of the stabilizing hyperbolic elements. We see that suﬃciently high positive powers of A send everything except for the half-plane contain- ing the repelling ﬁxed point of A into the half-plane containing containing the attracting ﬁxed point. It is easy to see that the ping-pong lemma now applies. As an application of hyperbolic geometry to group theory, we prove the following theorem which is a reﬂection of more general properties of so-called hyperbolic groups: Theorem 2.6. A free group F has the unique roots property: if g, h ∈ F and n ∈ Z, we consider the equation g n = h. If a solution exists, it is unique. Proof. We do the case where F has rank two and leave the general case to the reader. We can embed F as a subgroup of P SL2 (R). If h ∈ F , then as an element of P SL2 (R) it is either elliptic, parabolic or hyperbolic. Suppose that g exists. It is easy to check that p ∈ H2 ∪ R ∪ {∞} is a ﬁxed point of h if and only if it is a ﬁxed point of g. It follows that g has the same COURSE NOTES 7 classiﬁcation as h. An elliptic element is classiﬁed by its ﬁxed point in H2 and the degree by which it rotates (after we conjugate the ﬁxed point to the origin in the unit disk). A parabolic element is classiﬁed by its ﬁxed point at inﬁnity and its translation distance along the x-axis in the upper half-plane model (after we conjugate the ﬁxed point to be at inﬁnity). A hyperbolic element is classiﬁed by its ﬁxed points at inﬁnity and its translation distance along its ﬁxed geodesic. In all but the elliptic case, the claim is clear. On the other hand, we can always choose the embedding of F into P SL2 (R) to avoid elliptic elements. Next we would like to calculate the areas of some naturally occurring shapes in H2 . To deﬁne what these objects might be, we deﬁne a half-space in H2 to be determined by a geodesic γ in H2 . The complement of γ in H2 is a union of two regions. We deﬁne a convex polygon in H2 to be a ﬁnite intersection of half-spaces. A polygon in H2 is a ﬁnite union of convex polygons. The intersection of two geodesics bounding P is called a vertex of P . The angle at the vertex is the Euclidean angle between the two tangent vectors. If two geodesics bounding sides of P meet at a point at the circle at inﬁnity, then the point at inﬁnity where they meet is called an ideal vertex of P . The reader should verify that the angle at an ideal vertex is zero. If P is a polygon in H2 , then removing larger and larger balls centered at a point in the interior of H2 yields a ﬁnite number of ends of the polygon. Intuitively, they can be deﬁned for any topological space X and are the connected components of the complements of larger and larger compact sets K ⊂ X. Rigorously, the ends of X, denoted e(X), are deﬁned to by lim π0 (X \ K). ← − K⊂X If an end is a piece of the polygon bounded by two geodesics going oﬀ to the same point at inﬁnity, we say that the end is a cusp of P . Proposition 2.7. The area of a polygon P is ﬁnite if and only if one of the two following conditions hold: (1) P is bounded. (2) P is unbounded but all of its ends are cusps. Proof. In the ﬁrst case, P is compact and hence has ﬁnite area. For the second case, we will ﬁnd an upper bound for the area of a cusp in the sequel. If P is unbounded and has an end which is not a cusp, then two consecutive geodesics bounding sides of P (in the cyclic order about the circle at inﬁnity) do not meet at the same point at inﬁnity. It is easy to check that the resulting end must have inﬁnite area. A triangle in P is an intersection of 3 half-spaces. The following theorem is an amazing fact about hyperbolic space which is a reﬂection of its nonzero curvature. Note that is deﬁnitely false in Euclidean geometry. Theorem 2.8 (Gauss–Bonnet Theorem for hyperbolic triangles). Let T be a hyperbolic triangle with one ideal vertex and two vertices in the interior of H2 with angles α and β. Then the area of T is π − (α + β). In particular, the area of an ideal triangle is π. 8 T. KOBERDA Proof. We consider the upper half-plane model. Since P SL2 (R) acts by isometries, we may assume the ideal vertex is at inﬁnity and that the other two vertices sit on a unit radius circle centered at the origin over the points a and b on the x-axis. We have: b ∞ 1 1 Area = dx dy = √ dy dx. y2 a 1−x2 y2 Simplifying, we get b β 1 √ dx = −1 dθ, a 1 − x2 π−α where we have made the substitution x = cos θ in the last step. The claim follows. It is easy to deduce the following corollary, which we leave as an exercise. Corollary 2.9. A hyperbolic triangle with angles α, β, γ has area π − (α + β + γ). 