METU Department of Mathematics Math 543 Low-dimensional Topology Fall 2013 S.Finashin

Final

Part 1 (Take-home)

due to January 6, 13:30, 54 points in 6 problems

Name and the student number:

Q1. (6 pts) Determine which of the Lens spaces L(11, k), k = 1, …, 10, are homeomorphic to each other ? Which of them are chiral and which ones are achiral ? Which ones are homotopy equivalent to L(11, 1) ?

Solution:

L(11, k) is homeomorphic to L(11, n) if n = ±k^{±1} mod 11, moreover, a homeomorphism is orientation preserving if kn = 1 mod 11 and orientation reversing if kn = −1 mod 11. We have

2 × 6 = 3 × 4 = 5 × 9 = 7 × 8 = 10 × 10 = 1 mod 11,

so, there are orientation-preserving homeomorphisms

(1)  \begin{aligned}L(11, 2) & ≅ L(11, 6) \\ L(11, 3) & ≅ L(11, 4) \\ L(11, 5) & ≅ L(11, 9) \\ L(11, 7) & ≅ L(11, 8) \end{aligned}

and together with the orientation-reversing homeomorphisms

\begin{aligned}L(11, 1) & ≅ −L(11, 10) \\ L(11, 2) & ≅ −L(11, 9) \\ L(11, 3) & ≅ −L(11, 8) \\ L(11, 4) & ≅ −L(11, 7) \\ L(11, 5) & ≅ −L(11, 6) \end{aligned}

it gives the following three homeomorphism classes in the set of lens spaces L(11, k):

\begin{matrix}L(11, 1) & , L(11, 10) \\ L(11, 2) & , L(11, 6), L(11, 5), L(11, 9) \\ L(11, 3) & , L(11, 4), L(11, 7), L(11, 8). \end{matrix}

All of the lens spaces L(11, k) are chiral, that is, do not admit orientationreversing homeomorphisms, because such a homeomorphism would mean an orientation preserving homeomorphism L(11, k) ≅ −L(11, k) = L(11, 11 − k), but there is no such ones in the list (1) (6 ≠ 11 − 2, 4 ≠ 11 − 3, etc.).

Space L(11, 1) is homotopy equivalent to L(11, k), where k = ±n^{2} mod 11 (i.e., a quadratic residue or opposite to it). Letting n = 1, 2, … we obtain quadratic residues n^{2} = 1, 4, 9, 5, 3 mod 11, and −n^{2} = 10, 7, 2, 6, 8 mod 11 so, all the lens spaces L(11, k) are homotopy equivalent to L(11, 1).

(In the other words, we have only one homotopy class of the lens spaces L(11, k) because −1 is not a quadratic residue modulo 11.)

Q2. (10 pts) Show how to present L(11, 3) (pay attention to the correct orientation) by a framed link, L, so that

(a) link L has one component with a negative rational framing;

(b) link L has one component with a positive rational framing;

(c) all the components of L have negative integer framings;

(d) all the components of L have positive integer framings.

Q3. (7 pts) Let X denote a plumbing 4-manifold with the dual graph 2—3—4.

(a) Sketch a framed link presentation of X.

(b) Find the matrix of the intersection form of X.

(c) Show that the boundary ∂X is a lens space L(p, q), and find p, q ≤ 0 paying attention to the orientation. (A Hint: use slam-dank moves).

Q4. (6 pts) Let X be a compact 4-manifold described by the dual graph (-2)—(-1)—(-2).

(a) Show that the boundary ∂X is S^{1} × S^{2}.

(b) Which closed 4-manifold is obtained by filling X with S^{1} × D^{3} ?

Q5. (10 pts) Let X be the Milnor fiber of the singularity D_{4} and M = ∂X (the link of this singularity).

(a) Using Kirby moves of the D_{4}-link present M by a framed link with only two component.

(b) Find the order of the homology group H_{1}(M) and deduce that it is impossible to fill in M with S^{1} × D^{3} like in the previous example.

Q6. (15 pts) Find the Betti numbers b_{i}, the signature σ and the indices b_{2}^{±} of the following 4-manifolds X.

(a) X = ℂP^{2}#(ℂP^{1} × ℂP^{1})#\overline{ℂP^{2}}.

Solution: Since X is connected, we have b_{0} = b_{4} = 1. Since the summands ℂP^{2}, ℂP^{1} × ℂP^{1} and \overline{ℂP^{2}} are simply connected, so is X,

and thus, b_{1} = b_{3} = 0.

Next, σ(X) = σ(ℂP^{2}) + σ(ℂP^{1} × ℂP^{1}) + σ(\overline{ℂP^{2}}) = 1 + 0 + (−1) = 0. More precisely, b_{2}^{±}(X) = b_{2}^{±}(ℂP^{2}) + b_{2}^{±}(ℂP^{1} × ℂP^{1}) + b_{2}^{±}(\overline{ℂP^{2}}) implies that

b_{2}^{+}(X) = 1 + 1 + 0 = 2 and b_{2}^{−}(X) = 0 + 1 + 1 = 2.

(b) S^{1} × S^{1} × F, where F is an oriented surface of genus 3.

Solution: Using multiplicativity of the Poincare polynomial P_{X}(t) = ∑_{k≥0} b_{k}(X)t^{k}, for X = S^{1} × S^{1} × F, we obtain

P_{X}(t) = P_{S^{1}}(t)P_{S^{1}}(t)P_{F} (t) = (1 + t)^{2}(1 + 6t + t^{2}) = 1 + 8t + 14t^{2} + 8t^{3} + t^{4},

that is b_{0} = 1, b_{1} = 8, b_{2} = 14, b_{3} = 8, b_{4} = 1.

Since X has a direct factor S^{1}, we have σ(X) = 0. Since b_{2} = b_{2}^{+} + b_{2}^{−} = 14 and b_{2}^{+} − b_{2}^{−} = σ = 0, we conclude that b_{2}^{+} = b_{2}^{−} = 7.

(c) X → ℂP^{2} is the double covering branched along a non-singular curve of degree 8.

Solution: As usual, b_{0} = b_{4} = 1, moreover, the double branched covering X is simply

connected (Lefschetz theorem), and thus, b_{1} = b_{3} = 0. Thus, χ(X) = b_{0} − b_{1} + b_{2} − b_{3} + b_{4} = 2 + b_{2}. By the generalized Riemann-Hurwitz formula,

χ(X) = 2χ(ℂP^{2}) − χ(A),

where A is the branching locus.

As a non-singular curve of degree 8, algebraic curve A is a topological surface or genus g = \frac{1}{2}(8 − 1)(8 − 2) = 21.

Thus, χ(A) = 2 − 2g = −40 and b_{2} = χ(X) − 2 = (6 − χ(A)) − 2 = 44. By the index formula,

σ(X) = 2σ(ℂP^{2}) − \frac{1}{2}(A ∘ A) = 2 − 32 = −30,

since A ∘ A = 8^{2} by Bezout theorem. Finally,

b_{2}^{+} + b_{2}^{−} =b_{2} = 44
b_{2}^{+} − b_{2}^{−} =σ = −30

yield b_{2}^{+} = \frac{1}{2}(44 − 30) = 7 and b_{2}^{−} = \frac{1}{2}(44 + 30) = 37.