Saveliev

Question 1

Genus g Handlebody

Answer 1

An orientable 3-dim mfld obtained from B^3 by attaching g many 1-handles D^1 \times D^2

Question 2

Existence of Heegaard Splitting

Answer 2

Any closed orientable 3-mfld admits a Heegaard splitting

Question 3

Proof of Existence of Heegaard Splitting

Answer 3

1) Take triangulation of manifold
2) Take H1= nbhd of vertices and edges
3) H2 = nbhd of faces and tetrahedra

Question 4

Stabilization of Heegaard Splittings

Answer 4

Get a genus g+1 HS by adding an unknotted 1-handle to a genus g HS

Question 5

Equivalent Heegaard Splittings

Answer 5

Two HS of a manifold are equivalent if there is a homeomorphism taking one splitting to another

Question 6

Stably Equivalent Heegaard Splittings

Answer 6

Two HS’s of a mfld are stably equivalent if they are equivalent after some # of stabilizations

Question 7

Two Heegaard splittings are stably equivalent if?

Answer 7

They are Heegaard splittings of the same manifold

Question 8

Mapping Class Group

Answer 8

A discrete group of symmetries on the space M.
Homeo(M) = group of all orientation preserving homeomorphisms of M.
Homeo0(M) = normal subgroup of homeomorphisms isotopic to id.

H(M) = Homeo(M)/Homeo0(M) is called the mapping class group of the surface M.

Question 9

Isotopic gluing maps

Answer 9

If f and f’ are isotopic gluing maps, then they produce homeomorphic manifolds.

Question 10

Heegaard genus

Answer 10

A manifold has heegaard genus g if it admits a heegaard splitting of genus g, and does not admit one of a smaller genus.

Question 11

Only closed manifold of genus 0?

Answer 11

This is S^3.

Question 12

Mapping class group of torus

Answer 12

SL(2,Z).
This comes from looking at homeo’s of T^2 as automorphisms on pi_1 = Z \oplus Z. Auto’s of Z \oplus Z are given by integral 2 by 2 invertible matrices. Matrices over Z are invertible iff det = 1, -1. And a matrix is orientation preserving iff det = 1. Therefore get SL(2,Z).

Question 13

Mapping class group is generated by…?

Answer 13

Dehn twists along , meridinal, longitudinal, and other meridinal curves in a handle body.

Question 14

Where we send the meridian completely determines the manifold.

Answer 14

We glue in solid torus in two steps.

1) Glue in solid cylinder, this is determined by a 2 by 2 integral matrix, see mapping class group
2) Then glue in 3-ball, along boundary = S^2. All orientation preserving homeo’s of S^2 are isotopic to id.

Question 15

Lens Space

Answer 15

L(p,q) is the lens space given by sending meridian to pm + ql.
Fundamental group of L(p,q) = Z/p.

Question 16

Manifolds of Heegaard genus 1

Answer 16

Any 3-dim manifold of genus 1 is either L(0,1) = S^1 \times S^2, or L(p,q), with p and q relatively prime, p >= 2 and 1 <= q <= p-1

Question 17

When are two lens spaces homeomorphic?

Answer 17

L(p,q) and L(p,-q) are always homeomorphic (via orientation reversing homeo).
L(p,q) and L(p,q’) are homeo when qq’ = 1, -1 mod p.

Question 18

When are two lens space homotopy equivalent?

Answer 18

L(p,q) and L(p,q’) are homotopy equivalent iff qq’ = \pm m^2 mod p for some integer m.

Question 19

Seifert Manifold M((a1,b1), (a2,b2), …, (an,bn)) construction

Answer 19

1) Let M be a 2-sphere with n disjoint disks removed.
2) M \times S^1 is a compact orientable 3-manifold whose boundary consists of n tori.
3) Given n pairs of relatively prime integers (ai,bi) ai > 1
4) Glue in n solid tori so that the meridian of the its solid torus is glued to a curve on one of my boundary tori by ail + bil.
The ith singular fiber is the image of my central circle of each solid tori under this gluing.

