Thanks for your comment, Eitan! This is a nice case to remember. Here the situation is somewhat different since AP of length 6 (over (Z/mZ)^n when 6|m) are ( in view of the new result) surprisingly easy (as easy as 3) but this does not say that AP of length larger than 6 are easy. (So we do not have statements about AP of length 10 over Z/10Z or of length 15 over Z/15Z reducing these problems to questions about smaller length AP). Regarding AP of length 4,5 and 6 the results give exponential improvement over (Z/mZ)^n when m is 0 mod 6. Now we know exponential improvement in (Z/6Z)^n for the cases of 4- 5- and 6- terms AP. Are 5-terms AP over (Z/5Z)^n and 4-term AP over (Z/4Z)^n so much harder? (And by “harder” I refer to how hard it is to prove things). BTW I plan a series of posts based on our open problem session so I’ll probably ask you more details about your probךem.

]]>“Before Smale proved this theorem, mathematicians became stuck while trying to understand manifolds of dimension 3 or 4, and assumed that the higher-dimensional cases were even harder. The h-cobordism theorem showed that (simply connected) manifolds of dimension at least 5 are much easier than those of dimension 3 or 4. The proof of the theorem depends on the “Whitney trick” of Hassler Whitney, which geometrically untangles homologically-tangled spheres of complementary dimension in a manifold of dimension >4. An informal reason why manifolds of dimension 3 or 4 are unusually hard is that the trick fails to work in lower dimensions, which have no room for untanglement.” ]]>

Thank you for the clarification

]]>Dear Kodlu, the statement “much easier” refers to the difficulty of proving an upper bound (or even exponentially-small upper bound). We have reasons to think that this is considerably harder for 4 compared to 3. And that it is as hard or harder for 5 compared to 4. And until the new result we could have thought that this is at least as hard for 6 compared to 4 and 5. I don’t think the difficulty of finding a proof is very related to the precis value of the upper bound.

*(I also edited the wording of the post to make the point clearer.)*

*GK: thanks*

that $5.709/6\approx 0.95$ for P6 while the corresponding ratio is more like $0.9$ for P4 and P5? ]]>