3. Quasi-isometries In this section we develop one of the most fundamental notions in geo- metric group theory, namely that of quasi-isometry. One of the motivating problems behind quasi-isometries is the fact that a presentation for a group is not well-deﬁned. To illustrate the problems this causes, we let G = S | R be a ﬁnitely generated group. We deﬁne the word length :G→Z by taking all strings in S which represent a ﬁxed word g. We deﬁne (g) to be the shortest such string (it may not be unique, but there exists a string of shortest length representing g). Consider two presentations for F2 , the free group of rank two. One is given by x, y , and the other is given by a, b, c | c = an . The reader should verify that both of these are presentations for F2 , and an isomorphism is given by sending x → a and y → b. Notice that with respect to the ﬁrst presentation, the word xn has length n. Identifying xn with an via the isomorphism, it has length 1 with respect to the second presentation. Let X and Y be metric spaces. We say that a function f : X → Y is a quasi-isometry if there exist constants λ ≥ 1 and C ≥ 0 such that 1 dX (x, y) − C ≤ dY (f (x), f (y)) ≤ λdX (x, y) + C. λ We say that X and Y are quasi-isometric (via f ) if there is a D ≥ 0 such that every point of Y is within a D-neighborhood of a point of f (X). Quasi-isometry is a rigorous expression of the notion that two spaces “look the same from far away.” Therefore, the constant C is introduced to allow for tearing on a small scale. Quasi-isometries are generally not continuous. COURSE NOTES 9 The reader should verify that any ﬁnite diameter metric space is quasi- isometric to a point. We will be dealing with quasi-isometry often when either X or Y (or both) are the Cayley graph of a ﬁnitely generated group. By the geometry of a group we mean the geometry of the quasi-isometry type of its Cayley graph. Recall that if G = S | R is a group, we construct a graph Γ(G, S) whose vertices are the elements of G and two vertices are connected if they diﬀer by an element of S. Precisely, we have that g and h are connected is there is an s ∈ S such that g = s · h. We will generally assume that S is closed under taking inverses and does not contain the identity element. We will partition S into two subsets S and (S )−1 , and we will label the edges of Γ(G, S) by elements of S . There will be a positive orientation on the edge from h to g if g = s · h. Traversing the edge in the other direction is tantamount to multiplying on the left by s−1 . Note that loops in Γ(G, S) correspond precisely to strings in S which represent the identity in G. Note also that the word length (g) coincides with the graph distance dΓ (1, g) in the Cayley graph. As an exercise, the reader should draw part of the Cayley graph for F2 with respect to the two presentations given above. Theorem 3.1. Let G admit presentations S1 | R1 and S2 | R2 with both S1 and S2 ﬁnite. Then Γ(G, S1 ) and Γ(G, S2 ) are quasi-isometric. Proof. Exercise. We now wish to state what is often called the fundamental observation ˇ of geometric group theory, or the so-called Milnor-Svarc Lemma. Before stating it, we need a few deﬁnitions. Let X be a metric space. We say that X is proper if closed balls of ﬁnite radius are compact. We say that a path γ between two points x, y in a metric space X satisfying d(x, y) = L is a geodesic if γ is an isometric to [0, L]. A metric space is called geodesic if every pair of points can be joined by at least one geodesic segment. A group action G × X → X is said to be proper if for each compact K ⊂ X, the set {g ∈ G | g · K ∩ K = ∅} is ﬁnite. ˇ Theorem 3.2 (Milnor-Svarc). Let X be a metric space which is geodesic and proper, and let G × X → X be a proper group action by isometries. Suppose furthermore that X/G is compact. Then G is ﬁnitely generated, and for each x ∈ X, the map g → g · x is a quasi-isometry G → X. Proof. The proof follows de la Harpe, who follows Milnor. Let π : X → X/G be the canonical projection. There is a metric on X/G given by looking at the distance between preimages of two points x and y. One needs properness to show that this is a metric. Let R be the diameter of X/G. If x ∈ X is ﬁxed, let B be a closed R-ball about x. The totality of the translates of B is a covering of X. Let S = {s ∈ G | s = 1, s · B ∩ B = ∅}. 