## Polymath 3: The Polynomial Hirsch Conjecture 2

Here we start the second research thread about the polynomial Hirsch conjecture.  I hope that people will feel as comfortable as possible to offer ideas about the problem. The combinatorial problem looks simple and also everything that we know about it is rather simple: At this stage joining the project should be very easy. If you have an idea (and certainly a question or a request,) please don’t feel necessary to read all earlier comments to see if it is already there.

In the first post we described the combinatorial problem: Finding the largest possible number f(n) of disjoint families of subsets from an n-element set which satisfy a certain simple property (*).We denote by f(d,n) the largest possible number of families satisfying (*) of d-subsets from {1,2,…,n}.

The two principle questions we ask are:

Can the upper bounds be improved?

and

Can the lower bounds be improved?

What are the places that the upper bound argument is wasteful and how can we improve it? Can randomness help for constructions? How does a family for which the upper bound argument is rather sharp will look like?

We are also interested in the situation for small values of n and for small values of d. In particular, what is f(3,n)? Extending the problem to multisets (or monomials) instead of sets may be fruitful since there is a proposed suggestion for an answer.

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### 104 Responses to Polymath 3: The Polynomial Hirsch Conjecture 2

1. Terence Tao says:

To summarise some of the discussion from the previous thread, I think it would be a good idea to focus a fair bit of effort on understanding f^*(3,n), defined as the largest number t of families one can have consisting of 3-element multisets {a,b,c} obeying the convexity condition. We have two basic examples that illustrate that this quantity is at least 3n-2:

* Nicolai’s (and Klas’s) example $F_i := \{ \{ \lfloor \frac{i+2}{3} \rfloor, \lfloor \frac{i+3}{3} \rfloor, \lfloor \frac{i+4}{3} \rfloor \} \}$ for i=1,…,3n-2

* My example $F_i := \{ \{a,b,c\}: a+b+c = i+2 \}$.

Perhaps there is a one-parameter family of intermediate examples that interpolate between these two. But there does not yet seem to be any example that breaks the 3n-2 barrier; perhaps we can start working out f^*(3,n) for small values of n (From Nicolai’s first comment it appears that f^*(3,n) = 3n-2 has already been verified up to n=8).

On the other hand, the [AHRR] argument, a version of which is on the wiki, gives something like $f^*(3,n) \leq 4n-1$ (there seems to be a bit of scope to improve the -1 constant term though). The basic idea is to try to greedily break up the families into blocks with a common element and with a fairly small total support, so that one can exploit the existing bound $f^*(2,n) = 2n-1$. It would be nice to see if there were other arguments that also gave a linear bound on $f^*(3,n)$; the argument based on using m-restrictions (to use Tim’s notation) seems to lose a lot of information somehow.

I also like Tim’s idea of trying to exploit using the 2-shadows of F_i (the set of pairs {a,b} that are contained in one of the sets in F_i). All our arguments are currently relying only on the 1-shadow U_i and so one can hope that the 2-shadows are somehow carrying more information.

• Ryan O'Donnell says:

Although investigating f*(3,n) indeed looks like it will be more fruitful, I also still like the simpler question raised by Klas: what is f(2,n)? (I.e., what if we really insist that each F_i is a (non-multi-)set of cardinality 2?) The answer is somewhere between (3/2)n and 2n.

• Terence Tao says:

I conjecture the answer here to be f(2,n) = 2n – O(log n). My construction starts from my multiset example, say for n=10:

F_1 = { {1,1} }
F_2 = { {1,2} }
F_3 = { {2,2}, {1,3} }
F_4 = { {2,3}, {1,4} }

F_9 = { {5,5}, {4,6}, …, {1,10} }
F_10 = { {5,6}, {4,7}, …, {2,10} }

F_18 = { {9,10} }
F_19 = { {10,10} }

basically, what one has to do here is to replace the degenerate multisets {1,1}, {2,2}, …, {10,10} by genuine multisets. I can do this by adding a few more numbers 11, 12, 13, etc. to the domain. For instance, I can replace {1,1} in F_1 by {1,11}. Then I can replace {2,2} in F_3 by {2,12}, {3,3} in F_5 by {3,12}, {4,4} in F_7 by {4,12}, and {5,5} in F_9 by {5,12}, but I have to throw in some more sets in the intermediate families F_4, F_6,F_8 to maintain convexity, e.g. I could add {1,12} to F_4, {6,12} to F_6, and {7,12} to F_8. At this point I run out of room for 12 and have to move up to 13: I replace {6,6} in F_11 by {6,13}, {7,7} in F_13 by {7,13}, {8,8} in F_15 by {8,13}, and {9,9} in F_17 by {9,13}, throwing (say) {3,13} back into F_12, {5,13} back into F_14, and {10,13} back into F_16 to recover convexity. Finally I have to replace {10,10} in F_19 by {10,14}. This gives f(2,14) >= 19. I think in general, this procedure costs a logarithmic factor in n, but I have not worked out the details properly.

• It would be interesting to see if Terry’s construction can be made more systematic and what kind of bound it actually gives.

For N=14 it seems to give exactly the same bound as the 3/2n construction I gave so in order to really notice the 2n part of the conjecture larger values of n are needed.

• Terence Tao says:

The AHRR argument proceeds by greedily breaking up the t families F_1, …, F_t into blocks $F_{a_i},\ldots,F_{a_{i+1}-1}$, such that each of them are supported in a fairly small set $S_i$ (with $\sum_i |S_i| \leq 2n-1$, in fact, and such that the supports $U_{a_i},\ldots,U_{a_{i+1}-1}$ in that block have a common element $m_i$. In the d=3 case, this means that the length of each block is at most $f^*(2,|S_i|) = 2|S_i|-1$, leading to a total length of at most $\sum_i (2|S_i|-1)$. One can show easily that we may assume wlog that there are at least two blocks for n at least 2, and this leads to an upper bound of 4n-4 for f^*(2,n).

In Nicolai’s example, there are n-1 blocks, of the form

{{i,i,i}}, {{i,i,i+1}}, {{i,i+1,i+1}}

(so the common element here is i, and S_i = {i,i+1} has cardinality 2), plus a final block {{n,n,n}}. The bound on f^*(2,|S_i|) used here is tight; the loss is coming from all the -1’s in the sum $\sum_i (2|S_i|-1)$, because there are so many blocks.

In my example, there are two blocks, one in which the index j of F_j ranges from 1 to 2n-1 (with common element 1 and S_i = [n]), and one in which j ranges from 2n-2 to 3n-2 (with common element n and S_i = {2,…,n}). The bound on f^*(2,n) is tight for the first block but not for the second, which is where the loss comes from. One way to think about it is that the second block can in fact be extended to a much larger block (from j=n to j=3n-2) that still has a common element n, though the total support is now enlarged slightly to [n]). The big overlap between these two blocks can then be “blamed” for the loss from 4n-O(1) back down to 3n-O(1).

So to get close to 4n, one has to use very few blocks, and they have to not “overlap” each other too much. I don’t know how to control this rather fuzzy notion of “overlap” though.

• Terence Tao says:

To put it another way: at present it is conceivable that f^*(3,n) is equal to 4n-4 for some n, but this is only possible under a specific scenario: one in which the first 2n-1 families F_1, …, F_{2n-1} share a common element (say 1) in their supports, while the remaining 2n-3 families F_{2n}, …, F_{4n-4} do not have 1 in their supports (which are then contained in {2,..,n}) but have another common element (say n) in their supports. Then the two blocks F_1,…,F_{2n-1} and F_{2n},…,F_{4n-4} both saturate the bound f^*(2,n) = 2n-1 without contradiction. Nevertheless, this situation should presumably not be possible, somehow because the two blocks are “competing” for the “same resources” in some way.

2. Terence Tao says:

Regarding random constructions, one idea I had was to write things in the dual format that was mentioned in [AHRR]. Namely, for every subset A of [n], consider the (possibly empty) interval I_A of indices i such that A is contained in one of the sets in F_i. (The convexity condition ensures that this I_A is indeed an interval.) The various properties of the F_i can then be reformulated into a single axiom:

* (Layer axiom) For any A, $\bigcup_{B \supsetneq A} I_B$ consists of $I_A$ with at most one element removed.

The question is then to maximise the cardinality t of $\bigcup_{A \subset [n]} I_A$.

Anyway, with this formulation it is tempting to find some semi-random way to select the intervals I_A, maybe starting from small A first and then working upwards. I haven’t tried this seriously though.

3. gowers says:

I was just thinking about lower bounds and had an idea for a construction. Thinking what it should give, I thought it ought to give a quadratic lower bound. Because of that I thought I should look at [AHRR] more carefully, and sure enough my construction wasn’t new. In fact, given that their construction is more complicated than mine but based on almost identical ideas, my guess is that mine isn’t even completely correct.

But for that very reason it seems worth briefly sketching it, either to find a slight simplification or (far more likely) to show why they did what they did.

The underlying thought is this. Suppose you have a family F of d-sets such that every (d-1)-set is contained in a set in F. These, I see from [AHRR], are called (d-1,d,n)-covering designs. Then given any two disjoint families G, H that consist of sets of size at most d, the triple (G,F,H) satisfies condition (*), since the intersection of two distinct sets of size at most d has size at most d-1. It follows that if we have a sequence $F_1,F_2,\dots,F_m$ of disjoint families of sets such that for each $i$ there exists $d(i)$ such that $F_i$ is a $(d(i)-1,d(i),n)$-covering design …

… ah, I’ve found my mistake. I was going to say that if the $d(i)$ are non-decreasing then the sequence would be connected. I had even checked this, but I had been careless, since I had forgotten the case where $F_i$ and $F_j$ consist of sets of size $d$ and $F_k$ of sets of a bigger size. Then we can pick a set $A\in F_i$ and a set $C\supset A$ from $F_k,$ and $A\cap C=A$ will not be contained in any set in $F_j.$ So now let me go back to AHRR and try to understand how they deal with this apparently major difficulty (major because if later sets have to avoid containing earlier sets then it seems to kill the idea completely when you have different d around).

