Homology of loop space

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.

• Isn't taking homology of a covering space just a sneaky way of doing homology with local coefficients? – Dylan Wilson Nov 17 '11 at 5:02

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 $n-1$.

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) $$E_2^{i,j} := H^i(\pi_1(X),\mathbb Q)\otimes H^j(\tilde{X},\mathbb Q) \implies H^{i+j}(X,\mathbb Q).$$

• Thanks! Though I have to admit that I don't see how to use this to prove the above theorem, it is nevertheless a good thing to know. – Carlos Garcia Nov 17 '11 at 4:30

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.

• Yes, but as I tried to indicate in my answer the statement is not quite right in the non simply connected case. The rational $H^0$ of a circle is not exactly a polynomial ring. – Tom Goodwillie Nov 17 '11 at 23:00
• sorry, I don't follow. I don't see any issues in the non simply connected case in general and for a circle in particular. there is no claim anywhere that $H^0$ is a polynomial algebra. – Vitali Kapovitch Nov 18 '11 at 0:25
• I meant to say the loop space of the circle. The assertion seems to say, when $X=S^1$, that $H^0(\Omega S^1;\mathbb Q}$ is $\mathbb Q[s^{-1}V]$ where $s^{-1}V$ is concentrated in degree $0$ and $1$-dimensional. – Tom Goodwillie Nov 18 '11 at 0:38
• oh, I see. that's not exactly what it says though. when $X$ is a circle then $V$ has one generator in degree 1 so that $s^{-1}V$ has one generator $a$ in degree 0 (and none in degree 1). the loop space of a circle has $\mathbb Z$ worth of connected components (all contractible) so that $H^0(\Omega S^1,\mathbb Q)$ is still isomorphic to $\mathbb Q[a]$. I think everything is formally correct here. in any case this is easily dispensed with by looking only at one connected component and setting $s^{-1} V$ to be zero in degree $0$. – Vitali Kapovitch Nov 18 '11 at 1:01
• $H^0(\Omega S^1)$ is the dual of a vector space. Its dimension is uncountable. That's what's bothering me. – Tom Goodwillie Nov 18 '11 at 11:56

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 ...

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.

• You say several wrong things here. First, a minimal model of X is quasi-isomorphic to (AP⁢L⁢(X),⁢d), not to (H*⁢(X,ℚ),0) (unless the space is formal). the definition of a formal space is also wrong. a space is formal if (AP⁢L⁢(X),⁢d) is quasi-isomorphic to (H*⁢(X,ℚ),0) but the differentials on the minimal models are not usually zero even for formal spaces. Lastly, in general a model of X does not give a model of Ω⁢X by a shift as you claim. The generators are shifted by 1 but the differentials all become 0 which they are not in the model of $X$. – Vitali Kapovitch Nov 17 '11 at 18:48
• You're right and I could have edited my answer accordingly but it would end up looking more like your answer which is good enough. – Somnath Basu Nov 17 '11 at 19:44

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.