Saveliev Flashcards
Genus g Handlebody
An orientable 3-dim mfld obtained from B^3 by attaching g many 1-handles D^1 \times D^2
Existence of Heegaard Splitting
Any closed orientable 3-mfld admits a Heegaard splitting
Proof of Existence of Heegaard Splitting
1) Take triangulation of manifold
2) Take H1= nbhd of vertices and edges
3) H2 = nbhd of faces and tetrahedra
Stabilization of Heegaard Splittings
Get a genus g+1 HS by adding an unknotted 1-handle to a genus g HS
Equivalent Heegaard Splittings
Two HS of a manifold are equivalent if there is a homeomorphism taking one splitting to another
Stably Equivalent Heegaard Splittings
Two HS’s of a mfld are stably equivalent if they are equivalent after some # of stabilizations
Two Heegaard splittings are stably equivalent if?
They are Heegaard splittings of the same manifold
Mapping Class Group
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.
Isotopic gluing maps
If f and f’ are isotopic gluing maps, then they produce homeomorphic manifolds.
Heegaard genus
A manifold has heegaard genus g if it admits a heegaard splitting of genus g, and does not admit one of a smaller genus.
Only closed manifold of genus 0?
This is S^3.
Mapping class group of torus
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).
Mapping class group is generated by…?
Dehn twists along , meridinal, longitudinal, and other meridinal curves in a handle body.
Where we send the meridian completely determines the manifold.
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.
Lens Space
L(p,q) is the lens space given by sending meridian to pm + ql.
Fundamental group of L(p,q) = Z/p.
Manifolds of Heegaard genus 1
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
When are two lens spaces homeomorphic?
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.
When are two lens space homotopy equivalent?
L(p,q) and L(p,q’) are homotopy equivalent iff qq’ = \pm m^2 mod p for some integer m.
Seifert Manifold M((a1,b1), (a2,b2), …, (an,bn)) construction
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.
Seifert Manifolds with 1 and 2 singular fibers
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.
Link
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
Link Neighborhood
Thicken each link component to get a collection of smoothly embedded disjoint solid tori whose cores form the link
Knot Exterior
When we remove the knot nbhd, we are left with the knot exterior. It will have torus boundary.
How is gluing map determined
Determined entirely by where the meridian of the knot nbhd is sent.