10 T. KOBERDA Note that S is closed under taking inverses and is ﬁnite since the action of G is proper. Let r = inf{d(B, g · B) | g ∈ G \ S}, and let λ = max d(x, s · x). s∈S Firstly, observe that r > 0. Indeed, for any one g, the distance is positive, so let r be a distance that works for a particular g. Let T = {g ∈ G \ S | d(B, g · B) ≤ r }. Then T is ﬁnite by properness and nonempty by deﬁnition, whence the claim. Next, we claim that S generates G. Let g ∈ G \ S. Let k be deﬁned by R + (k − 1)r ≤ d(x, g · x) < R + kr. The reader should check that k ≥ 1. Since X is geodesic, we can connect x and g · x by a piece-wise geodesic path with endpoints {x, x1 , . . . , xn } such that d(x, x1 ) < R and d(xi , xi+1 ) < r. Since translates of B cover X, −1 there is a gi ∈ G such that gi · B contains xi . If si = gi−1 gi , we can write g = s1 · · · sn . The reader should verify that each si ∈ S. Finally, for all g ∈ G, we have 1 1 d(x, g · x) ≤ dS (1, g) ≤ d(x, g · x) + 1. λ r Here, dS is the word length with respect to S \ {1}. By what we did to prove that S generates G, we obtain 1 R dS (1, g) ≤ k ≤ d(x, g · x) + 1 − . r r The other inequality is clear. It follows easily that the map G → X given by g → g · x is a quasi-isometry. One of our goals is exhibit some non-obvious quasi-isometries. One such quasi-isometry will be that the fundamental group of a closed surface of genus g > 1 is quasi-isometric to H2 . Before we do this, we should know what the closed surfaces are. Theorem 3.3. Let Σ be a closed orientable surface. The Σ is homeomorphic to a surface of genus g ≥ 0. We call the surface of genus zero the sphere and the surface of genus one the torus. Proof. Suppose ﬁrst that if γ ⊂ Σ is a simple closed loop then Σ \ γ is disconnected. We claim that then Σ is homeomorphic to the sphere. Let ∆ be a ﬁnite cell decomposition of Σ and let T be a maximal tree in the 1-skeleton of ∆. Note that Σ \ T is connected. Let ∆∗ be the dual cell decomposition of Σ and let Γ be a maximal subgraph of the 1-skeleton of ∆∗ subject to the condition Γ∩T = ∅. Then Γ is a tree. Indeed, observe that Γ is connected. If Γ contains a loop then Γ separates Σ, which contradicts the maximality of T . It follows that Σ is a union of small neighborhoods of T and Γ and is hence homeomorphic to the sphere. Let Σ contain a simple loop γ which does not separate Σ which sits within the 1-skeleton of Σ. Cutting along such a loop and gluing in two disks gives COURSE NOTES 11 another surface Σ . Notice that the Euler characteristic of Σ is greater than that of Σ by two. If Σ still has nonseparating loops, repeat this process, which we call “cutting and patching”. Finally, we must show that this is a ﬁnite process. Note that Σ has a cell decomposition coming from Σ, except that Σ has two more faces and one extra vertex and edge for each vertex and edge appearing in γ. Since γ is a circle, the number of extra vertices and edges is equal. Suppose that the cell decomposition of Σ is not redundant, in the sense that each vertex has degree at least three. Then the ratio of edges to vertices must then be at least 3/2. Note that γ has two preimages in Σ . For each vertex appearing in γ, one of the two vertices in the two preimages of γ must be non-redundant. By deleting a redundant vertex, we decrease the number of edges and vertices by one. If one preimage of γ has at most one non-redundant vertex, then we can delete the entire preimage of γ collapse the “loose ends” of the 1-skeleton. This process will decrease the total number of edges and vertices in the 1- skeleton in Σ from the numbers in Σ. We call this process “simpliﬁcation”. Therefore, we take Σ and perform the two processes above. Either the number of edges goes oﬀ to inﬁnity or remains bounded. If it remains bounded then since some of the contributions to the edges in Σ are one of the preimages of γ and since γ separates Σ , after ﬁnitely many cuts and patches and simpliﬁcations we must obtain the sphere. If the number of edges goes oﬀ to inﬁnity, then the ratio of edges to vertices must tend to one, a contradiction. 