1 SMSTC Geometry and Topology 2011 2012 Lecture 5 The Seifert van Kampen Theorem Andrew Ranicki (Edinburgh) 10th November, 2011
2 Introduction Topology and groups are closely related via the fundamental group construction π 1 : {spaces} {groups} ; X π 1 (X ). The Seifert - van Kampen Theorem expresses the fundamental group of a union X = X 1 Y X 2 of path-connected spaces in terms of the fundamental groups of X 1, X 2, Y. The Theorem is used to compute the fundamental group of a space built up using spaces whose fundamental groups are known already. The Theorem is used to prove that every group G is the fundamental group G = π 1 (X ) of a space X. Lecture follows Section I.1.2 of Hatcher s Algebraic Topology, but not slavishly so.
Revision : the fundamental group The fundamental group π 1 (X, x) of a space X at x X is the group of based homotopy classes [ω] of pointed loops ω : (S 1, 1) (X, x). The group law is by the concatenation of loops: [ω 1 ][ω 2 ] = [ω 1 ω 2 ]. The inverse is by the reversal of loops: [ω] 1 = [ω] with ω(t) = ω(1 t). A path α : I X induces an isomorphism of groups α # : π 1 (X, α(0)) π 1 (X, α(1)) ; ω [α ω α]. Will mainly consider path-connected spaces X : the fundamental group π 1 (X, x) is independent of the base point x X, and may be denoted π 1 (X ). 3
Three ways of computing the fundamental group I. By geometry For an infinite space X there are far too many loops ω : S 1 X in order to compute π 1 (X ) from the definition. A space X is simply-connected, i.e. X is path connected and the fundamental group is trivial π 1 (X ) = {e}. Sometimes it is possible to prove that X is simply-connected by geometry. Example: If X is contractible then X is simply-connected. Example: If X = S n and n 2 then X is simply-connected. Example: Suppose that (X, d) is a metric space such that for any x, y X there is unique geodesic (= shortest path) α x,y : I X from α x,y (0) = x to α x,y (1) = y. If α x,y varies continuously with x, y then X is contractible. Trees. Many examples of such spaces in differentiable geometry. 4
If Three ways of computing the fundamental group II. From above X p X is a covering projection and X is simply-connected then π 1 (X ) is isomorphic to the group of covering translations Homeo p ( X ) = {homeomorphisms h : X X such that ph = p} (This will be proved in a later lecture). Example If then p : X = R X = S 1 ; x e 2πix π 1 (S 1 ) = Homeo p (R) = Z. 5
Three ways of computing the fundamental group III. From below 6 Seifert-van Kampen Theorem (preliminary version) X 1. Ẏ X 2 If a path-connected space X is a union X = X 1 Y X 2 with X 1, X 2 and Y = X 1 X 2 path-connected then the fundamental group of X is the free product with amalgamation π 1 (X ) = π 1 (X 1 ) π1 (Y ) π 1 (X 2 ). G 1 H G 2 defined for group morphisms H G 1, H G 2. First proved by van Kampen (1933) in the special case when Y is simply-connected, and then by Seifert (1934) in general.
7 Seifert and van Kampen Herbert Seifert Egbert van Kampen (1907-1996) (1908-1942)
8 The free product of groups Let G 1, G 2 be groups with units e 1 G 1, e 2 G 2. A reduced word g 1 g 2... g m is a finite sequence of length m 1 with gi G 1 \ {e 1 } or g i G 2 \ {e 2 }, g i, g i+1 not in the same G j. The free product of G 1 and G 2 is the group G 1 G 2 = {e} {reduced words} with multiplication by concatenation and reduction. The unit e = empty word of length 0. See p.42 of Hatcher for detailed proof that G 1 G 2 is a group. Exercise Prove that {e} G = G, G 1 G 2 = G2 G 1, (G 1 G 2 ) G 3 = G1 (G 2 G 3 )
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10 The free group F g For a set S let S = free group generated by S = Z. s S Let g 1. The free group on g generators is the free product of g copies of Z F g = a 1, a 2,..., a g = Z Z Z. Every element x F g has an expression as a word x = (a 1 ) m 11 (a 2 ) m 12... (a g ) m 1g a m 21 1... with (m ij ) an N g matrix (N large), m ij Z. F 1 = Z. For g 2 F g is nonabelian. F g F h = F g+h.
