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1月13日 ： Covering a 3-manifold with special subsets
We discuss the smallest number of open subsets U
needed to cover a closed 3-manifold M where U is
required to have a certain property. This property is
one of the following:
- U is homeomorphic to R^3.
- U is homeomorphic to S^1 x R^2.
- U is homeomorphic to an open subset of R^3.
- U is contractible in M.
- U is pi_1-contractible in M.
- U is H_1-contractible in M.
- U is S^1-contractible in M.
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1月20日 ： Three manifolds with S^1-category 2
A subspace U of a manifold is S^1-contractible (in M) if there are maps
f:U→S^1, α:S^1→M such that the inclusion i:U→M is homotopic to αf.
cat_S^1(M) is the smallest number n such that there exist n open
S^1-contractible subsets of M whose union is M.
Theorem. Let M^3 be a closed 3-manifold. cat_S^1(M^3)=2 if and only
if π_1 M^3 is cyclic.
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2月3日 ： On the homeotopy group of the nonorientable surface of genus 3
We will show that, if N=P^2 # P^2 # P^2, then the homeotopy group
Mod(N) of N is GL(2,Z).
Also we show that the natural epimorphism Homeo(N)→ Mod(N) has a section.
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2月24日 ： Knot-theoretic Equivalents of the Kervaire Conjecture
Kervaire's Conjecture: If G is a nontrivial group, then G*Z cannot be
normally generated by one element.
Think of a properly embedded, minimal genus Seifert surface (connected
2-manifold) in a knot exterior.
It is incompressible, compact, orientable, nonseparating (ICON).
But its boundary is connected.
Can you find an ICON surface F with disconnected boundary in a knot
exterior E?
If you can prove such surfaces do not exist you have proved Kervaire's
Conjecture.
For Kervaire's Conjecture is equivalent to
Conjecture Z: If F is an ICON surface in a knot exterior E,
then π_1(E/F)=Z.
And this conjecture is true if ∂F is connected.
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3月3日 ： Active Sums and the Homeotopy Group of Surfaces
Given a generating family $\cal{F}$ of subgroups of $G$, closed under
conjugation, and their mutual actions,
a new group $S$, called the active sum of $\cal{F}$, can be constructed.
This group, modulo centers, is isomorphic to $G$; more precisely
$S/Z(S)\approx G/Z(S)$ where $Z$ denotes center.
Two examples will be mentioned:
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$G$ is the homeotopy group of an orientable surface of genus $\geq
2$ and $\cal{F}$ is the family of (infinite cyclic)
Dehn twist subgroups.
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$G=\pi(S^{n+2}-M^{n})$ where $M^{n}$ is a closed orientable
$n$-manifold and $\cal{F}$ is the family of
(infinite cyclic) meridian subgroups.
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3月10日 ： Unsolvable problems about knots and related groups
(with C. McA. Gordon and J. Simon)
[Theorem.] If there is an $n$-knot whose group has unsolvable
word problem, then the problem of deciding
whether or not an $n$-knot is trivial is algorithmically unsolvable.
[Conjecture.] There is a $2$-knot group with unsolvable word
problem.
[Theorem.] There is a $3$-knot (and therefore a surface in
$S^{4}$) whose group contains all finitely
presented groups.
[Corollary.] There is a $2$-knot (and therefore a surface in
$S^{4}$) whose group has unsolvable word problem.
Let ${\cal K}_{n}=\{$groups of complements of $n$-spheres in
$S^{n+2}\}$ \\
${\cal S}=\{$groups of complements of orientable closed surfaces in
$S^{4}\}$ \\
${\cal M}=\{$groups of complements of $2$-spheres in
$\underbrace{S^{2}\times S^{2}\# \cdots \# S^{2}\times
S^{2}}_{n-times}$ for some $n\}$ \\
${\cal G}=\{$Finitely presented groups$\}$ \\
[Conjecture.] Let $\cal{A}$ and $\cal{B}$ be members of $\{
{\cal K}_{0}, {\cal K}_{1}, {\cal K}_{2},
{\cal K}_{3}, {\cal S}, {\cal M}, {\cal G} \}$ such that ${\cal
B}\subsetneqq {\cal A}$ and
${\cal A}\supset {\cal K}_{2}$, then there does not exist an algorithm
that can decide,
given a finite presentation of a group $G$ in ${\cal A}$, whether or
not $G$ is in ${\cal B}$.
[Theorem.] The conjecture is true if ${\cal A}\not={\cal
K}_{2}$.
[Theorem.] Let $I$ be a set of natural numbers containing a
number $>1$. Then there does not exist an algorithm
that can decide, given a finite presentation of a group $G$, whether
or not $H_{i}G=0$ for all $i\in I$.
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3月17日 ： A formulation of the Poincar\'{e} Conjecture using knot groups
An \textit{Artin presentation} is a group presentation
$$(x_{1}, \ldots, x_{n}: r_{1}, \ldots, r_{n})$$
such that
$$\prod_{i=1}^{n}r_{i} x_{i} r_{i}^{-1}=\prod_{i=1}^{n}x_{i}$$ in
$F(x_{1}, \ldots, x_{n})$.
[Definition.] If $(x_{1}, \ldots, x_{n}: r_{1}, \ldots, r_{n})$
is an Artin presentation of
the trivial group, then $\| (x_{1}, \ldots, x_{n}: r_{1}, \ldots,
r_{n-1}) \|$ is a \textit{RAT-group}.
[Definition.] A \textit{knot group} is a group that has a
presentation
$$(y_{1}, \ldots y_{n}: s_{1}y_{1}s_{1}^{-1}y_{2}^{-1}, \ldots,
s_{n-1}y_{n-1}s_{n-1}^{-1}y_{n}^{-1},
s_{n}y_{n}s_{n}^{-1}y_{1}^{-1})$$ such that
$$\prod_{i=1}^{n} s_{i}y_{i}s_{i}^{-1}=y_{2}y_{3}\cdots y_{n}y_{1}$$
in $F(y_{1}, \ldots, y_{n})$.
[Conjecture.] RAT-groups are knot groups.
[Theorem.] This conjecture is equivalent to Poincar\'{e}'s
Conjecture.
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