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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.

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1  
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
    
Thanks to everyone for the great answers! –  Carlos Garcia Nov 18 '11 at 3:29

5 Answers 5

up vote 9 down vote accepted

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 $n-1$.

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

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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.

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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 ...

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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.

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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

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