9

As I read an algebraic topology book, I felt I knew exactly what the fundamental group is geometrically! I thought it counts the number of independent cycles. (my definition of dependence cycles (that may be incorrect): $\alpha,\beta$ are two dependent cycles if $\exists m,n\in\Bbb Z$ s.t. $\alpha^n=\beta^m$.)

And according to this explanation it was always a serious question for me that why almost all authors compute the fundamental group of circle in very long way, some of them in a separate chapter and after proving many theorems instead of observing that there is only one independent cycle in circle! After these doubts, I read that $\pi_1(\Bbb RP^2)=\Bbb Z_2$. And here was the end of my dream. Since in term of my interpretation, the fundamental group is always free product of some $\Bbb Z$ but $\pi_1(\Bbb RP^2)=\Bbb Z_2$ is not in that form! So I want to know that

What does fundamental (homotopy) groups measure?

Isn't it true that if the space does not have a trivial loop, then it must contain at least one generator?

I think my interpretation is partially true but it ignores some information of the space. Or perhaps the missed point of my interpretation is that I don't consider the base point to be fixed.

The above interpretation also can be applied for $\pi_2$ (and $\pi_n$). i.e. $\pi_2$ counts the number of independent topological spheres (But in this case I don't know what is the independent topological spheres!!).

C.F.G
  • 8,523
  • Intuitively homotopy groups measure the same thing as homology groups, but formalized in a different way so that homology and homotopy groups are not exactly the same thing (there are certain relations between them however, such as the Hurewiz theorem). Therefore this question might help you. – Javi Oct 20 '20 at 11:42

1 Answers1

4

Maybe instead of general ideas, look at simple examples where the fundamental group is not a free product. Let's look at a torus, represented as a square with opposite sides identified. Each of the sides $a$ and $b$ represent incontractible loops, but if you go around the square, then the loop $aba^{-1} b^{-1}$ can be contracted (shrinking the boundary of the square into a mid-point).

Following picture illustrates a homotopy between $ab$ and $ba$ in a torus.

enter image description here

With $\Bbb{RP}^2$, represent it as a ball with lower half-circle $a$ identified with upper half-circle $a$. Then there is no way to contract $a$ but of course, $aa$ goes around the circle and can be contracted.

Blockquote

Nullhomotopy of $aa$:

enter image description here

Let's try to nullhomotop just $a$.. intuitively, you can probably see you don't succeed

enter image description here

To illustrate that fundamental group of $\mathbf{SO}(3)$ is $\Bbb Z_2$ is even more intuitive. Take a glass of beer, hold it in your right arm. Rotate it by 360 degrees, with your shoulder being "fixed" (and don't pour anything out). Your hand is "twisted" and represent a loop in $\mathbf{SO}(3)$. Move your hand up and continue twisting in the same rotation; after another 360 degrees, your hand is untwisted again. Great video.

If you want to think in terms of "independence" -- think of generators and relations and note that $\Bbb Z=\langle \alpha| \varnothing\rangle$ (empty relation) and $\Bbb Z_2=\langle \alpha| \alpha^2=1\rangle$. For $\pi_1$, you can restrict yourself to the 2-skeleton. Homotopy group of $1$-skeleton is indeed a free group of generating cycles. Every two-cell adds one relation.

