6
votes
0answers
79 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
37 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
35 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
156 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
118 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
76 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 ...
9
votes
1answer
239 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. ...