4. Geometrization of surfaces (INCOMPLETE) The ﬁrst result allows us to simplify the geometrization process for all surfaces to geometrization for one single surface. Theorem 4.1. Let G = π1 (Σg ) and let G < G have index n. Then G = π1 (Σn(g−1)+1 ). In particular, a closed surface of arbitrary genus (greater than two) covers a surface of genus 2. Analogously, let F be a free group of rank d and let F < F have index n. Then F has rank n(d − 1) + 1. In particular every ﬁnite rank free group is contained in a free group of rank 2, and it can be embedded with ﬁnite index when it is nonabelian. Proof. The Euler characteristic of Σg is 2 − 2g, and Euler characteristic is multiplicative under taking covers. It follows that G is the fundamental group of a surface whose Euler characteristic is n(2 − 2g). Writing the genus as g , we see that g − 1 = n(g − 1), whence the claim. The assertion for free groups is analogous. The claims about embeddings follow from the fact that a surface or free group has a normal subgroup of every index. 5. The geometry of H2 /P SL2 (Z) Let X be a metric space and G a group acting properly and discontin- uously on X. Let F be a closed domain (the closure of a nonempty open set). We call F a fundamental domain for the action of G if 12 T. KOBERDA (1) g · F = X. g∈G (2) F 0 ∩ g · F 0 = ∅ for all nonidentity elements of G. The boundary ∂F of F is F \ F 0 . The collection of translates of F by G is called a tesselation of X. As an exercise, the reader should check that the semi-annulus bounded by the circles |z| = 1 and |z| = 2 in the hyperbolic plane is a fundamental domain for the isometry z → 2z. Note that fundamental domains are not well-deﬁned. We will produce a fundamental domain for the action of P SL2 (Z) on H2 . Proposition 5.1. The matrices 1 1 1 0 , 0 1 1 1 generate P SL2 (Z). Proof. It suﬃces to show that these matrices generate SL2 (Z). Recall that SL2 (Z) acts naturally on Z2 . If a b A= ∈ SL2 (Z), c d then the greatest common divisor of the entries in any column or row is one. Running the Euclidean argument on any of these pairs of integers yields (1, 0) or (0, 1). The reader should verify that it follows that a there are matrices P and Q which are products of 1 1 1 0 , 0 1 1 1 and their inverses such that 1 0 P AQ = n 1 for some n. z It follows that the isometries z → z + 1 and z → z+1 generate P SL2 (Z). As an exercise, the reader should ﬁnd an expression for z → −1 in terms of z these isometries. A fundamental domain for P SL2 (Z) is bounded between −1/2 ≤ (z) ≤ 1/2 and above the circle |z| = 1. Proposition 5.2. The closed domain above is a fundamental domain for the action of P SL2 (Z). Proof. Let 1 = A be as above. We have: |cz + d|2 = c2 |z|2 + 2 (z)cd + d2 > c2 + d2 − |cd| = (|c| − |d|)2 + |cd|. The lower bound is at least one and is an integer. It follows that (A(z)) < (z). Replacing z and A by A(z) and A−1 gives a contradiction. Note that P SL2 (Z) acts on H2 with some ﬁxed points. Recall that any element of P SL2 (R) with a ﬁxed point is elliptic, and any elliptic element of P SL2 (Z) has ﬁnite order (why?) The reader should verify that the order of any torsion element of P SL2 (Z) has order at most 3. The reader should also COURSE NOTES 13 verify that z → z−1 has order 3 and preserves the cube roots of unity on the z unit circle. The images of the cube roots of unity and i in H2 /P SL2 (Z) are precisely the images of points of H2 which are ﬁxed by ﬁnite order elements of P SL2 (Z) (the so-called “orbifold points”). The reader should check that the volume of H2 /P SL2 (Z) is π/3. Proposition 5.3. The torsion elements of P SL2 (Z) generate it. Proof. We have 1 −1 0 −1 1 −1 = 1 0 1 0 0 1 and 0 −1 1 −1 1 0 = . 1 0 1 0 1 1 Note that if Γ < P SL2 (Z) is torsion free then its index is at least 6. Let Γ(n) be the kernel of the map P SL2 (Z) → P SL2 (Z/nZ). The reader should verify that P SL2 (Z/2Z) ∼ S3 . As a more diﬃcult = exercise, the reader should check that Γ(2) is free of rank 2. 