• gowers says:

From a brief glance it looks as though the basic idea is this. I have tried to build what one might call an “increasing” sequence of families, and that has created the problem. They build an “increasing” family and tie it together with a “decreasing” family, at only small cost to the bound. Then if you have a triple of sizes $c,c,d$ in the increasing family, which is the problematic case, the sizes in the decreasing family will be $n-c,n-c,n-d,$ which is not a problematic case. There are few details to check of course, but I think that is the basic idea. The tying process is simply this: you take two disjoint ground sets, put an increasing family $F_1,\dots,F_m$ in one, a decreasing family $G_1,\dots,G_m$ in the other, and then take $H_i$ to be the set of all unions of a set in $F_i$ with a set in $G_i.$

• gowers says:

Oops, no, that’s not it, since it doesn’t actually deal with the difficulty. There’s still something clever going on that I haven’t grasped. I think I’d better stop this sequence of comments and keep the rest of my flounderings to myself.

• Another comment on AHRR. They prove that the largest number of disjoint (n,r+1,r)-covering designs is at most n-r, and at least n-r(r+2).

For r=2 the upper bound is tight for large n since there exist sets of n-2 disjoint steiner triple systems, assuming that n is in the right class modulo 6, as proved in http://ams.math.uni-bielefeld.de/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=RT&pg7=ALLF&pg8=ET&review_format=html&s4=&s5=disjoint%20steiner%20triple&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=10&mx-pid=692824 and various papers finishing some remaining cases.

4. In this https://gilkalai.wordpress.com/2010/09/29/polymath-3-polynomial-hirsch-conjecture/#comment-3453 comment Tim was considering constructions for large families with d=2. If the graph used in the construction are all matchings this would lead to an “interval edge-colouring”, introduced by Asratyan in the late 80s. In these colourings the colours on the edges incident with any vertex must form an interval, and they must form a proper edge colouring.

For such colourings it is known that the number of colour classes can’t be more than 2n-3, see e.g. http://arxiv.org/abs/1007.1717v2

5. Nicolai Hähnle says:

As for f(2,n), you can get at least $2n - O(\sqrt{n})$ by slightly extending the construction in our paper. The high-level view of our construction for d=2 is:

B(A)
M(A,1;B,1)
B(B)

Here, A and B are sets of size m and n = 2m. The number of families in the B-blocks is m-c, the number of families in the M-block is m, which gives the roughly 3n/2.

Now, extend the construction to

B(A)
M(A,1;B,1)
B(B)
M(B,1;C,1)
B(C)
….
B(Z)

Taking k sets of size m for a total n=km, you get $k(m-c) + (k-1)m$ families. Setting k=m gives $2n - O(k+m)$ families.

I have to think about Terry’s approach of starting with the multiset construction and then nuking the “degenerates”.

If I recall correctly, a similar argument also gives $3n – O(\sqrt{n})$ for the case d=3 for the set case.

• While it would be interesting to see if the root term could be reduced I think these constructions put even more weight on Terry’s question regarding the upper bound for $f^*(3,n)$.

I’ll sit down later today and try to digest the EHR upper bound a bit better.

6. Nicolai Hähnle says:

Third and last attempt to get a comment through: For d=2, you can get 2n-O(sqrt(n)) by extending our construction. This is merely a short test comment, because my previous comments seem to have disappeared.

• Nicolai Hähnle says:

Okay, now we’re getting somewhere. The idea is as follows. In our paper we have a high-level construction like this:

B(A)
M(A,1;B,1)
B(B)

The B-blocks are called bridges because you can extend this construction.

• Nicolai Hähnle says:

Our construction in the paper is this:

Bridge(A)
Mesh(A,B)
Bridge(B)

This can be extended in a straightforward way to add more mesh and bridge blocks:

Mesh(B,C)
Bridge(C)

etc.
If you take k sets of m elements, you end up with k(m-c) + (k-1)m families (the first summand for the bridge blocks, the second summand for the mesh blocks). Set k=m and you get the 2n minus a square root term.

If I recall correctly, analogous constructions can be done for any fixed d.

7. gowers says:

It still feels to me as though the elementary argument can be squeezed to give more. An initial target would be to get a bound of $n^{c\log n}$ for a smaller constant $c,$ and then the hope would be to iterate the improvement to get a bound of a genuinely different type. I’m writing this comment not because I can actually do this, but because I want either to persuade people that it is possible or to understand why it is harder than it at first looks.

The basic idea is this. In the elementary argument we divide up the families into three subintervals as follows. We take the largest initial segment $F_1,\dots,F_r$ such that all families are supported in a set of size $n/2$ and note that $r$ is at most $f(n/2).$ We also take the largest final segment $F_{s+1},\dots,F_N$ such that all families are supported in a set of size $n/2$ and note that $N-s\leq f(n/2).$ Then the union of the supports up to $F_{r+1}$ intersects the union of the supports after $F_s,$ so there must be some element common to the supports of all of $F_{r+1},\dots,F_s,$ which leads to the bound of $f(n-1)$ for $s-r.$

Now let us think about the supports of $F_1,\dots,F_r,$ which I shall call $U_1,\dots,U_r.$ Suppose that $U_i$ contains an element that is not contained in a later $U_j.$ Then that element is banned from all $F_k$ beyond $j.$ But that means that one of two possibilities holds. Either the sets $U_1,\dots,U_r$ form an increasing sequence, in which case it is easy to find an element common to all the sets and do the next stage of the iteration far more efficiently, or there is an element that is banned from all sets after $U_r,$ in which case it seems as though we can improve the bound for the middle section from $f(n-1)$ to $f(n-2).$

This suggests to me that either I am making a stupid mistake or we can put a bit more effort into the elementary argument (note that if the above reasoning is correct then it is just the start of what one could do) and get a stronger bound.

• gowers says:

I think I may have seen my mistake. The trouble is that if we know that every intermediate family includes an element $m$ and we also know that every intermediate family includes a second element $r,$ we can’t necessarily restrict to sets that contain both $m$ and $r$ because the sets that contain $r$ could be different from the sets that contain $m.$ So we don’t get an easy improvement from $f(n-1)$ to $f(n-2).$

However, I still can’t rid myself of the feeling that more can be got out if we try to push the elementary argument.

• gowers says:

Oops, my mistake has a mistake. The improvement from f(n-1) to f(n-2) came from banning an element rather than insisting on it, so it looks as though it could be OK after all. I’ll think about this properly in about two hours’ time.

8. gowers says:

Just before I start this comment I want to make a quick meta-comment. I’m treating this polymath project as I treat ones that have taken place on my blog, in the sense that I am spilling out a lot of half-baked thoughts, or even thinking in real time as I write, rather than spending an hour or two with a piece of paper and reporting back if I find anything interesting. But I don’t mind modifying this practice if anyone (and particularly Gil) would rather I wrote less and thought harder first.

But in the absence of such an instruction for the time being, I’m going to make a preliminary attempt to improve the elementary argument. The basic idea will be to consider not just one iteration but two and see what happens. (A few minutes ago I did in fact sketch something out on a piece of paper over a cup of coffee, and my impression then was that I could save a factor of n, which is not that interesting on its own but it is vaguely promising.)

As in the elementary argument, and after permuting the ground set if necessary, we can assume that the sequence of families is divided into three intervals, one consisting of families that live in $[n/2],$ one of families that all use the element $m=n/2+1,$ and one of families that live in $[n/2+2,n]$ (or something along those lines — I don’t want to get too involved in parity considerations).

Let’s now consider the first interval $F_1,\dots,F_s$ of families. We know that $s\leq f(n/2),$ but now I want to say something more detailed. To do that I would like to divide into two cases. Let $U_i$ be the support of $F_i.$ In the first case let us assume that the $U_i$ up to $i=s$ are not nested upwards. In that case, some $U_i$ contains an element $a$ that is not contained in a later $U_j,$ From that it follows that $a$ can never be used again, so the number of families in the middle interval is at most $f(n-2)$ (as we see by restricting to sets that contain $m$ and exploiting the fact that they do not contain $a$). So in that case we get a bound of at most $2f(n/2)+f(n-2).$

In the second case, the $U_i$ are nested upwards in the initial interval and downwards in the final interval (or else we can run the previous argument but reflected in $n/2$). That means that any element used in $F_1$ is used in all of $F_2,\dots,F_m,$ so the number of families is at most $f(n/2-1)$ by the restriction argument. So in this case we get a bound of at most $2f(n/2-1)+f(n-1).$

Which of these two bounds is bigger? That’s equivalent to asking which is bigger out of $2(f(n/2)-f(n/2-1))$ and $f(n-1)-f(n-2).$ Assuming that $f$ grows reasonably fast, as we may, the second is much bigger, which implies that $2f(n/2-1)+f(n-1)$ is a new upper bound.

What effect does that very slightly improved recurrence have on the bound for $f$? One way of trying to estimate this is to apply the iteration to the inner $f(n-1)$ to get an upper bound of $2f((n-1)/2-1)+f(n-2),$ and again and again until the middle interval has got down to size $n/2.$ That gives us exactly what we would have got with the original iteration except that all the arguments have had 1 subtracted. Now if $f(n)$ is in the general ball park of $n^{\log_2n},$ then the difference we make by subtracting 1 from n is a ratio of around $(1+1/n)^{\log_2n},$ which is not that wonderful. In fact, it’s not that wonderful even when we iterate the observation, which we get to do about $\log_2n$ times. Indeed, it seems to give an improvement by a constant factor (which comes almost entirely from right at the beginning of the induction).

However, there is more to say, which I’ll say in a new comment.

• Gil Kalai says:

Dear Tim, regarding the meta-comment. Writing many half-baked thoughts even in real time is most welcome. (I am a bit slow contributing this time but I intend to catch up.) Other forms of operation are welcome as well.

