Tangles are decided by weighted vertex sets

We show that, given a $ k $-tangle $ \tau $ in a graph $ G $, there always exists a weight function $ w: V(G)\to\mathbb{N} $ such that a separation $ (A,B) $ of $ G $ of order $<k $ lies in $ \tau $ if and only if $ w(A)<w(B) $, where $ w(U):=\sum_{u\in U}w(u) $ for $ U\subseteq V(G) $.


February 5, 2019
We show that, given a k-tangle τ in a graph G, there always exists a weight Tangles in graphs have played a central role in graph minor theory ever since their introduction by Robertson and Seymour in [3]. Informally, a tangle in a graph G is an orientation of all low-order separations of G satisfying certain consistency assumptions. Tangles capture highly connected substructures in G in the sense that every such substructure defines a tangle in G by orienting each low-order separation of G towards the side containing most or all of that substructure. In view of this, if some tangle of G contains the separation (A, B), we think of A and B as the 'small' and the 'big' side of (A, B) in that tangle, respectively; our main result will confirm this intuition.
Formally, a separation of a graph G = (V, E) is a pair (A, B) with A ∪ B = V such that G contains no edge between A B and B A, and the order of a separation (A, B) is the size |A ∩ B| of its separator A ∩ B. Furthermore, for an integer k, a ktangle in G is a set consisting of exactly one of (A, B) and (B, A) for every separation (A, B) of G of order < k, with the additional property that no three 'small' sides of separations in τ cover G, that is, that there are no ( As a concrete example, if G contains an n × n-grid for n k, then the vertex set of that grid defines a k-tangle τ in G by letting (A, B) ∈ τ for a separation of order < k if and only if B is the unique side of (A, B) containing, say, 90% of the vertices of that grid. In this way, the vertex set of the n × n-grid 'defines τ by majority vote'.
In [1] Diestel raised the question whether all tangles in graphs arise in the above fashion, that is, whether all graph tangles are decided by majority vote: given a k-tangle τ in a graph G, is there always a set X of vertices such that a separation (A, B) of order < k lies in τ if and only if |A ∩ X| < |B ∩ X|? A partial answer to this was given in [2], where Elbracht showed that such a set X always exists if G is (k − 1)-connected and has at least 4(k − 1) vertices. The general problem appears to be hard.
In this paper, we consider a fractional version of Diestel's question and answer it affirmatively, making precise the notion that B is the 'big' side of a separation (A, B) ∈ τ : given a k-tangle τ in G, rather than finding a vertex set X which decides τ by majority vote, we find a weight function w : V → N on the vertices such that for all separations (A, B) of order < k we have (A, B) ∈ τ if and only if the vertices in B have higher total weight than those in A. The existence of a vertex set X as in Diestel's original question would then imply that there is such a weight function with values in {0, 1}.
For a graph G there is a partial order on the separations of G given by letting (A, B) ≤ (C, D) if and only if A ⊆ C and B ⊇ D. One of the main ingredients for the proof of Theorem 1 is the following observation about those separations in a tangle τ that are maximal in τ with respect to this partial order. It says, roughly, that they divide each other's separators so that, on average, those separators lie more on the big side of the separation than on the small side, according to the tangle. Additionally we shall use a result from linear programming. For a vector x ∈ R n we use the usual shorthand notation x ≥ 0 to indicate that all entries of x are non-negative, and similarly write x > 0 if all entries of x are strictly greater than zero.

Lemma 3 ([4]).
Let K ∈ R n×n be a skew-symmetric matrix, i.e. K T = −K. Then there exists a vector x ∈ R n such that Kx ≥ 0 and x ≥ 0 and x + Kx > 0.
We are now ready to prove Theorem 1.
Proof of Theorem 1. Let a finite graph G = (V, E) and a k-tangle τ in G be given. Since G is finite it suffices to find a weight function w : V → R ≥0 such that a separation (A, B) of order < k lies in τ precisely if w(A) < w(B); this can then be turned into such a function with values in N.

For this it is enough to find a function w : V → R ≥0 such that w(A) < w(B) for all maximal elements (A, B) of τ : for if w(A) < w(B) and (C, D) ≤ (A, B) then w(C) ≤ w(A) < w(B) ≤ w(D).
So let us show that such a weight function w exists.
To this end let (A 1 , B 1 ), . . . , (A n , B n ) be the maximal elements of τ and set Observe that, by Lemma 2, we have m ij + m ji > 0 for all i = j and hence the matrix M + M T has positive entries everywhere but on its diagonal (where it has zeros). We further define Then K is skew-symmetric, that is, K T = −K. Let x = (x 1 , . . . , x n ) T be the vector obtained by applying Lemma 3 to K. We define a weight function w : V → R by Note that w has its image in R ≥0 and observe further that, for Y ⊆ V , we have With this, for i ≤ n, we have where (M x) i denotes the i-th coordinate of M x. Thus w is the desired weight function if we can show that M x > 0, that is, if all entries of M x are positive. From x + Kx > 0 we know that at least one entry of x is positive. Let us first consider the case that x has two or more positive entries. Then K x > 0 since K has positive values everywhere but on the diagonal, and hence since Kx ≥ 0. Therefore, in this case, w is the desired weight function.
Consider now the case that exactly one entry of x, say x i , is positive, and that x is zero in all other coordinates. Then for j = i we have (M x) j ≥ (K x) j > 0 and thus w(B j ) − w(A j ) = (M x) j > 0. However (M x) i = 0 and thus w(A i ) = w(B i ), so w is not yet as claimed. To finish the proof it remains to modify w such that w(A i ) < w(B i ) while ensuring that we still have w(A j ) < w(B j ) for j = i. This can be achieved by picking a sufficiently small ε > 0 such that w(A j ) + ε < w(B j ) for all j = i, picking any v ∈ B i A i , and increasing the value of w(v) by ε.
We conclude with the remark that Theorem 1 and its proof extend to tangles in hypergraphs without any changes. Even more generally, the following version of Theorem 1, which is formulated in the language of [1], can be established with exactly the same proof as well: Theorem 4 then applies to tangles in graphs or hypergraphs by taking for U the universe of separations of a (hyper-)graph and for P the given k-tangle. (See [1] for more on the relation between graph tangles and profiles.) Theorem 4 holds with the same proof as Theorem 1, since Lemma 2 holds in this setting too, using the definition of profile rather than the tangle axioms.