Tverberg type theorems for matroids
Abstract
In this paper we show a variant of colorful Tverberg’s theorem which is valid in any matroid: Let be a sequence of nonloops in a matroid of finite rank with closure operator . Suppose that is colored in such a way that the first color does not appear more than times and each other color appears at most times. Then can be partitioned into rainbow subsequences such that . A subsequence is called rainbow if it contains each color at most once. . In particular,
The conclusion of our theorem is weaker than the conclusion of the original Tverberg’s theorem in , which states that , whereas we only claim that . On the other hand, our theorem strengthens the Tverberg’s theorem in several other ways: \@setpar it is applicable to any matroid (whereas Tverberg’s theorem can only be used in ), instead of we have the stronger condition we add a color constraints that are even stronger than the color constraints in the colorful version of Tverberg’s theorem.\@noitemerr Recently, we used the first property and applied the noncolorful version of this theorem to homology groups with coefficients to obtain several nonembeddability results, for details we refer to [GMP15]. , and
1 Introduction
Tverberg’s theorem [Tve66] states that given points^{1}^{1}1We allow repetitions among these points. in , it is possible to split these points into sets with intersecting convex hulls, that is with .
If one replaces convex hulls with affine hulls, one obtains a valid statement (Lemma 1), which has the advantage of being applicable to any field [GMP15, GMP16]. Lemma 1 is also easier to prove than the original Tverberg’s theorem. Since the proof only uses properties of closure operators, the statement does generalize to any matroid (Lemma 2). In both these cases the conclusion can be strengthened a bit: instead of , one can require .
In this paper we study the variant of Tverberg’s theorem for matroidal closures and show that it allows a colorful version – a generalization where the original points are colored and one furthermore requires that no resulting set , contains two or more points of the same color.
While the version without colors is straightforward [GMP16, Lemma 12] the proof of the colorful version is more subtle. Moreover, our proof method yields an efficient algorithm that finds the required sets in polynomial time.
1.1 Terminology
Before we state our results formally, let us introduce some notations and terminology which will allow us to nicely present the statements and proofs. We assume that the reader is acquainted with the basic matroid theory. We always use the symbols and to denote nonnegative integers. We use the symbols , , and for matroidal closure, affine closure, convex hull and rank function, respectively.
If is a set, we consider a sequence of elements from as a set of pairs . With this convention we can use the set theoretic terminology for sequences: is the length of the sequence, means that is a subsequence of , we know what it means for two subsequences to be disjoint, we can use the operation of (sequence) difference, etc.
If is a sequence and we need to refer to the set , we use the symbol .
If is a map from the subsets of (for example a closure operator, rank function), and is a sequence in , we use a shorthand . To make formulas and equations shorter, we leave out the parantheses after the operators , , and when there is no danger of confusion.
A coloring of a sequence is any map into some set of colors, that is, assigns to each pair a color from . The sequence is rainbow with respect to , if the restriction of to is injective.
1.2 Main results
Let us first state the noncolorful variant of Tverberg’s theorem for affine hulls and its easy generalization to matroidal closures.
Lemma 1 (Tverberg for affine hulls [Gmp15, Gmp16]).
Let be a sequence of points in an affine space of dimension . If^{2}^{2}2We do not require to be finite, therefore the slightly unusual formulation. , then there exist pairwise disjoint subsequences of with . In fact, there are pairwise disjoint subsequences satisfying .
Lemma 2 (Matroidal Tverberg).
Let be a (finitary^{3}^{3}3Finitary matroids are generalization of matroids to not necessary finite ground sets. They add the following axiom to the usual axioms for finite matroids: If , then there exists a finite set such that . With these addition, such terms as rank or basis can be correctly defined.) matroid of rank with closure operator and be a sequence of points in with . Then there exist pairwise disjoint subsequences of satisfying .
In [GMP16, Lemma 13] we only stated that there exists sets with . However, the proof there implies Lemma 1, and (if one replaces with the closure operator of a matroid) Lemma 2. In the case of matroids of finite rank, both lemmas can also be obtain as a direct consequence of Theorem 3.
