What is the rational homology of the unordered configuration space of points in the plane?

What is the rational homology of the unordered configuration space of $n$ points in the Euclidean plane?

In the unordered configuration space, I know there is torsion in homology, but I just want to know the Betti numbers, i.e. what is
$$\beta_j = \dim H_j (\mbox{Conf}_n(\mathbb{R}^2),\mathbb{Q} ),$$
as a function of $n$ and $j$?

For the ordered configuration space, there is no torsion in homology and we have the Poincaré polynomial
$$\beta_0 + \beta_1 t + \beta_2 t + \dots = (1+t)(1+2t) \dots (1+(n-1)t).$$

Answer

This is classical and in some form is due to Arnol’d (The cohomology ring of the colored braid group, which unfortunately is in Russian) but I am having trouble locating a precise statement and reference for the unordered as opposed to the ordered configuration spaces. Since $\text{Conf}_n(\mathbb{R}^2)$ is the classifying space of the braid groups $B_n$ you are equivalently asking about their (rational) group homology.

$\text{Conf}_n(\mathbb{R}^2)$ has a Galois cover by the ordered configuration space, which I will denote $\text{OConf}_n(\mathbb{R}^2)$, with Galois group $S_n$. (These spaces are classifying spaces of the pure braid groups $PB_n$.) It follows that the rational homology of $\text{Conf}_n(\mathbb{R}^2)$ satisfies

$$H_{\bullet}(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q}) \cong H_{\bullet}(\text{OConf}_n(\mathbb{R}^2), \mathbb{Q})^{S_n}$$

so we can do this computation by understanding not only the rational homology of the ordered configuration spaces but also the action of $S_n$ on them.

The ordered configuration spaces are homotopy equivalent to the spaces of the little $2$-disks operad $E_2$, and accordingly their rational homology is the rational homology $H_{\bullet}(E_2, \mathbb{Q})$. This rational homology forms another operad (in graded vector spaces) and it’s known exactly what this operad is:

Theorem: The rational homology of the little $d$-disks operad is the $d$Poisson operad $\text{Pois}_d$ of (graded) commutative algebras equipped with a Poisson bracket of homological degree $d-1$ (often stated as cohomological degree $1-d$; I hope I am negating this correctly).

You can check out Dev Sinha’s The homology of the little disks operad for more details here. When $d = 2$ we get the Gerstenhaber operad. The Poisson bracket comes from the generator of $H_{d-1}(\text{OConf}_2(\mathbb{R}^d), \mathbb{Q})$; here $\text{OConf}_2(\mathbb{R}^d)$ is homotopy equivalent to $S^{d-1}$. Besides being a compact way of describing all of the homology at once this description also tells us the action of $S_n$: it’s given by permuting the operations in the Poisson operad. For example we have the graded skew-symmetry $\{ x, y \} \mapsto \{ y, x \} = -(-1)^{d-1} \{ x, y \}$ of the Poisson bracket itself coming from the action of $S_2$ on $\text{OConf}_2(\mathbb{R}^d)$; the sign here corresponds to the action of the antipode of $S^{d-1}$ on $H_{d-1}(S^{d-1})$.

Now it suffices to determine for $d = 2$ which of the $n$-ary operations in the Poisson operad are invariant under the action of $S_n$. In degree $0$ the only operation (up to scale) is the $n$-ary commutative product $x_1 \dots x_n$ which is always invariant; this corresponds to $H_0 \cong \mathbb{Q}$. In degree $1$ we have the operation

$$\sum_{i < j} \{ x_i, x_j \} \prod_{k \neq i, j} x_k$$

which is also invariant (the Poisson bracket commutes and $\{ x_i, x_j \}$ has degree $1$ but the other $x_k$ still have degree $0$ so the action of $S_n$ on monomials is just by permutation with no signs) and which is also the unique invariant (up to scale).

In higher degrees the operations are spanned by “monomials” in the commutative product and the Poisson bracket which involve either products $\{ x_i, x_j \} \{ x_k, x_{\ell} \}$ of two or more Poisson brackets or nested brackets $\{ x_i, \{ x_j, x_k \} \}$. (By repeated application of the Leibniz rule we can at least assume WLOG that commutative products are never nested inside Poisson brackets.) Here I believe what happens is that “signs cancel”: that is, summing any monomial over its $S_n$-orbit gives $0$, which implies that there are no $S_n$-invariant operations. It is basically clear in my head how this works but I am having trouble writing down an explicit rigorous proof; in any case, assuming this (I will try to fix this gap later), we have:

Corollary: The rational homology of $\text{Conf}_n(\mathbb{R}^2)$, and hence of the braid groups, is given for $n \ge 2$ by

$$H_i(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q}) \cong \begin{cases} \mathbb{Q} & \text{ if } i = 0, 1 \\ 0 & \text{ otherwise}. \end{cases}.$$

(The cases $n = 0, 1$ can be handled separately; in both cases we get $\mathbb{Q}$ in degree $0$.) In other words, maybe somewhat surprisingly, the unordered configuration spaces have the rational homology of a circle.

