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

## Homology

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.

## Cohomology

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:

Succinctly:

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 http://www.mit.edu/~sanathd/(co)homology-of-omega-S%5En.pdf.