# The polymath blog

## March 6, 2017

### Rota’s Basis Conjecture: Polymath 12

Filed under: polymath proposals — tchow8 @ 11:18 pm

There has been enough interest that I think we can formally declare Rota’s Basis Conjecture to be Polymath 12. I am told that it is standard Polymath practice to start a new blog post whenever the number of comments reaches about 100, and we have reached that point, so that is one reason I am writing a second post at this time. I am also told that sometimes, separate “discussion” and “research” threads are created; I’m not seeing an immediate need for such a separation yet, and so I am not going to state a rule that comments of one type belong under the original post whereas comments of some other type belong under this new post. I will just say that if you are in doubt, I recommend posting new comments under this post rather than the old one, but if common sense clearly says that your comment belongs under the old post then you should use common sense.

The other reason to create a new post is to take stock of where we are and perhaps suggest some ways to go forward. Let me emphasize that the list below is not comprehensive, but is meant only to summarize the comments so far and to throw in a few ideas of my own. Assuming this project continues to gather steam, the plan is to populate the associated Polymath Wiki page with a more comprehensive list of references and statements of partial results. If you have an idea that does not seem to fit into any of the categories below, please consider that to be an invitation to leave a comment about your idea, not an indication that it is not of interest!

### Matroids with No Small Circuits

I want to start with an idea that I mentioned in my MathOverflow post but not in my previous Polymath Blog post. I think it is very promising, and I don’t think many people have looked at it. Geelen and Humphries proved that Rota’s Basis Conjecture is true for paving matroids. In the case of vector spaces, what this means is that they proved the conjecture in the case where every (n – 1)-element subset of the given set of n2 vectors is linearly independent. It is natural to ask if n – 1 can be reduced to n – 2. I have not digested the Geelen–Humphries paper so I do not know how easy or hard this might be, but it certainly could not hurt to have more people study this paper and make an attempt to extend its results. If an oracle were to tell me that Rota’s Basis Conjecture has a 10-page proof and were to ask me what I thought the method was, then at this point in time I would guess that the proof proceeds by induction on the size of the smallest circuit. Even if I am totally wrong, I think we will definitely learn something by understanding exactly why this approach cannot be extended.

### Independent Partial Transversals

Let me now review the progress on the three ideas I mentioned in my first blog post. In Idea 1, I asked if the n2 vectors could be partitioned into at most 2n – 2 independent partial transversals. A nice proof that the answer is yes was given by domotorp. Eli Berger then made a comment that suggested that the topological methods of Aharoni and Berger could push this bound lower, but there was either an error in his suggestion or we misunderstood it. It would be good to get this point clarified. I should also mention that Aharoni mentioned to me offline that he unfortunately could not participate actively in Polymath but that he did have an answer to my question about their topological methods, which is that the topological concepts they were using were intrinsically not strong enough to bring the bound down to n + 1, let alone n. It might nevertheless be valuable to understand exactly how far we can go by thinking about independent partial transversals. Ron Aharoni and Jonathan Farley both had interesting ideas along these lines; rather than reproduce them here, let me just say that you can find Aharoni’s comment (under the previous blog post) by searching for “Vizing” and Farley’s comment by searching for “Mirsky.”

### Local Obstructions

Idea 2 was to look for additional obstructions to natural strengthenings of Rota’s Basis Conjecture, by computationally searching for counterexamples that arise if the number of columns is smaller than the number of rows. Luke Pebody started such a search but reported a bug. I still believe that this computational search is worth doing, because I suspect that any proof that Rota’s Basis Conjecture holds for all matroids is going to have to come to grips with these counterexamples.

Note that if we are interested just in vector spaces, we could do some Gröbner basis calculations. I am not sure that this would be any less computationally intensive than exhausting over all small matroids, but it might reveal additional structure that is peculiar to the vector space case.

### Algebraic Geometry

There has been minimal progress in this (admittedly vague) direction. I will quote Ellenberg’s initial thoughts: “If you were going to degenerate, what you would need to do is say: is there any version of this question that makes sense when the basic object is, instead of a basis of an n-dimensional vector space V, a 0-dimensional subscheme of V of degree n which is not contained in any hyperplane? For instance, in 2-space you could have something which was totally supported at the point (0,1) but which was “fat” in the horizontal direction of degree 2. This is the scheme S such that what it means for a curve C to contain S is that S passes through (0,1) and has a horizontal tangent there.”

Let me also mention that Jan Draisma sent me email recently with the following remarks: “A possible idea would be to consider a counterexample as lying in some suitable equivariant Hilbert scheme in which being a counterexample is a closed condition, then degenerate to a counterexample stable under a Borel subgroup of GLn, and come to a contradiction. ‘Equivariant’ should reflect the action of GLn × (SnSnn). However, I have not managed to make this work myself, even in low dimensions. In fact, having a good algebro-geometric argument for the n = 3 case, rather than a case-by-case analysis, would already be very nice!”

### Alon–Tarsi Conjecture

Now let me move on to other ideas suggested in the comments. There were several thoughts about the Alon–Tarsi Conjecture that the Alon–Tarsi constant LnevenLnodd ≠ 0 when n is even. Rebecca Stones gave a formula that, as Gil Kalai observed, equated the Alon–Tarsi constant with the top Fourier–Walsh coefficient for the function detn; i.e., up to sign, the Alon–Tarsi constant is

ΣA (–1)σ(A) det(A)n,

where the sum is over all zero-one matrices and σ(A) is the number of zero entries in A. This formula suggests various possibilities. For example one could try to prove that LnevenLnodd ≢ 0 (mod p) where p = 2n + 1 is prime, because in this case, det(A)n must be 0, 1, or –1. This would already be a new result for n = 26, and the case n = 6 is small enough to compute explicitly and look for inspiration. Luke Pebody posted the results of some computations in this case.

Another possibility, suggested by Gil Kalai, is to consider a Gaussian analogue. Instead of random zero-one matrices, consider random Gaussian matrices and try to understand the Hermite expansion of detn, in particular showing that the coefficient corresponding to all ones is nonzero. This might be easier and might give some insight.

Note also that in the comments to my MathOverflow post, Abdelmalek Abdesselam proposed an analogue of the Alon–Tarsi conjecture for odd n. I do not think that many people have looked at this.

### Generalizations and Special Cases

Some generalizations and special cases of the conjecture were mentioned in the comments. Proving the conjecture for graphic matroids or binary matroids would be an enormous advance. There is a generalization due to Jeff Kahn, in which we have n2 bases Bij and we have to pick vijBij to form an n × n grid whose rows and columns are all bases. Another generalization was prompted by a remark by David Eppstein: Suppose we are given n bases B1, …, Bn of a vector space of dimension mn, and suppose we are given an n × n zero-one matrix with exactly m 1’s in every row and column. Can we replace each 1 in the matrix with a vector in such a way that the m vectors in row i are the elements of Bi and such that the m vectors in every column form a basis?

Juan Sebastian Lozano suggested the following reformulation: Does there exist a group G such that V is a representation of G and there exists giG such that gi Bi = Bi+1, and for every vector bB1,

span{g0b, …, gn – 1b} = V

where gi = gig1 and g0 is the identity?

### Other Ideas

Fedor Petrov mentioned a theorem by him and Roman Karasev that looks potentially relevant (or at least the method of proof might be useful). Let p be an odd prime, and let V be the Fp-vector space of dimension k. Denote V* = V \ {0} and put m = |V*|/2 = (pk – 1)/2. Suppose we are given m linear bases of the vector space V

(v11, …, v1k), (v21, …, v2k), …, (vm1, …, vmk).

Then there exist pairwise distinct x1, …, xm, y1, …, ymV* and a map g:[m] → [k] such that for every i ∈ {1, …, m} we have yixi = vig(i).

Gil Kalai notes that the Alon–Tarsi conjecture is related to the coloring polynomial of a graph and asks if we can learn anything by considering more general polynomials such as

Π {(xiλexj) : i < j, {i,j} = eE(G)},

where the λe are weights associated to the edges e.

1. Thank you for this wonderful summary!

Here are some very preliminary thoughts which might be of interest. I was thinking that a perhaps instructive structure might be the exterior algebra on $V$. There are three things that are attractive about this:

1) If we consider Rota’s basis conjecture to be a statement on the existence of certain non-zero elements in the $n^{th}$ power of the graded algebra, then prima facie, it looks like the idea of a paving matroid is exactly a vector space upon which you can induct on the dimension of $V$ when proving the existence of these non-zero elements. I don’t think that this will give new information about the matroid problem — in fact it seems to me that the obstruction to doing this more generally is exactly the combinatorial/independence problem which the more general matroid approach aims to solve. However, if there is some structure specific to the vector space case, as opposed to the matroid, it seems the exterior algebra would be the natural place to capture it.

2) I do not understand fully the discussion of the Alon–Tarsi Conjecture above, but it seems that if we are talking about the signs of permutations (i.e. looking for odd v even latin squares, where the elements in our squares are basis elements), then one might be able to recover the formula given by Rebecca Stone by considering the action of $S_n$ on the highest exterior power in the exterior algebra. This might lend itself to some generalization (it seems to me that this is roughly the context of the mathoverflow question you link on the Alon–Tarsi Conjecture for odd $n$, not with the exterior algebra, but with the tensor algebra, but that was just at a glance).

3) Speaking of my reformulation on the last post: one can probably make a representation theoretic reformulation on the exterior algebra (e.g. maybe using the ‘induced representation’, meaning the representation passed into the exterior algebra) in which one might get explicit conditions on the determinants of the things in the image of the homomorphism into $End(V)$.

Anyway, my general point is simply that the exterior algebra might be a good place to think about the intersection of some of the approaches, which might help in the solving the problem, even if I doubt it is anywhere near enough on its own.

Comment by Juan Sebastian Lozano — March 7, 2017 @ 12:24 am

2. Small comment on the algebraic geometry, still with very little content: Draisma’s comment is a much better way of saying what I was trying to say. Or rather, you might say: the content of my question is to ask what it would *mean* for a degenerate (e.g. Borel-fixed) object to be a “counterexample” to Rota (since it wouldn’t literally be subject to Rota’s conjecture in its original form.) I also want to endorse Draisma’s bringing up of the symmetry group GL_n × (S_n ⋉ S_n)^n, which acts on every vector space in this whole story and their symmetric powers too. What do we know about the ring of invariants?

Comment by JSE — March 7, 2017 @ 5:04 am

• I briefly discussed Draisma’s idea with Patrick Brosnan. One somewhat annoying fact is that Rota’s Basis Conjecture (or any variation or generalization that I have seen) hypothesizes that certain sets are bases, and being a basis is an open condition rather than a closed condition. I’m not sure how serious an obstacle this is?