11 The infinite dihedral group Warning The free product need not be free. Example Let G 1 = Z 2 = {e 1, a}, G 2 = Z 2 = {e 2, b} be cyclic groups of order 2, with generators a, b such that a 2 = e 1, b 2 = e 2. The infinite dihedral group is the free product D = Z 2 Z 2 = {e, a, b, ab, ba, aba, bab, abab,... }, with a 2 = b 2 = e. This is an infinite group with torsion, so not free.
12 The subgroups generated by a subset Needed for statement and proof the Seifert - van Kampen Theorem. Let G be a group. The subgroup generated by a subset S G S G is the smallest subgroup of G containing S. G consists of finite length words in elements of S and their inverses. Let S G be the subset of G consisting of the conjugates of S S G = {gsg 1 s S, g G} The normal subgroup generated by a subset S G S G G is the smallest normal subgroup of G containing S.
Group presentations Given a set S and a subset R S define the group S R = S / R S = S /normal subgroup generated by R. with generating set S and relations R. Example Let m 1. The function a a m = {e, a, a 2,..., a m 1 } Z m ; a n n is an isomorphism of groups, with Z m the finite cyclic group of order m. R can be empty, with S = S = S Z the free group generated by S. The free product of G 1 = S 1 R 1 and G 2 = S 2 R 2 is G 1 G 2 = S 1 S 2 R 1 R 2. 13
Every group has a presentation Every group G has a group presentation, i.e. is isomorphic to S R for some sets S, R. Proof Let S = G = G Z and let R = ker(φ) be the kernel of the surjection of groups Then Φ : S G ; (g n 1 1, g n 2 2,... ) (g 1) n 1 (g 2 ) n 2.... S R G ; [x] Φ(x) is an isomorphism of groups. It is a nontrivial theorem that R is a free subgroup of the free group S. But we are only interested in S and R as sets here. This presentation is too large to be of use in practice! But the principle has been established. While presentations are good for specifying groups, it is not always easy to work out what the group actually is. Word problem: when is S R = S R? Undecidable in general. 14
How to make a group abelian The commutator of g, h G is [g, h] = ghg 1 h 1 G. Let F = {[g, h]} G. G is abelian if and only if F = {e}. The abelianization of a group G is the abelian group G ab = G/ F G, with F G G the normal subgroup generated by F. If G = S R then G ab = S R F. Universal property G ab is the largest abelian quotient group of G, in the sense that for any group morphism f : G A to an abelian group A there is a unique group morphism f ab : G ab A such that f : G G ab f ab A. π 1 (X ) ab = H 1 (X ) is the first homology group of a space X about which more in a later lecture. 15
Free abelian at last Example Isomorphism of groups (F 2 ) ab = a, b aba 1 b 1 Z Z ; a m 1 b n 1 a m 2 b n 2... (m 1 + m 2 +..., n 1 + n 2 +... ) with Z Z the free abelian group on 2 generators. More generally, the abelianization of the free group on g generators is the free abelian group on g generators (F g ) ab = g Z for any g 1. 16 It is clear from linear algebra (Gaussian elimination) that Z is isomorphic to Z if and only if g = h. g h It follows that F g is isomorphic to F h if and only if g = h.
Amalgamated free products 17 The amalgamated free product of group morphisms is the group i 1 : H G 1, i 2 : H G 2 G 1 H G 2 = (G 1 G 2 )/N with N G 1 G 2 the normal subgroup generated by the elements i 1 (h)i 2 (h) 1 (h H). For any h H i 1 (h) = i 2 (h) G 1 H G 2. In general, the natural morphisms of groups j 1 : G 1 G 1 H G 2, j 2 : G 2 G 1 H G 2, j 1 i 1 = j 2 i 2 : H G 1 H G 2 are not injective.
18 Some examples of amalgamated free products Example G G G = G. Example {e} H {e} = {e}. Example For H = {e} the amalgamated free product is just the free product G 1 {e} G 2 = G 1 G 2. Example For any group morphism i : H G G H {e} = G/N with N = i(h) G G the normal subgroup generated by the subgroup i(h) G.
