This post will concern tuesday morning. Tuesday evening will be in a later post.

With the morning thus came, again, the pain in the legs. However, I'm told it'll be better if I keep on skiing.

The mathematics in this report will come sooner than in the last; mainly because the lectures start at 8.30 and not at 17.00. :)

First out is BĂȘnoit Fresse with

## Little cubes operad actions on the bar construction of algebras

The reduced bar construcion of augmented associative algebras is given by fixing a field [tex]F[/tex], and for an augmented associative algebra [tex]A[/tex] giving a chain complex [tex]B(A)[/tex] such that [tex]B_n(A)=\hat A^{\otimes n}[/tex] where [tex]\hat A[/tex] denotes the augmentation ideal of [tex]A[/tex] with the differential [tex]\partial(a_1\otimes\dots\otimes_n)=\sum_{i=1}^{n-1}a_\otimes\dots\otimes a_ia_{i+1}\otimes\dots\otimes a_n[/tex]

If the product of [tex]A[/tex] is commutative, then the shuffle product of tensors provides [tex]B(A)[/tex] with the structure of a differential commutative algebra. In the talk, Fresse starts looking at the algebraic structure of [tex]B(A)[/tex] for algebras with a homotopy commutative algebra:

### Results

- If [tex]A[/tex] is an [tex]E_\infty[/tex] algebra, then [tex]B(A)[/tex] can be equipped with the structure of an [tex]E_\infty[/tex] algebra.
- If [tex]A[/tex] is an [tex]E_n[/tex] algebra, then [tex]B(A)[/tex] can be equipped with the structure of an [tex]E_{n-1}[/tex] algebra.

1 can be motivated in arXiv:math.AT/0601085.

2 is a consequence of the assertion that the endomorphism prop of the bar construction is equivalent to the prop of co-associative and [tex]E_{n-1}[/tex]-multiplicative bialgebras.

Next up is Clemens Berger, talking about

## On the combinatorial structure of E_{n} operads

May et.al. studied the n-fold loop spaces [tex]\Omega^nX=\operator{Top}_*(S^n,X)[/tex] of maps of the n-sphere to the space, by considering the k-fold n-fold loop space, i.e. maps [tex]S^n\to S^n\wedge S^n\wedge\dots\wedge S^n=\mathcal O_n(k)[/tex]. It turns out that it isn't necessary to study the entire space of maps [tex]S^n\to\mathcal O_n(k)[/tex] but it's enough to look at the suboperad of "little n-disks" [tex]\mathcal D_n[/tex].

Theorem (Boardmann-Voft, May, Segal): Each n-fold loop space is canonically a [tex[\mathcal D_n[/tex] algebra.

[tex]\mathcal D_\infty[/tex] gives rise to an [tex]E_\infty[/tex] operad. [tex]\mathcal D_1[/tex] gives rise to an [tex]E_1[/tex] operad. If we can find a similar intrinsic characterisation of [tex]\mathcal D_n[/tex] operads for other n, it would yield a combinatorial structure of an n-fold loop space.

Fiedorowicz has been able to obtain such a classification for n=2 (with symmetric props replaced by braid groups).

The next talk was held by Muriel Livernet on

## The associative operad and the weak order on the symmetric groups

Livernet started out by defining symmetric and nonsymmetric operads, and gave a pair of adjoint functors - the forgetful functor from symmetric to non-symmetric operads, and tensoring with the group algebra to symmetrize a non-symmetric operad.

At this point, parts of the lighting in the room fell down on the audience. No serious consequences.

The weak Bruhat order can be defined as ordering permutations [tex]\sigma\leq\tau[/tex] if [tex]\operator{Inv}(\sigma)\subseteq\operator{Inv}(\tau)[/tex] where [tex]\operator{Inv}[/tex] is the inversion set of the permutation, i.e. the set of pairs i,j such that i\sigma_j[/tex].

Generating a new basis for the associative operad, using the MĂ¶bius inversion on the relevant Bruhat order, the composition rule ends up being a summation over a specific interval in the Bruhat order.

Interesting combinatorial question asked: How many permutations are there (of a fixed length) such that no proper substring of the permutation (written as a permuted string) is an entire interval. For instance, (2,4,1,3) has no proper substring that fills out an interval, whereas (2,4,3,1) has the substring 2,4,3 which fills out [2,4].

Shouldn't this question be easily answered by simply counting all permutation that DO have a subinterval? It's easy enough to count your way through the intervals - choose a length and a startingpoint. There are length! ways to embed that interval, and easily enough controlled number of points to insert the interval. The remaining parts are also easily distributed. However, double counting will occur - which messes up the counting quite a bit. Maybe using inclusion-exclusion will lead the way? For, for instance [2,4] to be counted a particular time, you'd want [2,4] to be embedded without contact to 3 and 5, as well as the remaining two sections to be interval-free. Maybe it's doable with some sort of recursion? There are, even then, cases of permutations with several intervals embedded, for instance (2,4,3,6,9,8,1,7) which contains [2,4] and [8,9].