In [GMP15, GMP16] we applied Lemma 1 to homology groups over finite fields. This enabled us to prove some inequalities for simplicial complexes embeddable into various manifolds. Our colorful matrodial Tverberg (Theorem 3) provides a control of the resulting sets, which enables us to further improve the bounds from [GMP15, GMP16]. For the details of the improvement, see the author’s thesis [Pat15].
We are now ready to state the main results of this paper.
Theorem 3.
Let be a matroid of a finite rank and be a sequence of nonloops in colored by some colors in such a way that at most elements of are colored by the first color, at most by the second color, at most by the third color, etc. If , then there exist pairwise disjoint rainbow subsequences of , such that .
Furthermore, if the time required to decide whether a point lies in the closure of a set is bounded by , then the subsequences can by found in time polynomial in , , and .
In the proof of Theorem 3 we encounter another version of colorful matroidal Tverberg’s theorem.
Theorem 4.
Let be a matroid of a finite rank and a sequence of nonloops in colored by colors in such a way that at least elements of are colored by the first color, at least by the second color, at least by the third, …, at least by the th color.
Then there exist pairwise disjoint rainbow subsequences of such that .
Furthermore, if the time required to decide whether a point lies in the closure of a set is bounded by , then the subsequences can by found in time polynomial in , , and .
Note the different conditions on the number of points of each color. In Theorem 3 these conditions are used to ensure that we have enough colors. In Theorem 4 we already have the right number of colors, but the conditions ensure that the length of is sufficient.
Moreover, these results are tight:
Proposition 5.
Tverbergtype theorems in
Let us now compare our main results with the related theorems valid in .
In this section denotes the dimensional simplex.
Tverberg’s theorem can be stated as follows: If is an affine map, there are pairwise disjoint faces of with . This is the reason why Tverberg’s theorem is also called affine Tverberg’s theorem. To avoid confusion, we have decided not to use the name “affine Tverberg” for Lemma 1.
If is a prime power, Özaydin [Öza87] showed that the same result holds for an arbitrary continuous map . The statement is known as topological Tverberg. It was a longstanding open problem, whether topological Tverberg can be extended to other values of . The negative answer came in 2015, when Frick (based on the previous work of Mabillard and Wagner [MW14]) constructed first counterexamples [Fri15]. Counterexamples for other values of and followed shortly afterwards. [MW15]
If is a prime, there is a colorful version of (topological) Tverberg’s theorem [BMZ15] as well: Suppose that the vertices of are colored in such a way, that no color is used more than times. Then for every continuous map , there are pairwise disjoint rainbow^{4}^{4}4Containing each color at most once. faces of with .
The colorful version provides more control over the resulting sets .
Even if is an affine map, the only known proof uses topological methods and needs the assumption that is prime. Whether this assumption can be relaxed in the affine situation is an open question. Moreover, the topological proof does not provide any way how to find the pairwise disjoint faces, it merely shows their existence.
We see that Theorem 3 does not require to be a prime number, it relaxes the conditions on the colors from topological version a bit and provides an efficient algorithm for finding the desired sets.
We also note that Bárány, Kalai and Meshulam proved another, very different Tverberg Type Theorem for Matroids [BGR15], they considered continuous maps from the matroidal complex and showed the following: If denotes the maximal number of disjoint bases in a matroid M of rank , then for any continuous map from the matroidal complex into there exists disjoint independent sets such that .
2 Tightness
We postpone the technical proofs of our main results, Theorems 3 and 4 to the end of the paper. First we prove Proposition 5 showing their tightness. The proof is a variant of the standard construction for showing that Tverberg’s theorem is tight.
We start with an auxiliary lemma.
Lemma 6.
Let be a matroid with finite basis . Then for any two sets
(2.1) 
Proof.
Since the operator is monotone, the inclusion is obvious. Let us now prove the opposite inclusion.
Let be an arbitrary element. We want to show that . If is a loop, . So assume that is not a loop.
Let and be inclusion minimal subsets with and , respectively. Since we assume that is not a loop, .
We will show by contradiction that , hence proving the claim. If , we may up to symmetry assume that there is an element which does not lie in .
From the inclusion minimality of follows that . The exchange principle yields . Similarly for an arbitrary .
The set is independent being a subset of a basis . By construction . Comparing the ranks of both sides and using the fact that , we see that has to belong to . Since was arbitrary, this implies – in contradiction with being minimal with . ∎
We can now finally prove Proposition 5.
Proof of Proposition 5.
