The (co)homology of the loop space of a n-sphere

Computing (co)homology can be rather hard without the appropriate tools. The Serre spectral sequence is one such tool. We compute the important example of H_\ast(\Omega S^n), and also provide a different proof of Proposition 3.22 in Hatcher’s Algebraic Topology, namely the computation of H^\ast(J(S^n))\cong H^\ast(\Omega S^{n+1}), where J(X) is the reduced James product (see section 4.J, for James). (Note that the cohomology rings are isomorphic since J(S^n)\simeq\Omega S^{n+1}.) This example will illustrate the usefulness of the Serre spectral sequence. Everything here has coefficients in \mathbf{Z}.

Recall that for a fibration p:E\to B such that B is simply connected, if F=p^{-1}(b), then the Serre spectral sequence of p says that E^2_{p,q}=H_p(B;H_q(F))\Rightarrow H_{p+q}(E). The universal coefficient theorem says that there is a natural short exact sequence 0\to H_p(B)\otimes H_q(F)\to H_p(B;H_q(F))\to\mathrm{Tor}^\mathbf{Z}_1(H_{p-1}(B),H_q(F))\to 0 that splits (albeit not naturally). Suppose H_q(F) or H_{p-1}(B) is torsionfree; then E^2_{p,q}\cong H_p(B;\mathbf{Z})\otimes H_q(F;\mathbf{Z})\Rightarrow H_{p+q}(E). This spectral sequence has a product, in the following sense. Suppose E_1\to B_1, E_2\to B_2, and E_3\to B_3 are Serre fibrations. Then a diagram:


Gives a product \left(E^r_{p,q}\right)_1\otimes \left(E^r_{p^\prime,q^\prime}\right)_2\to \left(E^r_{p+p^\prime,q+q^\prime}\right)_3.


Our first goal is to study H_\ast(\Omega S^n).

Theorem: H_\ast(\Omega S^n)\cong\mathbf{Z}[x_1], where the element x_1 is the image of 1 under H_n(S^n)\cong \pi_n(S^n)\cong \pi_{n-1}(\Omega S^n)\to H_{n-1}(\Omega S^n).

Proof. To prove this, consider the path-loop fibration \Omega S^n\to PS^n\to S^n. The path space PS^n is contractible. Hence the Serre spectral sequence says that E^2_{p,q}=H_p(S^n;H_q(\Omega S^n))\Rightarrow H_{p+q}(\Omega S^n). We know that E^2_{p,q}=H_q(\Omega S^n) if p=0,n, and is zero otherwise. Consider the generator \sigma of H_0(S^n); then H_0(\Omega S^n)\simeq\langle \sigma\rangle. Note that almost all differentials are zero, except d^n:E^n_{n,q}\to E^n_{0,q+n-1}. We therefore have:


We claim that all the nonvanishing maps are isomorphisms.

To see this, note that E^{n+1}_{p,q}\cong\cdots\cong E^\infty_{p,q}=F^p H_{p+q}(PS^n)/F^{p-1}H_{p+q}(PS^n). But PS^n is contractible, so E^\infty_{p,q}\cong 0 unless p=q=0. Therefore if d^n was not an isomorphism, we would get nontrivial elements in E^{n+1}_{p,q}, which is weird. Therefore, H_k(\Omega S^n)\cong\mathbf{Z} if k is a multiple of (n-1), and is zero otherwise. Say that x_\ell generates E^2_{0,\ell(n-1)}. Then the generator of E^2_{n,\ell(n-1)} is \sigma\otimes x_\ell, and thus the differential goes d(\sigma\otimes x_\ell)=x_{\ell+1}.

Now we will study the multiplicative structure on H_\ast(\Omega S^n). We have to choose three Hopf fibrations; indeed, we can pick the most obvious ones: \Omega S^n\to PS^n\to S^n, \Omega S^n\to PS^n\to S^n, and \Omega S^n\to \Omega S^n\to\ast. It can easily be checked that the desired diagram commutes, so that we have a product on spectral sequences. For the fibration \Omega S^n\to \Omega S^n\to \ast, it is rather obvious that E^k_{0,q}=H_q(\Omega S^n) for all q and E^k_{p,q}=0 for p>0, and all differentials are zero.

Therefore, if E^k now denotes the Serre spectral sequence for \Omega S^n\to PS^n\to S^n, the multiplication is H_q(\Omega S^n)\otimes E^r_{p^\prime,q^\prime}\xrightarrow{\times} E^r_{p^\prime,q+q^\prime}. Since x_\ell\times\sigma=\sigma\otimes x_\ell\in E^2_{n,\ell(n-1)} (under the UCT isomorphism E^2_{n,0}\otimes E^2_{0,\ell(n-1)}=H_n(S^n)\otimes H_{\ell(n-1)}(\Omega S^n)\cong H_n(S^n;H_{\ell(n-1)}(\Omega S^n))=E^2_{n,\ell(n-1)}), and since we know that d(\sigma\otimes x_\ell)=x_{\ell+1}, it follows that x_{\ell+1}=dx_\ell\times\sigma\pm x_\ell\times d\sigma=\pm x_\ell\times x_1 because dx_\ell=0 and d\sigma=x_1. Thus, by induction, x_{\ell}=\pm x_1^\ell. The desired result follows.