9. gowers says:

The extra that one can say — and I don’t know how significant it is — is that if the above argument can’t be improved then some pretty strong things can be said about the system. Suppose we have that the supports of the first $r$ families form a non-decreasing sequence with $U_r=[n/2].$ There are all sorts of ways that we might try to improve on the obvious bound of $f(n/2)-1$ for $r.$ To discuss this, let me assume WLOG that the $U_i$ are initial segments. Then instead of using the fact that $1\in U_i$ for every $i,$ we could use the fact that all but at most $f(t-1)$ families use the element $t.$ If for no $t$ does that give rise to a substantial improvement, then we know that several $t$-restrictions fail to be monotone.

If a $t$-restriction fails to be monotone, it tells us that we can find $i such that $at\in F_i\setminus F_j,$ which bans $at$ from the lower shadow of all subsequent families. However, this is not yet a very strong condition because we have only a linear number of 2-sets that we are able to ban. Or at least I think that is the case.

I think I need to do a bit of thinking away from a keyboard, but one problem that might be worth considering is what the effect is on the bound if we have a ground set of the form $A\cup B$ with $A$ and $B$ disjoint sets of the same size, and we have a matching $\phi$ between $A$ and $B,$ and we ban all sets that have a subset of the form $\{a,\phi(a)\}.$

10. Pingback: Polymath3 « Euclidean Ramsey Theory

11. Gil Kalai says:

I have a sort of meta comment: I presented the problem in a fairly “naked” way because I thought this will be most effective. If there are connections, equivalent formulations, or other things that I should elaborate on please let me know. (Often, I find it hard to juggle between equivalent formulations no matter how simple the equivalence are…) I do intend to repeat the simple connection between the formulation in terms of diameter of certain graphs and the formulation in terms of families because maybe it is useful to be familiar with both. Any other advice regarding running the project is most welcome.

• gowers says:

I rather like the naked formulation myself, because if I understood better how the combinatorial problem implied results about polytopes I think I’d be more afraid of thinking about it …

• Gil Kalai says:

Let me repeat the equivalence between the two EHRR formulations: One formulation was this:

You have a graph G so that every vertex v is labeled by a subset S(v) of {1,2,…,n} (We can further require that |S(v)|=d or that |S(v)|<=d.) We assume that all these subsets are distinct and that
(**) for every two vertices v and u there is a path so that all vertices w in the path satisfy $S(v)\cap S(u)\subset S(v)$.

f(n) (respectively f(d,n)) is the maximum diameter of this graph.

Starting with such a graph G and a vertex u we obtain families ${\cal F}_1,{\cal F}_2,\dots$ by letting ${\cal F}_i$ to be all the sets of labeles of vertices of $G$ which are of distance i+1 from u.

The condition (*) is satisfied because when you consider a set S(v) in ${\cal F}_i$ and S(w) in ${\cal F}_k$ then the path between v and w which satisfies (**) will necessary have a representative in ${\cal F}_j$ for every j strictly between i and k.

In the other direction, if you have families you can just associate to every set in every family a vertex and connect all vertices for the ith family to all vertices in the (i+1)th family.

• Eddie Kim says:

Gil,

With your express invitation to write out equivalent formulations, even if the equivalence is quite silly. Here is an equivalent formulation of the number of disjoint covering designs problem:

Color the (k+1)-faces of an (n+1)-simplex so that each k-face has all colors. How many colors are possible?

I find it a little easier to think of it in the polar formulation:

Color the (n-k-1)-faces of an (n+1)-simplex so that each (n-k)-face has all colors. How many colors are possible?

Question: is there a color assignment function that can be defined on each facet of the simplex, in such a way that it is well defined over the entire simplex?

12. Terence Tao says:

It may be good to have a “non-inductive” proof of f(2,n) <= 2n-1 (or f^*(2,n) <= 2n-1), as this may shed more light on how to bound f^*(3,n) accurately. The main thing here, I think, is to get the main term 2n; to begin with, one can be sloppy with the -1 error, viewing it as a lower order term.

The idea I had in this regard was to try to "factor" the problem of constructing families F_1,…,F_t of 2-element sets (or multi-sets) in to that of constructing the 1-shadows (aka supports) U_1,…,U_t first, and then selecting the 2-shadows (which, in this d=2 setting, are just F_1,…,F_t).

One obvious constraint on the U_i is convexity, which is a simple enough condition. Another constraint, if $U_a$ and $U_b$ have a common element m, then we know that $b-a+1 \leq |\bigcup_{a \leq i \leq b} U_i|$ from the $f^*(1,n)=n$ bound. Again, this looks like an easy condition to understand.

The trick is, once one fixes the 1-shadows U_i, how to fill in the 2-shadows. Here, there are two constraints. Firstly, no edge can be used more than once. Secondly, each F_i actually has to cover all of its support U_i.

This is sort of a packing problem. One can try to abstract away the mechanics of how one actually does the packing, by instead writing down basic bounds on the F_i that come from volume considerations. For instance, since the F_i have to cover U_i, we have

$|F_i| \geq |U_i|/2$

but then since they have to be disjoint, we have

$\sum_{a \leq i \leq b} |F_i| \leq \frac{1}{2} |\bigcup_{a \leq i \leq b} U_i|^2$.

Perhaps this (together with the information on the shadows U_i) is already enough to get a bound of $f^*(2,n) \leq 2n+O(1)$?

If we understand how this works then perhaps we can ramp it up to d=3, looking at how the 1-shadows, 2-shadows, and 3-shadows interact with each other.

p.s. I'll be traveling for the next few days and will probably post quite little to this thread for a while.

• Terence Tao says:

I should perhaps clarify: the [AHRR] argument gets $f^*(2,n) \leq 2n-1$ purely by using the information on the supports U_1,…,U_t, namely the convexity and the bound $b-a+1 \leq |\bigcup_{a \leq i \leq b} U_i|$ when $U_a \cap U_b \neq \emptyset$, using a covering argument (which, coincidentally, bears some resemblance to the Besicovitch covering lemma). But I’m hoping that some of the 2-shadow information can help reduce the dependence on these 1-shadow facts, because if one repeats the 1-shadow argument in d=3 one only gets the 4n-O(1) bound rather than the 3n-O(1) bound that we are shooting for.

• One thought on the structure of the $F_i$ for the d=2 case.

If we view each $F_i$ as a graph with vertex $U_i$ I believe that we can assume that no connected component contains a path on three edges. If it did we could delete the middle edge of the path without violating any of the conditions.

So every component of $F_i$ is either an edge or a star, which means that $|U_i|/2 \leq |F_i| \leq |U_i|-1$

• In general we should be able to delete any set in $F_i$ such that all its $(d-1)$-subsets are contained in other members of $F_i$.

13. Gil Kalai says:

Here are some meta-thoughts and later some thoughts of the problem.

1) The ideal mode of polymath in my eyes would allow a large number of people (Ok, by large I probably mean a few dozens) to follow a project comfortably along with their main research projects and other duties. I realize that in the successful polymath1 the mode of participation was more intensive (and, of course, since I truly want this problem get settled I would be happy to see such an intensive paste over here,) but in order to have useful projects of this kind, perhaps a few in parallel, it should be in a paste, and form (and in a software platform ), which allow people to follow it along with their main activities, and perhaps even to jump in and out.

2) Perhaps the most difficult thing in such a project is to take a step initiated by one participant to push it a little forward, preparing it for more participants . Perhaps this is a bit like in football, if you have to receive the ball from another player and than to dribble with it and move it to another player this is sort of complicated. (Here I mean European football, aka soccer).

3) As the host, I will be happy to do anything which you may think can be useful to the project. (Unfortunately, serving pizza or coffee is not supported by wordpress at this time).

14. Gil Kalai says:

1) I really like the plan to try to understand f*(3,n) (or f(3,n)). (Without neglecting also f(2,n)). I never thought about this avenue, but in hindsight it looks good.

There is a sister problem to PHC which is about effective (randomized) pivot rules for LP problems with d variables and n variables. (And various abstractions) and also there looking at fixed d (by a long list of people: Megiddo, Seidel, Klarkson, Sharir-Welzl, Matousek, Chazelle…) was an avaenue also for results when d is general.

2) What is really frustrating about the reccurence f(d,n)<=2f(d,n/2)+f(d-1,n-1) is that intuitively we may think that when we reach the common 'm' and delete it we dont have to start all over again; but rather the problem is already warm. Moreover by replacing f(d,n/2) by f(d,2n/3) we can reach one out of n/3 m's at our choice. We need an idea how to gain something inside the reccursion.

3) One idea I tried to play with (for which I need the graph with labelled set formulation) is this. Suppose that you give weights w_1,…,w_n to the vertices. And you give weights to sets which is the product of the weights of the vertices. And that when you compute the diameter you count every vertex by the weight of its label. It looks that maybe we can get a reccurence relation to f(d,w_1,…,w_n) the upper bound of such weighted diameter. I do not see how it can help us though.

4) trying to find examles which somehow behave as badly as the reccurence, perhaps with some randomization is certainly something to be tried.

5) I wonder for what class of graded lattices the upper bound argument work. It works for sets; it works perhaps for atomic graded lattices in which every interval is atomic; but somehow monomials are not such lattices so it may eextend further. (This is a side question except for the desperate thought that if we indeed the scope of the inequality sufficiently we may make it sharp.)

6) There are some pleasant inequalities for f(d,n) when n is a constant tines d. This suggests that the really bad case may be when n is a power of d.

Off topic: Shouldn’t [AHRR] be [EHRR]? {GK: right.}

16. I do not know if this is of any use, but there is a way to encode the problem that makes the convexity condition simple (but mess up the disjointness). Simply consider for a subset $A$ of $1,\ldots,n$ the sets $I(A)$ of indices $i$ such that there is a $B\in\mathcal{F}_i$ containing $A$. Then the convexity translates to the $I(A)$ being intervals, while the disjointness can be translated into the property

(1) that the set $i(A)$ of indices appearing in $I(A)$ but not in any $I(A’)$ where $A’$ runs over the successors of $A$, is empty or a singleton.

Of course, the definition of $I(A)$ make $I$ a decreasing map (with respect to inclusion) and the data of a decreasing map $I$ satisfying (1) defines a unique convex family of subsets.