Comment by tchow8 — March 28, 2017 @ 3:00 pm

3. Have people seen this paper that just came out? It should be checked.
Carlos Gamas, ON THE CONJECTURE OF ALON-TARSI, Gulf Journal of Mathematics Vol 4, Issue 4 (2016) 108-116

Abstract. The Alon-Tarsi Conjecture states that for even n, the number of even latin squares of order n differs from the number of odd latin squares. In this note we prove that this conjecture is true if and only if there exists a permutation ζ ∈ S_{n^2} and a spherical function, φ, such that φ(ζ)\ne 0.

Comment: it looks like there are a couple of misprints in this paper. Hopefully they are not serious. (Both the spherical function and the permutation are defined at the beginning of the paper. So the actual statement in the abstract could be misleading?)

Comment by David Glynn — March 7, 2017 @ 9:33 am

4. It seems that the Gaussian analog is actually equivalent to the Alon-Tarsi conjecture.

It can be formulated as follows:

Let $n$ be even and let $X=(x_{ij})$ be a random $n \times n$ matrix. Then the expectation

$\mathbb E (det (X)^n \prod_{1 \le i,j \le n} x_{ij})$ is non zero.

(Perhaps it is positive.)

We can take $X$ to be a random Boolean matrix with $\pm 1$ entries or a random Gaussian matrix with entries distributed $N(0,1)$.
There is maybe some hope that because we know a lot about random Gaussian matrices (and correlation inequalities for Gaussian measures) there will be some other way/formula showing that the expectation in question for the Gaussian case is positive.

Comment by Gil Kalai — March 7, 2017 @ 2:00 pm

• Sorry to be dense, but why is the Gaussian analog equivalent to the Alon-Tarsi conjecture?

Comment by tchow8 — March 10, 2017 @ 12:09 am

• Suppose the elements of X are independent, symmetric about 0 and each have non-0 finite variance. We do not assume identical distributions.

Express $\mathrm{det}(X)^n\Pi_{i,j}x_{i,j}$ as a polynomial in the $x_{i,j}$.

It is a homogeneous polynomial of degree $2n^2$ in $n^2$ variables, in which each monomial contains each variable at least once.

The expected value of any monomial is 0 if it has a variable raised to an odd power (since each variable is symmetric around 0 and they are all independent). Any monomial that is not of such a form must include every variable at least twice. By the degree condition, it must include each variable exactly twice.

Therefore the expectation of $\mathrm{det}(X)^n\Pi_{i,j}x_{i,j}$ is equal to $c_n$ times the product of the variances of the cells, where $c_n$ is independent of the distribution of the cells, and is equal to the $\Pi_{i,j}x_{i,j}$ coefficient in $\mathrm{det}(X)^n$.

Finally: $\mathrm{det}(X)^n$ can be written as the sum of elements $\sigma_1, \ldots, \sigma_n$ of the product of the signs of the sigmas times $X_{i, \sigma_ji}$, which is equal to $\Pi_{i, j}x_{i, j}$ precisely when the sigmas form the rows of a Latin square.

Therefore Alon-Tarsi is equivalent to the expectation of $\mathrm{det}(X)^n\Pi_{i,j}x_{i,j}$ being non-0.

Comment by Luke Pebody — March 28, 2017 @ 6:38 pm

• So, for n=2, I looked for a simple expression for $\mathbb{E}(\mathrm{det}(X)^k\Pi_{i,j}x_{i,j}$ and proved that for even k it is -((k+2)*k*k!)/8.

I then generated the same numbers for n=4, but haven’t been able to see what the formula could be:

The sequence goes:

576 for k = 4

14929920 for k = 6

534252257280 for k = 8

29855671566336000 for k = 10

2594739381460402176000 for k = 12

341103472674836305674240000 for k = 14

The quotients of consecutive terms are:

25920/1, 35784/1, 3967700/71, 313482312/3607, 444727920/3383

In the spirit of throwing pasta at the wall and seeing what sticks, I thought I would share these numbers here.

Comment by Luke Pebody — March 30, 2017 @ 6:25 am

5. Tim asked me to write something about online versions of Rota’s Basis
conjecture. Here, the n bases arrive one at a time, and after you receive
the i-th basis, you have to decide immediately in what order to write it
as i-th row of the array you are creating. Of course, the goal is that
the columns of the eventual array are again independent. Is
there an algorithm that does this for i up to and including n? There
are at least two versions of this question:

(1) The input consists of vectors in a vector space. In this case,
for _even_ n, the online version still follows from Alon-Tarsi (in
char 0), and for _odd_ n any online algorithm can be fooled in the
last steps (if the field contains sufficiently many roots of unity);
see https://arxiv.org/pdf/1312.5953.pdf

(2) The input consists of elements in an abstract matroid. In other
words, when the new basis arrives, you get told which subsets of
the elements so far are independent, but not more. In this case, an online algorithm
can be fooled regardless for any n≥3 (even with linearly realisable matroids); see page 9 of
https://arxiv.org/pdf/1312.5953.pdf

This leads to a question that I think might prove tractable: determine
the maximal number f(n)<n (n≥3) such that there exists an online
algorithm in scenario (2) for arranging f(n) bases, but not for arranging
f(n)+1 bases. E.g., does f(n) grow as a square root of n?

Comment by Jan Draisma — March 7, 2017 @ 3:57 pm

• No, I’m pretty sure $f(n)$ is linear, at least $n/4+c$ and at most $n/2+c$. Indeed you can do $k+1$ bases in $4k$ dimensions (at least for linearly realisable matroids), but you can’t do them in $2k-1$ dimensions.

For the lower bound, you simply use an extension of the argument from the linked paper above: There exist vectors $v_{i,j}\in\mathbb{R}^{2k-1}$ for $1\le i\le k$ and $1\le j\le 2k-1$ such that they are in general position (i.e. any $2k-1$ of them are linearly independent), but the vectors $v_j=\sum_{i=1}^kv_{i,j}$ for $1\le j\le 2k-1$ are the same (non-zero) vector. (Proof: fix a prime $p$ that is greater than
$\frac{((2k-1)k)!}{(2k-1)!((2k-1)(k-1))!}$, and choose the sum $v_1=v_2=\ldots=v_{2k-1}$ and the vectors $v_{i,j}$ with $i|A|+|B|,$ then at least one of $A$ and $B$ can be extended by some element of $D$.

This is true for linearly realisable matroids because the space generated by the union of $A$ and $B$ is of dimension at least $|C|$, so the intersection of the spaces generated by $A$ and by $B$ is of dimension at most $|A|+|B|-|C|$, so is less than $|D|$, so there is some element of $D$ that is not in the intersection of those spaces, so there is some element of $D$ that is either not in the space generated by $A$ or the space generated by $B$.

So our strategy in $4k$ dimension is: when the $j$th basis arrives (for $1\le j\le k+1$), make it so the first $k+1-j$ sets of 4 columns each form an independent set of size $4j$, and then use Hall’s Basis Theorem to make the other columns all be independent.

It is clear by the basis exchange axiom that we can make the first $k+1-j$ sets of 4 columns each form an independent set of size $4j$ (we start with an independent set of size $4j-4$ and have $4j$ independent vertices to choose from).

For the remainder, we have $4(j-1)$ columns (of $j-1$ vectors each) of which the first pair of pairs of columns both make a matroid of dimension at least $2(j-1)$, the next pair of pairs of columns both make a matroid of dimension at least $2(j-2)$, and so on until the final pair of pairs of columns both make a matroid of dimension at least 2.

Create a bipartite graph with partition $S_1$ of the $4(j-1)$ columns and $S_2$ of the $4(j-1)$ vectors we need to add in from the latest basis, where an edge exists if the column and the vector make an independent set (of size $j$).

Each vertex in $S_1$ makes a matroid of dimension $j-1$, so there are at least $3(j-1)$ of the vertices in $S_2$ adjacent. In particular if $T$ is a subset of $S_1$ is of size at most $3(j-1)$, the total number of neighbours of $T$ is at least $|T|$. If, on the other hand, $T$ is of size $4(j-1)-p$ where $p < (j-1)$, it includes both columns from one of the first $p+1$ pairs of columns, so it includes two pairs of columns which together make a matroid of dimension at most $2j-2-p$. So these two columns (let’s call them $A$ and $B$) are of size $j-1$, and their union contains a set $C$ of size $2j-1-p$, so if $D$ is any set of size more than $p$, either $A$ or $B$ is extendible by an element of $D$. In particular, there are at most $p - 1$ elements of $S_2$ which are not adjacent to either $A$ or $B$, so there are at least $4(j-1)-p=|T|$ neighbours of $T$.

So the conditions of Hall’s Marriage Theorem apply, and we can add the remaining vectors of the basis to the matrix and make each column independent.

My instincts are that this proof can be improved to actually show that you can do $2k$, solving the problem completely. I did this for $k=2$, but couldn’t quite do so for $k=3$.

Comment by Luke Pebody — March 25, 2017 @ 5:38 pm

• If someone can edit this to make the latex things work, that would be very welcome. It seems to have caused a huge gap including the end of the proof of the upper bound and the start of the proof of the lower bound. Is there a ‘preview my post’ feature?

Comment by Luke Pebody — March 25, 2017 @ 5:43 pm

• I tried to fix your LaTeX; hopefully this is what you wanted. I’m new to WordPress myself so I am not aware of a “preview” feature; I agree that that would be nice.

Comment by tchow8 — March 26, 2017 @ 9:16 pm

• Thanks Tim, that looks a lot better.

It has occurred to me the Lemma I needed (if A, B, C and D are independent sets in a matroid with C contained in $A\cup B$ and with |C|+|D|>|A|+|B| then either A or B can be extended by some element of D) is true by the submodularity of the rank function applied to $A\cup D$ and $B\cup D$.

The rank of the intersection of $A\cup D$ and $B\cup D$ is at least |D| clearly, and the rank of their union is at least |C|. It follows that the sum of their ranks is more than |A|+|B|, and hence either the rank of $A\cup D$ is more than |A| (and hence A is not a maximal independent set in $A\cup D$) or the rank of $B\cup D$ is more than |B| (and hence B is not a maximal independent set in $B\cup D$).

Comment by Luke Pebody — March 26, 2017 @ 9:57 pm

• Since this turned out pretty unreadable, I threw together some latex of the two proofs at https://www.overleaf.com/8773999gccdbdmfdgkm

Comment by Luke Pebody — March 26, 2017 @ 4:36 pm

• Pretty sure the same proof works to show you can do k+1 bases in 3k dimensions. What I’m using the groups of 4 for is that if you remove 0 or 1 columns, there’s still a pair remaining that works together. The same would be true with groups of 3.

Comment by Luke Pebody — March 26, 2017 @ 5:12 pm

• So I think it might be possible to extend this method to fully solve the question for online matroids.

I feel that the answer is that you can do k + 1 bases in rank 2k, but can’t necessarily do them in rank 2k-1. So far it is proved you can in rank 3k and can’t in rank 2k-1.

