Algebraic topology: Calculating the fundamental group

This lecture is part of an online course on algebraic topology.
We calculate the fundamental group of several spaces, such as a ficure 8, or the complement of a circle in R^3, or the group GL3(R).
For the other lectures in the course see • Algebraic topology

[@CodingDragon04]

"Reasonably obviously" is now my favorite combination of adjectives.

[@irrazionalex226]

Thank you for your amazing lectures Prof. Borcherds! I have a question: is there a reference where I can find the computation for the fundamental group of the space consisting in the two cones over the hawaian earrings joined at a point?

[@sachalucienmoserferreira2233]

Nice to find your channel on yt,its a great pleasure to be see a lecture
from a field medallist. Hugs from Brazil,keep going on your videos for
sheare with us a bit of your knowledgment.

[@veganmathematician2896]

Why are the fundamental groups of O(2) and O(3) the same as SO(2) and SO(3) respectively? Why does the Z/2Z (two connected components) not make a difference?

[@tofu-munchingCoalition.ofChaos]

SO(n) and O(n) have the same fundamental group if you use a basepoint A in SO(n). Because every loop starting at A in O(n) is contained in SO(n) as well as every homotopy between loops.
As the two components of O(n) are homeomorphic (multiplication with an element B in O(n) but not SO(n) permutes the components) hence the fundamental groups of the components are isomorphic.
Z/2Z in the calculation of the video is a discrete space with trivial fundamental group and not the fundamental group of a space. Perhaps this is where your confusion came from?
FYI, in general:
a) The fundamental group of a space or path component is not well-defined. It depends on a basepoint.
b) The isomorphism class of the fundamental group in the same path component is independent of the base point. The fundamental groups in one path component are canonically isomorphic iff the (isomorphism class of the) fundamental group of the path component is abelian. So only in the case of abelian fundamental groups makes it sense to say a group is *the* fundamental group of the path component.
c) Different path components do not need to have even isomorphic fundamental groups.

[@veganmathematician2896]

@@tofu-munchingCoalition.ofChaos Thank you, I think that makes sense!

[@vinbo2232]

The last demostration, the shoulder-elbow-twist is the 3 degree of freedom of SO(3), right? then how the (-1)*(-1)=1 like position transform can lead to the fundamental group Z/2Z exactly? Thanks.

[@forheuristiclifeksh7836]

17:48

[@migarsormrapophis2755]

yeeeeeeeeeeeeeeeeeee

[@brendawilliams8062]

Looks like a black hole and white hole and lots of interesting stuff

[@annaclarafenyo8185]

is pi_1 = Z an unknot detector? If so, it seems like it could be fast to check non-abelian nature of a presented group.

[@ay5960]

Checking whether a finitely presented group is abelian or not is undecidable meaning there are no algorithms let alone a fast one.

[@annaclarafenyo8185]

​@@ay5960 Noted, but that's for arbitary presentations and arbitarily complex abelian groups. In this particular case of knot complement, it's not an arbitary group, and it isn't an arbitary abelian group either. Every small loop (a loop that only crosses one part of the knot) must be a generating element, the abelianization of the group is the homology of the knot, and there is an explicit map to the abelianization. With these extra facts, I am not sure it is still difficult.

[@ay5960]

@@annaclarafenyo8185 The homology of knot complement is Z, so your question is equivalent to asking for an algorithm to detect whether the fundamental group is Z or not. Fundamental group being equal to Z is equivalent to the fact that the knot is unknot. So you question is equivalent to asking for an algorithm to detect unknot in polynomial time. This is a famous question and it is open. The major results in this direction indicate that unknot recognition is in NP (I think the algorithm is due to Haken). I don't know the state of the art for unknot detection algorithms but it seems that Lackenby has given a quasi-polynomial time algorithm.

[@brendawilliams8062]

Thankyou.

[@040_faraz9]

3:10 thts smash

[@diribigal]

No, he was right that it's the wedge sum. It just so happens that unfortunately, the LaTeX command for the smash product symbol is \wedge (and the LaTeX command for the wedge sum symbol he wrote is \vee ).

[@040_faraz9]

@@diribigal i see🙂

[@fbkintanar]

around 3:10, the descending wedge can be pronounced "Vel" which is Latin for or. en.wikipedia.org/wiki/Descending_wedge. Also see the comments of 040_Faraz and dirigibal

[@jusiphstolin8074]

first comment uwu

[@040_faraz9]

Congrats