Let be a basis of the matroid . It suffices to take
(2.2) 
Let be disjoint subsequences of . Then
where the first equality follows by inductive application of Lemma 6 and the second equality uses the fact that each element is missing in at least one sequence . ∎
We also note that the assumption in Theorem 3 that there are at most points of the first color is necessary. Otherwise, one can consider the sequence in where the first elements are red and the last element is blue. Then although the length of can be arbitrary, there are no three disjoint rainbow subsequences , , with . On the other hand, it is not true that this condition is necessary in every matroid. For example, consider the affine line over the field with two elements.
3 The proof
The reduction of Theorem 3 to Theorem 4 follows a well known pattern, a similar reduction previously appeared in the proof of the optimal colored Tverberg theorem [BMZ15] or in Sarkaria’s proof for the prime power Tverberg theorem [Sar00, 2.7.3], see also de Longueville’s exposition [dL02, Prop. 2.5]. Nevertheless, there are subtle differences because we are working in greater generality and because we need to take algorithmic aspects into consideration.
Theorem 4 implies Theorem 3..
Assume that the assumptions of Theorem 3 are satisfied. We show how to turn the sequence and the matroid with closure operator into a sequence and matroid with closure operator that satisfy the assumptions of Theorem 4. Moreover, we construct and the coloring of in a such way that the sets , , …, will satisfy will imply that is rainbow. and the rainbowness of iff and only if
Let be the rank of and the number of colors used in . From the conditions follows that .
If the length of is strictly larger than , we throw the superfluous elements of away. This does not add a point of any color, therefore all assumptions of Theorem 3 remain preserved. So we may assume that the length of is precisely .
We form from by adding new coloops^{5}^{5}5A coloop is an element that is independent on any set that does not contain . In other words, we form as the direct sum of with the uniform matroid . . Now we form the sequence by appending to .
Clearly we can color the new elements of so that in total there are exactly points of the first color, and exactly points of every other color.
We see that , satisfy the assumptions of Theorem 4. It follows that there are rainbow subsequences of satisfying .
Since the points are coloops and since each one of them was added exactly times, it follows that they cannot contribute to . Consequently, and .
We conclude that are the required subsequences of .
Observe that the reduction is polynomial in , , and . ∎
Now we can start with the proof of Theorem 4. Here we describe the main idea. We let be a rainbow independent subsequence of the maximal rank. In an ideal case and we may obtain the remaining subsequences by apply induction on the sequence inside .
However, we may be unlucky. It may happen that no such satisfies , see Fig. 1.
We see that in this case we could simply take the subsequence and unify colors blue and red into one color (say violet). Then lives in a submatroid of rank and satisfies the conditions of Theorem 4, so we may use induction. We obtain subsequences of satisfying . Moreover they are not only rainbow in the violetorange coloring, but also in the original blueorangered coloring. . These are clearly also subsequences of
In the proof we show that if , we may always resolve the situation by an analogous trick.
Let us now carry out the technical details. Since we promised an algorithmic solution, we describe an algorithm that finds the desired subsequences.
Proof.
First we compute . Since instead of we can consider the subsequence formed by the elements colored by the first colors (while preserving all assumptions of Theorem 4), we may assume that .
Now we find an inclusion maximal independent rainbow subsequence of . This can clearly be done in time polynomial in , , and .
We will proceed in the proof by induction on the triple (in lexicographical ordering). If or the statement is trivial, so assume .
If , then . Because is rainbow, and satisfy the assumptions of Theorem 4 for . By applying induction we obtain disjoint rainbow subsequences of with we see that are the desired disjoint rainbow subsequences with . If we now set
Therefore we may assume that
(3.1) 
We would like to increase by adding a point of a color that is not yet used in . Unfortunately, this is not possible without replacing some points of first. Our algorithm uses a cycle to find out which points to replace and how. Within the cycle we need to keep track of “replacement rules” which makes this part a bit technical. Moreover, there are three possibilities what can occur at one iteration of the cycle: \@setpar either we construct a larger independent rainbow set , we find the desired sets in a smaller submatroid, or we adjust the replacement rules.\@noitemerr
The cycle
In the th step () of the cycle the replacement rules consist of the following data:

