cohomology ring structure of conf($\mathbb{R}^m$, 3)

I am attempting to compute the (integral) cohomology ring structure of the 3 configuration space of $\mathbb{R}^m$ and have run into a few doubts.

Using a result of Fadell and Neuwirth, we have that conf($\mathbb{R}^m$, 3) is the total space of the fibration:

p: conf($\mathbb{R}^m$, 3) $\rightarrow$ conf($\mathbb{R}^m$, 2) given by the obvious map onto the first 2 factors with fibre homeomorphic to $\mathbb{R}^m - Q$ where $Q$ is two points.

Since $\mathbb{R}^m - Q$ is homotopy equivalent to a wedge of two $m-1$ spheres, it has cohomology groups $\mathbb{Z}$ and $\mathbb{Z}\bigoplus \mathbb{Z}$ in dimensions $0$ and $m-1$ respectfully (Mayer-Vietoris).

Similarly conf($\mathbb{R}^m$, 2) is homotopy equivalent to $S^{m-1}$ where a map is $(x,y) \rightarrow \frac{(x-y)}{|x-y]}$. Assuming trival local coefficients (where the only problem could arise when $m=2$ since otherwise the base is simply connected) we can apply the Serre spectral sequence with $E_2^{p,q} = H^p(S^{m-1}, H^q(\vee_2 S^{m-1} ))$.

An application of the UCT (since everything is free) gives that $E_2^{p,q} = H^p(S^{m-1}) \otimes H^q(\vee_2 S^{m-1} )$

We then have that $E_2 = E_{\infty}$ since we only have cohomology in degree $0$ and $m-1$ for the base and fibre and hence there can be no non-trivial differentials.
This leads me to a few questions.

1. In general the $E_{\infty}$ product structure doesn't determine the product structure on the total space (see for example p 29 SSAT ) but things work out in this instance since everything is a free $\mathbb{Z}$ module and of finite type? What is a necessary and sufficient condition on the $E_{\infty}$ page structure to guarantee that the product structures coincide?

2. I know that $H^*(S^{m-1}) = \mathbb{Z}[a_1]/(a_1^2)$ where $|a_1| = m-1$ and that $H^*(S^{m-1} \vee S^{m-1}) = \mathbb{Z}[a_2]/(a_2^2) \times \mathbb{Z}[a_3]/(a_3^2)$ where $|a_2|=|a_3| = m-1$.

Does this simply imply that $H^*($conf($\mathbb{R}^m$, 3) ) $\simeq \mathbb{Z}[a_1]/(a_1^2)$ $\otimes \mathbb{Z}[a_2]/(a_2^2) \times \mathbb{Z}[a_3]/(a_3^2)\simeq \mathbb{Z}[a_1,a_2,a_3]/(a_1^2=a_2^2=a_3^2)$? Am I missing something?

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try mathoverflow. –  john mangual Sep 15 '11 at 4:43

First, Juan S is correct in assessing the final state of the Serre spectral sequence.

Second, with regard to your point (1) the cohomology ring structure is, in this case, not determined by the multiplication on the $E_\infty$ term, even though everything is of finite type over $\mathbb{Z}$. Specifically, you have the two classes $\alpha, \beta \in (0,m-1)$, and in the spectral sequence you have that $\alpha \cup \beta = 0$ because it lands in the zero group. However, this only guarantees that the product is zero "up to filtration." You have the possibility that $\alpha \cup \beta$ becomes some linear combination of $\alpha \cup \gamma$ and $\beta \cup\gamma$ in the actual cohomology ring.

One of the biggest red flags that this actually happens is that the relations you're getting don't look symmetric. The action of the symmetric group on this configuration space permutes $\alpha$, $\beta$, and $\gamma$, and so you can't have $\alpha \beta = 0$ when neither of $\alpha \gamma$ and $\beta \gamma$ are.