In order to place the (k+1)’th basis, it is sufficient to achieve that some 2 columns so far together have rank 2k, that if you cover up any column there are 2 columns remaining that together have rank at least 2k-1, that if you cover up any 2 columns there are 2 columns remaining that together have rank at least 2k-2 and so on (up to if you cover up any k-2 columns there are 2 columns remaining that together have rank at least k+2).

We stop at k-2, because it is trivial that if you cover up any k-1 columns there are 2 columns remaining that together have rank at least k+1: otherwise in the remaining k x (k+1) grid all the columns are maximal independent sets and all the rows are larger independent sets.

For k=2, you want to achieve that some 2 columns have rank 4, which is trivial to achieve in rank 4 by extending independent sets..

For k=3, you want to achieve both that some 2 columns have dimension 6 (trivial in rank 6 by extending independent sets) and also that if you cover up any column, some 2 uncovered columns have rank at least 5. Wondering if maybe someone can see how to achieve this in rank 6? (Or prove it can’t always be done)

In particular, for any k, we would be done if we could achieve that after the k’th basis, we can find k disjoint pairs of columns such that one pair combined has rank 2k, one pair has rank 2k-1, one pair has rank 2k-2, …, one pair has rank k+1. Maybe you can even arrange it so you can find k disjoint pairs of columns such that each pair combined has rank 2k?

Comment by Luke Pebody — March 28, 2017 @ 6:14 am

• I haven’t gotten around to studying your argument yet, but I just wanted to say that fully solving the question for online matroids would be an excellent partial result and I hope that others will jump in to help.

Comment by tchow8 — March 28, 2017 @ 3:08 pm

• So far I could only digest the lower bound. I think it might be useful to summarize it as follows. In an n-dimensional space, we pick a line, and n subspaces of dimension n/2 that all contain it, but are otherwise generic. For each subspace, take n/2 generating vectors. Then any n of these n x n/2 vectors are independent. So if we get them in a matrix as n/2 rows of length n, we can do nothing but randomly put them in columns. If we are unlucky, we might just group them exactly such that each column generates one of the n/2-dimensional subspaces. But then if the next row we get contains a vector from the line they have in common, we are #\$&!ed.

Comment by domotorp — March 29, 2017 @ 11:25 am

• Yes, except that the dimension of the subspaces must be less than n/2, in order for the statement ‘any n of these n by subspaces are independent’ to be true.

Comment by Luke Pebody — March 31, 2017 @ 6:25 am

• Regarding the upper bound, I think one could use some form of Hall’s theorem. For example, suppose that we could arrange the first k rows (of length 2k) such that any line is contained in at most k of the 2k k-dimensional subspaces generated by the columns. Construct a bipartite graph where one class is the collection of these 2k subspaces, while the other class is the 2k vectors of the (k+1)-th basis, and a vector is connected to a subspace if it’s not contained in it. Then the above condition would be sufficient to find a perfect matching. Unfortunately this condition is not always true, but probably we can guarantee it with some care when arranging the first k rows.

Comment by domotorp — March 30, 2017 @ 7:40 am

6. This is coming to an earlier question of Jordan-in what way could this conjecture made more general. I propose the following: Take any $n$ by $n$ array of nonzero vectors such that for any $1\leq k \leq n-1$ no matter how we pick $k$ vectors from distinct rows on each of the remaining rows we can find a vector not in their span. This is easily checked out by having bases on each of the rows or having the example of a vectors linearly independent on the first column and linear multiple of each of them on the rows.
One motivation for this is: we draw an $n$-partite graph on the rows and draw an edge between the vertices in different parts if they are independent. Rota would say that each vertex is connected to at least $n-1$ vertices in any other part and then we want to find n disjoints paths through the parts of the graph. Jordan’s example is even more rich in edges, every vertex has degree $n$.
The condition stated above seems very reminiscent of the marriage lemma condition.

Comment by VladM — March 7, 2017 @ 7:19 pm

• Once again, this fails on Colin McDiarmid’s example. Denote the edges of $K_4$ by $12, 13, 14, 23, 24, 34$ and create duplicates $14', 24', 34'$ of three of the edges. Then the sets $\{14, 14', 23\}$, $\{24, 24', 13\}$, and $\{34, 34', 12\}$ satisfy your condition but the conclusion of Rota’s Basis Conjecture fails.

I think it is going to be difficult to formulate a satisfactory generalization of this type without coming to grips with the “local obstructions” that I mentioned above.

Comment by tchow8 — March 7, 2017 @ 7:30 pm

7. On Matroids with No Small Circuits.

A paving matroid is one in which the sets of size less than the rank are
all independent. With Peter Humphries I proved that:

Theorem.
Let $B_1,\ldots ,B_n$ be disjoint sets of size $n \ge 3$, and let
$M_1 \ldots ,M_n$ be rank $n$ paving matroids on $\cup_i B_i$
such that $B_i$ is a basis of $M_i$ for each $i \in \{1,\ldots,n\}$.
Then there exist $n$ disjoint transversals $A_1,\ldots,A_n$ of
$B_1,\ldots ,B_n$ such that $A_i$ is a basis of $M_i$ for each
$i \in \{1,\ldots ,n\}$.

This strengthening of Rota’s Basis Conjecture fails for matroids
in general and for $n=2$ for paving matroids. After an unpleasant
case analysis establishing the result for the $n=3$ case, the result
is proved by an easy inductive argument.

Tim is proposing extending the results in our paper, for any fixed
$k>1$, to rank-$n$ matroids in which the $(n-k)$-element sets are
all independent. It could well be true that the above theorem itself
would hold for such matroids so long as $n$ is large relative
to $k$. To prove such a result one would likely need to do a lot
of work to establish it for some value of $n$, but then the induction
should be easy.

I don’t see this leading to a full proof of Rota’s Basis Conjecture,
but it is a concrete approach toward interesting partial results.

On a related note, it is conjectured that the proportion of $n$-element
matroids that are paving tends to one as $n$ goes to infinity. Given the
exciting new results of Jorn van der Pol and Rudi Pendavingh on the
structure of almost all matroids one might be able to prove that
Rota’s Basis Conjecture (for matroids) is almost always true.

Comment by Jim Geelen — March 7, 2017 @ 8:56 pm

• I have been reading the Geelen-Humphries paper and I would like to explain here the idea behind their induction step, which is the part of their argument that I think is most valuable if we are thinking of extending the results.

The starting point is the observation that in Rota’s Basis Conjecture, it is easy to arrange for the first column of the grid to be a basis. It is then natural to imagine covering up the first row and column and applying some kind of induction hypothesis to the remaining $(n-1)\times (n-1)$ grid. However, several issues arise if we try to do this naively. First of all, the rows of the $(n-1)\times (n-1)$ grid aren’t bases of an $(n-1)$-dimensional vector space; they’re still sitting inside an $n$-dimensional vector space. Secondly, and more seriously, even if we were to get the columns of the $(n-1)\times (n-1)$ grid to all be linearly independent, there’s no guarantee that when we return to the original grid and add back the vector $v_{1j}$ from the first row, the $j$th column will still be linearly independent.

The first idea that Geelen and Humphries have to try to address these problems is contraction. For those not familiar with matroid theory, contraction may be thought of as orthogonal projection: If $M$ is a set of nonzero vectors in a real vector space and $v\in M$, then the contraction of $M$ by $v$ (written $M/v$) is obtained by orthogonally projecting all the vectors of $M$ onto the hyperplane perpendicular to $v$, and deleting $v$ itself (which has been projected to zero, of course). This has the effect of reducing the rank by one, and also has the nice property that if we take any basis of the contraction and add back the original vector $v$, we get a basis for the original vector space. In Rota’s Basis Conjecture, the idea is that for each column $j$, we contract by $v_{1j}$. This (seemingly) addresses both issues in the previous paragraph.

However, in the process of seemingly solving these issues, we’ve introduced new issues. The vectors in the first row are all distinct (in fact, linearly independent) so contracting by $v_{1j}$ can give a totally different result from contracting by $v_{1j'}$ if $j\ne j'$. This is why Geelen and Humphries end up proving something about $n$ different matroid structures on the same ground set. Another issue is, if we contract by $v_{1j}$, how do we know that we’ll get a basis (for the contraction) in the $j$th row? We somehow need to ensure that if we combine $v_{1j}$ with the vectors $v_{j2}, v_{j2}, \ldots, v_{jn}$, the result is a basis of the original space.

This latter condition is something that Geelen and Humphries state explicitly as a lemma, and prove using the assumption that we are dealing with paving matroids. Namely, they prove that if there are no linearly dependent sets of size less than $n$ (actually, as mentioned in the previous paragraph, they prove this for $n$ different paving matroid structures on the ground set, but I’ll ignore this for simplicity), then it is possible to find $v_{j1} \in B_j$ for all $j\in \{2, 3, \ldots, n\}$ so that $\{v_{11}, v_{21}, v_{31}, \ldots, v_{n1}\}$ is a basis and also $\{v_{1j}, v_{j2}, v_{j2}, \ldots, v_{jn}\}$ is a basis for all $j\in \{2, 3, \ldots, n\}$.

The interesting thing about this lemma is that it is a type of basis exchange property. We’re starting with $n$ bases and then we’re simultaneously swapping out $n-1$ elements from the first basis with one element from each of the other bases in a way that leaves everything a basis at the end. As indicated above, this particular basis exchange property does not hold in full generality, but relies on the assumption that there are no small linearly dependent sets. Geelen and Humphries prove this lemma by induction. Here I want to focus not on the induction step of this argument, but on the argument that lets them prove the base case, because it illustrates something that I think is of general interest for Rota’s Basis Conjecture.

The key claim is the following: Suppose we have a matroid with no dependent sets of size less than $k$, where $k \ge 3$. Let $I_1$ and $I_2$ be two disjoint independent sets of size $k$, and let $S_1$ be any subset of $I_1$ of size 2 and let $S_2$ be any subset of $I_2$ of size 3. Then there exists $x \in S_1$ and $y\in S_2$ such that if we “swap” $x$ and $y$, then the resulting sets $I_1 \backslash x \cup y$ and $I_2 \backslash y \cup x$ are independent.