• Here is an illustration of the preceding formulation. I propose to recover the fact that if {1,…,n} is in some convex family, then this family has cardinal at most 2n.

Under the hypothesis, we have that I(1,2,…, n) is some singleton (the alternative would be that it is empty). But then, all the I(A) with A of cardinal n-1 must be concentrated in a size 3 interval (each one of them have at most one point that is not in I(1…n) ). An immediate induction shows that the I(A) with card A=n-k are concentrated on an interval of size 1+2k. Last, $I(\emptyset)$ must be concentrated on an interval of size at most 2n.

• We can also get the same kind of information on the lower lever. Assuming that I(1,…,n) is empty, I claim that there are at most 3 sets A of cardinal n-1 such that I(A) is not empty.

Such sets satisfy that I(A) is a singleton, and if there were at least 4 of them, there would be 2 such sets A,B whose indices differ by at least 3. Let C be the intersection of A and B. Then I(C) contains I(A) and I(B), so that it has cardinal at least 4. But A, B and {1,…,n} are the only successors of C, a contradiction with the disjointness condition.

• gowers says:

Is that basically the dual formulation mentioned by Terry above?

17. gowers says:

I just wanted to mention a slight modification to how one can describe the problem. It isn’t enough of a modification to qualify as a reformulation, since it is too trivially equivalent to the original problem, but it is a slight change of language that I find I prefer. (Also, it is the kind of observation that anyone who has thought seriously about the problem will be aware of already I’m sure. This comment is therefore mainly for the benefit of people who are dipping into the discussion and hoping to follow some of it.)

Recall that a \textit{down-set} is a collection of subsets of $[n]$ that is closed under taking subsets. We would like to find an upper bound for the length of a sequence of down-sets $D_1,D_2,\dots,D_m$ if it satisfies the following two conditions.

1. For every set $A\subset[n]$ the set of $i$ such that $A\in D_i$ is an interval.

This can be rephrased as follows, which I find useful because it tells us when certain sets are banned from all future $D_k.$

1′. If $i and $A\in D_i\setminus D_j,$ then $A\notin D_k.$

2. No two of the $D_i$ share a maximal element.

Let me prove that this is equivalent to the original formulation. Given a sequence of families $F_i$ satisfying the disjointness and convexity conditions, let $D_i$ be the down-set generated by $F_i$ (that is, the collection of all subsets of sets in $F_i.$ Then every maximal element of $D_i$ is an element of $F_i,$ so the disjointness condition for the $F_i$ implies the disjointness condition for the $D_i.$ Also, if $i and $E\in D_i\cap D_k,$ then we can find sets $A\in F_i$ and $C\in F_k$ that contain $E.$ It follows from the convexity condition that we can find a set $B\in F_j$ that contains $A\cap C$ and hence $E.$ Therefore $E\in D_j.$

In the other direction, given the down-sets $D_i$ we can define $F_i$ to be the set of maximal elements of $D_i.$ Then we get the disjointness condition trivially. Also, if $i and $A\in F_i$ and $C\in F_k,$ then $A\cap C\in D_i\cap D_k$ and hence, by condition 1, $A\cap C\in D_j,$ which implies that there is some set in $F_j$ that contains $A\cap C.$

Now why should this formulation be even slightly advantageous? I don’t know if I yet have a precise answer to this — it’s more like a hunch that it encourages one to think about the problem in “the correct” way. One aspect of it that makes me feel more optimistic is that down-sets are quite structured objects, so it somehow feels more plausible that it should be hard to create sequences of down-sets with constraints.

What are these constraints? Well, as we define a longer and longer sequence of down-sets, we find that more and more sets become forbidden. And building a down-set that is forced to avoid a whole lot of sets can be quite hard.

Let me try to justify the first of those two statements in a small way. Suppose we tried to be very extreme and produce a sequence of down-sets with each one contained in the next, but still satisfying the distincnt-maximal-elements condition. Then each maximal element in $D_1$ would have to be contained in a strictly larger maximal element in $D_2,$ and so on. So trivially we can’t have an increasing sequence of down-sets satisfying the disjointness condition of length greater than $n+1.$

And now let me think very briefly about the second condition by considering what effect it would have if we knew that our down-sets were not allowed to contain the set $12.$ That means that WLOG the supports start off not containing 1 or 2, then they contain 1, then they contain neither, then they contain 2, and finally they contain neither again. (Some of these intervals could be empty.) This isn’t getting me anywhere.

One final brief and speculative thought. Perhaps the problem would yield to some kind of very clever induction that takes into account far more than just the size of the ground set. For example, we might try to generalize the problem vastly by giving an upper bound for the maximum length $f(n,\mathcal{A})$ of a sequence of down-sets satisfying the given condition, given that all sets in $\mathcal{A}$ are forbidden. And perhaps we could express that in terms of some other pairs (integer, set-system) that are in some sense simpler.

• gowers says:

Sorry, a better way of putting that very last point is to fix $n$ once and for all and just do an induction on the set systems $\mathcal{A}.$ Then $f(n-1)$ is just the length of the longest sequence of down-sets when the set $\{n\}$ is banned. If we define $f(\mathcal{A})$ to be the length of the longest sequence when no sets in $\mathcal{A}$ are allowed, then the hope would be to show that $f(\mathcal{A})$ can be bounded above in terms of the values at $f(\mathcal{B}_i)$ for some “worse” set-systems $\mathcal{B}_i,$ where that means that they contain more sets, or smaller sets, or something like that. (I think what we really want is that they have bigger upper shadows.)

• gowers says:

Let me have another go at saying, still in vague terms, what it might be possible to do.

For each $i$ and each $A\subset[n]$ let $D_i(A)$ be the trace of $D_i$ on $A,$ that is, the collection of all sets $A\cap B$ such that $B\in D_i.$ Then $D_i(A)$ is of course still a down-set. Also, if the down-sets $D_i$ satisfy the convexity condition (by which I mean conditions 1 and 1′ above) then so do the down-sets $D_i(A).$ This follows trivially from the fact that $D_i(A)$ is the intersection of $D_i$ with the power-set of $A.$

A simple consequence of this is that if $A\in D_i,$ then the sets $D_j(A)$ for $j>i$ are monotone decreasing. Indeed, if $j and $C\in D_k(A),$ then since $C\in D_i(A)$ (as $D_i$ is a down-set) it follows that $C\in D_j(A)$ (by the convexity condition).

Now we know that monotone families of down-sets can’t be too big, so it seems at first as though this “downward pressure” that $D_i$ places on the subsequent down-sets should be rather strong. Unfortunately, that doesn’t follow because we lose the distinct-maximal-elements condition.

However, there may be some pressure in this direction too. After all, the number of families that contain a given set $B$ is at most $f(n-|B|),$ which is pretty small if $B$ is at all large and $f$ grows superpolynomially. So even if we don’t have a distinct-maximal-elements condition we might hope for a don’t-use-maximal-elements-too-often condition. Whether anything like this can be developed into an argument that gives an interesting bound is of course quite another matter.

18. I think that f(4)=8. What follows is a case-by-case argument; I am not sure that it gives any real insight. I use the dual formulation notations.

First, the case when I(1234) is not empty is settled so we assume the contrary. Moreover, at most three of the I(abc) can be non empty. First consider the case when I(123), I(124), I(134) are non empty. Then by looking for example at the I(1a) we see that the 3 singletons are confined in an interval of length 3. It follows that the I(ab) are confined in an interval of length 5, that the I(a) are confined in an interval of length 7, and we are done because $I(\emptyset)$ contains at most one more element.

The second case is when exactly two I(abc) are non-empty, I(123) and I(124) say. If their values differ by 2 (which is maximum possible), then I(13) and I(23) are confined in an interval E of length 3, I(14) and I(24) are confined in an interval F of length 3, and F intersects E in only one point (between I(123) and I(124)). Now I(34) can be empty, in which case the I(ab) are confined in an interval $E\cup F$ of length 5, or I(34) is a singleton and we only have to show that it is in that same interval.
But if it where, e.g., on the right of this interval, then I(3) would have at least 2 points not lying in any of I(13), I(23), I(123) nor I(34), a contradiction. If I(123) and I(124) are adjacent or equal, then all of I(ab) where ab is not 34 are confined in a interval G of length 4, which is a union of two intervals of length 3, the first one containing I(13) and I(23), the second one I(14) and I(24). A similar argument than above proves that I(34) must be adjacent to G, so that once again the I(ab) are confined in an interval of length 5.

The third case is when exactly one of the I(abc) is not empty, I(123) say. Then I(12), I(23) and I(13) are confined in a interval E of length 3 around I(123), and I(14), I(24), I(34) are either empty or singletons. Looking at I(1) shows that (if not empty) I(14) is in the 2-neighborhood of E. But looking at I(4) shows that all non-empty I(a4) are at distance at most 2, so that all I(ab) are confined in some interval of length 5. Some of them could be empty, but in any case it is easily checked that $I(\emptyset)$ is confined in an interval of length 8.

For the fourth case, assume that all I(abc) are empty but some I(ab) are not. Any two I(ab), I(ac) must lie at distance at most 2 (look at I(a) ). If for example I(12) and I(34) where at distance 5 or more, then all other I(ab) would be empty (otherwise they should be close to both I(12) and I(34)), so that I(1), I(2) would be confined in an interval E and I(3), I(4) to an interval F, such that E and F are separated by a gap of length 2. We then get a contradiction by looking at $I(\emptyset)$. We conclude that all non-empty I(ab) are confined in a length 5 interval if all 1234 are represented, or confined in an interval of length 3 otherwise. In both cases, we get the desired conclusion.

Last, consider the case when all I(ab) are empty. Then we have at most 4 points covered by the I(a), and at most 5 by $I(\emptyset)$, and we are (finally) done.

• I moved this to the wiki.

• Paco Santos says:

The sentence “at most three of the I(abc) can be non empty”, has the following counter-example:
0
1
12
123
124,134,23
234
34
4
… but the conclusion f(4)=8 is still correct since the sentence “all the I(abc) are confined in an interval of length 3” holds when the four I(abc) are non-empty.

