Gil Kalai has officially started the Polymath3 project (Polynomial Hirsch conjecture) with a research thread at his blog.
The original aim of this project is to prove the polynomial Hirsch conjecture, which is a conjecture in the combinatorial geometry of polytopes. However, there is a reduction due to Eisenbrand, Hahnle, Razborov, and Rothvoss that would deduce this conjecture from a purely combinatorial conjecture, which can be stated as follows.
Combinatorial polynomial Hirsch conjecture. Let
be non-empty collections of subsets of
with the following properties:
- (Disjointness)
for every
.
- (Connectedness) If
and
, there exists
such that
.
Then t is of polynomial size in n (i.e.
).
For instance, when n=3, one can obtain such a family with t=6, e.g.
one can show that this is the best possible value of t for this choice of n. The best possible value of t for n=4 is still not worked out; it is between 8 and 11.
One appealing thing about this problem is that there is a simple elementary argument that gives the bound for all
; and so in some sense one is “only a logarithm away” from proving the conjecture. Anyway, the project is just starting, and does not require any particularly specialised background, so anyone who may be interested in this problem one may want to take a look at the research thread.

7 comments
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1 October, 2010 at 2:29 am
Américo Tavares
Typo: missing ‘latex’ in $A \cap B \subset C$,
. [Corrected, thanks - T.]
1 October, 2010 at 5:18 am
Scott Carnahan
I think the Disjointness axiom needs an empty set. [Corrected, thanks - T.]
2 October, 2010 at 6:27 am
Johan
There is one bracket too much in the definition of F_5. [Not sure I see this... - T.]
2 October, 2010 at 8:15 am
Anonymous
I think you’ve got one too many bracket also, should be F_5 = {{2,3},{3}}, rather than {{2,3}},{{3}}
[Ah, I see the problem now: F_6 had run off beyond the column width of the blog and so was not visible. Fixed now - T.]
2 October, 2010 at 7:05 pm
David Hansen
Should “polytops” be “polytopes”?
[Corrected, thanks - T.]
12 October, 2010 at 5:09 pm
Eric
Why is
{\mathcal F}_1 = \{ \emptyset\}, {\mathcal F}_2 = \{ \{1\}\}, {\mathcal F}_3 = \{\{1,2\}\}, {\mathcal F}_4 = \{\{1,2,3\}\}, {\mathcal F}_5 = \{\{2,3\}\}, {\mathcal F}_6 = \{\{3\}\};
optimal?
Why cant I shove in {{2}} to get
{\mathcal F}_1 = \{ \emptyset\}, {\mathcal F}_2 = \{ \{1\}\}, {\mathcal F}_3 = \{\{2\}\}, {\mathcal F}_4 = \{\{1,2\}\}, {\mathcal F}_5 = \{\{1,2,3\}\}, {\mathcal F}_6 = \{\{2,3\}\}, {\mathcal F}_7 = \{\{3\}\};
(I may be making a silly error, but I cannot see it)
12 October, 2010 at 5:15 pm
Eric
Nevermind!! Please ignore my post, I see that {1}\cap {1,2,3}={1} not subset of {2}. I should think about things more before posting next time.