Showing the Fundamental Group of S^1 is isomorphic to the integers

trmukerji14

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(1-t)}[/itex]
H(s,t)={e[itex]^{2\pi in2st}[/itex]e[itex]^{2\pi im2s(1-t)}[/itex] for s in [0,1/2]
{e[itex]^{2\pi im(2s-1)t}[/itex]e[itex]^{2\pi in(2s-1)(1-t)}[/itex] for s in [1/2,1]

Any help would be very much appreciated on my part.

rasmhop

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.

trmukerji14

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(1-t))}[/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(2s-1)(1-t))}[/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.

rasmhop

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 (1-t)} & \textrm{for }s \in [0,1/2] \\
e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)} & \textrm{for }s \in [1/2,1]
\end{cases}[/tex]
Let us focus on fixing the fact that this is not well-defined 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 (1-t)} & \textrm{for }s \in [0,1/2] \\
e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)}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 well-chosen 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}(1-t) = (m+n)\frac{1}{2} t + 2(1-t) + \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 well-defined homotopy of loops.

trmukerji14

I got it! This is awesome.

I got [itex]\alpha(t)[/itex]=m(1-t) 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 (1-t)} & \textrm{for }s \in [0,1/2] \\
e^{2\pi i(m+n)st}e^{2\pi i n 2 (2s-1)(1-t)}e^{2\pi i m(1-t)} & \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 (1-t)} & \textrm{for }s \in [0,1/2] \\
e^{2\pi i(m+n)st}e^{2\pi i n (2s-1)(1-t)}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.