• You’re right, I have been a little sloppy here. Thanks for the correction.

• Paco Santos says:

I have rewritten the section “on small n” in the wiki. Apart of showing that f(n)=2n for n=1,2,3,4, I included the remark that the function f'(n) (the one obtained when every F_i is a singleton) is always 2n.

19. (continuing with the dual notation introduced earlier). If we define a summit as a subset A of [n] such that I(A) is a singleton and no superset of A has this property, then we get immediately that the I(B) where B is a subset of A are all confined in an interval of length at most 2|A| centered at I(A).

Moreover, if B is another summit intersecting A, then the singletons I(A) and I(B) are $|A\cup B|-|A\cap B|$-close, since $I(A\cap B)$ is confined by both A and B. Unless I am mistaken, we can deduce that f(n) is bounded above by $\max \sum |A_{i+1}\cup A_i|-|A_{i+1}\cap A_i|$ where the max is on all antichains $A_1,\ldots, A_k$ such that $A_{i+2}\cap A_i=\emptyset$ for all i (simply consider the left-wise summit, then reading right the last summit not disjoint to him if it exists, otherwise the next summit whatever it is).

• If the bound on the distance between two intersecting summits where correct, we could deduce a very good bound (quadratic or linear) on f(n), but it is not. The summits cannot be used to confine the I(B) so easily, so the argument crumbles.

• I tend to think that there is something to get from looking at the summits. The case when 12…n appears in one of the families should be somehow generalized.

Maybe I should translate the definition of a summit I gave above in the main definition: it is simply a subset A of [n] that appears in one of the families, but such no superset of A does. In other words, they simply are the maximal sets appearing in the families.

One way one could try something, would be to design a procedure that takes two adjacent summits (appearing in families $\mathcal{F}_i$ and $\mathcal{F}_j$ such that no $\mathcal{F}_k$ with i < k < j contains a summit) and modify the families so that their union becomes a summit, without decreasing (two much) the number of families.

Let me give a very weak incentive to do that. If there is only one summit then we are done (by which I mean that m is at most 2n), since by definition all other sets appearing in the families are subsets of this one, and all goes as if 12…n appeared.

I also feel that when there are only two summits A and B, we can still prove that m is at most 2n by controlling the distance between their indices, and add to the intermediate families between them a sequence of sets going up and down from A to $A\cup B$ to B, without loosing the convexity condition.

But maybe this is a bit speculative.

20. Steve Klee says:

Hi everyone,

I made this comment on the previous instantiation of the blog, so I just wanted to add it again here. I apologize for spamming if this is a stupid idea, but I feel like it is applicable to the d=2 case that has been the subject of much discussion of late.

My thought is that the set families that have been defined describe the facets of a simplicial complex with the property that all of its faces have connected links. But really, in studying the Hirsch conjecture, we are dealing with simplicial polytopes, which have tons of additional structure. With this thought in mind, suppose we add the following condition on our collection of set families F_i: (I hope the tex-ing works, I don’t really know how to make the blog use math mode)

(**) Each (d-1)-element subset of [n] is active on at most two of the families F_i.

Here, I mean that a set S is active on F_i if there is a d-set T \in F_i that contains S

This condition must be satisfied for polytopes — any codimension 1 face of a simplicial polytope is contained in exactly two facets.

So how does this help us in the case that d=2? Suppose F_1,\ldots,F_t is a family of disjoint 2-sets on ground set [n] that satisfies Gil’s property (*) and the above property (**)

Let’s count the number of pairs (j,k) \in [n] \times [t] for which j is active on F_k. Each j \in [n] is active on at most two of the families F_i, and hence the number of such pairs is at most 2n. On the other hand, since each family F_k is nonempty, there are at least two vertices j \in [n] that are active on each family F_k. Thus the number of such ordered pairs is at least 2t.

Thus t \leq n, giving an upper bound of f(n,2) \leq n when we assume the additional condition (**).

This technique doesn’t seem to generalize to higher dimensions without some sort of analysis of how the sizes of the set families F_i grow as i ranges from 1 to t.

• For d=3 I think you have a lower bound of at least 5/3n for this version.

Partition 1…n into disjoint triples {1,2,3}, {4,5,6},….

Create a complete 3-partite 3-uniform hypergraph H with vertex classes (1,4)(2,5)(3,6)
Partition the edge set of this hypergraph into 4 perfect matchings. Each class of H has only two vertices so each 2-tuple will be in exactly two of the prefect matchings.
Use each matching as an F_i between {1,2,3}{4,5,6}

Repeat this for each consecutive pair of triples in the first list.

This gives 5(n/3-1)+1 families.

• Actually you have to delete one of the matchings in order to have a correct family, so the bound is 4/3n instead of 5/3n

• Paco Santos says:

I do not quite understand this construction. If I am not wrong, your “complete hypergraph” is the set of faces of an octahedron, the (1,4) (2,5) (3,6) being the pairs of opposite vertices. Similarly, your “four matchings” are the pairs of opposite faces of the octahedron. In particular, {1,2,3}{4,5,6} is one of them.

What do you mean then by “Use each matching as an F_i between {1,2,3}{4,5,6}”?

• Paco Santos says:

… sorry, I had not read Klas’ and Nikolai’s posts below …

• Gil Kalai says:

Dear Steve

Your comment is certainly very good and thanks for repeating it. You are correct that we consider a very abstract form of the problem. Our setting is even more general than the class of simplicial complexes with all links connected (where we do not know anything better than a linear lower bound.) I am not sure what happens when you add (**). It is certainly a good question.

• Nicolai Hähnle says:

Klas, I don’t think your construction works as is. If you want to start with {123} and end with {456}, then what would your second family be? Wlog it could be something like {126,453}, but then already 12 and 45 are the only “surviving” pairs that you can use.

• Nicolai, there are many short examples for this e.g.
{{1,2,3}} {{1,2,4}} {{1,4,5}} {{4,5,6}}

I’m in a hurry right now but I can post the explicit example from my earlier post a little later today.

• Nicolai Hähnle says:

Thank you, that would be much appreciated.

• Nicolai you are quite correct. When I tired to write down my example properly I found that I had miscounted an intersection, it got cover three times instead of 2.

Using the example I gave above works, but that only gives the rather dull bound of n-2

• After trying out a few constructions for this problem with d=3 I have not been able to get much of a improvement. My first, failed, construction gave n-2. I have another construction which gives a lower bound of n-1.

{{{1, n-4, n-2}, {n-5, n-4, n-1}}, {{1, n-3, n-2}, {n-5, n-3, n-1}}, {{1, n-3, n-1}, {n-5, n-3, n-2}}, {{1, n-1, n}, {n-5, n-2, n}}, {{n-2, n-1, n}}}

• n has to be at least 7 in order for all sets at the end to be distinct.

21. Gil Kalai says:

Let me try to address Tim’s question about a half systematic way to translate Nicolai’s conjecture to a linear algebra statement.

One think I am a bit confused in Nicolai’s question is what precisely replace condition (*) in the multiset version and definition of f*(d,n). Dear Nicolai can you reexplain?

Anyway we can think about Nicolai’s conjecture but just for sets for now (when we dont know it to be sharp).

Here is one possible suggestion. Consider the space of monomials of degree d in n variables. $x_1,...,x_n$ mod away the monomials which involve two non adjacent variables. This will be the vector space. (But it is all negotiable, of course; you can replace the relation $x_ix_j=0$ for j-i>1 by a more complicated relation if needed.)

Now how to associate to families vectors? Here are a few suggestions

1) Suppose that the order in which the elements of {1,2,…,n} appears in sets represented in the families is compatible with the usual ordering 1,2,…,n (namely if i appears in a set in a family F_j(i) but not in a set of a smaller index family then j(1)<=j(2)<=…<=j(n)). Then associate to a

1 a) Associate to each i the variable x_i to each set (or later multiset) the monomial and to each family the sum of monomials.

1 b) The same except you associate to a family a generic linear combination of the monomials associates to sets in the family.

2)a;b the same as you take associate to each element 'j' a genaric linear combination of the variables x_1,…,x_n rather then the rule in 1)

The hope is that for some such vector space (where essentially you built the vector space based on Nicolai's most simple example) you can use (*) (and the facts that the families are disjoint) to imply linear independence.

• Gil Kalai says:

1) is cannot work since monomials supposrted on more than 2 variables will vanish. so you need something towards 2). Of course, the more complicated you make the vector space the harder it is to apply the condition (*).

• Nicolai Hähnle says:

The choice of basis which is essentially powers of $x_i$ is tempting but it’s not clear to me how one would map a single set onto that. The easiest way that avoids the zero-ing is just to treat it as a sum, i.e. a (multi-)set {1,1,2,3} gets mapped to $x_1^2 + x_2 + x_3$.

Ultimately though we want to assign each family a vector, and just taking sums would destroy information about which 2-sets, 3-sets, etc. appear in the family.

Hmm. I’ll read up on some of the linked examples on the Tricki again, maybe inspiration will come then.

22. Nicolai Hähnle says:

It may be easier to think of the multiset version in terms of monomials over the variables x_1, …, x_n. Then f^*(d,n) is the largest number of disjoint families F_1, …, F_t of degree d monomials such that

(*) for i < j < k, whenever $m_1 \in F_i$ and $m_2 \in F_k$, then there exists a monomial $u \in F_j$ such that $gcd(m_1, m_2) | u$.

To get from this formulation back to the equivalent set case, all you do is replace "monomial" by "square-free monomial".

I'll return with more later.

23. Nicolai Hähnle says:

Another idea that goes in the algebraic direction would be to look at algebras of the form $k[x_1,...,x_n]/(J_i)$, where $J_i$ is some ideal corresponding to the family $F_i$. Then perhaps (*) can be translated into a statement about homorphisms between these algebras, and related to that maybe one can say something about the length of chains of some modules.

