Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Showing that the infinite dimensional sphere $S^{\infty}$ is contractible is rather easy by constructing an explicit contraction (Hatcher gives a nice one). I thought it might be a nice exercise to try and show this using loop spaces and the fact that $\pi_1(S^{\infty})$ is trivial.

Let the loopspace $\Omega X$ be the space of base pointed loops in $X$. For now, assume that the loopspace $ΩS^{\infty}$ is homotopy equivalent to $S^{\infty}$. Then, $\pi_2(S^{\infty})=\pi_1(\Omega S^{\infty})=\pi_1(S^{\infty})=0$. Continuing by induction, we get that every homotopy group of $S^{\infty}$ is trivial and so $S^{\infty}$ is contractible because $S^{\infty}$ is a CW-complex (by Whitehead). As far as I can tell, this argument has no holes in it.

My problem now is in proving the assumption that $\Omega S$ is homotopy equivlanet to $S$. I haven't had too much experience with loopspaces or function spaces in general so I'm struggling somewhat.

One approach might be to find a fibration $F\rightarrow\Omega S^{\infty}\rightarrow S^{\infty}$ and then playing about with long exact sequences in homotopy, but I'm not sure if such a fibration exists. Maybe something like 'If $f\colon [0,1]\rightarrow S^{\infty}$ is a loop, let $p\colon \Omega S^{\infty}\rightarrow S^{\infty}$ be given by $p(f)=f(1/2)$' would work? Is p a fibration? And if so, what do the fibers look like? The best case scenario would be for some fibration to exist with contractible fiber. However, I'm worried that showing such a fiber is contractible would amount to showing that $S^{\infty}$ is contractible, which defeats the object of the exercise.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

You might be able to get something using the Freudenthal theorem (though this is a bit overkill admittedly), since it studies the connectedness of the natural map to $\Omega \circ S$. Supposing the assertion that the suspension of $S^\infty$ is again $S^\infty$ (we have to use the limiting behavior of this space somewhere), then if $S^\infty$ is $n$-connected, the map

$\pi_k(S^\infty) \to \pi_k(\Omega S^\infty) = \pi_{k+1}(S^\infty)$

is an isomorphism for $k \le 2n$. Then this is certainly true for $k = n$, which gives that $\pi_{n+1}(S^\infty)$ is trivial as well, and so $S^\infty$ is at least $(n+1)$-connected -- we can continue on to hit all the homotopy groups.

share|improve this answer
    
What a lovely little proof. Very much the kind of thing I was looking for. I'm new to the site; is this the point that I accept an answer, or should I wait in case there are more answers? –  Daniel Rust Apr 16 '12 at 17:11

Although I do not know an answer to your question I wanted to remark the following.

It seems to me as if you would like to have a "nicer" proof of the fact that the infinite sphere is contractible. What you try to do is, prove that it is weakly contractible and use Whiteheads theorem (which I too think is nicer than an explicit contraction).

But the fact that the infinite sphere is weakly contractible may easily seen from the fact that given $n \geq 0$ it may be equipped with a CW structure such that its n-skelleton is a point and using the cellular approximation theorem.

share|improve this answer
    
I think I would mostly agree that I am trying to find a 'nicer' proof for some definition of nicer. A homology argument works well you're right (together with Hurewicz).For purely selfish reasons however, I was hoping to use homotopy groups and loopspaces in order to give a novel proof of a rather standard result. The part of the proof I have given seems so neat that it would be a shame if the assumption I made couldn't be proved without avoiding the standard proof. –  Daniel Rust Apr 15 '12 at 0:30

Since $\pi_1(S^\infty)=0$, it is rather easy to prove that $S^\infty$ and $\Omega S^\infty$ are homotopy equivalent : set $\rho : S^\infty \to \Omega S^\infty$, $\rho(p)(t)=p$ for every $t\in[0,1]$ (since $\rho(p)$ is a path, it makes sense to write $\rho(p)(t)=p$) and $\tau : \Omega S^\infty \to S^\infty$, $\tau(\gamma)=\gamma(0)$. Now $\tau\circ\rho$ is the identity, and $\rho\circ\tau$ is isotopic to the identity, since $\pi_1(S^\infty)=0$, and so every loop is homotopy equivalent to the constant loop on its base point : choose an isotopy $F_\gamma$ from each loop $\gamma$ to its base point, then set $F:\Omega S^\infty \times [0,1]\to\Omega S^\infty$ by $F(\gamma,t)=F_\gamma(t)$, and this is a homotopy between the identity and $\rho\circ\tau$.

share|improve this answer
2  
This proof cannot work, otherwise it would give a proof that any simply connected space is homotopy equivalent to its loopspace. It breaks in the last step: why should $F$ be continuous? –  Dylan Wilson Apr 16 '12 at 14:53
    
Regardless of Dylan's point, the functions $\rho$ and $\tau$ aren't well defined, although this may be my fault in omitting to define terms. The notation $\Omega X$ was meant to refer to the space of base pointed loops in $X$. I think you may have confused $\Omega X$ with $\Lambda X$ the space of all loops on X (the free loopspace). I've edited the question to reflect that we are working in the pointed topological category. –  Daniel Rust Apr 16 '12 at 16:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.