Showing the Fundamental Group of S^1 is isomorphic to the integersby trmukerji14 Tags: fundamental, integers, isomorphic, showing 

#1
Dec2712, 09:22 PM

P: 4

Hi,
I am reading J.P. May's book on "A Concise Course in Algebraic Topology" and have approached the calculation where [itex]\pi[/itex][itex]_{1}[/itex](S[itex]^{1})[/itex][itex]\cong[/itex]Z He defines a loop f[itex]_{n}[/itex] by e[itex]^{2\pi ins}[/itex] I want to show that [f[itex]_{n}[/itex]][f[itex]_{m}[/itex]]=[f[itex]_{m+n}[/itex]] I understand this as trying to find a homotopy between f[itex]_{n}[/itex]*f[itex]_{m}[/itex] and f[itex]_{m+n}[/itex] I have some attempts some attempts which have been unsuccessful are H(s,t)= f[itex]_{n+mt}[/itex]*f[itex]_{m(1t)}[/itex] H(s,t)={e[itex]^{2\pi in2st}[/itex]e[itex]^{2\pi im2s(1t)}[/itex] for s in [0,1/2] {e[itex]^{2\pi im(2s1)t}[/itex]e[itex]^{2\pi in(2s1)(1t)}[/itex] for s in [1/2,1] Any help would be very much appreciated on my part. 



#2
Dec2812, 05:18 PM

P: 418

Your idea seems correct. To separate into cases depending on whether s is smaller or larger than 1/2 is a good idea in this problem (as you have done).
However try to think about what you are doing. [itex]f_{m+n}[/itex] loops with a speed of (m+n) for the whole interval [0,1]. [itex]f_n * f_m[/itex] instead first loops with speed 2m for [0,1/2], and then loops with speed 2n for [1/2,1]. Let us first focus on what our homotopy should be for [itex]s \in [0, 1/2][/itex]. In this case we start with a speed of 2m and would like to end up with a speed of m+n. In other words for the part [itex]s\in[0,1/2][/itex] we would like to have [tex]e^{2\pi i 2 m s}[/tex] when t = 0 and [tex]e^{2\pi i (m+n) s}[/tex] when t = 1. In your case when we plug in t=0 we get [tex]e^{2\pi i m 2s}[/tex] as we wanted, but when we plug in t=1 we get [tex]e^{2\pi i n 2 s}[/tex] but we wanted [tex]e^{2 \pi i (m+n) s}[/tex] Can you figure out how to correct this? The same problem happens when [itex]s \in [1/2,1][/itex], so you must also correct your homotopy here. If you think you have done this, then all that remains is to check that this is a homotopy between the correct loops and that your two expressions agree on the intersection [itex]s = 1/2[/itex]. As a side remark the homotopy you constructed instead shows [tex][f_n][f_m] = [f_m][f_n][/tex] which will follow from what you want to prove eventually, but it is also interesting if you couldn't figure out how to prove the general result. 



#3
Dec3012, 03:16 PM

P: 4

Thank you for the help.
So far what I have is for s in [0, 1/2] H(s,t)= e[itex]^{2\pi i(m+n)st}[/itex]e[itex]^{2\pi im2s(1t))}[/itex] and for s in [1/2, 1], I have: H(s,t)= e[itex]^{2\pi i(m+n)st}[/itex]e[itex]^{2\pi in(2s1)(1t))}[/itex] This satisfies our conditions for a homotopy for t=0 and for t=1. Unfortunately, when we consider s it fails miserably. For a basepoint of 1. We have that H(0,t)=1 and that is good but when we try to plug in s=1/2 or s=1 we do not get what we want. So I'm scratching my head at the moment trying to fix this. As for the side remark, I understand that if we show this equivalence we can define a homomorphism from the integers to the fundamental group of the circle which will give us the abelian property. But I don't think I showed the abelian property with the homotopy I had since that one fails also when s=1. Thanks for the help again. 



#4
Dec3012, 06:00 PM

P: 418

Showing the Fundamental Group of S^1 is isomorphic to the integers
There are basically still 3 issues with your homotopy:
1) You forgot an extra coefficient of 2 in front of n in your H for s >= 1/2. 2) It doesn't line up on s=1/2, i.e. the two expressions you gave are different when s=1/2 3) It isn't a homotopy of loops (i.e. not a based homotopy). 1 is easily fixable and I assume you can easily do this. If you solve 2, then 3 should be solved automatically since your speeds gives you whole revolutions (this is obviously not a rigorous argument and you will need to check, but intuition never hurts). Fixing 1 for you I assume you wrote down: [tex]H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s1)(1t)} & \textrm{for }s \in [1/2,1] \end{cases}[/tex] Let us focus on fixing the fact that this is not welldefined for s=1/2. Ideally we would like some continuous function [itex]\alpha : [0,1] \to \mathbb{R}[/itex] such that [tex]H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s1)(1t)}e^{2\pi i \alpha(t)} & \textrm{for }s \in [1/2,1] \end{cases}[/tex] would work as a homotopy. Note that this does not change the "speeds" we work with, but just where around the circle the second case takes over when s reaches 1/2, so as long as [itex]\alpha[/itex] is wellchosen this should let us fix the problem at s=1/2. Clearly for this to work you must have (up to addition by an integer) [tex](m+n)\frac{1}{2}t + 2m\frac{1}{2}(1t) = (m+n)\frac{1}{2} t + 2(1t) + \alpha(t)[/tex] This should give you an expression for [itex]\alpha[/itex] which will let you fix your homotopy, and then you should get a welldefined homotopy of loops. 



#5
Dec3112, 12:58 PM

P: 4

I got it! This is awesome.
I got [itex]\alpha(t)[/itex]=m(1t) which gave the well defined homotopy [tex]H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s1)(1t)}e^{2\pi i m(1t)} & \textrm{for }s \in [1/2,1] \end{cases}[/tex] This almost worked except when s=1, we did not get our desired result of 1. So if we use this: [tex]H(s,t) = \begin{cases} e^{2\pi i (m+n) st}e^{2\pi i m 2 s (1t)} & \textrm{for }s \in [0,1/2] \\ e^{2\pi i(m+n)st}e^{2\pi i n (2s1)(1t)}e^{2\pi i \alpha(t)} & \textrm{for }s \in [1/2,1] \end{cases}[/tex] We have our homotopy of loops. Thanks again for the help.I was really stuck out there. 


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