6. HNN extensions and embedding theorems for finitely generated groups In this section we will discuss some basics concerning Higman-Neumann- Neumann extensions (HNN extensions) and their applicability to embed- dings of countable groups in certain ﬁnitely generated groups. Let G be a group and A, B < G subgroups together with an isomorphism φ : A → B. The HNN extension of G given by φ is a group with the presentation Gφ = G, t | t−1 at = φ(a), a ∈ A . We call G the base of the HNN extension. Let i = ±1. We say that the sequence g0 , t 1 , . . . , t n , gn , n ≥ 0, is reduced if there is no sequence t−1 gi t with gi ∈ A or tgi t−1 with gi ∈ B. Lemma 6.1 (Britton’s Lemma). If n ≥ 1 and the sequence g0 , t 1 , . . . , t n , gn is reduced, then g0 t 1 · · · t n gn = 1 in Gφ . An easy consequence of Britton’s Lemma is the following: Lemma 6.2. Let Gφ be an HNN extension with base G. If g ∈ Gφ has ﬁnite order then g is conjugate to an element of ﬁnite order in G. 14 T. KOBERDA A more diﬃcult generalization of Britton’s Lemma gives a way of dis- tinguishing two elements in an HNN extension based on a so-called normal form. Choose coset representatives for A and B in G, with the identity representing A and B. A sequence g0 , t 1 , . . . , t n , gn , is called a normal form if: (1) g0 ∈ G is arbitrary. (2) i = −1 implies gi is a representative of a coset of A in G. (3) i = 1 implies gi is a representative of a coset of B in G. (4) no subsequence of the form t , 1, t− occurs. Lemma 6.3 (Normal Form Theorem for HNN extensions). The base group G embeds in Gφ and any element of Gφ has a unique representative in normal form. The easy construction of HNN extensions yields some amazing conse- quences. Theorem 6.4. Every countable group C can be embedded in a quotient G of F2 . We may assume that the two generators of G have inﬁnite order. G will have n-torsion if and only if C does. Proof. Let C = c1 , . . . | s1 , . . . . Write F = C ∗ a, b . The set {a, b−1 ab, b−2 ab2 , . . .} generates a free subgroup of a, b (why?), as does {b, c1 a−1 ba, . . . , cn a−n ban , . . .}. Write G = F, t | t−1 at = b, t−1 b−i abi t = ci a−i bai . Then C embeds in G by the Normal Form Theorem, and the claim about torsion follows from Britton’s Lemma. Corollary 6.5. There exist uncountably many non-isomorphic quotients of F2 . Proof. For every nonempty set S of primes, take the direct sum of cyclic groups of order p for p ∈ S. Applying the previous theorem gives the claim. Corollary 6.6. There exists a ﬁnitely generated group which contains each ﬁnitely presented group as a subgroup. Proof. The direct sum of all ﬁnitely presented groups is a countable group. Even more bizarre embedding theorems are true. For instance: Theorem 6.7. Every countable group can be embedded in a six-generator simple group. Certain HNN extensions have very strange properties. A group G is called Hopﬁan if any epimorphism G → G is an isomorphism. COURSE NOTES 15 Proposition 6.8. The (2, 3)–Baumslag-Solitar group G = b, t | t−1 b2 t = b3 is not Hopﬁan. Proof. Deﬁne φ : G → G by φ(t) = t and φ(b) = b2 . It is easy to check that φ is well-deﬁned and surjective. Furthermore, [t−1 bt, b] ∈ ker(φ) and is nontrivial by Britton’s Lemma. A group is called residually ﬁnite if the intersection of all its ﬁnite index subgroups is trivial. Theorem 6.9. Any ﬁnitely generated residually ﬁnite group G is Hopﬁan. It is easy to see that the ﬁnitely generated hypothesis cannot be omitted. Proof. Let 1 = g ∈ G be in the kernel of an epimorphism φ from G to itself. Let n > 0. There exist only ﬁnitely many subgroups of index n since G is ﬁnitely generated. Pulling these subgroups back by φ, we get the same list of index n subgroups, all of which contain g. It follows that g is in their intersection. If G is residually ﬁnite, we obtain a contradiction. 7. Unsolvability of certain decision problems in group theory The ﬁrst goal of this section is to construct an explicit example of a group which has an unsolvable word problem. What this means precisely is that we have a group G= S|R such that no algorithm can take an element of F (S) and determine whether or not it represents the trivial word in G. An algorithm, on the other hand, is a set of instructions which can be executed by a Turing machine. A Turing machine consists of: • A ﬁnite set Q of states. • A ﬁnite alphabet Γ. • A distinguished blank symbol b ∈ Γ. • A transition function Q×Γ → Q×Γ×{L, R} which may be partially deﬁned. • An initial state q0 ∈ Q. • A set F ⊂ Q of ﬁnal states. Less obscurely, a Turing machine takes an inﬁnite tape which has all but ﬁnitely many places marked with b, and ﬁnitely many symbols in Γ. The Turing machine starts reading the tape in the initial state. After reading an entry, the