7
$\begingroup$

Here was a question about resonance arrangement. It is defined as follows.

Let $x_i$ be the standard coordinates on $\mathbb{C}^n$. For each nonempty $I\subseteq\{1,\dots,n\}$, define the hyperplane $H_I$ to be the hyperplane given by $l_I:=\sum _{i∈I}x_i=0$.

My question is about the top degree (co)homology of the complement to this arrangement.

I have the following conjecture: the space of top degree forms, which are wedge products of forms $d \log l_I$, is generated by tree-forms.

A tree form is defined by a rooted tree with $n$ edges labeled by $x_i$'s. Any (non-root) vertex gives a linear function on $\mathbb{C}^n$, which is the sum of all labels on edges, lying on the shortest path from the root to the vertex. The wedge product of $d\log$'s of these linear functions for all vertices is the tree form. By the way, is it a known object?

I would deeply appreciate any comments.

$\endgroup$
9
  • 6
    $\begingroup$ The resonance arrangement is notoriously difficult to understand in any nice way. One of the most recent papers about it I know of is arxiv.org/abs/1903.06595. (I don't think it talks about cohomology of complement of complex arrangement at all- but of course # of regions of complement of real arrangement is a closely related thing, which again is not well understood.) $\endgroup$ Sep 23, 2020 at 16:57
  • 1
    $\begingroup$ For something even more recent (and in line with Sam Hopkins's comment), see arxiv.org/pdf/2008.10553.pdf $\endgroup$
    – user35313
    Sep 24, 2020 at 2:07
  • $\begingroup$ @SamHopkins Thank you for your comment and the reference. $\endgroup$ Sep 24, 2020 at 16:56
  • $\begingroup$ @user61318 Thank you for the reference. $\endgroup$ Sep 24, 2020 at 16:57
  • $\begingroup$ The cohomology of hyperplane arrangements is described by the Orlik-Solomon algebra which has exactly the dlog forms as basis. Section 3 of the book of Orlik and Terao "Arrangements of hyperplanes" has a description of the "no broken circuits" basis of the top degree. I suppose that means the question is about expressing the "no broken circuits" basis in terms of your "tree form" basis. $\endgroup$ Oct 11, 2020 at 19:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.