How do we prove this claim? First we note the following: Denote the two elements of $S_1$ by $x$ and $x'$, and let $y$ be arbitrary; then we claim that either $I_1 \backslash x \cup y$ or $I_1 \backslash x' \cup y$ is independent. For if not, they are both linearly dependent, and since there are no dependent sets of size less than $k$, they are both circuits (minimal dependent sets). They are distinct and both contain $y$, so by the circuit elimination axiom of matroid theory, their union, minus $y$, must contain a circuit. But their union minus $y$ is a subset of $I_1$ itself, which is independent; contradiction. So, either $I_1 \backslash x \cup y$ or $I_1 \backslash x' \cup y$ is independent. Now apply this argument to each element $y \in S_2$ in turn; by the pigeonhole principle, there must be some $y_1, y_2\in S_2$ such that both $I_1 \backslash x \cup y_1$ and $I_1 \backslash x \cup y_2$ are independent (or else this claim holds for $x'$, but without loss of generality we may assume that it is $x$). Then, applying the argument at the beginning of this paragraph with $I_2$ in place of $I_1$ and $y_1, y_2$ in place of $x, x'$, we can infer that for one of $y_1$ or $y_2$—and without loss of generality we may assume it is $y_1$—we must have $I_2 \backslash y_1 \cup x$ is independent as well.

The reason I think the argument in the previous paragraph is interesting is that it can be interpreted as saying that the absence of small circuits implies a weak form of base-orderability. Base-orderability is, I think, a crucial concept for the conjecture; it is known that strongly base-orderable matroids not only satisfy Rota’s Basis Conjecture but also have no local obstructions, so they satisfy various strengthenings of the conjecture. I may say something more about this later, but this comment is already long enough!

Comment by tchow8 — March 9, 2017 @ 5:16 pm

• The argument seems correct, except that it seems that half of the time you wrote independent instead of dependent; so in the claim, we need to suppose that there is no dependent set of size less than $k$. Also, let me remark that it might be better to suppose from the beginning that $I_1$ and $I_2$ are disjoint (otherwise the statement is not necessarily true).

Comment by domotorp — March 9, 2017 @ 9:06 pm

• I fixed the dependent/independent typos; thanks!

What’s a counterexample when $I_1$ and $I_2$ are not disjoint?

Comment by tchow8 — March 9, 2017 @ 10:54 pm

• If $I_1=I_2$ and $S_1 \cap S_2=\emptyset$.

Comment by domotorp — March 10, 2017 @ 9:11 am

• Ah, I see. Technically, $I_1 \backslash x \cup y$ and $I_2 \backslash y \cup x$ are independent in this case as well, but they now have smaller cardinality than the original independent sets, and the term “swap” is misleading, so I agree that it is better to specify that $I_1$ and $I_2$ are disjoint.

Comment by tchow8 — March 20, 2017 @ 4:59 pm

• I think that the prospects for generalizing Geelen–Humphries are good. Here are some thoughts.

Definition. Let $k$ be a non-negative integer. Say that a matroid of rank $n$ is $k$-paving if it has no circuits of size less than $n-k+1$.

Thus a 0-paving matroid is a uniform matroid and a 1-paving matroid is a paving matroid in the usual sense. I believe that it is straightforward to prove the following Proposition (though it would be good for someone to verify):

Proposition. For any fixed $k$, the class of $k$-paving matroids is minor-closed; i.e., if you delete or contract an element of a $k$-paving matroid, the result is a $k$-paving matroid.

My first thought is that it is natural to try to generalize the Geelen–Humphries result to $k$-paving matroids.

A second observation is that, as I alluded to an in earlier comment, a key lemma in their paper is the following:

Key Lemma. Let $B_1, \ldots, B_n$ be disjoint sets of size $n \ge 3$, and let $M_1, \ldots, M_n$ be rank-$n$ paving matroids on $\bigcup_i B_i$ such that $B_i$ is a basis of $M_i$ for each $i\in \{1, \ldots, n\}$. Then there is an ordering of the elements of $B_1$ as $a_1, \ldots, a_n$ and a transversal $\{b_2, \ldots, b_n\}$ of $\{B_2, \ldots, B_n\}$ such that the set $(B_1 \backslash \{a_2, \ldots, a_n\}) \cup \{b_2, \ldots, b_n\}$ is a basis of $M_1$, and for all $j \in \{2, \ldots, n\}$, $(B_j \backslash b_j) \cup a_j$ is a basis of $M_j$.

One thing that I realized recently is that if all the matroids $M_i$ are identical, with $M_1 = \cdots = M_n = M$ say, then the Key Lemma holds for any matroid $M$, not just paving matroids. This follows from the following result, which I believe was first proved in “Comments on bases in dependence structures” by Richard Brualdi, Bull. Austral. Math. Soc. 1 (1969), 161–167.

Strong Basis Exchange. If $B_1$ and $B_2$ are bases of a matroid, then for any $e \in B_1$ there exists $f\in B_2$ such that both $(B_1 \backslash e) \cup f$ and $(B_2 \backslash f) \cup e$ are bases.

To prove the Key Lemma when $M_1 = \cdots = M_n = M$, use Strong Basis Exchange to swap each $a_j$ in turn with some $b_j \in B_j$, as $j$ runs from 2 to $n$.

I think that the Key Lemma is false in general if the $M_i$ are not all identical and are not all paving matroids (though I don’t have a counterexample offhand; maybe someone else can construct one). Certainly Brualdi’s proof breaks down. However, I suggest the following plan: Tinker with the statement of the Key Lemma, as well as the statement of the Theorem in Geelen’s comment above, to get the induction step of Geelen–Humphries to work for a more general class of matroids than paving matroids (perhaps $k$-paving matroids, or perhaps an even more general class of matroids). This seems eminently doable to me. Of course, even if this plan succeeds, it doesn’t prove much by itself in the absence of a proof of the base case. But I think it would be an important step forwards.

Comment by tchow8 — March 20, 2017 @ 6:16 pm

• Here are some further thoughts about the “tinkering” I mentioned above.

Suppose that we modify the Theorem in Geelen’s comment by replacing “paving matroid” with “2-paving matroid.” I don’t know if this modified assertion is true, but suppose we try to prove it using similar methods. If I haven’t made a mistake, the class of 2-paving matroids is minor-closed, so the same strategy, of using contraction to reduce to a smaller instance, has a fighting chance of working. In fact, I think that if the Key Lemma holds for 2-paving matroids then the inductive step (though not necessarily the base case) of the proof of the Theorem should go through.

However, some trouble arises when we try to prove the Key Lemma by mimicking the argument of Geelen and Humphries. I think you can only conclude that if you pick any three elements of a basis then you can swap one of them out with an element of a disjoint basis of another 2-paving matroid (provided that the rank is at least seven), and this lets you get $(B_1 \backslash \{a_2, \ldots, a_{n-1}\}) \cup \{b_2, \ldots, b_{n-1}\}$ to be a basis as well as $(B_j\backslash b_j) \cup a_j$ to be a basis for $j \in \{2, \ldots, n-1\}$, but then you get stuck at the last step $j = n$.

Now maybe the Key Lemma is still true, and one just needs a different argument. On the other hand, here’s another possible diagnosis of what’s going wrong: We know that the Key Lemma is true for all matroids when all the matroid structures are equal, and what’s causing us trouble is that we’re trying to prove it for a whole sequence of distinct matroid structures $M_1, \ldots, M_n$. It seems difficult to avoid considering different matroid structures on the same ground set if we stick to the strategy of contracting a different vector for each column. Nevertheless, from the point of view of Rota’s Basis Conjecture, we don’t necessarily need to prove the Key Lemma for arbitrary $M_i$. For example, the $M_i$ that arise in the (purported) inductive proof of Rota’s Basis Conjecture are all minors of the same initial matroid. So maybe what we need to do is to narrow our sights a little and only try to prove the Key Lemma for matroids $M_i$ that enjoy some degree of similarity, that is preserved when we perform certain sequences of deletions and contractions from the same starting matroid. By the same token, the Theorem itself might need to be similarly narrowed. These narrowings might make the Key Lemma (and the Theorem) easier to prove, yet still general enough to imply Rota’s Basis Conjecture.

Comment by tchow8 — March 21, 2017 @ 8:44 pm

• Question. Let $M$ be a base-orderable matroid with ground set $E$, and let $E' \subseteq E$. Let $M_1$ and $M_2$ be matroid structures on $E'$, each of which is obtained by a sequence of deletions and contractions of the elements of $M \backslash E'$ (but not necessarily the same sequence of deletions and contractions; so although $M_1$ and $M_2$ are both minors of $M$ and have the same ground set $E'$, they are not necessarily isomorphic). Suppose further that $M_1$ and $M_2$ have the same rank. Is it necessarily the case that if $B_1$ is a basis of $M_1$ and $B_2$ is a basis of $M_2$, then there exists a bijection $f : B_1 \to B_2$ such that for all $e\in B_1$, $(B_1 \backslash e) \cup f(e)$ is a basis of $M_1$ and $(B_2 \backslash f(e)) \cup e$ is a basis of $M_2$?

It is plausible to me that the answer is yes. For example, note that an independent set of a minor is independent in the original matroid, so both $B_1$ and $B_2$ are necessarily independent in $M$. It seems plausible that we could extend both of them to bases in $M$, use the base-orderability of $M$ to get a bijection, and then restrict the bijection. Or something like that.

If the answer is yes, then I think that the Geelen–Humphries induction argument goes through for base-orderable matroids. In the Theorem of Geelen’s comment above, we replace “paving matroid” by “base-orderable matroid,” and we also require that all $n$ matroid structures arise from a sequence of deletions and contractions from the same starting matroid. We modify the Key Lemma similarly. The Key Lemma then follows from a positive answer to the Question, because we can swap each $a_j$ in turn with some $b_j \in B_j$ as $j$ runs from 2 to $n$. Then we can apply the induction hypothesis to the contractions $M_i/a_i$ restricted to the set $(B_2 \cup \cdots \cup B_n) \backslash \{b_2, \ldots, b_n\}$.

Comment by tchow8 — March 29, 2017 @ 1:23 am

• Unfortunately, the answer to the Question is no. Let $M$ be the matroid associated with the graph that is a disjoint union of two triangles $T_1$ and $T_2$. Choose $e_1 \in T_1$ and $e_2 \in T_2$, and let $M_1 = M / e_1 \backslash e_2$ and let $M_2 = M / e_2 \backslash e_1$. Then all these matroids are base-orderable, and $M_1$ and $M_2$ are even abstractly isomorphic (but not equal), but it is easy to check that the desired bijection $f$ does not necessarily exist.

Comment by tchow8 — March 29, 2017 @ 5:51 pm

• The Key Lemma is not true for all “2 paving matroids”. In particular, the rank of $B_2\cup B_3\cup\ldots B_n$ could be n-2 in $M_1$.

Comment by Luke Pebody — April 2, 2017 @ 7:33 am

8. Let me say a bit more about base-orderability. A matroid is strongly base-orderable if, for any two bases $B_1$ and $B_2$, there exists a bijection $f:B_1\to B_2$ such that for any subset $S\subseteq B_1$, the set $B_1 \backslash S \cup f(S)$ is a basis. It is not hard to show that this axiom implies that $B_2 \backslash f(S) \cup S$ is also a basis. Strong base-orderability is a very strong condition. We have the following result:

Proposition. Let $M$ be a strongly base-orderable matroid of rank $n \ge 2$ with $2n$ elements, and suppose that there exists a partition of $M$ into $2$ disjoint bases. Given any $n$ disjoint independent sets $I_1, \ldots, I_n$ with $0 \le |I_i| \le 2$, there exists an $n\times 2$ grid such that the elements of $I_i$ appear in row $i$ and such that both columns are bases of $M$.

