## Prologue

Consider the following problems:

P3: What is the maximum density of a set A in $(\mathbb Z/3\mathbb Z)^n$ without a 3-term AP? (AP=arithmetic progression.)

This is the celebrated Cap Set problem and we reported here in 2016 the breakthrough results by Croot-Lev-Pach-Ellenberg-Gijswijt (CLPEG) asserting that $|A| \le 2.755..^n$.

P4: What is the maximum density of a set A in  $(\mathbb Z/4\mathbb Z)^n$ without a 4-term AP?

P5: What is the maximum density of a set A in  $(\mathbb Z/5\mathbb Z)^n$ without a 5-term AP?

Moving from 3 term AP to 4 term AP represents a major leap for such problems. In some cases moving from 4 to 5 also represents a major leap. Usually the techniques that work for $k=5$ continue to work for $k>5$, that is, I am not aware of cases where moving from 5 to 6 also represent a major leap. In other words, we do not have reasons to think, so far,  that progress on P6 (below) will be considerably harder than progress on P5.

P6: What is the maximum density of a set A in $(\mathbb Z/6\mathbb Z)^n$ without a 6-term AP?

And, of course, we can move on to P7, P8, P9,…. ,

It is known that the density of a set $A \subset (\mathbb Z_m)^n$ without $m$-term AP for  every fixed m goes to zero with n. This was proved by  Furstenberg and Katznelson using ergodic theoretical tools. Following the CLPEG breakthrough we can certainly believe that the density that guarantees m-term AP in $\mathbb Z_m^n$ for every fixed m is exponentially small in n. I vaguely remember that the best known bounds for P4 are polylogarithmically small with n and were achieved by Green and Tao using tools from Gowers’ work on 4AP-free sets over the integers.

### A little more before we move on

To have some general notation let $r_k(\mathbb Z_m^n)$ denote the maximal size of a set $A \subset \mathbb Z_m^n$ with no k distinct elements in arithmetic progression. It is reasonable to believe that $r_m(\mathbb Z_m^n) \le \delta_m^n m^n$ for some $\delta_m < 1$, that is that sets in $\mathbb Z_m^n$ avoiding $m$ terms AP are exponentially small. To be more precise let

$\delta_m= \frac {1}{m} \lim r_k(\mathbb Z_m^n)^{1/n}$

We can believe that the limits exist,  that $\delta_m<1$, and we may also guess (wildly but safely) that they satisfy ${\delta_3} < {\delta_4} < {\delta_5} \cdots$.

These problems, as well as the analogous problems over the integers, (Roth’s and Szemeredi’s theorems and later developments), and the related density Hales Jewett problem,  form a  holy grail of additive combinatorics and they attracted a lot of attention also on this blog.

Here is a partial list of posts: 2008 Pushing the Behrend bound; 2009A, 2009B 2009C Attempts to relate the problem to the Frankl-Wilson Theorem (and the polynomial method) and to Frankl-Rodl theorem and also to find counterexamples;  2009D polymath1;  2009P public opinion poll about the true behavior;  2010 The logarithmic bound for Roth is reached; 2011 cap sets and matrix multiplications; 2016a, 2016b The CLPEG breakthrough; 2016W A polynomial method workshop;  2020 The logarithmic bound for Roth is broken.

Let’s move on to the new result by Péter Pál Pach and Richárd Palincza

## Theabsolutely stunningresult by Péter Pál Pach and Richárd Palincza

As we already mentioned, so far we do not have reasons to expect that proving density results for 6-terms AP in $\mathbb Z_6^n$ (or $\mathbb Z$) will be much harder than the question over $\mathbb Z_5^n$. However, what Pach and Palincza proved is as follows:

Theorem:  sets avoiding 6-term arithmetic progressions in $\mathbb Z_6^n$ have size at most $5.709^n$.

The big surprise is that the problem P6 over integers modulo six is much easier than P4 and P5. The proof is based on a simple combinatorial reasoning showing that

$r_6(\mathbb Z_6^n) \le 2^{n+1} \sqrt {3^nr_3(\mathbb Z_3^n).}$

The paper arXived on September 2020 is

Click and enjoy this beautiful, stunning and very simple proof.

Congratulations to Péter and Richárd!

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### 6 Responses to Péter Pál Pach and Richárd Palincza: a Glimpse Beyond the Horizon

1. kodlu says:

Very nice. Is your statement of “much easier” simply referring to the fact
that $5.709/6\approx 0.95$ for P6 while the corresponding ratio is more like $0.9$ for P4 and P5?

• Gil Kalai says:

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.)

• kodlu says:

Thank you for the clarification

2. Spelling says:

Tau –> Tao ……

GK: thanks

3. eitan bachmat says:

Kind of reminds of the situation in topology, the following is quoted from the wikipedia entry on the h-cobordism theorem:
“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.”

• Gil Kalai says:

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.