A first candidate for $J_i$ would be the ideal created by all monomials that do not divide an element of $F_i$.

I don’t have a good candidate for the homomorphisms though, and I have to admit that I’m a bit out of my depth in this field.

24. Yury Volvovskiy says:

I was thinking about padding the sequence without changing its length. For instance if U and V are supports of two consecutive families and there’s an element $u$ which is in U but not in V and an element $v$ which is in V but not in U, one can add the subset {u,v} to either of the families.

Moreover, if we have a triple of consecutive supports, U, V, and W and V is a proper subset of both U and W, one can add the subset {u,w} to the middle family (here u is an element of U\V and w of W\V).

The supports of padded sequence are somewhat restricted. For instance, we can’t have a support of size 1 in the middle of such a sequence.

25. gowers says:

I’ve been trying unsuccessfully to prove a bound of the form $f(n)\leq Cf(n/2)$ (and hence $f$ is polynomial) in the following way. I’ll stick with the down-sets formulation from this comment. Suppose that we have a sequence of down-sets $D_1,D_2,\dots,D_m$ satisfying the convexity and disjointness conditions. (Recall that in this formulation the convexity condition says that if $i and $A\in D_i\cap D_k$ then $A\in D_j,$ and the disjointness condition says that no two $D_i$ share a maximal element.) Now let $X$ and $Y$ be two sets that partition the ground set, and just to be more specific let’s assume that $n$ is even and they both have size $n/2.$ Given any down-set $D,$ define $D(X)$ to be the set of all $A\cap X$ such that $A\in D,$ or equivalently to be the set of all $A\in D$ such that $A\subset X.$ (These are equivalent because $D$ is a down-set.) Then trivially if $D_1,\dots,D_m$ satisfy the convexity condition, then so do $D_1(X),\dots,D_m(X)$ and $D_1(Y),\dots,D_m(Y).$

However, these two sequences do not satisfy the disjointness condition (obviously, or else we would have proved the bound $f(n)\leq f(n/2)$). So one way we could think about the problem is this. Suppose we are given the down-sets $D_1(X),\dots,D_m(X)$ and $D_1(Y),\dots,D_m(Y)$ that satisfy the convexity condition. How can we find down-sets $D_1,\dots,D_m$ supported in $X\cup Y$ that project correctly and satisfy the disjointness condition as well?

In principle, this can be used to investigate either the lower bound or the upper bound. In the case of the lower bound we might try to take an example for $n$ that satisfies both conditions, “stretch” it in two different ways, at the expense of losing the disjointness condition, and then put the two resulting sequences of down-sets together in some way to recover the disjointness condition with a new ground set of size $2n.$ In the case of the upper bound, we might try to prove that if $m$ is at least 10 times as big as $f(n/2)$ then the disjointness condition fails so badly for $D_1(X),\dots,D_m(X)$ and $D_1(Y),\dots,D_m(Y)$ that there is no way to put the two sequences together to get a sequence with ground set $X\cup Y$ that satisfies the disjointness condition.

Note that trying to prove a lower bound this way would be trying to prove something rather strong: it is far from obvious that you can afford to forget all about what the down-sets actually are once you’ve proved a lower bound for $n/2.$ And something similar is true of the upper bound. It could be that some sequences of down-sets are easy to stretch, but then give rise to sequences that are no longer easy to stretch, or something like that.

Just to end, here are two very simple remarks about stretching. Here are two different ways of taking the sequences $E_1,\dots,E_m$ supported in $X$ and $F_1,\dots,F_m$ supported in $Y$ and producing a sequence $D_1,\dots,D_{2m}$ supported in $X\cup Y.$ (This is supposed to be the trivial bound that one is trying to beat.)

The first method is simply to take the sequence $E_1,\dots,E_m,F_1,\dots,F_m,$ regarded as a sequence of down-sets with ground set $X\cup Y.$ (For convenience let us assume that all the down-sets contain at least one non-empty set.) In other words, the stretching consists in converting $E_1,\dots,E_m$ into the sequence $E_1,\dots,E_m,\{\emptyset\},\dots,\{\emptyset\}$ and converting $F_1,\dots,F_m$ into the sequence $\{\emptyset\},\dots,\{\emptyset\},F_1,\dots,F_m$ and we then put the new sequences together by simply taking $D_i$ to be the union of the ith down-set in the first sequence with the ith down-set in the second sequence.

The second method is more complicated but still doesn’t give anything interesting. Following [EHHR], write $E\otimes F$ for the set of all $A\cup B$ such that $A\in E$ and $B\in F.$ If $E$ and $F$ are down-sets, then the maximal elements of $E\otimes F$ are unions of maximal elements of $E$ with maximal elements of $F.$ Therefore, if we take any sequence of the form $D_i=E_{j(i)}\otimes F_{k(i)}$ with the property that the sequences $(j(i))$ and $(k(i))$ are increasing, and for every $i$ either $j(i+1)>j(i)$ or $k(i+1)>k(i),$ then we satisfy the disjointness and convexity conditions. Obviously since at least one of $j(i)$ and $k(i)$ is changing each time, such a sequence has length at most $2m,$ so we don’t get anything non-trivial.

I now see that these two methods are not interestingly different — the first is (almost) the special case of the second where you increase the $j(i)$ first and then the $k(i).$

A final remark — in the above construction the “stretching” was simply a duplicating of the ground sets. But it is not necessary to do it that way. One might, for example, duplicate some maximal elements and change others. It is also not necessary to take each $D_i$ to be the maximal down-set that projects to $D_i(X)$ and $D_i(Y)$ (which is $D_i(X)\otimes D_i(Y)$).

26. (Re-corrected version of the previous comment; sorry I have been trapped by the > symbols. To Gil Kalai: would you be kind enough to erase my two previous comment please?)

I would like to propose an argument that gives a reformulation of the problem with less data. At first, I thought it would lead to $f(n)=2n$, but of course there where a mistake.

I continue to think in terms of summits, but for simplicity I won’t use the dual formalism. Let recall that a summit is a subset $A\subset [n]$ that appears in one of the families $\mathcal{F}_i$ and is maximal with this property with respect to inclusion.

It has been explained already that if $[n]$ itself appears (and is, therefore, a summit), then $m\leq 2n$. The key point is to adapt the argument to bound the distance between any two summits.

Lemma: if $A\in \mathcal{F}_i$ and $B\in \mathcal{F}_j$ are any two summits, then
$|i-j|\leq |A|+|B|-2|A\cap B|=|A\cup B|-|A\cap B|$ or there is a third summit $C\in\mathcal{F}_k$ such that i< k< j and $A\cap B\subset C$.

Note: if one gets rid of the second possibility, then $f(n)=2n$.

We prove the lemma by contradiction. Assume that $A\in \mathcal{F}_i$ and $B\in \mathcal{F}_j$ are two summits not satisfying the conclusion. There is some $k$ between $i$ and $j$ so
that $|k-i|\geq |A|-|A\cap B|$ and $|j-k|\geq |B|-|A\cap B|$, and one of the inequality is strict. By convexity, there is some set $X\in\mathcal{F}_k$ containing $A\cap B$. Moreover, there is no summit such that $X\subset C \in \mathcal{F}_l$ with $l$ between $i$ and $j$.

Assume for example that $k-i$>$|A|-|A\cap B|$. Then there is a set $X_1\in\mathcal{F}_{k-1}$ such that $X\subset X_1\subset A$. Inductively, we get a sequence of sets $X_\alpha\in\mathcal{F}_{k-\alpha}$ such that $X_\alpha\subset X_{\alpha+1}\subset A$. This sequence has length $k-i$>$|A|-|A\cap B|$, a contradiction. QED.

Note: if there is a third summit such that $X\subset C \in \mathcal{F}_l$ with i<l<j,
then assuming that $(i,j)$ where chosen minimal in the lexicographic order among possible conterexample indices, we get that $|l-i|\leq |A|+|C|-|A\cap C|$ and $|l-j|\leq |B|+|C|-|B\cap C|$.
This does not lead to the desired bound on $|i-j|$ because $C$ can contain a lot of elements that do not lie in $A\cup B$. This gives at least indication on how to try to design a long convex sequence of families.

Reformulation: let $g(n)=\max |A_1|+p(A_k)-p(A_1)+|A_k|$ where the max is on the following data:
an antichain $A_1,\ldots,A_k\subset [n]$ and an non-decreasing integer-valued function $p$ defined on that antichain, such that $|p(A_i)-p(A_j)|\leq |A_i|+|A_j|-2|A_i\cap A_j|$ whenever there is no $k$ between $i$ and $j$ such that $A_k\supset A_i\cap A_j$.
Then $f(n)\leq g(n)$.

• Sorry the formating is still not ok. I Give up my attempts to get it right since the comment is readable nonetheless.

27. Arne Müller says:

I have a slight different perspective to the problem (dunno if it leads to somewhere). It actually simply combines the notion of downsets from gowers with this interval perspective.

Consider the graph G over the vertices $\mathcal{P}([n])$ and edges $E = \{\{u,v\} : u \subset v \}$. In this view downsets are just sets D where holds for every $x \in D$ that any shortest path from $\emptyset$ to x is entirely contained in D.

Then you extrude the graph over some time axis. That will somehow correspond to the chain, but lets don’t care on the how for now.
Next you consider downsets to be active over some time interval.

The basic idea is that you actually only need to stack active downsets on top of each other. (stack on top = the upper downset must contain the lower one). This means if downset D is active, all downsets $T \subset D$ must be active, too.

Important note: These downsets do not directly correspond to the downsets generated by the $F_i$. They rather correspond to the intersections of downsets of the downsets generated by $F_i$.

None the less you can build a chain from such a stacking (just consider the time points, where a new downset becomes active). And also in reverse get a stacking from each chain (you get intervals and stacking property by convexity).

I think this may be a useful perspective, because it gives you the disjointness property of the chains implicitly (downsets may only become active once and hence the summits of each downset will only be used once).

28. Arne Müller says:

How does the notion of my above post to apply the method described in [EHRR] which gives you this $\Omega(n^2/log n)$ bound?

