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


by trmukerji14
Tags: fundamental, integers, isomorphic, showing
trmukerji14
trmukerji14 is offline
#1
Dec27-12, 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(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.
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rasmhop
rasmhop is offline
#2
Dec28-12, 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.
trmukerji14
trmukerji14 is offline
#3
Dec30-12, 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(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
rasmhop is offline
#4
Dec30-12, 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 (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
trmukerji14 is offline
#5
Dec31-12, 12:58 PM
P: 4
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.


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