C.F.G
  • 8,523
Peter Franek
  • 11,522
  • I just can't figure out what does $\Bbb Z_2$ means geometrically and how it construct in practice (that seems to be a dream)!! – C.F.G Oct 27 '20 at 15:18
  • I also couldn't relate your SO(3) explanation with its fundamental group. – C.F.G Oct 27 '20 at 15:19
  • There is a bit more explanation here but it could be better. Maybe simplest is the $RP^2$ example. Can you see, from my picture, that if you fix the endpoint of the circle $a$, however you transform this loop, it will never be trivial? Can you think of a way how to shring $aa$ (fixing the endpoint) to a trivial loop? – Peter Franek Oct 27 '20 at 15:33
  • With the hand and beer.. parametrize your arm with $[0,1]$ and consider each point in your arm to represent one rotation (one element in $SO(3)$). Then the state of your arm (from shoulder to palm) represent one curve in $SO(3)$. If you fix the shoulder and palm and do any movement with your arm, it represent a homotopy between two curves. The fact that a 360 degree rotated hand (= a curve starting with trivial rotation and ending with 360 degree rotation) cannot be untwisted shows it is homotopically nontrivial. But doing the same twist twice (curve from 0 to 720 degrees) is un-twisted. – Peter Franek Oct 27 '20 at 15:47
  • In regards to your first one of two last comments: No. I can't see from your picture. Isn't it true that if the space hasn't a trivial loop then it should contain at least one generator? – C.F.G Oct 27 '20 at 16:13
  • 1
    And I don't see any relation between your torus example and my question. Your example shows that the fundamental group of torus is abelian that I knew this. – C.F.G Oct 27 '20 at 16:18
  • and I realized the SO(3) double rotation (=720) using the YouTube link you provided. and again I don't see any relation between fundamental group of SO(3) in your example and my question. – C.F.G Oct 27 '20 at 16:21
  • You say "I know that $\pi_1$ of torus is abelian" and you write "(I thought that...) the fundamental group is always free product of some ℤ", how does this go together? – Peter Franek Oct 27 '20 at 16:46
  • 1
    I added another picture with nullhomotopy of $aa$ in $RP^2$.. but maybe you will write me that you know that... – Peter Franek Oct 27 '20 at 16:51
  • You are right! Perhaps I convinced myself in this specials case because I knew $ab=ba$ in torus. and $\pi_1(T^2)=Z *Z/C$ – C.F.G Oct 27 '20 at 16:52
  • Dear Peter: First, Can I work with disc with two boundary identified instead of $RP^2$ and shouldn't I be care about? Sorry for my poor imagination, but in your last image i can see a is not nullhomotopic because I can't imagine $a$ in real $RP^2$ instead of disc. Aside of all of this, by $a$ is not trivial, then it must contain at least one generator? – C.F.G Oct 27 '20 at 17:05
  • What is your definition of $RP^2$? Usually it is defined as this, as a disc with opposite points identified. Or a set of lines in a 3-space. You can identify a space of lines with the upper hemisphere; then clearly lines that are in $xy$ plane correspond to the boundary circle. The curve $a$ corresponds to a set of lines so that you start with the $x$-axis, move it around in the $xy$ plane, and do a rotation by $90$ degrees (and end up with the same line). – Peter Franek Oct 27 '20 at 17:13
  • I meant working with disc before identifying its opposite points as you (and all experts) did. Sorry that I wasted your time. – C.F.G Oct 27 '20 at 17:33
  • Not sure now if I understand. Whatever this space is.. even if you don't see it is $RP^2$, if you agree that it is some topological space, and see that $a$ is nontrivial and $aa$ is trivial, it might answer your question. No wasted time, it's fine, all the best – Peter Franek Oct 27 '20 at 17:47
  • Now your last change is meaningful for me. that $x$ is much helpful. – C.F.G Oct 27 '20 at 18:08
  • So by $a$ is nontrivial, then is it this means that $\pi_1$ must contain at least one generator? – C.F.G Oct 27 '20 at 18:22
  • Yes, each group can be represented by generators and relations. Nontrivial group = at least one generator. Here $\pi_1 = {a, a^2 = 1}\simeq \mathbb{Z}_2$ – Peter Franek Oct 27 '20 at 19:34
  • OMG, I always forget to think about "relation"! Thanks. I think that is the the point I missed it. Now it is so easy. Now due to your good comments I don't know why my post got +7 upvote! – C.F.G Oct 27 '20 at 20:39
  • An small question: Can orientability affect to $\pi_1$? – C.F.G Oct 27 '20 at 20:44
  • 2
    There is not a straight-forward relation between orientability and $\pi_1$ (AFAIK). For instance $SO(3)=\mathbb{RP}^3$ is orientable (moreover, a Lie group) and the fundamental group is still $\mathbb{Z}_2$. – Peter Franek Oct 27 '20 at 21:06
  • Thanks again a lot. For your time. I forget that $\Bbb Z$ is itself a loop with empty relation. pls add these last comments to your answer for further readers. – C.F.G Oct 27 '20 at 21:13
  • But I tried to write it from the beginning, in the last section "If you want to think in terms of "independence....". Any suggestion what should be added? Feel free to edit it, if you can – Peter Franek Oct 27 '20 at 21:15
  • You are right, I am so careless!! – C.F.G Oct 27 '20 at 21:18
  • 2
    There is a small relationship between orientability and $\pi_1$. Namely, for a non-orientable manifold, $\pi_1$ must contain an index two subgroup (which corresponds to the orientation double cover). In particular, if $\pi_1 \cong \mathbb{Z}_n$ with $n$ odd, then the manifold must be orientable. – Jason DeVito - on hiatus Oct 28 '20 at 13:29
  • @JasonDeVito: Thank you. So $\pi_1 \cong \Bbb Z^l*\mathbb{Z}{2k}$ or $\pi_1 \cong \mathbb{Z}{2}$ says nothing about orientability? – C.F.G Oct 29 '20 at 14:05
  • 2
    @C.F.G: Correct. As Peter mentions, both $\mathbb{R}P^2$ and $\mathbb{R}P^3$ have $\pi_1 = \mathbb{Z}2$, but one is orientable while the other is non-orientable. One can also construct orientable and non-orientable spaces with $\pi_1 = \mathbb{Z}^\ell\times \mathbb{Z}{2k}$ or $\pi_1 =\mathbb{Z}^\ell \ast \mathbb{Z}{2k}$. For example, to make a non-orientable thing with $\pi_1 = \mathbb{Z}{2k}$, you can use $(L_k \times S^2)/(\mathbb{Z}2)$ with $\mathbb{Z}_2$ acting on the Lens space $L_k$ by the usual thing with quotient $L{2k}$, and acting on $S^2$ via the antipodal map. – Jason DeVito - on hiatus Oct 29 '20 at 14:11