# Fundamenal Group

1. Sep 14, 2006

### waht

I'm sort of confused about combining loops under this group's operation.

If you have two loops with a basepoint x, would you travel along one loop then comeback to where you started at x, then start traveling along the second loop? Or would the group operation combine those loops into one giant loop?

2. Sep 14, 2006

### mathwonk

just traverse both loops, one after the other. that in efect does combine them.

3. Sep 14, 2006

### mathwonk

if your definition of a loop is a map from [0,1], then you go over each one twice as fast as before, so you then run over both in one unit of time.

4. Sep 14, 2006

### mathwonk

but it doesn't become a group until you impose the equivalence relation of homotopy. until then i think it is just a monoid.

5. Sep 15, 2006

### waht

$\alpha : \pi_1(X,x_0) \rightarrow \pi_1(X,x_1)$

so under such map, is alpha still a loop?

6. Sep 16, 2006

### matt grime

alpha is a map, not a loop, in what you wrote. And elements of pi_1(X,x) are homotopy classes of loops.

7. Sep 17, 2006

### waht

Are the homotopy classes just a collection of equivalent loops?

and what kind of map is alpha exactly between 2 homotopy loop classes. I studied the proof that

$\alpha : \pi_1(X,x_0) \rightarrow \pi_1(X,x_1)$

is an isomorphism. But I don't understand how they treat alpha as a path and not a loop.

8. Sep 17, 2006

### matt grime

yes a homotopy class of loops is an equivalence class, the equivalence relation beign homotopy, obviously.

alpha is a group homomorphism the way you write it. It seems there is something you're not telling us.

9. Sep 17, 2006

### waht

yea, I'm self-studying this stuff. And it can get difficult.

So guess I will start from the beggining.

Ok so, two paths are path homotopic if they have the same initial and final points. And

$f \simeq_p f'$

is an equivalence ralation. The path homotopy equivalence class is [f]

What I'm not sure right now, is [f] a set of all possible homotopic paths?

10. Sep 17, 2006

### mathwonk

11. Sep 18, 2006

### matt grime

No, definitely not. Two paths are homotopic if there exists a homotopy between them. So, what is a homotopy?

12. Sep 18, 2006

### homology

Some thoughts

You've got lots of paths. Some of them have the same initial and terminal points. Some of those can be continously deformed into each other. Those are path homotopic. If they are path homotopic. For the purposes of homotopy theory, they are considered the same. Consider the punctured plane. That is, R^2 minus the origin. Now, any paths in the punctured plane that have the same initial and terminal points form a "loop" if you look at "the picture" If the origin (which is "missing" in the punctured plane) is inside this loop then the paths are not path homotopic, because you couldn't continously deform one of them into the other without dragging it over the missing origin. The fact that they're not homotopic tells you that there is a "hole" in the space, that its not simply connected.

Now there are a lot of possible paths that could be in any space and so you organize them. From the example above it should be clear that loops can tell you about the "holes" in a space. So why not study those paths that start and end at the same point, i.e. loops. The particular point where they start and end is called the basepoint. They're kind of like lassos. So you plaster your topological space with all these loops and you cinch them tight. Anything that you can "cinch" all the way back to your basepoint isn't very interesting. If you can "cinch" (and by "cinch" I mean continously deform) the loop back to the basepoint then that loop is homotopic to the constant loop which starts at the basepoint, goes nowhere, and finishes at the basepoint. But if your space is interesting, some loops will not be null homotopic (won't be cinched back to the basepoint). These will get "caught" or "hung up on" holes in the space. Of course there will be lots of different loops around the same point. More technically, there will be lots of path homotopic loops around the same point. Now there's no need to repeat information. If you found a hole, you found a hole, you don't need to find it over and over again. So you make equivalence classes based on path homotopy of loops.

Now because these are loops, and because they are all set at the same basepoint (so they all start and end at the same place) you can "add" them by doing one loop and then doing another. Just like Mathwonk said, you add them by doing each one twice as fast so that you make a new loop, one that starts at the basepoint, goes out around path number 1 comes back to the basepoint and then heads out again for a trip around path number 2, finally returning to the basepoint. So this loop looks like a pair of petals on a flower, but its a loop nontheless.

Now, as you'll know from reading whatever you're reading (I like Massey, personally) "adding" equivalence classes of the loops in this manner is well defined (if you don't know you should prove it). And you've got inverses (going around the loop in the other direction), an identity (the null loop which is the basepoint) and so you've got a group.

now the alpha you've been talking about is a group homomorphism as Matt Grime said. Look at what its doing. Its mapping one fundamental group to another fundamental group. Both groups are concerned with the same topological space X, but have different basepoints. What your book is saying is that it doesn't matter what basepoint you choose (as long as your space is path connected). This is good, it means we don't have to worry about missing information if we pick a particular point to base our loops at. This what it means for alpha to be an isomorphism. For a path connected space, the fundmental groups at each basepoint are all isomorphic.

Well, that ought to keep you busy for a while. Try to organize your facts and theorems and distinguish between notations. As the other contributors have mentioned, alpha is not a path, it is not an equivalence class, it is a function between groups that preserve the group structure, i.e. a homorphism.

cheers,

Kevin

13. Sep 18, 2006

### waht

Wow, thanks for the reply, that's better than intro to topology by Gamlin, and Munkres I'm studying from.

Just have another quick question, for a loop as denoted by pi, is it required there exists a hole in space?

14. Sep 19, 2006

### matt grime

pi, as in capital pi is not a loop. It is a group. If you mean 'when is Pi_1(X,x) not trivial', then it is approximately correct to say this is when there are 'holes'.

A loop is just a map from S^1 the circle (or the unit interval with the end points going to the same place), and there is no requirement for there to exist any holes for that to make sense, or exist.

15. Sep 23, 2006

### mathwonk

here isa cute manifestation iof the fundamental group, using the crucial homotopy lifting proeprty:

consider the field C(X) of rational functions with complex coeeficients, and a finite galois extension K, where K the fraction field of the domain C(X)[Y]/(f(X,Y)), where f is an irreducible polynomial.

let G be the galois group of the extension K over C(X). we are going to compute this using ∏1.

set f=0 in the complex plane C^2, with zero locus Z, and project Z onto a suitable axis, whichw e will asume is the X axis.

this projection allows us to pull back rational functions from the X axis to Z, and induces exactly the field extension C(X) in K.

moreover the degree of the projection, i.,e. the number of points of Z over a general point of the X axis, equals the degree n of this field extension.

now remove from the X axis all points lacking n preimages on Z, and pick a general base point P in the X axis from the remaining points. what remains is a connected covering space (Z--->X axis) of degree n.

Then for each choice of a point Q of Z lying over P, every loop in the punctured X axis based at P has a unique lift to an arc in Z beginning at Q. The end point of this arc defines a permutation of Q,a s a point of the fiber over P.

I.e. we obtain a group homomorphism from the fundamental group of the ounctured X axis, into the eprmutations group S(n) of the fiber over P. The image of this hoomomorphism is the GALOIS GROUP OF THE FIELD EXTENSION OF K OVER C(X).

emphasis unintended but appropriate.

what about them apples?

Last edited: Sep 23, 2006