set of colors (this set corresponds to colors that we may use while replacing some points),

subsequence of (eventually we would like to replace the subsequence of by another sequence ),

for each element whose color is in and which does not lie in a subsequence of (we want to replace with , hence increasing the length of our subsequence by one)
To simplify the terminology, if is a subsequence of , let denote the set of all the colors used by elements of . If is a set of colors, let be the subsequence of formed by all elements with color from .
We want the data to satisfy the following conditions:

[label=()]

,

for some ,

,

and

and
Note that conditions 2 and 3 imply that only contains elements that have the same colors as points in plus one additional point that has color , which is not yet present in .
The first step () is easy. We set and let be all the colors of except for those already used in . No element is contained in^{6}^{6}6 are the elements of whose color lies in and we assume that contains only nonloop elements. , so we need to define the set for every such . We simply put .
Now we check that the above defined sets satisfy all the prescribed conditions. Note that by (3.1), . This together with the fact that is independent implies that . Since we have colors, there is a color that is not used in . In other words, is nonempty.
So suppose that the sets , and are already constructed. Since there are three cases that may occur:

[label=)]

,

or

and .
We deal with the particular cases separately:
Case 1:
In this case, we may apply the trick we used for Fig. 1. Let us describe it formally.
We set and . has rank and by (3.1) we know that . It follows that and since , we also have .
Condition 1 implies , so there is a point .
Because is rainbow and independent and , has distinct elements, say . We define . In we recolor and all points of color by a new color .
Because (we evaluate with respect to the original coloring), the assumption (Case 1) implies that is a sequence of elements from . Also in there are colors, at least elements of color and at least elements of all the remaining colors. Therefore, the assumptions of Theorem 4 are satisfied for . By induction we obtain the desired disjoint rainbow subsequences of (which itself is a subsequence of ) with . By the construction of the new coloring these subsequences are also rainbow in the original coloring of . . These subsequences are rainbow with respect to the new coloring of
Case 2:
In this case, we construct a new independent rainbow subsequence with : We pick a point with and set .
Before we show that such is a rainbow independent subsequence of size , we prove the following auxiliary equality:
(3.2) 
Indeed,
where the last equality uses the fact that from condition 4. Because any closure operator satisfies
(3.3) 
we may rewrite the expression further to
By condition 4 , which reduces the equality to:
Using (3.3) again, we obtain
Since , Equation (3.2) follows.
Using the fact that , we are now ready to verify that is a rainbow independent subsequence with .

is rainbow: contains one element of color that is not used in , otherwise it uses the same colors as . Because , we see that uses exactly colors. This, together with the previous item, yields that is rainbow.

is independent: From the equality (3.2) we get Moreover, we have chosen a point which satisfies , so Since was independent and has exactly one element more, the independence of follows.
Let be an inclusion maximal independent rainbow subsequence of that contains . We may now start our algorithm again but this time we replace the maximal independent rainbow subset by . We have decreased the quantity and preserved and . By induction we obtain the desired disjoint rainbow subsequences with .
Case 3: and
In this case, we show how to construct sets , and for every with we construct a subsequence .
We choose to be any inclusion minimal subsequence satisfying
(3.4) 
Because we assume that , such set does exist. We further define
(3.5) 
Before we construct , we prove the following auxiliary claim:
Claim 6.1.
(3.6) 
Proof.
Now we construct sets for all points satisfying . Let be such a point. By definition of , , so cannot lie in . Equation (3.5) implies . Because is a rainbow set^{8}^{8}8 is rainbow!, there exists a unique element with . Since we assume , we have , where the last inclusion follows from condition 1. In particular, , hence
(3.7) 
Since is an inclusion minimal subsequence of for which , there exists an element such that . Since , the exchange principle implies .
It easily follows that
(3.8) 
Claim 6.1 together with (3.7) imply that . Since was chosen to satisfy , we have as well. Together with , this implies that is defined. We set^{9}^{9}9We note that does depend on the choice of , i.e., if we choose another that satisfies , we obtain a different set .
(3.9) 

Condition 3: By definition . Because is a subset of the rainbow set , is itself rainbow. Together with , where , this implies that the sets and are disjoint. Since (Equation (3.7)), and (conditions 2 and 5), we have and . From follows . Since , we have , where the last equality uses the induction hypothesis for . Claim 6.1 then yields as desired.