Homology of loop space

Question

I've been reading Galatius's Park City notes on the Madsen-Weiss theorem (available here).

On page 8, he states the following theorem. Let $X$ be a space such that $\pi_1(X)$ is abelian and acts trivially on the rational cohomology of the universal cover of $X$. Let $V = \oplus_{n \geq 1} V_n$ be a graded vector space and $V \rightarrow H^{\ast}(X;\mathbb{Q})$ a linear map such that the induced map $\mathbb{Q}[V] \rightarrow H^{\ast}(X;\mathbb{Q})$ is an isomorphism in degrees $\leq n$. Then there is a map $\mathbb{Q}[s^{-1} V] \rightarrow H^{\ast}(\Omega X;\mathbb{Q})$ which is an isomorphism in degrees $\leq n-1$. Here $s^{-1} V$ is the result of shifting the gradings on $V$ down by one.

He claims that this can be proved by "This can be proved using the Serre spectral sequence for the path-loop fibration over the universal cover $\tilde{X}$".

Despite a lot of work, I have been unable to fill in the details of this use of the Serre spectral sequence. It seems weird -- how on earth is the homology of the universal cover at all related to the homology of $X$?

Can anyone help me? Thank you very much.

Answer 1

I believe the argument Galatius had in mind is the following. Let us write $\bar{V} = \oplus_{n \geq 2} V_n$, so $V = V_1 \oplus \bar{V}$. All cohomology will be rational, and we write $G := \pi_1(X)$. Lets go ahead and suppose $G$ is finitely-generated and $X$ has rational cohomology of finite type.

By assumption $V_1 \to H^1(X) = H^1(G)$ is an isomorphism. As $G$ is an abelian group the extension to a free commutative graded algebra $\mathbb{Q}[V_1] \to H^*(G)$ is an isomorphism. We also have the composition $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$, and I claim this is an isomorphism in degrees $\leq n$.

To see this, we apply the spectral sequence described by Emerton in his answer, which goes $$E_2^{p,q} = H^p(G ; H^q(\widetilde{X})) \Longrightarrow H^{p+q}(X).$$ As the action of $G$ on $H^q(\widetilde{X})$ is trivial, we can re-write $E_2$ as $H^*(G) \otimes H^*(\widetilde{X}) = \mathbb{Q}[V_1] \otimes H^*(\widetilde{X})$. The image of $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$ can support no differentials (as these classes are pulled back from $X$, by construction). The map $\mathbb{Q}[\bar{V}] \hookrightarrow H^*(X) \to H^*(\widetilde{X})$ is an isomorphism in degree 0 and 1 (this is obvious, as there is nothing in degree 1 on both sides). In the minimal degree in which it is not as isomorphism (below $n$), we see we get a contradiction. If it is not surjective, there are extra classes in $H^*(\widetilde{X})$. These must support a differential, but we know everything in lower degrees is correct, so there is nothing for such a differential to hit. Otherwise, the map is surjective but not injective. But then we do not find enough cohomology in this degree to produce the result on $E_\infty$, which we know to be $\mathbb{Q}[V]$.

So, we have that $\mathbb{Q}[\bar{V}] \to H^*(\widetilde{X})$ is an isomorphism in degrees $\leq n$. Now we use the path fibration $$\Omega_\bullet X \simeq \Omega \widetilde{X} \longrightarrow P\widetilde{X} \longrightarrow \widetilde{X},$$ where the fibre is identified with the basepoint component of the loop space of $X$. We are now in the situation of (1) in Goodwillie's answer. We have the map $c: \mathbb{Q}[s^{-1} \bar{V}] \to H^*(\Omega_\bullet X)$ and want to see that it is an isomorphism. In low degrees the first differential is $$d_2 : H^1(\Omega_\bullet X) \longrightarrow H^2(\widetilde{X}) = V_2$$ and this must be an isomorphism as the spectral sequence converges to zero. This shows that $c$ is an isomorphism in degree 1. Now we argue in a similar way as above by thinking about the minimal degree in which it is not an isomorphism: this argument takes us up to degree $n-1$.

Answer 2

This is a response to your final question: how can the cohomology of $\tilde{X}$ be related to that of $X$?

There is a Hochschild--Serre spectal sequence for the quotient of a space by a group action. If $G$ acts freely on $Y$, with quotient $X$, it says that $$E_2^{i,j} := H^i(G,H^j(Y,A)) \implies H^{i+j}(X,A).$$ (Here $A$ can be any abelian group of coefficients, and $H^i(G, H^j(Y,A) )$ denotes group cohomology; note that since $G$ acts on $Y$, the cohomology $H^j(Y,A)$ is naturally a $G$-module.)

