cohomology of free loop space on $S^3$

Question

Let $LS^3=\text{maps}(S^1,S^3)$ denote the free loop space on $S^3$. I want to compute the cohomology of this space. I think that it should be $\mathbb{Z}$ in all degrees except degree one where it should be zero. I have computed this using the Serre spectral sequence for the fiber sequence $\Omega S^3\to LS^3\to S^3$. However i am not 100% that this is true since for the computation I used that $H^2(LS^3)=\mathbb{Z}$ but I have no way right now of seeing that this is true. So I was wondering if somebody can confirm my result and if there is a nice argument for $H^2(LS^3)=\mathbb{Z}$.

Answer 1

The fiber sequence $\Omega S^3\to LS^3\to S^3$ leads to the long exact sequence of homotopy groups. Since any map $S^1\to S^3$ is nullhomotopic, $LS^3$ is path connected and the long exact sequence then shows that $LS^3$ is simply connected. I claim that $\pi_2(LS^3)=\mathbb{Z}$. It then follows from Hurewicz that $H_2(LS^3) = \mathbb{Z}$ and from the universal coefficient theorem that $H^2(LS^3)=\mathbb{Z}$. In low degrees the long exact sequence of homotopy groups takes the form $$\cdots\to\pi_3LS^3\to \pi_3S^3\to \pi_2\Omega S^3\to \pi_2LS^3\to 0\to\cdots.$$ Now $LS^3\to S^3$ has a section (send $x\in S^3$ to the constant loop at $x$), hence $\pi_3S^3\to \pi_2\Omega S^3$ is zero, which in turn implies that $\pi_2\Omega S^3\cong \pi_2LS^3$. Finally, note that $\pi_2\Omega S^3\cong \pi_3S^3 = \mathbb{Z}$.

Addendum: You actually don't need to know that $H^2(LS^3)=\mathbb{Z}$. In the Serre spectral sequence the only possible non-trivial differentials are $$d_3^{0,2q}\colon H^0(S^3;H^{2q}(\Omega S^3))\to H^3(S^3;H^{2(q-1)}(\Omega S^3)).$$ Both source and target are $\mathbb{Z}$ and to show that these are zero it suffices to show that $d_3^{0,2}$ vanishes, since the spectral sequence is multiplicative and $H^\ast(\Omega S^3)$ is a divided power algebra. Now the edge homomorphism $$H^3(S^3) = H^3(S^3;H^0(\Omega S^3))=E_2^{3,0}\to E_\infty^{3,0}\subset H^3(LS^3)$$ is the map induced by $LS^3\to S^3$, hence is injective since there is a section $S^3\to LS^3$. It follows that also $E_2^{3,0}\to E_\infty^{3,0}$ is injective, which means that $d_3^{0,2}$ vanishes as claimed.

Answer 2

Here is an alternative approach. $S^3$ is actually a Lie group, namely $SU(2)$, and in general if $G$ is a topological group then the fiber sequence $\Omega G \to LG \to G$ splits in the sense that we have a homeomorphism $LG \cong \Omega G \times G$ given by

$$LG \ni f(t) \mapsto (f(0)^{-1} f(t), f(0)) \in \Omega G \times G$$

where $f$ is thought of as a periodic function $\mathbb{R} \to G$ (of period $1$, say) and we take the basepoint for defining $\Omega G$ to be the identity. This means that to compute the cohomology of the free loop space (as a ring, even), instead of applying the Serre spectral sequence you can just apply the Kunneth formula.