Fun aside: this computation turns out to be related, via the Grothendieck trace formula, to the fact that the probability that a random monic polynomial over $\mathbb{F}_q$ is squarefree is $1 – \frac{1}{q}$, which implies that the computation can actually be redone (using the Grothendieck trace formula, together with a comparison theorem between $\ell$-adic and singular cohomology) by counting that there are $q^n – q^{n-1}$ monic squarefree polynomials of degree $n$; I learned this from Jordan Ellenberg’s blog post The braid group, analytic number theory, and Weil’s three columns.

Edit: Here is the promised calculation. I still believe the “signs cancel” approach above will work, but since I’m not sure how to organize the calculation nicely, let’s do the following instead. Since the action of a finite group on a rational vector space is completely reducible, invariants are canonically isomorphic to coinvariants. And we have a trick available for calculating coinvariants: namely, if $O(n)$ is any operad in graded vector spaces, the free $O$-algebra on a graded vector space is $V$ is

$$\bigoplus_{n \ge 0} V^{\otimes n} \otimes_{S_n} O(n).$$

When $V$ is the $1$-dimensional vector space in degree $0$, $V^{\otimes n}$ is the trivial $1$-dimensional representation of $S_n$ in degree $0$, so this calculates exactly the $S_n$-coinvariants $O(n)_{S_n}$. To keep track of the answers for different values of $n$ we will introduce a second grading for $V$; let’s call it the “vector grading,” and we’ll call the other grading the homological grading.

The upshot is that we’ve reduced to calculating the free Gerstenhaber algebra on a generator $x$ in homological degree $0$ (and vector degree $1$). This is actually quite straightforward. First, we get the Poisson (really I should say “Gerstenhaber” here) bracket $\{ x, x \}$, which lives in homological degree $1$ and vector degree $2$. The graded Jacobi identity gives $\{ x, \{ x, x \} \} = 0$, so no nested brackets appear. And since $\{ x, x \}$ is odd, it anticommutes with itself with respect to ordinary multiplication, so $\{ x, x \} \{ x, x \} = 0$. So all we can do is take the product of at most $1$ copy of $\{ x, x \}$ with any number of copies of $x$, and that’s all.

So the free Gerstenhaber algebra on $x$ has a basis given by $x^i$ (which lives in homological degree $0$ and vector degree $i$) and $x^i \{ x, x \}$ (which lives in homological degree $1$ and vector degree $i + 2$), for $i \in \mathbb{Z}_{\ge 0}$. The correspondence with the rational homology of the configuration spaces is that the subspace in homological degree $i$ and vector degree $n$ is $H_i(\text{Conf}_n(\mathbb{R}^2), \mathbb{Q})$. This exactly reproduces the previous calculation (and as a bonus corrects an error I had previously made, which is to forget that the cases $n = 0, 1$ behave differently).

Fun aside #2: this free Gerstenhaber algebra on a generator in degree $0$ occurs “in nature” as the rational homology $H_{\bullet}(\Omega^2 S^2, \mathbb{Q})$ of the double loop space of the $2$-sphere, which is in turn the “free grouplike $E_2$ space” on a point. The generator $x$ itself corresponds to the generator of $\pi_2(S^2) \cong \pi_0(\Omega^2 S^2)$, embedded into $H_0(\Omega^2 S^2, \mathbb{Q})$, while the Poisson bracket $\{ x, x \}$ corresponds to the Whitehead bracket of this generator with itself, which I believe is something like twice the Hopf fibration as an element of (the rationalization of) $\pi_3(S^2) \cong \pi_1(\Omega^2 S^2)$, embedded into $H_1(\Omega^2 S^2, \mathbb{Q})$. The connection between double loop spaces and braid groups is also classical (I think) but I don’t know what a good reference is. The nLab page on iterated loop spaces has some references.

Attribution
Source : Link , Question Author : Matthew Kahle , Answer Author : Qiaochu Yuan

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