Our next goal is to compute H^\ast(\Omega S^n). Let \Lambda_\mathbf{Z}[x] denote the exterior algebra on one generator, i.e., \mathbf{Z}[x]/x^2, and let \Gamma_\mathbf{Z}[x] denote the divided polynomial algebra.

Theorem: If n is odd, then H^\ast(\Omega S^n)\cong\Gamma_\mathbf{Z}[x] where |x|=n-1. If n is even, then H^\ast(\Omega S^n)\cong H^\ast(S^{n-1})\otimes_\mathbf{Z} H^\ast(\Omega S^{2n-2})\cong\Lambda_\mathbf{Z}[x]\otimes_\mathbf{Z} \Gamma_\mathbf{Z}[y] where |x|=n-1 and |y|=2(n-1).

Proof. From the cohomological spectral sequence, we compute (like above) that E^{p,q}_2 is H^q(\Omega S^n) if p=0,n, and is zero otherwise. The only nontrivial differential is d_n:E^{0,q}_2\to E^{n,q-n+1}_2. Arguing as above, we find that H^q(\Omega S^n)\cong\mathbf{Z} if q is a multiple of (n-1), and is 0 otherwise.

Let \sigma generate H^n(S^n); then \sigma generates H^0(\Omega S^n). Consider a generator x_\ell\in H^{\ell(n-1)}(\Omega S^n). Then d_n:E^{0,\ell(n-1)}_n=H^{\ell(n-1)}(\Omega S^n)\to H^{(\ell-1)(n-1)}(\Omega S^n)=E^{n,(\ell-1)(n-1)}_n sends x_\ell\mapsto x_{\ell-1}\sigma, which is also (clearly) a generator of H^{(\ell-1)(n-1)}(\Omega S^n) since \sigma generates H^0(\Omega S^n).

Let’s consider first the case that n is even. Then x_1\in H^{n-1}(\Omega S^n) satisfies x_1^2=0 (by graded commutativity, since n-1 is odd). Therefore, x_1x_k\in H^{(k+1)(n-1)}(\Omega S^n) can be written as N_k x_{k+1} for some integer N_k. This means that d_n(x_1x_k)=d_n(N_k x_{k+1})=N_k d_n(x_{k+1})=N_k x_k\sigma. But from the Leibniz formula, we also know that d(x_1x_k)=d(x_1)x_k-x_1d(x_k)=\sigma x_k-x_1x_{k-1}\sigma=\sigma x_k-N_{k-1}x_k\sigma=(1-N_{k-1})\sigma x_{k-1}. This is good, because N_1=0 (again because x_1^2=0). Thus N_k is (k+1)\bmod 2, i.e., x_1x_k=x_{k+1} if k is even, else x_1x_k=0 if k is odd.

Also, x_2\in H^{2n-2}(\Omega S^n) commutes with everything, i.e., d_n(x_2^k)=d_n(x_2)x_2^{k-1}+x_2d_n(x_2^{k-1})=\cdots kx_2^{k-1}d_n(x_2)=kx_2^{k-1}x_1\sigma. Now, x_2^k\in H^{2k(n-1)}(\Omega S^n), so x_2^k=M_kx_{2k} for some integer M_k. It then follows that d_n(x_2^k)=M_kd_n(x_{2k})=M_kx_{2k-1}\sigma, so that M_kx_{2k-1}\sigma=kx_2^{k-1}x_1\sigma=kM_{k-1}x_{2(k-1)}x_1. We determined that x_{2k-1}=x_1x_{2(k-1)} above (since 2(k-1)+1\equiv 1\bmod 2), so M_kx_1x_{2(k-1)}\sigma=kM_{k-1}x_{2(k-1)}x_1, i.e., M_k=k! by induction. To summarize:




Let’s now consider the case that n is odd. Then x_1\in H^{n-1}(\Omega S^n) commutes with everything since n-1 is even. Consequently, d_n(x_1^k)=kx_1^{k-1}d_n(x_1)=kx_1^{k-1}\sigma. Now, x_1^k=N_kx_k\in H^{k(n-1)}(\Omega S^n) for some N_k. This means that d_n(x_1^k)=N_kd_n(x_k)=N_kx_{k-1}\sigma, i.e., x_k=k!x_1. This therefore means that:


We’re done.

All of this is also at


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