3
votes
1answer
106 views

Relation between the braid group and the mapping class group of the plane

According to the following link, page 248, the braid group modulo its center is isomorphic to the mapping class group of the $N$-times punctured plane, i.e. $B_N/Z(B_N)\cong M_N(\mathcal(R)^2)$. Could ...
5
votes
2answers
135 views

Framed Cobordism Classes of links in $\mathbb R^3$

We know that every link in $S^3$ is framed cobordant to the unknot with some framing. The idea is to study smooth homotopy classes of maps from $S^3$ to $S^2$. Actually in the title I have given ...
14
votes
1answer
176 views

Visualize Fourth Homotopy Group of $S^2$

I know $\pi_4(S^2)$ is $\mathbb{Z}_2$. However, I don't know how to visualize it. For example, it is well known that $\pi_3(S^2)=\mathbb{Z}$ can be understood by Hopf Fibration. Elements in ...
0
votes
1answer
62 views

Is it true that any two tame knots are homotopic?

My understanding is that if the embeddings $f_0,f_1$ are tame knots then $H(t,\theta) = (1-t)f_0(\theta) + t f_1(\theta)$ is a homotopy between them, thus all tame knots are homotopic. Is this the ...
3
votes
0answers
40 views

some help on the group of unknotted

Show that the group of the unknotted $K=\{(z_z,z_2)\in \mathbb{S^3} : |z_1|=1 \}$ is infinite cyclic. where $\mathbb{S^3}$ is to be considering as the unit vectors in $\mathbb{C^2}\cong \mathbb{R^4}$. ...
4
votes
2answers
194 views

A puzzle on knotted surfaces

Only after having learned that the somehow only notion of equivalence of knots is definitely "ambient isotopy" I stumbled over this blog entry on ambient isotopy. (Had it been earlier!) What bothers ...
1
vote
1answer
167 views

Equivalence of knots

It's intuitively clear what it means that two knots $K,K'$ are essentially the same, but it can be termed and defined more precisely in different ways. Are all of them equivalent? $K, K'$ are ...
0
votes
1answer
103 views

Pure braid group, stabilizer

From group theory we know that a homomorphism $\phi: G \to \operatorname{Sym}(S)$, where S is a set, then $\operatorname{Sym}(S) \cong \Sigma_n $. Its kernel is given as $\bigcap_{s \in S}G_s$, which ...
10
votes
1answer
303 views

Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?

This question has been cross posted on MathOverflow with some very interesting answers and discussion. I'm currently writing a project on the braid groups and their analogues on closed surfaces. ...