Sketch of proof: Let $B_1$ and $B_2$ be disjoint bases of $M$. Denote the elements of $B_1$ by $1, 2, \ldots, n$. Let $f$ be the the bijection guaranteed by strong base-orderability, and write $1'$ for $f(1)$, $2'$ for $f(2)$, etc. Construct any $n\times 2$ grid such that the elements of $I_i$ appear in row $i$. Then a Hall’s marriage theorem argument along the lines of Proof 2 of Theorem 1 in this paper shows that we can permute the elements within each row so that both columns contain the numbers $1$ through $n$ if we erase the primes. Strong base-orderability implies that the two columns must be bases.

Theorem 4 of this paper then implies that the generalization of the above Proposition with $2$ replaced by any larger number is also true, and so this yields a proof of Wild’s theorem that Rota’s Basis Conjecture holds for strongly base-orderable matroids.

As already discussed in earlier comments, there are local obstructions (such as $K_4$ and $J$) that prevent the above Proposition from being true in general. The point is that these local obstructions vanish in the presence of strong base-orderability. Thus if we wish to search for local obstructions, we should look at non-base-orderable matroids. Another way to think about it is that if we want to generalize Wild’s theorem, we might want to look at weaker base-orderability hypotheses and see how far we can get with them. A very optimistic, but not inconceivable, route to proving Rota’s Basis Conjecture would be to formulate a generalization to $n\times k$ grids that has some kind of weak base-orderability hypothesis $H$ with two properties: (1) when $k < n$, $H$ blocks annoying counterexamples like $K_4$ and $J$, and (2) when $k = n$, $H$ is automatically satisfied by all matroids.

Administrative note: I will be unavailable to participate in Polymath for the next week or so. Obviously, this should not prevent others from continuing to work.

Comment by tchow8 — March 10, 2017 @ 12:00 am

• Here is another thought about base-orderability. A matroid is said to be base-orderable (note that I have dropped the adverb “strongly”) if for any two bases $B_1$ and $B_2$, there exists a bijection $f : B_1 \to B_2$ such that for every $x \in B_1$, the sets $(B_1 \backslash x) \cup f(x)$ and $(B_2 \backslash f(x)) \cup x$ are both bases. The following theorem is proved in J. de Sousa and D. J. A. Welsh, “A characterisation of binary transversal structures,” J. Math. Anal. Appl. 40 (1972), 55–59.

Theorem. A binary matroid (i.e., a matroid representable by vectors in a vector space over the field with two elements) is base-orderable if and only if it contains no $K_4$ minor.

As we have seen, base-orderability (or at least strong base-orderability) makes life easier for the would-be prover of Rota’s Basis Conjecture. So one could imagine a strategy for proving Rota’s Basis Conjecture for binary matroids that proceeds along the following lines: When base-orderability seems to be useful, split into two cases. In Case A, go ahead and assume base-orderability. In Case B, exploit the above theorem to conclude that there must be a $K_4$ minor lying around somewhere, and do something special to take care of it.

Comment by tchow8 — March 21, 2017 @ 9:01 pm

• After studying the literature a bit more, I learned that binary matroids that contain no $K_4$ minor are the same as series-parallel matroids, and in particular are strongly base-orderable (because strong base-orderability is closed under series and parallel connection). Thus Rota’s Basis Conjecture is known to be true for binary matroids with no $K_4$ minor.

This suggests to me that binary matroids are a good testing ground for proof strategies that involve grappling with what I’ve been calling “local obstructions.” In some sense, $K_4$ is the only local obstruction that we have to worry about for binary matroids; we don’t have to worry about $J$ (which is not binary) or any other unknown local obstructions.

Comment by tchow8 — April 8, 2017 @ 3:56 am

• To be a little less vague, here is an outline of a possible strategy for proving Rota’s Basis Conjecture for binary matroids. Recall that strongly base-orderable matroids satisfy an $n\times k$ version of the conjecture: If we have $n$ disjoint independent sets of size $k$, whose union can be partitioned into $k$ disjoint bases, then there is an $n\times k$ grid whose rows are the given independent sets and whose columns are all bases. So for binary matroids, we could imagine constructing the first $j$ columns of our desired $n\times n$ grid (for some hopefully not-too-large value of $j$) in such a way that the remaining $n(n-j)$ elements contain no $K_4$ minor (and therefore comprise a strongly base-orderable matroid); then as long as these $n(n-j)$ elements can be partitioned into $n-j$ disjoint bases, we will be done.

Of course, this plan requires us to get a handle on how many $K_4$ minors there can be in a binary matroid. So, as a preliminary step, let me pose this question. If $M$ is a binary matroid, let $k(M)$ denote the minimum number of elements we have to delete from $M$ to produce a $K_4$-minor-free matroid. Abusing notation slightly, for every positive integer $n$, let $k(n) = \max_M k(M)$ where the maximum is over all binary matroids $M$ with $n$ elements. What can we say about how big $k(n)$ is?

Comment by tchow8 — April 8, 2017 @ 3:58 pm

• I think that $k(M)$ can be as large as $n-\sqrt{8n}$ already for binary matroids. This is because a $K_4$ minor free graph can have at most $2|V(G)|-3$ edges. So if the $n$ elements are arranged to form all edges of a complete graph on $\sqrt{2n}$ vertices, we get $k(M)\ge n-\sqrt{8n}$.

Comment by domotorp — April 8, 2017 @ 8:45 pm

• Can you please explain why a $K_4$-minor-free (simple?) graph can have at most $2|V(G)| - 3$ edges?

Comment by tchow8 — April 10, 2017 @ 2:19 pm

• Hadwiger proved minimum degree is at most 2.

Comment by Luke Pebody — April 10, 2017 @ 2:26 pm

• Sorry, I don’t understand this comment. What is the “minimum degree”? It sounds like the smallest degree of a vertex but with that definition, $K_n$ plus an isolated vertex has minimum degree zero, which is certainly at most 2. Can you provide a reference?

Comment by tchow8 — April 10, 2017 @ 5:52 pm

• Ah, good! Thanks. So it looks like my naive plan is unlikely to succeed. Still, I am wondering if perhaps it is possible to characterize “benign” versus “malignant” occurrences of $K_4$ minors and to figure out how to construct columns inductively in a way that avoids painting oneself into a malignant corner. The paper by Harvey, Kiraly, and Lau should give some important clues about what kinds of malignancies can arise.

By the way, I think it’s also worth mentioning that Marcel Wild has another paper that may be relevant: Base exchange properties of graphic matroids, Discrete Math. 148 (1996), 253-264. Although graphic matroids were mentioned early on in the Polymath project as a possible special case of interest, it hasn’t been clear to me until now what would make the graphic matroid case easier than the general case. But since $K_4$-minor-free graphs are strongly base orderable, and since graphic matroids are known to have special basis exchange properties that aren’t true in general, I now feel more hopeful that graphic matroids may be more tractable.

Comment by tchow8 — April 11, 2017 @ 2:28 pm

• I’m having trouble figuring out how to prove that if you delete an element $e$ from a base-orderable matroid $M$ then the resulting matroid $M \backslash e$ is still base-orderable. Of course if the rank $r(M \backslash e)$ equals the rank $r(M)$ then this is obvious, but if $r(M \backslash e) = r(M) - 1$ then the obvious line of argument is to extend the given bases $B_1$ and $B_2$ of $M\backslash e$ to bases $B_1'$ and $B_2'$ of $M$, then restrict the bijection $f : B_1' \to B_2'$ to $B_1$. The thing I’m worried about is that maybe $e \in B_2$ and $f$ sends some element of $B_1$ to $e$, in which case $f$ does not restrict to a bijection between $B_1$ and $B_2$.

I’m confident the result is true because I see it stated everywhere. Welsh’s book even gives a proof but it’s essentially the above argument and so it looks to me as though it has a gap. Oxley’s book states it as an exercise and cites the literature, but when I looked up the papers the terminology was sufficiently different that I couldn’t figure out how to extract what I wanted. Can someone else help?

Comment by tchow8 — March 23, 2017 @ 1:58 am

• O.K., I see what I was missing now:

Fact. Let $M$ be base-orderable. Let $B_1$ and $B_2$ be bases of $M$, and let $f : B_1 \to B_2$ be a bijection whose existence is guaranteed by the base-orderability axiom. Then $f$ restricted to $B_1 \cap B_2$ must be the identity map.

The proof is obvious as soon as you think about it: If $e \in B_1 \cap B_2$ and $f$ is not the identity map, then $(B_2 \backslash f(e)) \cup e$ contains “two copies of $e$” and hence cannot be a basis.

So now returning to the “obvious line of argument” in my comment above, if $r(M\backslash e) = r(M)-1$ then that must mean that every basis of $M$ contains $e$. In particular, $B_1'$ and $B_2'$ must both contain $e$, and by the above Fact, $f$ is the identity on $e$. So $f$ does indeed restrict to a bijection between $B_1$ and $B_2$, and we’re done.

I am now hopeful about proving Rota’s Basis Conjecture for base-orderable matroids, which would generalize Wild’s result that the conjecture holds for strongly base-orderable matroids.

Comment by tchow8 — March 28, 2017 @ 2:52 pm

9. Some remarks on Alon-Tarsi conjecture as a correlation inequality (that Tim asked about). (I inquired a few random matrices friends and this comment reflects also a discussion with Ofer Zeitouni and remarks by Krzysztof Oleszkiewicz and Ron Peled).
Start with any symmetric (around zero) real probability distribution $D$ and consider an $n \times n$ random matrix $X=(x_{ij})$ with entries i.i.d. from $D$ and consider the expectation

${\bf E}=\mathbb E (det (X)^n (\prod_{1 \le i,j \le n} x_{ij}))$.

When you evaluate this expectation in terms of monomials in the $X_{ij}$ the contribution of monomials that appears with degree one is zero. (Because $D$ is symmetric around zero). The contribution to the monomial
$\prod_{i,j} x_{ij}^2$ is precisely a signed summation of Latin squares. This comes from the product of $n$ disjoint generalized diaginals in the expansion of $det(X)^n$,
The sign is the product of signs of all $n$ permutations representing these generalized diagonals.

This expression does not depend on the distribution, it is zero when $n$ is odd, and the common belief that for even values of $n$ there are more even than odd Latin squares gives (I think) that

${\bf E}$ is negative when $n=2~(\mod 4)$ and ${\bf E}$ is positive if $n=0~(\mod 4)$. (This should be double checked).