In your case, taking $Y = \tilde{X}$, $G = \pi_1(X)$, and $A = \mathbb Q$, your hypothesis is that $G$ acts trivially on the cohomology of $Y$, and so you get (at least under some finiteness conditions) $$E_2^{i,j} := H^i(\pi_1(X),\mathbb Q)\otimes H^j(\tilde{X},\mathbb Q) \implies H^{i+j}(X,\mathbb Q).$$

Answer 3

The result the OP wants easily follows from rational homotopy theory. Let $(\Lambda W, d)$ be the minimal model of $X$, i.e it's a minimal Sullivan algebra over $\mathbb Q$ with a quasi-isomorphism $(\Lambda W, d)\to (A_{PL}(X),d)$. It exists since $X$ is a nilpotent (in fact, simple) space by assumption. If you don't know what $(A_{PL}(X),d)$ is, just work over $\mathbb R$ and for a smooth manifold $X$ think of the algebra of exterior differential forms on $X$. Choose a basis of $V$ and map it to closed elements of $\Lambda W$ in the corresponding cohomology classes. Extend this map multiplicatively to $\phi:(\mathbb Q[V],0)\to (\Lambda W, d)$. By assumption this map is a quasi-isomorphism up to dimension $n$. Since $(\mathbb Q[V], 0)$ is also minimal this means that $\phi$ is an isomorphism up to dimension $n$. In particular $V_i\cong \pi_i(X)\otimes \mathbb Q$ for $1\le i\le n$. This gives homotopy groups of $\Omega X$ with degree shift by 1 and the claim follows since $\Omega X$ is an H-space and thus all the differentials in its minimal model are 0. The argument by Oscar Randal-Williams above makes all of this more explicit without relying on rational homotopy theory which has this stuff baked in. Note that by the argument above the assumptions on $X$ imply that $X$ is intrinsically formal up to dimension $n$, i.e. given its rational cohomology ring its minimal model is uniquely determined up to degree $n$. That's what makes the computation of $\pi_i(X)\otimes \mathbb Q$ particularly easy here.

Answer 4

Next, I suggest that you think about two special cases:

(1) When $\pi_1(X)$ is trivial. This will involve using the Serre spectral sequence for the contractible path fibration over $X$ with fiber $\Omega X$, including the multiplicative structure.

(2) When $\pi_k(X)$ is trivial for all $k>1$, i.e. when $\tilde X$ is contractible. For example, when $X$ is a circle. Oh, actually Galatius is saying something wrong in this case. He probably should be using homology instead of cohomology ...

Answer 5

This may not be exactly what you're looking for but rational homotopy theory provides an answer.

Step $1$ : Let $X$ be a nilpotent space, i.e., $\pi_1(X)$ is nilpotent and $\pi_i(X)$ is a nilpotent module over $\pi_1(X)$ for $i\geq 2$. Note that this is satisfied in your case. We also assume that $X$ has finite (rational) cohomology in each dimension.

Step $2$ : Methods from Sullivan's rational homotopy theory tells us that one can find a model for the rational cohomology of $X$, i.e., find a differential graded algebra $(\mathcal{A},d)$ and a map of algebras $\varphi:(\mathcal{A},d)\to \big(H^\ast(X;\mathbb{Q}),0\big)$ which is a quasi-isomorphism, i.e., an isomorphism when we pass to cohomology. This algebra $\mathcal{A}$ is usually written as $\Lambda V$ for a graded vector space $V$. Elements in $V$ are called indecomposables and one can take $V$ as the direct sum of the duals of $V_i:=\pi_i(X)\otimes\mathbb{Q}$ and hence the name rational homotopy theory.

Remark : A space $X$ is called formal if there is a model (these are often called Sullivan models) for $H^\ast(X;\mathbb{Q})$ with vanishing differential. H-spaces ans spheres are examples of such spaces.

Step $3$ : It can be shown (refer Rational Homotopy Theory by Felix, Halperin and Thomas) that a model $\Lambda V$ for $X$ defines a model for $\Omega X$ by shifting the underlying vector space $V$ down by $1$. It turns out that the differential for $\Lambda(s^{-1} V)$ can be taken to be zero as $\Omega X$ is an $H$-space.

Answer 6

Here's an argument that works in the non simply-connected case and avoids the universal cover and spectral sequences assuming the existence of rationalizations.

Assume that $X$ has rational cohomology of finite type up to degree $n$. Choosing a basis for $\bigoplus_{i=1}^{n}V_i$ corresponds via the map $V\rightarrow H^*(X,\mathbb{Q})$ to a map of spaces $$f\colon X\rightarrow \Pi_{i=1}^{n}K(\mathbb{Q},i)^{dim(V_i)}.$$ The induced map in rational cohomology $$\mathbb{Q}[\bigoplus_{i=1}^{n}V_i]\cong H^*(\Pi_{i=1}^{n}K(\mathbb{Q},i)^{dim(V_i)},\mathbb{\mathbb{Q}})\rightarrow H^*(X,\mathbb{Q})$$ is an iso up to degree $n$ by assumption. Therefore, it is also an iso on rational homology in these degrees. Since $X$ is a simple space, it's rationalization $X_\mathbb{Q}$ exists and we conclude that $$f_\mathbb{Q}\colon X_\mathbb{Q}\rightarrow\Pi_{i=1}^{n}K(\mathbb{Q},i)^{dim(V_i)}$$ is an iso on integer homology up to degree $n$, hence on homotopy groups up to degree $n$, hence $f$ induces an iso on rational homotopy groups up to degree $n$. Therefore $$\Omega f\colon\Omega X\rightarrow \Pi_{i=1}^{n}K(\mathbb{Q},i-1)^{dim(V_i)}$$ induces an iso on rational homotopy groups up to degree $n-1$. So does $$\Omega_\bullet X\rightarrow \Pi_{i=2}^{n}K(\mathbb{Q},i-1)^{dim(V_i)}$$ by restricting to path components. By doing the same reasoning as before backwards, this map induces an iso on rational cohomology up to degree $n-1$, hence $$\mathbb{Q}[s^{-1}(V)]\cong H^*(\Omega_\bullet X,\mathbb{Q})$$ up to degree $n-1$ as desired.