Question 20

Seifert Manifolds with 1 and 2 singular fibers

Answer 20

A Seifert manifold with one or 2 singular fibers is a lens space.
A Seifert manifold with >2 singular fibers is not homeomorphic to a lens space.

Question 21

Link

Answer 21

A finite collection of smoothly embedded disjoint closed curves in a closed orientable 3-mfld M is called a link.
One component link is a knot

Question 22

Link Neighborhood

Answer 22

Thicken each link component to get a collection of smoothly embedded disjoint solid tori whose cores form the link

Question 23

Knot Exterior

Answer 23

When we remove the knot nbhd, we are left with the knot exterior. It will have torus boundary.

Question 24

How is gluing map determined

Answer 24

Determined entirely by where the meridian of the knot nbhd is sent.

Question 25

Integral Surgery

Answer 25

p/q surgery is integral if q = 1, -1

Question 26

Lickerish Wallace Theorem

Answer 26

Every closed orientable 3-mfld can be obtained from S3 by an integral surgery on a link in S3.

Question 27

Continued Fraction Decomposition

Answer 27

p/q can be written as x1 - 1/(x2 - 1/(x3 - 1/(….))))

Question 28

Plumbing Graph

Answer 28

Each vertex is unknot, labeled with surgery coefficient
Adjacent vertices represent linked unknots
Every Lens space and Seifert 3-mfld has such a description

Question 29

Cobordism

Answer 29

compact smooth 4-mfld W is cobordism between M1 and M2 if boundary W = M1 cup M2.
If M1 empty, we say M2 is cobordant to 0.
Any closed oriented 3-mfld is cobordant to 0

Question 30

Surgery on 4 manifolds

Answer 30

Take M * (0,1) and N(k) in M*1, then attach D^2 * D^2 along N(k). This gives us 4-mfld W. This is called the trace of surgery on k.
W is a cobordism between M and surgery on M

Question 31

Linking Number

Answer 31

lk(L1,L2) is given by summing all crossings of L1 under L2 with the appropriate sign

Question 32

Canonical longitude of a knot

Answer 32

This canonical longitude satisfies lk(k,l) = 0.

This is because we can be picture l living in the seifert surface of k, which means lk = 0

Question 33

Kirby Moves

Answer 33

K1: Add or delete unlinked unknot with framing +- 1
K2: Slide one component of Link over number with proper twisting considerations
Can only be done on integral surgeries!

Question 34

Kirby moves and homeomorphic manifolds

Answer 34

3-Manifolds by surgery on L1 and L2 are homeomorphic IFF L1 and L2 can be connection through sequence of Kirby moves

Question 35

Blow up/down move

Answer 35

Move away unknot with framing +- 1 from rest of link by giving any strands through this unknot a left/right twist

Question 36

Fundamental Class

Answer 36

Let M be a closed oriented connected, simply-connected 4-mfld. H_4(M) = H^0(M) = Z. Choosing a generator in H_4(M) corresponds to choosing orientation of M. Once M is oriented, the generator of H_4(M) is called the fundamental class of M and is denoted [M].

Question 37

Intersection Form

Answer 37

Let M be a closed oriented connected, simply-connected 4-mfld. Consider the bilinear form on the free Abelian group H^2(M)
Q_M: H^2(M) * H^2(M) –> Z
(a,b) -> (a cup b, [M])
This is a symmetric integral form called the intersection form of the manifold M.

Question 38

Seifert Surface for a knot

Answer 38

A connected compact oriented surface embedded in S3 with the knot as boundary.
Every oriented link in S^3 bounds a seifert surface.
Any two seifert surfaces for an oriented link are stably equivalent.

Question 39

Four ball genus

Answer 39

Find a connected compact oriented surface embedded in B^4, with the knot as genus. The minimal genus of all the surfaces is the four ball genus.

Question 40

Seifert Matrix

Answer 40

Take a curve x on a Seifert surface for k, and do the positive push off, x+. The Seifert matrix is the n * n matrix lk(xi, xj+). This will be genus by genus in dimension.
The Alexander polynomial is the determinant of [t^(1/2)S - t^(-1/2)S^T], where S is the seifert matrix.