Along this line, the cohomology ring is of strictly smaller rank than the candidate that you list in item (2).

One way to get the actual cohomology ring is as follows. There is a map $$conf(\mathbb{R}^m,3) \to conf(\mathbb{R}^m,2)^3 \simeq (S^{m-1})^3$$ that sends $(x,y,z)$ to $((y,z), (x,z), (x,y))$. This is a map of bundles and respects the action of the symmetric group; over any point $(x,y)$, it is homotopy equivalent to the inclusion $S^{m-1} \vee S^{m-1} \to S^{m-1} \times S^{m-1}$. This gives you a map of Serre spectral sequence in the opposite direction which is surjective. Examining this has the following consequences.

• The map $H^*((S^{m-1})^3) \to H^*(conf(\mathbb{R}^m,3))$ is surjective, where the former is the exterior algebra $\mathbb{Z}[a_1,a_2,a_3]/(a_i^2)$ you list in (2).

• This map is an isomorphism in degree $m-1$, and the action of $\Sigma_3$ is dual to a permutation action on $\mathbb{Z}^3$. You can choose generators so that the permutation $\sigma$ sends $a_i$ to $a_{\sigma^{-1} i}$. (Note: Cohomology is contravariant so this is a right action.)

• The kernel in degree $2(m-1)$ is $\Sigma_3$-equivariant of rank $1$, and the image is free of rank $2$. This forces the kernel to be generated by $(a_1 a_2 + a_1 a_3 + a_2 a_3)$.

So the cohomology ring is actually $$\mathbb{Z}[a_1,a_2,a_3]/(a_1^2, a_2^2, a_3^2, a_1 a_2 + a_1 a_3 + a_2 a_3).$$ (This is actually the hard one. You can use this result to get a grip on the configuration spaces of more points in $\mathbb{R}^m$.)

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Thank you for the clarification. I realise that I got confused by the semantics on p. 139 in McCleary where he says that the spectral sequence converges to $H^*(E,R)$ as $algebras$. –  confusedmath Oct 20 '11 at 10:02

This is really a comment, but I ran out of room (and I give no gurantee it is correct)

Shouldn't there be some relation between the generators? If I've done the spectral sequence correctly there is a $\mathbb{Z}$ on the $q=0$ row in at $p=0$ and $p=m-1$. On the $q=m-1$ row there is a $\mathbb{Z} \oplus \mathbb{Z}$ at $p=0$ and $p=m-1$. I'm not sure there can be any extension problems here, and so we know the cohomology as abelian groups:

$$H^p(\text{Conf}(\mathbb{R}^3,3);\mathbb{Z}) \simeq \begin{cases} \mathbb{Z} &\text{ if } p=0 \\ \mathbb{Z}\oplus\mathbb{Z}\oplus \mathbb{Z} &\text{ if } p=m-1 \\ \mathbb{Z} \oplus \mathbb{Z} &\text{ if } p=2(m-1) \end{cases}$$

By multiplicity of the spectral sequence I think we should we should have $E_2^{0,m-1} \otimes E_2^{m-1,0} \simeq E_2^{m-1,m-1}$ So if we denote the generators of the $\mathbb{Z} \oplus \mathbb{Z}$ in position $(0,m-1)$ as $\alpha,\beta$ and the $\mathbb{Z}$ in $(m-1,0)$ as $\gamma$ shouldn't we get that the $\mathbb{Z}\oplus\mathbb{Z}$ in $(m-1,m-1)$ is generated by $(\alpha \otimes \gamma,\beta \otimes \gamma)$?

This paper looks useful. I note the comment there that $H^{2(m-1)}(\text{Conf}(\mathbb{R}^3,3);\mathbb{Z})$ is a free $\mathbb{Z}$ module spanned by $A_{21}A_{31}$ and $A_{31}A_{32}$.

One final place that might help is this monograph by Cohen,Lada and May. See specifically page 250.

Hopefully this is of some help!

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