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Let $X_n$ be the space of $n$ distinct labeled points in $\mathbb{R}^3$, which is equipped with an action of the symmetric group $S_n$. It is well known that the total cohomology of $X_n$ is isomorphic to the regular representation of $S_n$, but I would like to know what representation one gets in each degree.

When $\mathbb{R}^3$ is replaced with $\mathbb{R}^2$, this question has been answered by Lehrer and Solomon, and Getzler expresses this answer in a beautiful form in Equation (2.5) of http://arxiv.org/pdf/alg-geom/9510018.pdf. If we let $f_{n,i}$ be the degree $n$ symmetric function associated with the action of $S_n$ on the degree $i$ cohomology of the configuration space, then Getzler gives a nice infinite product expansion for the power series $\sum_{n\geq 0}\sum_{i\geq 0}(-x)^if_{n,i}$. If there were a formula similar to this one for the configuration space of points in $\mathbb{R}^3$, that would be great.

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  • $\begingroup$ Probably you know this, but: the question is equivalent to asking what representation of the symmetric group you get when you consider the subspace of the $3$-Poisson operad (like the Poisson operad, but the Poisson bracket is in degree $-2$) involving $n$ operands and some number of Poisson brackets. This is reasonably explicit but maybe you want something even more explicit than this. $\endgroup$ – Qiaochu Yuan Sep 14 '15 at 21:19
  • $\begingroup$ @QiaochuYuan Once you start keeping track of number of Poisson brackets, the 3-Poisson and 1-Poisson operads are the same. In either case the Leibniz rule provides a "distributivity" that says that this is the space of forests of Lie words whose leaves are the letters $a,\dots,n$. (Where, of course, "$n$" is the $n$th letter.) I say "forest" because a Lie word looks a bit like a tree. This space has a basis consisting of alphabetized lists of Lyndon words. But, of course, the description of Lie words in terms of Lyndon words is not good for extracting $S_n$ actions. $\endgroup$ – Theo Johnson-Freyd Sep 15 '15 at 1:14
  • $\begingroup$ The number of trees in the forest, of course, is (the same data as) the number of Poisson brackets, i.e. the homological grading in Nick's question. $\endgroup$ – Theo Johnson-Freyd Sep 15 '15 at 1:16
  • $\begingroup$ So by Qiaochu's remark your question is the same as asking for the $S_n$ representation on $k$-fold commutative products of Lie/Lyndon words with a total of $n$ variables participating (each once). This should break up as a direct sum over partitions (the number of leaves on the different trees) of a product wreaths product of (1) permuting the trees with $m$ leaves and (2) for each tree, the $S_m$ representation thereon, assuming I'm tracking things correctly. That latter contribution is the $S_m$-action on the linear-in-every-variable component of the free Lie algebra on $m$ letters. $\endgroup$ – Theo Johnson-Freyd Sep 15 '15 at 1:22
  • $\begingroup$ I.e. the $S_m$ action on the arity-$m$ part of the Lie operad. I don't know much about this representation. The best I can say is that the $L_\infty$ operad gives a resolution of this representation in terms of (usual, as opposed to Lie) rooted trees (perhaps twisted by some sign representations). $\endgroup$ – Theo Johnson-Freyd Sep 15 '15 at 1:27
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One can indeed modify the formulas in the paper of Getzler to get an answer for any $\mathbf R^d$, by judiciously inserting minus signs and making substitutions $x \mapsto x^{d-1}$ in various places, but if I try I'll probably get it wrong. So let me explain how it works instead, and hopefully there are no sign errors and everything ends up in the right cohomological degree.

The cohomology ring of $F(\mathbf R^d,k)$ was computed by Cohen. It is the graded commutative algebra generated by variables $\omega_{ij}$ of degree $(d-1)$ for distinct $i,j \in \{1,\ldots,k\}$, modulo the relations

  • $\omega_{ij}^2 =0$,
  • $\omega_{ij} = (-1)^d \omega_{ji}$,
  • $\omega_{ij}\omega_{ik} + \omega_{jk}\omega_{ji} + \omega_{ki}\omega_{kj} = 0$.

One can associate a graph to a monomial in these generators, with vertices $\{1,\ldots,k\}$ and an edge between $i$ and $j$ if $\omega_{ij}$ occurs in the monomial. The relations imply that the monomial is zero unless this graph is a forest.

