Fundamental theorem of Morse theory for $\Omega(S^n )$

Question

Using the Fundamental theorem of Morse Theory we can prove that $\Omega(S^n)$ is homotopically equivalent to a CW complex with one cell each in dimensions $o,n-1,2(n-1), \cdots$ and so on. But how can I attack these cells? For example $\Omega(S^2) \simeq e^0 \cup e^1 \cup^2 \cup \cdots $. But I don't think that $\Omega(S^2)$ is homotopically equivalent to $\mathbb{R}P^\infty$...

Answer 1

The answer below is repeated from the comment thread for my answer at: Path space of $S^n$

Of course, $\Omega(\mathbb{S}^2)$ isn't homotopy equivalent to $\mathbb{RP}^{\infty}$. In fact, they don't even have isomorphic homotopy groups. The reason is that the universal cover of $\mathbb{RP}^{\infty}$ is the contractible infinite dimensional sphere $\mathbb{S}^{\infty}$. It's a (easy to prove) theorem that for $n\geq 2$, $\pi_n$ of the universal cover of a (path-connected) space agrees with $\pi_n$ of the space. Therefore, $\pi_n(\mathbb{RP}^{\infty})=0$ for $n\geq 2$. Finally, $\pi_1(\mathbb{RP}^{\infty})=\mathbb{Z}/2$ because the universal cover has fiber of cardinality two.

In general it's a hard problem to explicitly compute the attaching maps of cells in a CW complex. The homotopy type of $\Omega(\mathbb{S}^2)$ is equally as complicated as the homotopy type of $\mathbb{S}^2$ because $\pi_n(\Omega(\mathbb{S}^2))\cong \pi_{n+1}(\mathbb{S}^2)$ (this is easy to prove using either the pathspace fibration or the loopspace/suspension adjunction).

If you're interested in computing homology, then the situation is nice for $n>2$. In this case, $\Omega(\mathbb{S}^2)$ has exactly one cell in each nonnegative integer multiple of $n-1>1$. Therefore, in cellular homology, the differentials are all zero and $H_i(\Omega(\mathbb{S}^n))=\mathbb{Z}$ if $i$ is a nonnegative integer multiple of $n-1$ and zero otherwise.

The situation is more subtle in the case $n=2$. In the cellular complex for $\Omega(\mathbb{S}^2)$, there is a $\mathbb{Z}$ in each degree but the differentials are (a priori) unknown. So, an alternative approach is needed unless one is given explicit information about the homotopy types of the attaching maps.

The approach that immediately comes to mind is the Serre spectral sequence applied to the pathspace fibration of $\mathbb{S}^2$. All the differentials on the $E_2$ page must be isomorphisms (except the ones that originate from the zero group and enter a nonzero group and vice-versa) because the $E_2$ page equals the $E_{\infty}$ page (and the $E_{\infty}$ page just has a $\mathbb{Z}$ in the $(0,0)$ lattice point because the pathspace of $\mathbb{S}^2$ is contractible). Since the differentials on the $E_2$ page go two units to the left and one unit up, it follows that the homology groups of $\Omega(\mathbb{S}^2)$ are all isomorphic and and the common isomorphism class (given by, e.g., the zeroth homology group!) is $\mathbb{Z}$. Therefore, $H_i(\Omega(\mathbb{S}^2))\cong \mathbb{Z}$ for all nonnegative integers $i$. The same result holds for cohomology by the universal coefficient theorem. However, the homotopy groups of $\Omega(\mathbb{S}^2)$ are unknown because they're unknown for $\mathbb{S}^2$!

I hope this helps!