1) I am less optimistic now that this probabilistic interpretation is useful. For Gaussian matrices there are nice expressions for the joint eigenvalues distribution but it does not seem to mix well with the product of entries. Of course the key would be to give another expression for this correlation or a wider context but I dont see any.

2) It would be nice to examine the case for complex distributions and some familiar families of random matrices like random Hermitian matrices.

3) This observation is essentially Theorem 1.9(e) of Kumar and Landsberg’s paper CONNECTIONS BETWEEN CONJECTURES OF ALON-TARSI, HADAMARD-HOWE, AND INTEGRALS OVER THE SPECIAL UNITARY GROUP.

Comment by Gil Kalai — March 10, 2017 @ 8:07 am

• There are very relevant papers by Jeannette Janssen. One classic paper The Dinitz problem solved for rectangles contains a proof based on Alon-Tarsi theorem of a rectangle case of Dinitz conjecture. It is very interesting if the polynomial used by Janssen can beused to prove a “rectangular” (this a bit weaker) case of the Rota conjecture. (Note that a similar slightly weaker form of the conclusion of Rota conjecture is conjectured by Ron Aharoni in much greater generality.) The other paper is On even and odd Latin squares. It contains the relation between the various definitions of signs of matrices (mentioned in my previous comment). Another remark is that the conjecture that there are more even Latin squares than odd ones may well be a wishful thinking based on a few very small cases. I wonder if somebody can offer a heuristic explanation why this should be true.

Comment by Gil Kalai — March 14, 2017 @ 8:07 am

• As a starting point for a heuristic argument that there are more even Latin squares than odd ones, we might note that the number of even derangements of $n$ minus the number of odd derangements of $n$ is $(-1)^{n-1}(n-1)$. Admittedly, I’m not quite sure what to do with this observation…

Comment by tchow8 — March 28, 2017 @ 3:11 am

• This means that the number of row-even partial $2\times n$ Latin squares minus the number of row-odd partial $2\times n$ Latin squares is
$(-1)^{n-1}(n-1)n!$.

Clearly for $3\times n$ this difference will be 0, so maybe we should instead consider partial $m\times n$ matrices where the first row is $1, 2, \ldots, n$.

Then for $m=1$ the difference is 1, for $m=2$, it is $(-1)^{n-1}(n-1)$,
and for $m=3$, it is 0, 0, 2, 0, 72, -320, 3600, -32256, …, which is at http://oeis.org/A098276. For $m=4$, the sequence starts 0,0,0,24,-576,12960.

Comment by Luke Pebody — March 28, 2017 @ 4:59 am

• It may be worth mentioning that Levent Alpoge has used the Kumar–Landsberg result to estimate the size of the Alon–Tarsi constant. Unfortunately, from the point of view of Rota’s Basis Conjecture, he provides only an upper bound rather than a lower bound, but I think it is still an interesting result.

Comment by tchow8 — March 20, 2017 @ 6:29 pm

10. […] the occasion of Polymath 12 devoted to the Rota basis conjecture let me remind you about the Alon-Tarsi conjecture and test […]

Pingback by Test Your Intuition about the Alon-Tarsi Conjecture | Combinatorics and more — March 15, 2017 @ 10:29 am

11. Let me provide another possible inductive approach:
Let $S$ be a sequence split into some subsequences $S_1, \ldots, S_r$ (called color classes).
We say that another subsequence is rainbow, if it contains at most one element from each $S_i$.

Now the lemma: (from https://arxiv.org/abs/1702.08170)
Let $M$ be a matroid of rank $r$ and $S$ be a sequence of $k\cdot r$ elements from $M$,
split into $r$ subsequences, each of length at most $k$.
Then any largest independent rainbow subsequence of $S$ is a basis of $M$
if and only if (*) there does not exist an integer $s and set of $s+1$ color classes, such that the union of these color classes has rank $s$.

The idea would be to remove the largest independent rainbow set, leaving $(k-1)r$ elements and use recursion. This of course does not always work, in many cases the remaining elements will not satisfy (*),
but it seems to me that in this case one can try to use some backtracking (going several steps back if necessary) and exchange axiom to force (*).

Comment by Pavel Paták — March 23, 2017 @ 9:13 am

• This is an interesting idea. My guess is that a more promising approach than backtracking is to try to modify/generalize the lemma to make the induction/recursion go through without backtracking. Once you start backtracking, it becomes difficult to keep track of all the different cases that could arise.

More specifically, can you give examples of how (*) can fail after a basis is removed? Then we could think about what conditions are needed to ensure that this scenario can’t happen. That will require modifying the induction hypothesis and maybe force more conditions, but hopefully the modification process will eventually converge.

Comment by tchow8 — March 28, 2017 @ 1:27 am

12. Is Rota’s basis conjecture]implied by the same statement restricted to vector spaces over a fixed field (say $\mathbb{R}$, or even $\mathbb{Z}_2$)?

Comment by dominiczypen — March 28, 2017 @ 6:17 am

• This is not known.

Comment by tchow8 — March 28, 2017 @ 2:41 pm

13. I wonder if the following version of colorful Tverberg theorem is correct. Consider $d+1$ (affinely independent) subsets of size $d+1$ of $\mathbb R^d$ such that the origin belongs to the interior of the convex hull of each set. Is it possible to find $d+1$ sets of size $d+1$ such that each set is a rainbow set and the interior of the convex hulls of all these sets have a point in common.

Comment by Gil Kalai — March 31, 2017 @ 10:36 am

• I would have a couple of questions regarding this very interesting conjecture.
1, Why is affinely independent needed?
2, In the conclusion, why are you asking for an arbitrary point in common instead of the origin?
3, Is the condition that the origin is in the convex hulls known to be needed for the conclusion? (I know that a topological version of Tverberg’s theorem has been disproved, but the current statement is about convex hulls.)

Comment by domotorp — April 1, 2017 @ 9:12 am

• Well, there are few versions of this conjecture that can be asked. For the condition you can ask that

a) The points in each set are affinely independent and
b) That the origin is in the interior of their convex hull.

And for the conclusion you can ask that

1) There are d+1 rainbow sets so that there is a point in their intersection and
2) There are d+1 rainbow sets such that the origin is in the interior of their intersection.

(Of course, b is stronger than a and 2 is stronger than 1.)

I asked if b implies 1. We can also make two stronger conjectures that a implies 1 and that b implies 2.

Now there is a conjecture by Reay that says that $(d+1)^2$ points in general conjecture have a partition into $d+1$ sets (necessarily all of size $d+1$) such that the interior of their convex hulls intersect. The conjecture a implies 1 is even stronger. Also b implies 2 seems very strong. In particular, b implies 2 give a very strong form of Barany’s colorful Caratheodory theorem which might be too good to be true.

Comment by Gil Kalai — April 1, 2017 @ 8:28 pm

• “b implies 2” is indeed too good to be true. Consider the following sets of points in the plane:
{(-2,2),(1,-2),(2,-1)}, {(-1,-2),(0,2),(1,-2)}, {(-2,-1),(-1,-2),(2,2)}.
Any set containing the origin in its interior must contain a point with positive second coordinate,
so at least one of (-2,2),(0,2),(2,2). If there are 3 disjoint rainbow sets, then each must contain exactly one of (-2,2),(0,2),(2,2).
But there is no rainbow set containing only (-2,2) from this set of points.

Comment by Tamás Király — April 6, 2017 @ 9:08 am

• Dear Tamas, Indeed I think that also a implies 1) is not true. Does your example satisfies 2)?

Comment by Gil Kalai — April 6, 2017 @ 4:22 pm

• Nice! This example seems to suggest that b) doesn’t seem to have much advantage compared to a).

Comment by domotorp — April 6, 2017 @ 7:18 pm

• @domotorp Why is that? is b implies 1 still open?

Comment by Gil Kalai — April 18, 2017 @ 8:43 pm

• It is still open of course, since a) implies 1 is also open. But I wonder if for example it is possible to construct from a d-dim input S satisfying a) a (d+1)-dim input S’ satisfying b) so that if S’ doesn’t satisfy 1, nor does S.

Comment by domotorp — April 19, 2017 @ 4:29 am

• I think I had an example against a) implies 1) but I have to find it.

Comment by Gil Kalai — April 19, 2017 @ 6:48 am

• Wouldn’t that give an example showing that $t(d,r)>r$ for $r=d+1$ in the (non-topological) colorful Tverberg problem?

Comment by domotorp — April 19, 2017 @ 7:59 am

• Ah I meant another 1) (from the first comment) that there is a point in common to the interior of the simplices. Sorry for the confusion.

Comment by Gil Kalai — April 19, 2017 @ 3:20 pm

14. I want to comment on the prospects for generalizing the Geelen–Humphries induction argument to a proof of the full conjecture.

First, as they point out in their paper, the following statement is FALSE:

False claim. Let $B_1,\ldots ,B_n$ be disjoint sets of size $n \ge 3$, and let $M_1, \ldots ,M_n$ be arbitrary rank $n$ matroids on $\cup_i B_i$ such that $B_i$ is a basis of $M_i$ for each $i \in \{1,\ldots,n\}$. Then there exist $n$ disjoint transversals $A_1,\ldots,A_n$ of $B_1,\ldots ,B_n$ such that $A_i$ is a basis of $M_i$ for each $i \in \{1,\ldots ,n\}$.

This is false for a rather trivial reason. The only control you have over the structure of $M_i$ is that $B_i$ is a basis, and the elements of the other $B_j$ could even be loops in $M_i$ (in the context of vector spaces, a “loop” is a copy of the zero vector). So if we want to prove Rota’s Basis Conjecture for all matroids this way, we need to impose some additional conditions on the $M_i$—strong enough to block this kind of counterexample, but weak enough that we can specialize to the case $M = M_1 = \cdots = M_n$ and infer Rota’s Basis Conjecture for an arbitrary rank $n$ matroid.

At first I thought that insisting that all the $M_i$ be minors of the same matroid should buy us something, but Jim Geelen pointed out to me by email that given any two matroid structures $M_1$ and $M_2$ on a common ground set $E$, there exists some matroid $M$ that has both $M_1$ and $M_2$ as minors. So that’s not enough.

Here is another remark. The Geelen–Humphries induction is based on the following construction: We start by figuring out how to arrange the first row and the first column of the desired grid, and (this is the Key Lemma) we do this in such a way that if we swap the first element in row $i$ with the first element in column $i$, then row $i$ remains a basis. Then we apply induction to arrange the remaining $(n-1) \times (n-1)$ elements, without disturbing the first row and column. It is therefore natural to ask if this general framework has any hope of proving the full conjecture. That is, suppose we are given $n$ bases of the same matroid, as in the original Rota Basis Conjecture. Suppose the first column and the first row are arranged so that the first column is a basis, and that for every $i$, swapping the first element in row $i$ with the first element in column $i$ leaves row $i$ a basis. Then does there necessarily exist a rearrangement of the remaining $(n-1) \times (n-1)$ elements (without disturbing the first row and column) that satisfies the conclusion of Rota’s Basis Conjecture?

