I understand that finding higher homotopy groups of spaces is difficult because there's no equivalent to the van-kampen theorem for n>1 and so building homotopy groups can't be done easily by deconstructing a space in to simpler spaces which we know the homotopy groups of. I was wondering though if this could be done for special cases such as the one-point union (wedge product) of two spaces. I'm asking because I want to be able to find the homotopy groups of bouquets of n-spheres in terms of the homotopy groups on a single n-sphere; or explicitly if possible, but I know how hard it is to work out the homotopy groups of spheres so I'm not getting my hopes up.

One approach I've thought about is trying to find some nice fibrations from the bouquets to the single spheres and then using the long exact sequence of homotopy groups for a fibration to deduce some properties of the groups; however I'm pretty new to the concept of fibrations so I'm not sure how to go about constructing such a map or even if it's possible.

So I guess my question is 'Is there any relatively simple method for deducing properties about the wedge product of two spaces, specifically spheres'. Even any special cases like the 1-dimensional case would be useful for my work. Any help would be greatly appreciated.

## homotopy groups of wedge products

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- MartianInvader
**Posts:**705**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: homotopy groups of wedge products

It's been a while since I've thought about this, but my understanding is that nothing nice happens. It's not necessarily the direct sum of the homotopy groups or anything else like that.

Except in the 1-dimensional case; in this case the first homotopy group of a wedge of n circles is the free group on n generators, and all the others are zero (for the same reason that the higher homotopy groups of a circle are zero).

Except in the 1-dimensional case; in this case the first homotopy group of a wedge of n circles is the free group on n generators, and all the others are zero (for the same reason that the higher homotopy groups of a circle are zero).

Last edited by gmalivuk on Mon Jan 02, 2012 3:56 pm UTC, edited 1 time in total.

**Reason:***You don't need jsMath to write the letter "n"*Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

- Talith
- Proved the Goldbach Conjecture
**Posts:**848**Joined:**Sat Nov 29, 2008 1:28 am UTC**Location:**Manchester - UK

### Re: homotopy groups of wedge products

Thanks for the reply. Yeh, I kind of guessed it wouldn't be nice. Is it at least possible to tell which of these groups are going to be non-trivial? How about finite or infinite? Also, is the natural map from a wedge of n m-spheres to a single m-sphere (the 'fold over' map for want of a better phrase) a fibration? It's not a covering map because of the identification point of the total space making the fibers different at different points, although the fact that the fibers aren't even homotopic at every point makes me think the map can't even be a fibration.

### Re: homotopy groups of wedge products

There's the Blaker's-Massey theorem (or "homotopy excision"). Wikipedia has a basic description. The answer is that you can figure out some of the low-dimensional data.

- Talith
- Proved the Goldbach Conjecture
**Posts:**848**Joined:**Sat Nov 29, 2008 1:28 am UTC**Location:**Manchester - UK

### Re: homotopy groups of wedge products

Thanks freikugel. It's definitely a partial answer and very useful. I found it quite funny that I used the Cayley graph in MartianInvader's avatar to prove that the higher homotopy groups of the wedge of two circles (and similarly for n circles) are all 0 due to its universal cover (the Cayley graph of its fundamental group) being contractible .

It does seem like covering spaces (and fibrations in general) and the long exact sequence of homotopy groups give a powerful method for finding a lot of higher homotopy groups for spaces. So far I've used the method to show that π

I wonder if anyone could suggest some other examples of spaces for which the higher homotopy groups can be worked out this way. Just the spaces of course (and possibly the fibration if it's a non-trivial map like the Hopf fibration) as this work will hopefully lead to an assessed project.

It does seem like covering spaces (and fibrations in general) and the long exact sequence of homotopy groups give a powerful method for finding a lot of higher homotopy groups for spaces. So far I've used the method to show that π

_{n}(RP^{n})=Z for n>1, to show that π_{n+k}(RP^{n})~π_{n+k}(S^{n}) for n>1, and that π_{k}(T_{g})=0 for k>1. I've also used the Hopf fibration to show that π_{n}(S^{2})~π_{n}(S^{3}) for n>2 and a few other simple results.I wonder if anyone could suggest some other examples of spaces for which the higher homotopy groups can be worked out this way. Just the spaces of course (and possibly the fibration if it's a non-trivial map like the Hopf fibration) as this work will hopefully lead to an assessed project.

### Re: homotopy groups of wedge products

The homotopy groups of a wedge of spaces (which are themselves suspensions) was worked out by Hilton and Milnor. Google "Hilton-Milnor Theorem". For a wedge of spheres the homotopy groups (mod torsion) are a type of free Lie algebra on the homotopy groups of the wedge summands, where the Lie bracket is the Whitehead bracket.

### Re: homotopy groups of wedge products

delooper wrote:The homotopy groups of a wedge of spaces (which are themselves suspensions) was worked out by Hilton and Milnor. Google "Hilton-Milnor Theorem". For a wedge of spheres the homotopy groups (mod torsion) are a type of free Lie algebra on the homotopy groups of the wedge summands, where the Lie bracket is the Whitehead bracket.

I think this is overstating the Hilton+Milnor result, which is a representation of Ω(SA v SB) as something less than a product of Ω(S A^[m] ^ B^[n]) for various repeated ms and ns; that can only be called a determination of the homotopy groups of SA v SB if the homotopy groups of all those smash products are known; but generally they're not. It's still pretty awesome, though!

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