The top degree $H^{k(d-1)}(F(\mathbf R^d,k))$ is spanned by monomials such that the corresponding graph is a tree. One can interpret such a monomial as a bracketing of $\{1,\ldots,k\}$: if $d$ is odd then the second and third relation correspond to anticommutativity and Jacobi identity for this bracketing, so the $S_k$-representation is given by the space $\mathrm{Lie}(k)$ spanned by Lie words on $k$ letters, and if $d$ is even we get instead $\mathrm{Lie}(k) \otimes \mathrm{sgn}_k$.

In lower degrees we get instead a sum over partitions of $\{1,\ldots,k\}$, and for each block $B$ in the partition a monomial such that the corresponding graph is a tree on the vertex set $B$. This tells us that we are considering a plethysm of symmetric functions.

Let $\widehat\Lambda$ be the degree completion of the ring of symmetric functions in infinitely many variables and let $\widehat\Lambda[[t]]$ be the power series ring in one variable over it. If $V$ is a representation of $S_k$, let $\mathrm{ch}_k(V)$ be the corresponding degree $k$ symmetric function. Define $$ C = \sum_{k \geq 0} h_k \in \widehat\Lambda,$$ the sum of the trivial representation in each arity. Let $$ L_d = \begin{cases} \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) \otimes \mathrm{sgn}_k t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ even} \\ \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ odd} \end{cases}$$ Then $$ \sum_{k \geq 0} \sum_{i \geq 0} t^i \mathrm{ch}_k H^i(F(\mathbf R^d,k)) = C \circ L_d.$$ The plethysm here satisfies $p_n \circ f(t,p_1,p_2,\ldots) = f(t^n,p_n,p_{2n},\ldots)$. You can also write this plethysm in terms of an infinite product or the "plethystic exponential", like Getzler does. One can describe $\mathrm{ch}_k\mathrm{Lie}(k)$ explicitly in terms of symmetric functions, and Getzler is using this in his paper. Namely, there is an isomorphism $\mathrm{Lie}(k) = \mathrm{Ind}_{C_k}^{S_k} \chi$ where $\chi$ is a primitive character of the cyclic group on $k$ elements, and $$ \mathrm{Ind}_{C_k}^{S_k} \chi = \frac 1 k \sum_{{d \mid k}} \mu(d)p_k^{k/d}.$$

Lurking in the background is an operadic perspective. The operad $e_d$ is given by the homology of the little $d$-dimensional disks and what we are trying to do is calculate the characteristic of this operad. The operad $e_d$ can be constructed as the product of the commutative operad and a $(d-1)$-fold suspension of the Lie operad by a distributive law. This tells us that the characteristic of $e_d$ is the plethysm of the characteristic of $\mathrm{Com}$ (which is $C$) and the characteristic of this $(d-1)$-fold suspension, which is $L_d$.

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  • $\begingroup$ Thanks! I knew the presentation and I was aware that the top degree part is isomorphic to $\operatorname{Lie}(k)$, but I didn't know the plethysm formula that you wrote down, which is very elegant and just the kind of answer I wanted. (In particular, SAGE is good with plethysms!) The only problem that I still have is that I can't seem to fine the formula for $\operatorname{ch}_k\operatorname{Lie}(k)$, in Getzler's paper or elsewhere. Maybe I'm just not looking in the right place? $\endgroup$ – Nicholas Proudfoot Sep 15 '15 at 15:44
  • $\begingroup$ Okay, I found a formula for $\sum_{k>0}\operatorname{ch}_k\operatorname{Lie}(k)$ in Proposition 7.11 of arxiv.org/pdf/1408.5415.pdf. Surely not the original source, but it will do! $\endgroup$ – Nicholas Proudfoot Sep 15 '15 at 15:53
  • $\begingroup$ I see now that product formulas also appear in arxiv.org/pdf/1505.04196.pdf (Theorems 2.11 and 2.22). $\endgroup$ – Nicholas Proudfoot Sep 15 '15 at 16:18
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    $\begingroup$ I see now that Getzler didn't state the formula in terms of the Lie operad in this paper, so I added this to the answer (and corrected a dumb mistake). $\endgroup$ – Dan Petersen Sep 15 '15 at 16:21

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