Unfortunately, the answer is no, and a simple counterexample is again provided by our old friend $K_4 = \{12, 13, 14, 23, 24, 34\}$. Let’s take the rows to be the bases $\{14, 12, 34\}$, $\{12, 13, 24\}$, and $\{34, 14, 23\}$, in that order. The first column and the first row are equal, so the swapping condition is trivially satisfied. On the other hand, it is easily checked that there is no way to make all three columns be bases without disturbing the first column.

I had been hoping that because the Geelen–Humphries induction goes from an $(n-1)\times (n-1)$ square to an $n \times n$ square without passing explicitly through non-square rectangles, we might be able to sidestep the $K_4$ obstruction. It looks like that is not possible. It’s still conceivable to me that if we choose the first column more delicately, we can still make this kind of induction argument work, but once again it looks like we have to come to grips with “local obstructions” like $K_4$.

Comment by tchow8 — March 31, 2017 @ 3:54 pm

• I have a more interesting example to show that all sets of size k being independent in all the matroids is not sufficient to prove the many-matroids version of RBC for rank 2k (even when all rows are independent in all the spaces).

In my example, the first 2k-1 matroids are all the same, and are the union of the k-uniform matroid on the left half of the grid and the k-uniform matroid on the right half of the grid. (By k-uniform I mean that all k element sets are independent, but no k+1 element sets are)

It is clear that this is of rank 2k and that a set of size 2k is independent if exactly k elements are in the left half. In particular, all the rows are independent.

Now if we have k-1 independent disjoint transversals from this matroid, the remaining elements will also be an independent transversal from this matroid.

For the final matroid, we take the same matroid except we switch the left elements and right elements on one row. Any set of size 2k that had exactly one element on that row could not be independent in both this matroid and the original matroid, so no complete rainbow exists.

Comment by Luke Pebody — March 31, 2017 @ 6:27 pm

15. I’m still hoping that someone will do a computational search for counterexamples consisting of five sets of independent sets of size 2. If nobody else does it then I guess I will eventually do it, although I’m not that great a programmer. Let me mention that Mayhew and Royle have found all matroids with nine elements, so that should simplify the calculation; using their database, we can quickly populate 9 of the 10 slots in the grid, and then run through all the possibilities for adding a 10th element. The ideas described in Mayhew and Royle’s paper should be helpful, although they focus a lot of attention on eliminating isomorphic copies, and that’s not so important for our calculation (presumably there won’t be that many counterexamples, especially if we don’t bother with configurations that contain $K_4$ among the first 9 elements, and so isomorphism testing can be carried out as a post-processing step).

Comment by tchow8 — April 1, 2017 @ 12:21 am

• That seems like it would be really useful, but the link to the db at http://people.csse.uwa.edu.au/gordon/small-matroids.html no longer works.

Comment by Luke Pebody — April 1, 2017 @ 7:11 am

• I wrote to Gordon Royle and his reply appears below.

Yes, my site was lost when I moved department…

https://cloudstor.aarnet.edu.au/plus/index.php/s/GBZYfixUsmaRzbO

Unzip the file and it lists the bases of each matroid, one per line in a format similar to the following:

63 5 3 10 ‘012’ ‘013’ ‘023’ ‘123’ ‘014’ ‘024’ ‘124’ ‘034’ ‘134’ ‘234′

This means

“Matroid number 63 has size 5, rank 3 and has 10 bases, which are {0,1,2}, {0,2,3}, {1,2,3}, …, {2,3,4}”

Enjoy…

Gordon

Comment by tchow8 — April 3, 2017 @ 3:01 pm

• The index numbers of the matroids on at most 9 elements which are of rank at most 5, are the union of two independent sets and can have some disjoint pairs such that whenever you express the matroid as a union of two independent sets, some pair is contained in one of the two parts are:

ground set of size 6: [130]

ground set of size 7: [381]

ground set of size 8: [1009, 1042, 1089, 1129, 1135, 1144, 1157, 1196, 1207, 1213, 1216, 1217, 1218, 1232, 1245, 1915]

ground set of size 9: [194069, 194127, 194232, 194452, 194459, 194475, 194488, 195069, 195070, 195086, 195095, 195099, 195100, 195101, 195103, 195104, 195105, 195106, 195107, 195108, 195135, 195157]

It’s possible there are more of size 9. Will confirm later.

Comment by Luke Pebody — April 4, 2017 @ 4:28 am

• Confirmed. This list is complete.

Comment by Luke Pebody — April 4, 2017 @ 4:50 am

16. Michael Nielsen and Terry Tao created a page on the Polymath Wiki about Rota’s Basis Conjecture, and I have been fleshing it out. It is still unfinished, particularly the section on partial results, but the References section should be pretty complete. If anyone sees a reference that I have missed, please comment here and I will add it.

Let me highlight two references that I personally feel deserve extra attention. The first is the paper by Marcel Wild. I have already mentioned it but I have never fully digested it, and glancing at it now, I think that it deserves closer scrutiny. The second is an unpublished manuscript of Michael Cheung that claims to computationally check Rota’s Basis Conjecture for $n = 4$ for all matroids. I’m a little surprised that this computation is feasible, given how many 16-element matroids there are, so it would be good for someone to verify it.

Comment by tchow8 — April 6, 2017 @ 3:21 am

17. I’ve started reading Marcel Wild’s paper more carefully and one thing I noticed, which I had either not noticed before or totally forgotten about, was that he has an interesting partial result about graphic matroids. If I’m reading it correctly, Theorem 3 says that if $G$ has maximum degree $d < n/3$ then it is possible to construct $\lceil (n+3d)/2d\rceil - 1$ columns of the desired grid. (Wild says "rows" instead of "columns" because his notation is the transpose of ours.)

This seems to be a pretty strong result for small $d$. Now if $d$ is small, then the bases necessarily overlap a lot, so it's certainly a special case, but still interesting I think. I will try to understand the proof technique better and describe it in a future comment.

Comment by tchow8 — April 17, 2017 @ 3:29 pm

• I believe I understand Wild’s proof now. He uses Rado’s theorem to construct the columns one at a time, and uses the low degree of the graph to show that the process can be continued for the stated number of columns.

I don’t think we’ve mentioned Rado’s theorem explicitly in Polymath 12 yet. It’s a generalization of Hall’s marriage theorem. If you have sets $S_1, S_2, \ldots, S_n$ of elements from a matroid such that for all $r \le n$, the rank of the union of any $r$ of the sets is at least $r$, then there is an independent set $\{x_1, x_2, \ldots, x_n\}$ such that $x_i \in S_i$ for all $i$.

Suppose now that we have a simple graph $G$ (i.e., no loops or multiple edges) with no isolated vertices, and let $n$ be the rank of the associated matroid. Rota’s Basis Conjecture hypothesizes that we have $n$ bases $B_1, \ldots, B_n$. Constructing the first column of the desired grid is easy, but let us do it by applying Rado’s theorem. We have to check that for all $k\le n$, the rank of the union of any $k$ of the $B_i$ is at least $k$. But even one of the $B_i$ has rank $n \ge k$ by definition, so everything is O.K.

Now let $B_i'$ denote the $(n-1)$-element set obtained from $B_i$ by deleting the element that was chosen to be in the first column. To construct the second column, Wild’s key observation is that if the maximum degree $d$ of $G$ is sufficiently small, then the sets $B_i'$ will necessarily satisfy Rado’s theorem. If $k\le n-1$, then we are done because even a single $B_i'$ will have rank at least $k$, so we just need to check that the union of all the $B_i'$ has rank $n$. Here is where the structure of the graph comes in.

Wild defines the “capacity” of a vertex to be the number of edges among the $B_i$ that are incident to it, counting multiplicities. Now, if $v$ is any vertex of $G$, then every one of the original bases $B_i$ contains an edge that is incident to $v$, because a basis by definition comprises a spanning tree for each connected component of $G$; thus the capacity of $v$ is at least $n$. When we delete the first column, the capacity of $v$ drops by at most $d$, since the first column is a basis and cannot contain multiple copies of an edge that is incident to $v$. Thus if $n > d$, then the union of the $B_i'$ must contain some edge incident to $v$. Since $v$ was arbitrary, this means that the union of the $B_i'$ contains a spanning subgraph of $G$, and hence its rank is $n$, as desired.

For the third column, we let $B_i''$ be $B_i'$ minus the element in the second column. We need to check the hypothesis of Rado’s theorem for the $B_i''$ with $k = n-1$ and $k = n$. For $k = n$, the same argument as before shows that the rank of the union of all the $B_i''$ is at least $n$ as long as $n > 2d$. For $k = n-1$, we can think of the union of $k$ of the $B_i''$ as being obtained by removing, from the original $n^2$ elements, three sets: the first column, the second column, and one of the rows (i.e., one of the $B_i''$). Each of these three sets is independent and hence can reduce the capacity of a vertex by at most $d$, so if $n > 3d$ then the union of $n-1$ of the $B_i''$ must contain a spanning subgraph of $G$, and therefore has rank $n$, which is more than enough (we only need rank $n-1$ to invoke Rado’s theorem).

In general, to get the $j$th column for $j\ge 3$, we need to verify the hypothesis of Rado’s theorem for $k\ge n-j+2$. Deleting $j-1$ columns and $j-2$ rows reduces the capacity of a vertex by at most $(2j-3)d$, so as long as $n - (2j-3)d > 0$ (i.e., as long as $j < (n+3d)/2d$) then Rado's theorem does the trick. This proves Wild's result.

Comment by tchow8 — April 20, 2017 @ 3:11 am

18. On the topic of graphs, I learned recently that there are some graph-theoretic conjectures with the same flavor as Rota’s Basis Conjecture. For example, a recent arXiv preprint by Horn and Nelsen proves a new partial result towards an old conjecture of Brualdi and Hollingsworth, which says that if the complete graph $K_{2m}$ (for $m \ge 3$) is edge-colored in such a way that every color class is a perfect matching, then there is a decomposition of the edges into $m$ edge-disjoint rainbow spanning trees (“rainbow” means that no two edges in the spanning tree have the same color). While I don’t see any direct connection with Rota’s Basis Conjecture, perhaps some of the proof techniques carry over from one setting to the other.

Comment by tchow8 — April 17, 2017 @ 8:19 pm

19. Jeff Kahn mentioned to me the beautiful “Wide Partition Conjecture” from the paper: T. Chow, C.K. Fan, M.X. Goemans and J. Vondrak, Wide Partitions, Latin Tableaux, and Rota’s Basis Conjecture, Advances in Applied Mathematics, 31, 334–358, 2003. pdf . (I think this conjecture might be interesting also in the Tverberg context we mensioned above.) Perhaps, Tim, you can explain and motivate this conjecture and its relation to Rota’s conjecture?