machine can change the entry on the tape, change its state, and move the reader to the left or to the right. To say that a group G has a solvable word problem means that we give a Turing machine a tape which has a word in the free group on the generators written on it, and the machine tells us if the word was trivial or not. For a free group on a ﬁnite set X, a Turing machine looks roughly as follows: the alphabet consists of X ∪ X −1 ∪ b. We write a word on a tape without inserting any blanks and feed it to the machine. If there are only blanks, the machine moves to a terminal state which declares the word to be trivial. Otherwise the machine moves to one of 2|X| states labeled by the elements 16 T. KOBERDA of X and X −1 . Every time the machine reads a generator, it moves to the state labeled by the generator unless it is in a state qx and reads x−1 . If this does not happen, the machine declares the word nontrivial. Otherwise, the machine moves into a reduction loop. It replaces xx−1 by two blank spots and then moves the letters it has not read over two spots until it reaches another blank spot, which indicates that the word is over. The machine then returns to the initial state and starts over. Let A be a countable subset of N. We say that A is computable if there is a Turing machine which decides if n ∈ A or n ∈ A. We say that A is / computably enumerable or c.e. if there is a surjective computable function f : N → A. A set is computable if and only if it and its complement are c.e. The reader not familiar with this fact should check it. There is an easy corollary to this observation: Corollary 7.1. A ﬁnitely presented residually ﬁnite group G has a solvable word problem. Proof. Since G is residually ﬁnite, we can computably enumerate all el- ements of G which are not the identity by enumerating homomorphisms G → Sn as n → ∞. Since G is ﬁnitely presented, we can eﬀectively enumer- ate which elements represent the identity. Thus the set of elements which represent the identity and those which do not are both c.e. Since there are only countably many computable sets (why?) and un- countably many subsets of N, there exist noncomputable subsets of N. Theorem 7.2. There exists a ﬁnitely generated group with an unsolvable word problem. Proof. Let S be a noncomputable subset of N. Let G = a, b, c, d | a−i bai = c−i dci for i ∈ S . Then solving the word problem in G allows us to compute S. A group is called computably presented if the set of relations is a c.e. subset of F (S). The following is a diﬃcult theorem of Higman: Theorem 7.3. A group G embeds in a ﬁnitely presented group H if and only if it is computably presented. It follows that if there exists a c.e., noncomputable subset of N, then there exists a ﬁnitely presented group with an unsolvable word problem. Lemma 7.4. There exists a c.e. set which is not computable. Sketch. There exists a universal Turing machine F : N × N → N which encodes all other Turing machines. Precisely, we can uniquely and eﬀectively code every Turing machine as a natural number. One way to do this is to associate to each symbol or instruction an unambiguous number to get a tuple (n1 , . . . , nk ) associated to a Turing machine Te . We then enumerate the primes in increasing order and associate to Te the number e = pn1 · · · pek . 1 k COURSE NOTES 17 Let K ⊂ N × N be the set of pairs (e, n) such that Te converges on the input n. Then K is c.e. by deﬁnition. K is not computable by a standard diagonalization argument. Proposition 7.5. There exist subgroups of Fn × Fn with unsolvable mem- bership problems for n suﬃciently large. Proof. Let H = x1 , . . . , xn | r1 , . . . , rm be a group with an unsolvable word problem. Let LH < Fn ×Fn be generated by elements (xi , xi ) and (1, rj ) for all i and j. It is an easy exercise to show that (u, v) ∈ LH if and only if u = v in H. With a little more work, it is possible to show that there is no algorithm which determines if a particular ﬁnite set of elements of Fn × Fn generates a proper subgroup. The membership and generator problem for Fn × Fn have been of interest historically because there are statements concerning maps from π1 (Σg ) → Fg × Fg which are equivalent to the Poincar´ conjecture. e Department of Mathematics, Harvard University, 1 Oxford St., Cambridge, MA 02138 E-mail address: koberda@math.harvard.edu