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Is it true that the loop space of a circle is contractible? Consider the long exact sequence in homotopy for the path fibration $\Omega S^1 \rightarrow \ast \rightarrow S^1$ shows all homotopy groups of the loop space to be zero, and then Whitehead's theorem kicks in and tells us that $\Omega S^1$ is contractible.

I can't see any flaw in this argument, but it's really rankling against my intuition of what a loop space "is". For instance, we know that $S^1$ has $\mathbb{Z}$ many loops up to homotopy, so surely it should have a fairly complicated loop space? So I suppose that there are two possible approaches to answering this question: either point out where I'm going wrong in my mathematics, or help rectify my intuition.

Secondly, does this generalise to the $n$-fold loop space of an $n$-sphere? I think it does; I've used the Serre spectral sequence to compute the cohomology rings of loop spaces of spheres and they all die at $\Omega^n S^n$.

Sorry for the "please check my working" nature of the question; it's sufficiently early in my mathematical life that I'm never convinced that there's no errors in my working (especially not with spectral sequences, which still feel a bit like doing magic to me). It was a lot easier to be (mathematically) confident as an undergraduate, when you could check your working against solutions sheets...

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up vote 8 down vote accepted

$\Omega S^1\cong \mathbb Z$ — your argument is almost correct: each connected component is contractible, but there are countably many such components ($\pi_0(\Omega S^1)=\pi_1(S^1)=\mathbb Z$).

As for $\Omega^n S^n$, these spaces have very non-trivial homotopy type: $\pi_k(\Omega^n S^n)=\pi_{n+k}S^n$ and there are a lot of non-trivial homotopy groups of spheres (say, $\pi_1(\Omega^2 S^2)=\pi_3(S^2)=\mathbb Z/2\mathbb Z$ already).

As for the Serre spectral sequence, maybe you made the standard mistake: when computing, say, $H(\Omega S^3)$ one sees that there is an element $x$ s.t. $d_2(x)=[S^3]$ and concludes that (since powers of $x$ "obviously" kill all cohomology classes from $E_2$) $H(\Omega S^3)$ "is" $\mathbb Z[x]$. But actually $d_2(x^n)=\mathbf{n}x[S^3]$, so powers of $x$ don't kill all classes: there also should exist classes $x^n/n!$ and the answer is not $\mathbb Z[X]$ but $\Gamma[x]$.

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Yes, I see now: the homotopy groups see it as zero, since the space is discrete, but we can use the isomorphism $\pi_n(X) \cong \pi_{n-1}(X)$ to spot that it has $\Z$ many such connected components. Cheers. Can we say anything about the cohomology of the spaces $\Omega^n S^n$? I have their $n$th cohomology groups dying, but my working is probably very shaky. – Alex Jun 16 '11 at 10:26
I have that the loop space is the divided polynomial algebra (or the divided polynomial algebra tensor an exterior algebra for even spheres), and then I considered the fibration $\Omega^2 X \rightarrow \ast \rightarrow \Omega X$ (where $X$ is a sphere). This has a Serre spectral sequence with blank stable page and $E_2$ page consisting of columns only in multiples of $n-1$. Everything on the $E_2$ page has to disappear, so the first $n-1$ $d_{n-1}$ differentials in the first column have to be isomorphims, so as to kill $E^{n-1, 0}, \ldots, E^{n-1, n-1}$, whose "outward" differentials all map – Alex Jun 16 '11 at 10:37
to zero. And continuing through the sequence like this, we get isomorphims $H^i \Omega^2 S^n \cong H^{i - n + 2} \Omega^2 S^n$, which give us that the cohomology of $\Omega^2 S^n$ is $\mathbb{Z}$ in dim $0, n-2, 2n-2, \ldots $ and zero elsewhere. We can then move onto the next path fibration in the sequence, where it'll be the $n-2$ differentials we need to look at and the cohomology will be nontrivial at multiples of $n-3$, and so on. That's my working in computing the cohomology modules of $\Omega^j S^n$, at any rate. With my working, when $j=n$, the cohomology must be zero. – Alex Jun 16 '11 at 10:41
And then from there we can compute the ring structure... but I figure it's best to sort out my mistakes one step at a time. Also, comments are terrible for trying to do actual maths in. – Alex Jun 16 '11 at 10:43
@Alex ...and in general the question about cohomology of $\Omega^n S^n$ is not easy, AFAIK. One should, perhaps, start from classical paper of Dyer and Lashof – Grigory M Jun 16 '11 at 10:46

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