Comment by Gil Kalai — April 17, 2017 @ 8:39 pm

• Brian Taylor and I noticed that in the context of bracket algebras, which were Rota’s original motivation, there was nothing special about square shapes; one could consider any Young diagram. We thought that generalizing Rota’s Basis Conjecture to Young diagrams might open up the possibility of proofs by induction that add a single box at a time. We did find a successful generalization to Young diagrams that we called the “wide partition conjecture”; unfortunately, it seems that this generalization makes life harder, not easier, so I don’t think that this is a promising avenue to proving Rota’s Basis Conjecture. However, there is a certain special case of the wide partition conjecture that is interesting and tantalizing in its own right, which I will now describe.

Given a Young diagram $\lambda$ with row lengths $\lambda_1, \ldots, \lambda_k$, we can ask, is there a way to put a number in each box in such a way that (A) in row $i$, each number from $1$ to $\lambda_i$ appears exactly once, and (B) in every column, the numbers that appear are all distinct? In general the answer is certainly no; for example, in a Young diagram that consists entirely of a single column, condition (A) forces you to put a 1 in each box, and this violates condition (B) as soon as there is more than one box in the column. If you study the problem for a bit, you will soon see that a necessary condition for a positive answer to the question is that $\lambda \ge \lambda^T$, where $\ge$ denotes dominance (a.k.a. majorization) order, and $\lambda^T$ denotes the conjugate (a.k.a. transpose) of $\lambda$. Moreover, if $\mu$ is any Young diagram whose multiset of rows is a submultiset of the multiset of rows of $\lambda$, then we must have $\mu \ge \mu^T$ as well. The wide partition conjecture states that this necessary condition is also sufficient.

By the way, this special case of the wide partition conjecture would follow from another conjecture that I described on MathOverflow a couple of years ago.

Comment by tchow8 — April 17, 2017 @ 9:30 pm

• Tim, three quick questions: For this conjecture (in the special case you mentioned) what do we get for the square case? Is there an analogue of Galvin’s theorem, and is there an analog of the Alon-Tarsi connection? (The paper said that this is likely but not yet verified.)

Comment by Gil Kalai — April 18, 2017 @ 8:50 pm

• For square shapes, the conjecture reduces to the trivial statement that “Latin squares exist.” As for Galvin, the theorem he proved was about the list chromatic index of an arbitrary bipartite multigraph, so it already says something about Young diagrams, but I’m not sure if that’s what you’re asking. Maybe you’re asking for an analogue of the Dinitz conjecture for Young diagrams? The closest thing I know is Theorem 1 in our paper, but that’s probably not quite what you want.

Finally, as you noted, in the paper we already said that it ought to be easy to formulate an Alon-Tarsi conjecture for Young diagrams, but I don’t think anybody has bothered to work out the details. Maybe now is a good time to work this out. Towards this end, I would consider shapes that are not only wide (i.e., satisfy the necessary conditions I mentioned above) but are also self-conjugate, because then you can ask for every column, as well as every row, to be a permutation, so that it is clear what “even” and “odd” mean. The analogue of Alon-Tarsi would state that for any wide self-conjugate shape, the number of even tableaux is not equal to the number of odd tableaux. Then the only thing left to check is that this analogue of Alon-Tarsi would imply the analogue of Rota’s Basis Conjecture for self-conjugate wide shapes. What requires some thought is how to deal with the fact that the rows and columns do not all have the same length. Presumably, we should switch from talking about “bases” to talking about “independent sets”; that’s easy enough, but then we’re going to have to deal with non-square matrices, for which linear independence is more complicated to express algebraically (it’s not just the nonvanishing of a single determinant).

Comment by tchow8 — April 18, 2017 @ 10:03 pm

20. Continuing the graph-theory angle, here is a hopefully tractable problem that I think would help us make some headway. Consider the two-column version of Rota’s Basis Conjecture. That is, suppose we have $n$ disjoint independent sets $I_1, \ldots, I_n$, each of size 2, and suppose that their union $\bigcup_j I_j$ can be written as the disjoint union of two bases. Then is there an $n\times 2$ grid whose $i$th row is $I_i$ and whose two columns are bases? We know that the answer in general is no because $K_4$ is a counterexample already for $n=3$, and we also know that the answer is yes for strongly base-orderable matroids (which, in the case of graphs, coincide with series-parallel graphs). My question is, if we restrict our attention to graphic matroids, can we characterize the counterexamples? We know that there has to be a $K_4$ minor somewhere. But do there have to exist three sets $I_{j_1}, I_{j_2}, I_{j_3}$ that are an explicit copy of $K_4$? My guess is no, but I would like to see some explicit examples.

While the case of two columns may seem very special, as I showed in one of my papers, in some situations one can reduce Rota’s Basis Conjecture to the case of two columns.

Comment by tchow8 — April 20, 2017 @ 3:31 pm

• Just as a trivial way of making a more complicated example:

a1x vs a2x a2x Vs a1x a1b Vs cd a1c Vs bd a2d Vs BC

Comment by Luke Pebody — April 20, 2017 @ 5:12 pm

• Could you please decipher this notation?

Comment by domotorp — April 20, 2017 @ 7:36 pm

• Yeah, that did turn out more cryptic than I had hoped.

The blue and green edges at the top mean that the top three edges must be connected in blue and green, so we can contract those edges and then the remaining coloured edges make a $K_4$

Comment by Luke Pebody — April 21, 2017 @ 7:16 am

• This is a nice observation. Thinking about where to go from here, I was led to the following thought. Given a graph, let us define the operation of “pulling out an edge” to mean deleting the edge, and then adding two new vertices to the graph with a new edge between them. Question: Given an $n\times 2$ graphic counterexample to the two-column version of Rota’s Basis Conjecture, can we always fix it by pulling out at most $n/3$ edges?

I expect that the answer is probably no, but the counterexamples should be more interesting, because I don’t think you can get them just by stringing together disjoint copies of $K_4$ and “uncontracting” edges as in Luke Pebody’s example.

Should the answer unexpectedly be yes, then that gives hope that we can control the copies of $K_4$. Pulling out edges can be thought of as a proxy for making sure that those edges get used in other columns, in some kind of induction argument.

Comment by tchow8 — April 24, 2017 @ 5:12 pm

• Could you be more precise? By putting out, you really want to add an independent edge to the graph? What is “fixing it?” Does it mean converting it to something that’s not a counterexample?

Comment by domotorp — April 24, 2017 @ 7:33 pm

• By “fixing it” I do mean converting it to something that’s not a counterexample. And yes, when I pull out an edge $e$, I first delete $e$, and then I add a new independent edge $f$ to the graph. If $e$ originally appeared in the two-element independent set $I_i$, then $f$ replaces $e$ in $I_i$. Another way to put it is that I alter the matroid by making all sets consisting of an independent set plus the given edge independent, and keeping all the other independent sets. Basically I want to delete the edge from the graph (because I’m imagining that this is just an inductive step in the ordinary Rota conjecture, and the edge is appearing in some other column) but technically I have to leave it in there because the two-column version of the conjecture involves exactly $n$ sets of size exactly 2.

Comment by tchow8 — April 25, 2017 @ 12:05 am

• But what happens to the $2n$ elements’ partitioning to two bases? You forget about that condition? I might be really misunderstanding something, but consider taking the original $2\times n$ matrix ordered such that the first column forms a basis. The second column can be covered by two independent sets (the respective parts of the two original bases). One of these independent sets will have at most $\frac n4$ elements. Why can’t we pull these out?

Comment by domotorp — April 25, 2017 @ 8:48 am

• I’m afraid I don’t understand your sentence, “One of these independent sets will have at most $n/4$ elements.” Of course, by the pigeonhole principle, one of these independent sets will have at most $n/2$ elements, but how are you getting $n/4$? In fact if we look at our favorite example of $K_4$ with $n=3$, then we see that $n/4$ cannot be achieved.

Comment by tchow8 — April 25, 2017 @ 5:01 pm

• Oops, right, somewhere halfway through my argument I forgot that $n$ was not the total number of elements.

Comment by domotorp — April 25, 2017 @ 6:11 pm

21. Given $n$ sets of size $n$ in an $n$-dimensional vector space $V$ it would be interesting to find various sufficient and necessary conditions for finding $n$ rainbow disjoint basis $B_1,B_2,\dots B_n$. So here is a tentative question: If this is not possible can you always find disjoint $n+1$ sets, $C_1,\dots,C_{n+1}$ such that each set contains at most one member from each $B_j$ and the intersection of all linear spans of the $C_i$s is non trivial. (Remark: we identify members of the $B_i$‘s by their index, and in case of repeated vectors for the purpose of “at most one” and “disjoint” we refer to this indexing.)

Comment by Gil Kalai — April 26, 2017 @ 12:05 pm

• This sounds like an interesting idea, but can you provide some motivation for the condition that you state? I’m having a hard time understanding why one would expect this to be true.

Comment by tchow8 — April 28, 2017 @ 2:15 pm

22. I’ve been continuing to read Wild’s 1994 paper and I think that the most interesting thing in it that I’ve not mentioned already is Lemma 6, which states the following.

Let $M$ be a matroid on an $n^2$-element set $E$ that is a disjoint union of $n$ independent sets $B_1, \ldots, B_n$ of size $n$. Assume that there exists another matroid $M'$ on the same ground set $E$ with the following properties:

(1) $M'$ is strongly base orderable.

(2) $r(X) \ge |X|/n$ for all $X \subseteq E$, where $r$ is the rank function of $M'$.

(3) All circuits $C$ of $M$ satisfying $\forall j: |C\cap B_j| \le 1$ remain dependent in $M'$.

Then there is an $n\times n$ grid whose $i$th row comprises $B_i$ and whose columns are independent in $M$.

That all strongly base-orderable matroids satisfy Rota’s Basis Conjecture follows at once if we take $M = M'$ since condition (2) is automatic by pigeonhole. However, there is a lot more that we might extract out of Lemma 6. Wild asks whether a suitable $M'$ always exists. As Wild recognizes, this is probably too optimistic, but he doesn’t have a counterexample. Maybe a suitable $M'$ always exists for graphic matroids?

Comment by tchow8 — April 28, 2017 @ 2:28 pm

23. I have started a new blog post: https://polymathprojects.org/2017/05/05/rotas-basis-conjecture-polymath-12-post-3/ Most new comments should probably be posted under there rather than here.

Comment by tchow8 — May 5, 2017 @ 3:55 am

24. Thanks for your good content.

Comment by Indresh — August 23, 2019 @ 6:14 pm

Blog at WordPress.com.