What [EHRR] does is the following: They build $d-1$ foundation downsets. On top of each such downset they add $l$ little peaks. Because $l = \Omega(n / \log n)$ and $d = \Omega(n)$, they get their lower bound.

The foundation downsets again form a valid chain (of length $d-1$) and the peaks only add points on top of the foundation downset and not next to it. Hence the peaks on different foundation downsets do not interfere. Peaks on the same foundation downset, do also not interfere, because they contain just sets containing one element more than the summits of the foundation downsets. This makes the construction work.

PS: This interference analysis is important, because each downset must only be active in one interval (and each peak or foundation downset may activate subset-downsets).

So can we extend the construction of [EHRR] with this perspective?

• Arne Müller says:

The formula that does not parse was supposed to be
$l = \Omega(n / \log n)$

The rest should be readable 😉

• gowers says:

I’m still having trouble understanding the [EHRR] construction. Can you give slightly more detail about the foundation downsets and the little peaks?

• Arne Müller says:

Each peak is generated by one layer $\mathcal{L}_i$ in the definition of a mesh.

For the foundation down-sets see the proof of Lemma 3:
$f$ is active on any given layer of the mesh $\mathcal{M}(A,i;B,j)$ if and only if $i \geq |f \cap A|$ and $j \geq |f \cap B|$.
Each foundation down-set corresponds to one mesh $\mathcal{M}(A,j;B,d-j)$.
So these $f$ (with $|f \cap A| \leq j$, $|f \cap B| \leq d-j$) are the elements of the j.th foundation down-set.

By the statement above follows directly that each peak (layer) covers its foundation down-set (mesh).

Actually Lemma 2 (from which the above statement of Lemma 3 is derived), says rather directly that the peaks do not cover more than their foundation-down-set. The peaks (layers) are made out of these $(A,i;B,j)$ sets.

29. gowers says:

I was going to suggest something similar: indeed, I thought it was exactly the same until you said that the downsets were not the downsets generated by the $F_i.$

If you do take $D_i$ to be the down-set generated by $F_i,$ and for each $A\subset[n]$ you let $I_A$ be the interval of all $i$ such that $A\in D_i,$ then the properties the $I_A$ must have are as follows.

(i) If $A\subset B$ then $I_A\supset I_B.$

(ii) For each $A$ there is at most one $i$ such that $A$ is maximal such that $i\in I_A.$ (That is, each $A$ is allowed to be maximal at most once.)

This makes sense for multisets too (with the obvious definition of $\subset$).

• Arne Müller says:

I guess this is pretty the same idea with the only difference that I had dropped condition (ii).
Instead of looking if $A$ is maximal for every $i \in I_A = [s_A, e_A]$ my idea was to simply look whether $A$ is maximal at $s_A$. If I’m not mistaken this should be possible wlog. Yes, because if you have the scenario where $A$ only becomes maximal somewhere in the middle (my chain-conversion would not contain $A$ which is bad, because smaller) you can also let $A$ start a bit earlier and then it is maximal at $s_A$, too.

30. gowers says:

One thought I had for a way to prove the exact bound for multisets was to try to show that in a sequence of down-multisets satisfying the necessary conditions it must always be possible to pick a sequence of maximal elements that “increase” in a certain sense. Roughly speaking, the meaning of “increase” would be that there is some ordering of the ground set such that you only ever replace elements with larger elements. Note that if you have a sequence of multisets of size d that is increasing in this sense then it will not be longer than the sequence in Nicolai’s example.

Let’s suppose that all the maximal elements have size d, and let’s suppose that the order in which new elements of the ground set are added is the usual order on $[n].$ Is it possible that we could have a maximal element of $D_i$ that is lexicographically larger than all the maximal elements of $D_{i+1}$? Let me try to produce an example with d=5, say.

I’ll start with $D_1=\langle 11111\rangle$ (meaning the down-multiset generated by the singleton 11111), then jump to $D_2=\langle 12345\rangle.$ There’s nothing now to stop us taking $D_3=\langle 12222\rangle,$ but this isn’t really a counterexample because after renaming the ground set we can think of this as $\langle 15555\rangle.$ If we do indeed think of it that way, then from now on we will be forbidden to put 1 with any of 2,3,4, so it will be hard to take a step backwards.

OK, although I have so far made only the feeblest of attempts to find a counterexample, let me make not exactly a conjecture but more of a statement that would I think prove that Nicolai’s bound is best possible but that is probably way too optimistic. It should be fairly easy to decide whether it is true. The statement is this: if $D_1,\dots,D_m$ are down-multisets satisfying the given condition then one can select from them a sequence of elements that are lexicographically increasing with respect to some ordering of the ground set.

Sorry, I realize I don’t quite mean lexicographically increasing. I want something a bit stronger than that. I define a partial order (with respect to the usual ordering of the ground set) as follows: given two submultisets $A,B\subset [n]$ I say that $A if there is a “bijection” $\phi$ from $A\setminus B$ to $B\setminus A$ such that $\phi(x)>x$ for every $x\in A\setminus B.$

For example, $1135<1236$ but it is not the case that $1135<1234.$ I think if we have multisets of different sizes then we might also say that a multiset increases if you remove one of its smallest elements and increases if you add an element that is maximal once added. (I’m not sure if this is the right definition — I need to mess around with trying to disprove the conjecture.) For instance, $246$ would be less than $2467$ and $46$ by this definition but would not be less than $2456.$

Even if none of this quite works, I don’t see an immediate reason that there couldn’t be a proof that follows this general strategy, which I’ll spell out once more.

1. Define a partial order on all multisets.

2. Prove that no strictly increasing sequence in this partial order can be longer than Nicolai’s example.

3. Prove that if $D_1,D_2,\dots,D_m$ satisfy the conditions of the conjecture, then we can select a strictly increasing sequence from them.

31. gowers says:

Let’s see what can be proved if all maximal elements have size 2. If I write something like {12,34} I’ll mean the down-set generated by 12 and 34.

WLOG the first down-set is {12} (since making the first one smaller cannot harm anything). Now let’s do a case analysis. If the next down-set does not use either 1 or 2, then we are done by induction. If it uses 1 but not 2, then WLOG the next down-set is {13}, since the only reason we need it to be large is so that it will cover intersections with 12, and if 2 is not used we just have to cover 1.

Hmm. Suppose we keep going for as long as possible choosing a set of the form 1r from each down-set. We can reorder the ground set so that we choose 12, then 13, then 14, etc. Eventually we reach 1t and can no longer find a set of the form 1r that has not been used already. But the disjointness condition means that we cannot find any maximal element that involves 1, so at this point we need to find a maximal element that is bigger (in the strange ordering) than 1t. If there is any maximal element in $D_{t+1}$ that involves a number $s$ that is at least $t,$ then we are done, since its other element is bigger than 1. If we can’t, then … well, to see what happens let me try it out with some small numbers.

Suppose, then, that we’ve chosen 12, 13, 14, 15 and have then found that the next down-set is supported in {2,3,4}. If it contains 34, say, then $D_3$ and $D_4$ must contain something that involves 3. Let us suppose they contain something like 37 and 39. So what we know so far is that the $D_i$ contain {12}, {13}, {14,37}, {15,39}, {34}. But we also know that $D_4$ must involve 4, so I’ll add something to that too, so that now we have {12}, {13}, {14,37}, {15,39,48}, {34}.

Oh dear, I think that is a counterexample to my hypothesis. However we order things we must have 1<3<4. Ah, but I have the choice of making 7<4. So let me rewrite this example with such a reordering: {12}, {13}, {15,34}, {16,39,58}, {35}. No, that doesn’t help. The problem is that all of 16, 39 and 58 are provably not smaller than 35 (since 5 appears before 6 and 9, and 3 appears before 8).

It still feels as though something is increasing too fast here, but I’m not sure what it is.

32. gowers says:

Another thought occurs to me, which is that perhaps compressions could help.

As a complete guess, let me see what happens if we try a so-called ij-compression. The basic idea of an ij-compression is to take some pair $i and replace every set $A$ in a set-system by $(A\cup\{i\})\setminus\{j\}$ unless that already belongs to the set-system, in which case you do nothing. For example, if you apply a 12-compression to the set-system $\{23,14,24,25\}$ you get the set-system $\{13,14,24,15\}.$

Typically what you want to prove is that some quantity of interest goes down when you compress, so if you want to minimize that quantity you can assume that your set-system is unaltered by any ij-compression. But perhaps we could find a way of simultaneously compressing the families $F_i$ and preserving the convexity and disjointness properties.

This is incredibly unlikely to work, but let me see whether we preserve these properties if we compress the families by taking their union, doing an ij-compression on that, and keeping track of which set belonged to which family. That is, we replace $A\in F_r$ only if that doesn’t destroy the disjointness property by hitting some $B\in F_s$ (where $s$ could equal $r$).

Since this is set up so that the disjointness property is trivially preserved, the problem will presumably be with the convexity property. So let $r and let $A\in F_r$ and $C\in F_t.$ Then we know that there is $B\in F_s$ such that $A\cap C\subset B.$ Given any set $E\in\bigcup F_h,$ let $\phi_{ij}(E)$ stand for the set you get by replacing $j$ by $i$ if $j\in E$ and the resulting set does not already belong to $\bigcup F_h.$

If $j\notin B$ then $\phi_{ij}(B)=B$ and $\phi_{ij}(A)\cap\phi_{ij}(C)=A\cap C,$ so we are fine. In fact, we can say more: if $j\notin A\cap C,$ then removing … actually, there might be a problem. OK, here’s an example: $A=\{1,2,4\}, C=\{3,5,6\}, i=3, j=4.$ Then if we do an ij-compression as described above, the intersection of $A$ and $C$ changes from being empty to being the singleton $\{3\},$ and there is absolutely no reason for any set in $F_s$ to contain either 3 or 4.

Nevertheless, it might be worth trying to come up with ways to transform existing examples into more canonical ones of the same length.

