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 Bênoit 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 was obtained by Smirnov, by Justin Smith, ...
2 was obtained for n=2 by an explicit construction by Hans Baues.
For the cochain algebra of a space [tex]A=C^*(X)[/tex] similar
structure results have been obtained by Kadeishvili-Saneblidze and by
Hess-Parent-Scott.
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 En 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 Möbius
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].