19 Injective amalgamated free products If i 1 : H G 1, i 2 : H G 2 are injective then G 1, G 2, H are subgroups of G 1 H G 2 with G 1 G 2 = G, G 1 G 2 = H. Conversely, suppose that G is a group and that G 1, G 2 G are subgroups such that G 1 G 2 = G. This condition is equivalent to the group morphism being surjective. Φ : G 1 G 2 G ; g k g k (g k G k ) Then G = G 1 H G 2 with H = G 1 G 2 G and i 1 : H G 1, i 2 : H G 2 the inclusions.
The Seifert - van Kampen Theorem Let X = X 1 Y X 2 with X 1, X 2 and Y = X 1 X 2 open in X and path-connected. Let i 1 : π 1 (Y ) π 1 (X 1 ), i 2 : π 1 (Y ) π 1 (X 2 ) be the group morphisms induced by Y X 1, Y X 2, and let j 1 : π 1 (X 1 ) π 1 (X ), j 2 : π 1 (X 2 ) π 1 (X ) be the group morphisms induced by X 1 X, X 2 X. Then Φ : π 1 (X 1 ) π 1 (X 2 ) π 1 (X ) ; x k j k (x k ) (x k π 1 (X k ), k = 1 or 2) is a surjective group morphism with ker Φ = N = the normal subgroup of π 1 (X 1 ) π 1 (X 2 ) generated by i 1 (y)i 2 (y) 1 (y π 1 (Y )). Theorem Φ induces an isomorphism of groups Φ : π 1 (X 1 ) π1 (Y ) π 1 (X 2 ) = π 1 (X ). 20
Φ is surjective I. 21 Will only prove the easy part, that Φ is surjective. For the hard part, that Φ is injective, see pp.45-46 of Hatcher. Choose the base point x Y X. Regard a loop ω : (S 1, 1) (X, x) as a closed path f : I = [0, 1] X = X 1 X X 2 such that f (0) = f (1) = x X, with ω(e 2πis ) = f (s) X. By the compactness of I there exist 0 = s 0 < s 1 < s 2 < < s m = 1 such that f [s i, s i+1 ] X 1 or X 2, written f [s i, s i+1 ] X i. Then f = f 1 f 2 f m : I X is the concatenation of paths f i : I X i with f i (1) = f i+1 (0) = f (s i ) Y (1 i m).
Φ is surjective II. 22 Since Y is path-connected there exists paths g i : I Y from g i (0) = x to g i (1) = f (s i ) X. The loop (f 1 g 1 ) (g 1 f 2 g 2 ) (g m 2 f m 1 g m 1 ) (g m 1 f m ) : I X is homotopic to f rel {0, 1}, with so that [g i f i g i+1 ] im(π 1 (X i )) π 1 (X ), [f ] = [f 1 g 1 ][g 1 f 2 g 2 ]... [g m 2 f m 1 g m 1 ][g m 1 f m ] im(φ) π 1 (X ). g 1 f 1 f 2 x 0 Hatcher diagram: A α = X 1, A β = X 2 g 2 f 3 A α A β
The universal property I. 23 An amalgamated free product G 1 H G 2 defines a a commutative square of groups and morphisms H i 1 G 1 j 1 i 2 j 2 G 2 G 1 H G 2 with the universal property that for any commutative square H i 1 G 1 i 2 k 1 k 2 G 2 G there is a unique group morphism Φ : G 1 H G 2 G such that k 1 = Φj 1 : G 1 G and k 2 = Φj 2 : G 2 G.
24 The universal property II. By the Seifert - van Kampen Theorem for X = X 1 Y X 2 the commutative square π 1 (Y ) i 1 π 1 (X 1 ) i 2 π 1 (X 2 ) j 1 j 2 π 1 (X ) has the universal property of an amalgamated free product, with an isomorphism Φ : π 1 (X 1 ) π1 (Y ) π 1 (X 2 ) = π 1 (X ).
The one-point union 25 Let X 1, X 2 be spaces with base points x 1 X 1, x 2 X 2. The one-point union is. X 1 X 2 = X 1 {x 2 } {x 1 } X 2 X 1 X 2. X 1. X 2 The Seifert - van Kampen Theorem for X 1 X 2 If X 1, X 2 are path connected then so is X 1 X 2, with fundamental group the free product π 1 (X 1 X 2 ) = π 1 (X 1 ) π 1 (X 2 ).
The fundamental group of the figure eight 26 The figure eight is the one-point union of two circles X = S 1 S 1 The fundamental group is the free nonabelian group on two generators: An element π 1 (X ) = π 1 (S 1 ) π 1 (S 1 ) = a, b = Z Z. a m 1 b n 1 a m 2 b n2 π 1 (X ) can be regarded as the loop traced out by an iceskater who traces out a figure 8, going round the first circle m 1 times, then round the second circles n 1 times, then round the first circle m 2 times, then round the second circle n 2 times,....
A famous Edinburgh iceskater 27
28 The fundamental group of a graph Let X be the path-connected space defined by a connected finite graph with V vertices and E edges, and let g = 1 V + E. Exercise Prove that X is homotopy equivalent to the one-point union S 1 S 1 S 1 of g circles, and hence that π 1 (X ) = F g, the free group on g generators. Prove that X is a tree if and only if it is contractible, if and only if g = 0. Example I.1.22 of Hatcher is a special case with V = 8, E = 12, g = 5..
A knot is an embedding The complement of K is Knots K : S 1 S 3. X K = S 3 \K(S 1 ) S 3. Two knots K 1, K 2 : S 1 S 3 are equivalent if there exists a homeomorphism of their complements h : X K1 = XK2. The fundamental group π 1 (X K ) of the complement of a knot K is an invariant of the equivalence class of K. Fact: π 1 (X K ) ab = H 1 (X K ) = Z. 29 The Seifert - van Kampen Theorem can be applied to obtain the Wirtinger presentation of π 1 (X K ) from a knot projection Exercise I.1.2.22 of Hatcher.
30 The unknot The unknot has complement with group K 0 : S 1 S 3 ; z (z, 0, 0) X K0 = S 1 R 2 π 1 (X K0 ) = Z K 0 (S 1 ) S 3 \K 0 (S 1 )
Knotting and unknotting 31 A knot K : S 1 S 3 is unknotted if it is equivalent to K 0. A knot K : S 1 S 3 is knotted if it is not equivalent to K 0. It is easy to prove that if the surjection π 1 (X K ) π 1 (X K ) ab = Z is not an isomorphism then K is knotted, since π 1 (X K ) = π1 (X K0 ) = Z. It is hard to prove (but true) that if the surjection π 1 (X K ) π 1 (X K ) ab = Z is an isomorphism then K is unknotted. This is Dehn s Lemma, originally stated in 1911, and only finally proved by Papakyriakopoulos in 1957. Musical interlude: Get knotted! (Fascinating Aida)
32 The trefoil knot The trefoil knot K 1 : S 1 S 3 has group π 1 (X K1 ) = {a, b aba = bab}. a K 1 (S 1 ) S 3 \K 1 (S 1 ) b The groups of K 0, K 1 are not isomorphic (since one is abelian and the other one is not abelian), so that K 0, K 1 are not equivalent: the algebra shows that the trefoil knot is knotted.
The torus knots I. For coprime m, n 1 use the embedding to define the torus knots S 1 S 1 S 1 ; z (z m, z n ) T m,n : S 1 S 1 S 1 S 3. 186 VII. THE FUNDAMENTAL GROUP 33 FIG. 97 Every torus knot S 1 S 1 S 1 S 3 is equivalent to one of T m,n (Brauner, 1928). into two subsets whose common boundary is the torus on which the knot lies. The tube will then be split along its length to form two half-tubes. The complex Q will be decomposed to form two solid tori, from Problem 4 of $14, such that a winding groove of semicircular profile has been milled out of each torus. We shall designate the finite solid torus as the subcomplex R and we
34 The torus knots II. See Example I.1.24 of Hatcher for the application of the Seifert - van Kampen Theorem to compute π 1 (X Tm,n ) = x, y x m = y n. The centre (= the elements which commute with all others) C = x m = y n π 1 (T m,n ) is an infinite cyclic normal subgroup such that Two torus knots T m,n, T m,n (m, n) = (m, n ) or (n, m ). The trefoil knot is K 1 = T 2,3. π 1 (X Tm,n )/C = Z m Z n. are equivalent if and only if
35 Links Let µ 1. A µ-component link is an embedding L : S 1 = S 1 S 1 S 1 S 3. A knot is a 1-component link. The complement of L is µ X L = S 3 \L( µ S 1 ) S 3. The fundamental group π 1 (X L ) detects the linking of the circles among each other. (And much else besides).
The unlink and the Hopf link 36 The unlink L 0 : S 1 S 1 S 3 has X L0 S 1 S 1 S 2, π 1 (X L0 ) = F 2 = Z Z. The Hopf link L 1 : S 1 S 1 S 3 has X L1 S 1 S 1, π 1 (X L1 ) = Z Z. Hatcher Example I.1.2, noting difference between S 3 and R 3.
Cell attachment 37 Let n 0. Given a space W and a map f : S n 1 W let X = W f D n be the space obtained from W by attaching an n-cell. X is the quotient of the disjoint union W D n by the equivalence relation generated by (x S n 1 ) (f (x) W ). An n-dimensional cell complex is a space obtained from by successively attaching k-cells, with k = 0, 1, 2,..., n Example A graph is a 1-dimensional cell complex. Example S n is the n-dimensional cell complex obtained from by attaching a 0-cell and an n-cell S n = D 0 f D n with f : S n 1 D 0 the unique map.
38 The effect on π 1 of a cell attachment I. Let n 1. If W is path-connected then so is X = W f D n. What is π 1 (X )? If n = 1 then X is homotopy equivalent to W S 1, so that π 1 (X ) = π 1 (W S 1 ) = π 1 (W ) Z. For n 2 apply the Seifert - van Kampen Theorem to the decomposition X = X 1 Y X 2 with X 1 = W f {x D n x 1/2}, X 2 = {x D n x 1/2}, Y = X 1 X 2 = {x D n x = 1/2} = S n 1
The effect on π 1 of a cell attachment II. The inclusion W X 1 is a homotopy equivalence, and X 2 = D n is simply-connected, so that π 1 (X ) = π 1 (X 1 ) π1 (Y ) π 1 (X 2 ) 39 = π 1 (W ) π1 (S n 1 ) π 1 (D n ) = π 1 (W ) π1 (S n 1 ) {e}. If n 3 then π 1 (S n 1 ) = {e}, so that If n = 2 then π 1 (X ) = π 1 (W ) {e} {e} = π 1 (W ). π 1 (X ) = π 1 (W ) Z {e} = π 1 (W )/N the quotient of π 1 (W ) by the normal subgroup N π 1 (W ) generated by the homotopy class [f ] π 1 (W ) of f : S 1 W. See Hatcher s Proposition I.1.26 for detailed exposition. If X = S 1 D 2 D n is a cell complex with a single S R n 3 0-cell, S 1-cells and R 2-cells then π 1 (X ) = S R.
Every group is a fundamental group 40 Let G = S R be a group with a presentation. Realize the generators S by the 1-dimensional cell complex W = S S 1 with π 1 (W ) = S the free group generated by S. Realize each relation r R π 1 (W ) by a map r : S 1 W. Attach a 2-cell to W for each relation, to obtain a 2-dimensional cell complex X = W D 2 such that R π 1 (X ) = S R = G.
Realizing the cyclic groups topologically 41 Example Let m 1. The cyclic group Z m = a a m of order m is the fundamental group π 1 (X m ) = Z m of the 2-dimensional cell complex X m = S 1 m D 2, with the 2-cell attached to S 1 = {z C z = 1} by m : S 1 S 1 ; z z m. X 1 = D 2 is contractible, with π 1 (X 1 ) = {e} X 2 is homeomorphic to the real projective plane RP 2 = S 2 /{v v v S 2 } with π 1 (X 2 ) = π 1 (RP 2 ) = Z 2. = D 2 /{w w w S 1 }
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