33. Woett says:

Dear Gil and others,

I’ll probably make a fool out of myself with this comment, but this problem is way too cute to not be wrong about at least once.

We are trying to maximize $f(2,n)$ by just making as many families as possible, family by family; Assume, WLOG, that $F_1 = {{1,2}}$. Then:

1) $F_2$ ‘introduces’ exactly $1$ new element, say $3$. Why? Because, unfortunately, it’s not possible to _not_ introduce a new element and it’s not necessary (for the moment I can’t make ‘it’s not necessary’ completely rigorous, but it feels so supertrue that I can’t imagine the proof of it being anything but trivial)

Furthermore, 1 of the following is true:

2a) We kill off a number, say $1$ and, WLOG, we may assume $F_2 = {{2,3}}$
2b) We do not kill off a number, but in effect we have to make $F_2$ a family of more subsets; $F_2 = {{1, 3}, {2, 3}}$.

Now we repeat; $F_3$ also introduces exactly 1 new element, and we again have the choice (assuming we went with 2b); do we want to kill off 1 (or 2) numbers (implying that $|F_3| - |F_2| = 0$ (or $1$)), or do we want to keep every element alive which demands $F_3$ to consist of 3 sets.

In general: $F_n$ introduces exactly 1 new element. Further we can kill off $i$ numbers, with $i$ being non-negative and smaller than $|F_{n-1}|$ implying $|F_n| - |F_{n-1}| = i - 1$

So after $n-2$ of these ‘introduce an element and kill some others’-procedures we killed, say, k of them and we may see that:

$F_{n-1} = {{k+1, k+2}, {k+2, k+3}, .., {n-1, n}}$

And we can no longer do anything. So $f(2, n)$ should equal $n-1$.

• Woett, if you check the earlier posts you will find that there are several constructions which give lower bounds larger than $n$.

• Paco Santos says:

Sorry again, the parameters in my first post were all wrong. Let me start over. (Gil, feel free to delete my previous two posts):

For example, I have found the following easy construction showing that f(2,2k) is at least 3k -2. (I have not seen this posted before; sorry if I missed it):

By inductive hypothesis, f(2,2k-2) is at least 3k-5 (induction starts with the trivial statement f(2,2)=1). Let G be a graph that achieves that bound, and let uv be an edge labelled with 2k-5 in that graph. We introduce two new vertices x and y and five new edges: the cycle uxvy with edges alternating the labels 2k-4 and 2k-3, and the edge xy with label 2k-2.

Needless to say, what this construction really shows is the formula

f(2, n+2) \ge f(2,n) + 3

To prove the statement f(2,n) \ge 2n – c for a constant c it would suffice a similar argument showing that you can increase f by 2k introducing only k new vertices…

• Paco Santos says:

Now I see my construction was already posted by Klas in the first thread (although he stated it gives 3n/2 and I still think it gives 3n/2 -2): https://gilkalai.wordpress.com/2010/09/29/polymath-3-polynomial-hirsch-conjecture/#comment-3485

• Klas markström says:

Paco, i stated 1+3(n/2-1), which simplifies to 3n/2-2

• Woett says:

Ah, I see it now. The reason I couldn´t prove the crucial part was that it wasn´t true 🙂 Still, I feel that the idea makes sense in some way. Let me ask something;

Can anyone give me an example where it is necessary to ‘introduce’ 3 new elements?

Because if this is never a necessity, we immediately have the bound $f(2,n) \leq 2n$. Although this might be wishful thinking of course

• Paco Santos says:

(To Klas) I see. sorry.

• Paco, nothing to be worry about. It had been worse if there was a real disagreement 🙂

34. Woett says:

Alright, I tried to correct my post a badrillion times now, hopefully wordpress allows me this time. Corrections:

$F_1 = \{\{1, 2\}\}$
(at 2a) $F_2 = \{\{2, 3\}\}$
(at 2b) $F_2 = \{\{1, 3\}, \{2, 3\}\}$
$F_{n-1} = \{\{k+1, k+1\}, \{k+2, k+3\},.., \{n-1, n\}\}$

The following sentence:

Why? Because, unfortunately, it’s not possible to _not_ introduce a new element and it’s not necessary (for the moment I can’t make ‘it’s not necessary’ completely rigorous, but it feels so supertrue that I can’t imagine the proof of it being anything but trivial)

should become

Why? Because, unfortunately, it’s not possible to _not_ introduce a new element and it’s not necessary (for the moment I can’t make ‘it’s not necessary’ completely rigorous, but it feels so supertrue that I can’t imagine the proof of it being anything but trivial) to introduce more than 1 new element. For example, there is no reason to add, say, $\{7, 10\}$ to $F_2$ as we can just as well put that set in the first family which contains $7$ or $10$

35. gowers says:

I had a thought about a statement that generalizes what we are trying to prove and that might serve as a better inductive hypothesis (if we are trying to bound $f(d,n)$ and know all about $f(c,n)$ whenever $c).

What I propose is that we get rid of the disjointness condition, but measure the length of the sequence differently. Let us say that $J$ is a maximality interval if there is some $A\subset[n]$ such that $A$ is maximal in $D_i$ whenever $i\in J.$ (Alternatively, in terms of the $F_i,$ we could define it to be an interval of the form $\{i:A\in F_i\}.$) Now define a sequence $D_1,\dots,D_M$ to have “length” at most $m$ if it is possible to write $\{1,2,\dots,M\}$ as a union of at most $m$ maximality intervals. Now I’d like to conjecture (without having thought about it at all, so there might be a small counterexample, but then I’d want to modify the conjecture rather than abandon it completely) that in the multisets case this modified length is still at most the length of Nicolai’s example.

This is a stronger statement, and becomes equivalent if we insist that any two maximality intervals are either the same or disjoint. The reason it might be useful is that if we construct an example with $d$-sets, we know that the $(d-1)$-shadow satisfies the convexity condition, but it doesn’t satisfy the disjointness condition. If we could nevertheless have a bound for its modified length, we would then have a much clearer idea of what we had to do with the $d$-sets to increase the modified length. Note that the overlapping of the maximality intervals seems very much to happen with the shadows, so this is an attempt to say something strong about them anyway.

• gowers says:

To continue this line of thought, let me discuss some extreme cases when we go from d=1 to d=2. If we want to maximize the number of maximality intervals needed when d=1, then the best we can do, writing $I_r$ for the interval of all $j$ such that $\{r\}$ is an element (and hence a maximal element) of $D_j,$ is to have intervals $I_1,I_2,\dots,I_n$ such that for every $r$ there is at least one $j$ that belongs only to $I_r.$ So in order to be as general as possible, let us assume that there are a few $j$ that belong to $I_1$ only, then some that belong to $I_1$ and $I_2,$ then some that belong to $I_2$ only, and so on.

Now let’s assume that these intervals are telling us the lower shadow of a system formed by 2-multisets. How can we increase the number of intervals?

For each $j$ we find that we either have one available element of the ground set or two consecutive available elements. So the only possible multisets involve two equal or consecutive elements. So the most general thing we can do is have an interval of 11s, then a (possibly overlapping) interval of 12s, then a (possibly overlapping, but not with 11 if we want to maximize the minimum cover) interval of 22s, and so on.

In other words, Nicolai’s example arises if we build up d and are as greedy as possible at each stage.

Now let’s go to the opposite extreme and see what happens if we can cover everything with just one interval at the d=1 stage. That is, let’s suppose that $\{1\}$ is contained in the lower shadow of every single family. Now if we want to cover with as few intervals as possible, we might as well restrict attention to sets of the form $1j.$ But then we clearly get a bound of $f(1,n).$

That wasn’t very interesting, so let’s try a situation where we have a large interval of families (with d=2) that use 1 but not n, then a large number that use both 1 and n, and finally a large number that use n but not 1. Can we cover with a small number of intervals?

Yes we can. Let $I$ be the interval of families that use 1 and let $J$ be the interval of families that use $n.$ By looking at the d=1 case we see that we can cover $I$ with at most $n$ maximality intervals, and the same for $J.$ But we cannot use $1n$ twice (except consecutively) by the convexity condition, so we get a bound of $2n-1,$ which is the correct bound. Note that this situation can occur, and does occur in Terry’s example where you take the collection of all pairs with sums $2,3,\dots,2n.$ Also, for what it’s worth, it’s not hard to prove that if this situation occurs then it’s actually forced to be Terry’s example.

I’m starting to feel as though this is a fruitful way to think about the problem. I’m going to think a little bit about what happens if the 1-shadow can be covered by three overlapping intervals but not two, and I’ll report back.

• gowers says:

I did indeed go away and think about it, but it soon led to something not too interesting: the $2n-1$ upper bound by an argument that was basically the usual elementary argument. So I cannot really claim that this is a fruitful way of thinking about the problem unless something comes out for d=3, which so far it hasn’t.

36. I can’t stop thinking that $f(n)=2n$; here is one bold conjecture that would imply it, and to which I have no counter example.

Conjecture: if one defines $I_k$ to be the interval of indices i such that k is contained in at least one of the elements of $\mathcal{F}_i$, then every $\ell\in[m]$ is either the first element of one of this intervals, or the last element of one of this intervals.

Of course, since there are only n intervals one gets $m\leq 2n$ if the conjecture holds.

Can anyone produce a convex sequence of families such that the conjecture does not hold?

• Benoit,
Take a complete bipartite graph G with $n/2$ vertices in each part. Let $F_1,...,F_{n/2}$ be a sequence of edge disjoint perfect matchings in $G$.

The sequence of families $F_1,...,F_{n/2}$ is convex, as has been observed earlier, and they all have the same support. However they are not a maximal family for d=2.

• Ok, thanks for the example, I think I now know what was wrong in my way to think about the problem.

• gowers says:

If we’re talking about the same $f(n),$ then the [EHRR] quadratic lower bound will certainly be a counterexample.

• Hum, sure it will. I was not aware of this bound; I will try to have a look at [EHRR].

Maybe this bound could be included in the wiki? I feel uncomfortable writting in the wiki myself until I understand all this a bit better.

37. Gil Kalai says: