Algebraic Topology Homework 7: Due Wednesday, October 14 Recall That a Meridian of a Solid Torus D × S 1 Is a Curve That Bounds

Home , Solid torus

Algebraic Topology Homework 7: Due Wednesday, October 14 Recall that a meridian of a solid torus D × S1 is a curve that bounds D×(point in S1), while a longitude is a curve of the form (point in D) ×S1. a b Problem 1 Let A = be an integer matrix. The linear transformation L(x) = c d 2 2 2 Ax maps IR to IR , and maps Z2 to Z2. It therefore maps the torus T 2 = IR /Z2 to itself. 2 What is the induced map on π1(T )? (We will eventually show that every continuous map between tori is homotopic to such a linear map.) 2 The generators of π1(T ) are a path γ1 that moves a distance 1 in the horizontal direction and a path γ2 that moves 1 in the vertical direction. The path L∗γ1 clearly moves a in the horizontal direction and b in the vertical direction, and so is homotopic 2 2 to aγ1 + cγ2 (where we are using additive notation for the Abelian group π1(T ) = Z . 2 Likewise, L∗γ2 is homotopic to bγ1 + dγ2. In other words, the induced map L∗ on Z is just multiplication by the integer matrix A.

Problem 2. Let X1 and X2 be two solid tori, and identify the boundary of each 2 one with IR /Z2, such that the ﬁrst coordinate refers to the meridian and the second to the longitude. Let A and L be as in problem 1, only with the extra condition that the determinant of A is ±1. Identify x in the boundary of X1 with L(x) in the boundary of X2, and let Y be the disjoint union of X1 and X2, modulo this identiﬁcation. Compute π1(Y ) in terms of the matrix elements of A.

Let α be the generator of π1(X1) and let β be the generator of π1(X2). Let U be a tubular neighborhood of X1, which is homotopy equivalent to X1, and let V be a tubular neighborhood of X2. The intersection U ∩V deformation retracts onto the common boundary of X1 and X2, and we can take ℓ1 and m1 (the longitude and the meridian of the boundary of X1) to be the generators of π1(U ∩ V ). By van Kampen, π1(Y ) is generated by α,β,ℓ1, m1 with relations from how π1(U ∩ V ) maps to π1(U) and to π1(V ). From the inclusion of U ∩ V into U, we see that ℓ1 is identified with α and m1 is identified with the d identity. From the inclusion of U ∩ V into V we see that ℓ1 is identified with β and m1 c d c is identified with β . We thus have α = β and β = 1, so our group is Zc if c 6= 0 and Z if c = 0. In general Y is the lens space L(c, a), except when c = 0, in which case we get S2 ×S1. 2 2 To see this, recall our construction of L(p, q) in class. In the 3-sphere |z1| + |z2| = 1, 2 2 we had two solid tori, one where |z1| ≤ 1/2, and the other where |z2| ≤ 1/2. After 2πi/p 2πqi/p modding out by our group action, which sends (z1,z2) to (z1e ,z2e ), we still get two solid tori. The longitude ℓ1 and meridian m1 of the first solid torus increase the ′ arguments of z1 and z2 by (2πq /p, 2π/p) and (2π, 0), respectively, while the longitude and meridian of the second increase these values by (2π/p, 2πq/p) and (0, 2π). It’s not ′ ′ hard to see that ℓ1 = q ℓ2 + (1 − qq )/pm2, while m1 = pℓ2 − qm2, so our matrix A a b −q (1 − qq′)/p is = ′ . The fact that we have −q and rather than q is an c d p q artifact of our convention of the direction of our meridian m2. With a different sign choice

1 q (qq′ − 1)/p we would have the matrix ′ . p q 3 Problem 3. Let K be a smoothly embedded knot in S3 (which you can view as IR plus a point at infinity), and let X be a closed ǫ-neighborhood of K. For ǫ small enough, X is a solid torus. Let X′ be another solid torus (viewed abstractly, not as a subset of p r Euclidean space) and let A = be an integer matrix of determinant 1 that maps the q s boundary of X′ to the boundary of X (not the other way around!), where we take the first coordinate to be the longitude and the second to be the meridian. Let X0 be the interior of X, and let Y =(S3 − X) ∪ X′, where we identify the boundary of X and the boundary of X′ via A. Show that, for fixed p and q, different choices of r and s yield the same space Y , up to homeomorphism. This procedure of creating Y from K, p and q is called “p/q Dehn surgery on the knot K”. [Hint: First see if you can prove independence of r for (p, q)=(1, 0).] [The same process can be applied successively to a knot in Y to get Z, to a knot in Z to get another space, and so on. Alternately, we can speak of doing surgery on a link, which is a disjoint collection of knots in S3. For each component of the link, drill out a neighborhood and glue in another solid torus, where the fractions p/q can be different for different components of the link. The Lickorish-Wallace theorem states that every compact, connected oriented 3-manifold can be obtained from S3 by Dehn surgery on an appropriate link. This gives a strong connection (dare I say a link?) between knot theory and the topology of 3-manifolds.] The meridian of a solid torus, viewed as an element of the fundamental group of the boundary, is uniquely defined up to sign. It’s the only generator of π1 of the boundary that bound a disk in the interior. However, there is no canonical choice of the longitude, 1 as ℓ1 + km1 is also a longitude for any integer k. Indeed, on the solid torus D × S with ikθ coordinates (z,θ), the automorphism (z,θ) → (ze ,θ) maps the meridian m1 to m1, but maps ℓ1 to ℓ1 + km1. 1 k Now apply this to Dehn surgery on the unknot. If A = , then a longitude of 0 1 X′ is identified with a longitude of X plus k times a meridian (in other words a longitude), while a meridian of X′ is identified with a meridian of X. In other words, replacing X with X′ doesn’t change anything, and we still have a 3-sphere. ′ p r ′ p r Finally, suppose that we have two matrices A = and A = ′ . Since q s q s both matrices have determinant 1, there must be an integer k such that r′ = r + kp and ′ ′ 1 k 1 k s = s + kq, so A = A . We just saw that was just a change in choice of 0 1 0 1 longitude for X′, so the action of A′ is, up to this choice, the same as the action of A. ′ 3 ′ More explicitly, if we get Y1 by gluing X1 to S − X via A and Y2 by gluing X2 to 3 ′ 3 S − X via A , then there is a homeomorphism Y2 → Y1 that is the identity on S − X ′ 1 ′ 1 ikθ and maps X2 = D × S to X1 = D × S by (z,θ) → (ze ,θ).

2 Problem 4. Let Y be obtained from S3 by p/q surgery on the unknot. What is Y when p/q = 0/1? What is Y when p/q = 1/0? For general p/q, compute π1(Y ) in terms of p and q. (This is another way to visualize the Lens space L(p, q), as is the construction of Problem 2) See the solution to problem 2. Recall that S3−X is itself a solid torus, whose longitude is the meridian of X, and whose meridian is the longitude of X. When p/q = 0/1, we are just gluing two solid tori by identifying their longitudes and meridians. This is the union of two disks, all times a circle, in other words Y = S2 × S1. When p/q = 1/0, X and X′ are the same so we are filling in exactly what we removed, and Y = S3. In general, we are q s gluing in two tori as in problem 2, only with the matrix , and π1(Y ) = Z . p r p Page 123, problems 2.1, 2.2, 2.3 and 2.4. For problem 2.2, think about which condition of being a local homeomorphism can fail when restricted to A. 2.1. There was some confusion about what a “basic family” of sets is. This isn’t a standard term, and I can think of two possible meanings: (1) It could mean that the family forms a basis for a topology (although not necessarily the topology that we’re using). In other words, that every point is in some set in the family, and that if U1 and U2 are sets in the family and x ∈ U1 ∩U2, then there is a family member U3 with x ∈ U3 and U3 ⊂ U1 ∩ U2. (2) It could mean that, given any point x and any neighborhood N of x, there is a family member U with x ∈ U ⊂ N. For open sets, these conditions are equivalent. For non-open sets they aren’t. I’ll write solutions based on the second definition, but anybody who used the first will get full credit.

Assuming (a), consider a basic family of open sets around a point x0, and restrict each member of this family to the arc component containing x0. This gives a basic family of path-connected open neighborhoods. Given (b), (c) is trivial, since a basic family of open neighborhoods is a basic family of neighborhoods. (d) is just a restatement of (c), if we use the second deﬁnition of “basic family”. All that remains is to show that (c) implies (a). Let x be an arbitrary point, let N be an arbitrary neighborhood and let V ⊂ N be a (not necessarily open) arc-connected neighborhood of x, which exists by (c). By deﬁnition, x is not on the boundary of V , so V 0 (the interior of V ) is an open neighborhood of x. Let W be the arc-component of x in V 0. I claim that W is open, and hence is an arc-connected neighborhood of x. To see this, let x′ be an arbitrary point in W . By assumption, there is a path from x to x′ in V 0. Furthermore, there is a path-connected (not necessarily open) neighborhood V ′ of x′ that sits inside V 0, so all of V ′ is connected to x by paths in V 0, so all of V ′ is in W . Since V ′ contains an open neighborhood of x′, an open neighborhood of x′ is in W . 2.2: Let X = IR, let Y be the unit circle, and let f(x) = eix. Now let A be the integers. Note that f is 1–1 on A. Since f(A) is dense in Y , the image of a bounded subset of A is not open in f(A), and f −1 is not continuous on the image of an unbounded subset

3 of A. 2.3: If f −1(y) is infinite and X is compact, then f −1(y) has an accumulation point x0. But then f, restricted to a neighborhood of x0, is not 1–1, and in particular can’t be a homeomorphism. 2.4: Since X is compact, f(X) is compact, hence closed (since Y is Hausdorff). How- ever, since f is a local homeomorphism, f(X) is open. Since Y is connected, this implies that f is onto. Now let y be an arbitrary point in Y . By the previous exercise, −1 f (y) = {x1,...,xk} is finite. Pick open neighborhoods Ui of xi that map homeomorphically to their images. Let V be a path-connected open neighborhood of y that sits −1 inside the intersection of all the f(Ui)’s. The path-components of f (V ) are precisely −1 f (V ) ∩ Ui (each of which is path-connected since it maps homeomorphically to V ), so V is an elementary neighborhood. This shows that f is a covering map.