--- title: Proof Attempt and the Strongly Base Orderable Case --- # Status I do not have a proof of the full conjecture. I do have a clean proof of a strong special case: > If the matroid is strongly base orderable, then two rainbow bases already suffice. So in that class the conjectured bound $3$ is true with room to spare. # Setup Let $M$ be a rank-$r$ matroid on $E$, and suppose - $E = A \sqcup B$, - $A$ and $B$ are bases, - the ground set is partitioned into $r$ color classes of size $2$. Write - $A = \{a_1,\dots,a_r\}$, - $B = \{b_1,\dots,b_r\}$. Later I will assume that $b_i = \varphi(a_i)$ for a bijection $\varphi : A \to B$. # A Useful Reformulation Fix any bijection $\varphi : A \to B$, and write $b_i = \varphi(a_i)$. For each subset $I \subseteq [r]$, define the transversal \[ T_I := \{a_i : i \notin I\} \cup \{b_i : i \in I\}. \] So $T_I$ picks exactly one element from each pair $\{a_i,b_i\}$. The question is: - when is $T_I$ rainbow? - when is $T_I$ a basis? The second question depends on the matroid. The first depends only on the coloring. ## The Color Graph From the coloring, build a multigraph $G$ on vertex set $[r]$ as follows. For each color pair: - if the color pair is $\{a_i,b_i\}$, add a loop at $i$; - if it is $\{a_i,b_j\}$ or $\{b_i,a_j\}$ with $i \neq j$, add an edge $ij$ with label $0$; - if it is $\{a_i,a_j\}$ or $\{b_i,b_j\}$ with $i \neq j$, add an edge $ij$ with label $1$. Because each element of $E$ appears in exactly one color pair, each vertex $i$ contributes exactly two half-edges, one coming from $a_i$ and one from $b_i$. Hence every vertex has degree $2$, so each connected component of $G$ is a cycle or a loop. ## Lemma: The Parity System Is Always Consistent For the graph $G$, consider the system over $\mathbf F_2$ \[ x_i + x_j = \lambda(ij) \] for every non-loop edge $ij$, where $\lambda(ij) \in \{0,1\}$ is its label, and for a loop the equation is $x_i + x_i = 0$. ### Proof It is enough to show that on every cycle, the sum of the labels is $0$ mod $2$. Take one cycle component $C$. Let - $p$ be the number of $A$-$A$ color pairs on $C$, - $q$ be the number of $B$-$B$ color pairs on $C$, - $m$ be the number of cross pairs ($A$-$B$ or $B$-$A$) on $C$. Then: - the number of $A$-elements used by the color pairs of $C$ is $2p + m$, - the number of $B$-elements used by the color pairs of $C$ is $2q + m$. But each vertex in $C$ contributes exactly one element from $A$ and one from $B$, so these two numbers are equal. Therefore $2p + m = 2q + m$, hence $p=q$. The label-$1$ edges are exactly the $A$-$A$ and $B$-$B$ pairs, so the number of label-$1$ edges on $C$ is \[ p+q = 2p, \] which is even. So every cycle has even total label, and the system is consistent. # The Strongly Base Orderable Case Recall the definition. ## Definition A matroid $M$ is strongly base orderable if for every two bases $B_1,B_2$ there exists a bijection $\varphi : B_1 \to B_2$ such that for every subset $X \subseteq B_1$, \[ (B_1 \setminus X) \cup \varphi(X) \] is again a basis. ## Theorem Let $M$ be a rank-$r$ matroid on $E$, and suppose - $E = A \sqcup B$ with $A,B$ bases, - $\varphi : A \to B$ is a bijection such that $(A \setminus X) \cup \varphi(X)$ is a basis for every $X \subseteq A$, - the elements of $E$ are colored by $r$ colors, each used exactly twice. Then there exist two rainbow bases whose union is $E$. In particular, every strongly base orderable matroid satisfies the rainbow-cover conjecture, and in fact needs at most $2$ rainbow bases. ## Proof Write $b_i = \varphi(a_i)$ and build the labeled multigraph $G$ above. By the lemma, the parity system \[ x_i + x_j = \lambda(ij) \] has a solution $x = (x_1,\dots,x_r) \in \{0,1\}^r$. Let \[ I := \{i \in [r] : x_i = 1\} \] Define \[ R := (A \setminus \{a_i : i \in I\}) \cup \varphi(\{a_i : i \in I\}) = \{a_i : x_i = 0\} \cup \{b_i : x_i = 1\}. \] By hypothesis, $R$ is a basis of $M$. I claim that $R$ is rainbow. Take any color pair. 1. If the pair is $\{a_i,b_i\}$, then $R$ contains exactly one of them by construction. 2. If the pair is $\{a_i,b_j\}$ or $\{b_i,a_j\}$, then the edge $ij$ has label $0$, so $x_i = x_j$. Hence exactly one of $a_i$ and $b_j$ lies in $R$. 3. If the pair is $\{a_i,a_j\}$ or $\{b_i,b_j\}$, then the edge $ij$ has label $1$, so $x_i \neq x_j$. Hence exactly one of the two same-side elements lies in $R$. Thus $R$ contains exactly one element of each color, so $R$ is rainbow. Now take the complementary solution $x' = 1-x$. It also satisfies the same parity system, because the equations are linear over $\mathbf F_2$. The corresponding set is \[ R' := \{a_i : x_i = 1\} \cup \{b_i : x_i = 0\} = E \setminus R. \] Again, by hypothesis, $R'$ is a basis, and by the same argument it is rainbow. Finally, \[ R \cup R' = E. \] So $R$ and $R'$ are two rainbow bases covering the whole ground set. # Corollary for the Double-Cover Variant In the strongly base orderable case, the partition-matroid special case of the double-cover conjecture also holds: - the two complementary rainbow bases $R$ and $R'$ are common bases of $M$ and the color partition matroid; - therefore $R,R,R',R'$ are four common bases covering each element exactly twice. So this special case is stronger than the original $3$-cover statement. # Why This Does Not Yet Prove the Full Conjecture The proof above splits the problem into two parts: 1. the coloring contributes a parity system on a 2-regular graph; 2. the matroid decides which transversals $T_I$ are actually bases. The first part is completely benign: for every bijection $\varphi : A \to B$, the parity system is always solvable. The real difficulty is the second part. In the strongly base orderable case, every $T_I$ is a basis, so the parity solution immediately gives a rainbow basis. In a general matroid, the parity-compatible sets $T_I$ need not be bases at all. # Failed General Approaches ## Attempt 1: Replace strong base orderability by multiple symmetric exchange Take disjoint bases $A,B$. A standard exchange theorem says that for each subset $X \subseteq A$, there exists some subset $Y \subseteq B$ with $|Y|=|X|$ such that \[ (A \setminus X) \cup Y \] is a basis. This is not enough here. The coloring prescribes a very specific subset of $B$ once one chooses $X$; call it $Y_{\mathrm{col}}(X)$. The exchange theorem only gives existence of some $Y$, not the prescribed $Y_{\mathrm{col}}(X)$. That is exactly the gap. ## Attempt 2: Work component-by-component on the color graph The color graph is a disjoint union of cycles. Each cycle has two natural local rainbow states. This suggests an induction on the cycle components. I could not make this work, because matroid basis choices do not decompose independently over those color components. Local feasible choices on two different color cycles need not glue to a global basis. ## Attempt 3: Use the common-base polytope Let $P$ be the partition matroid of the coloring. Then both $M$ and $P$ have rank $r$, and $E$ decomposes into two bases in both. Hence the vector $(1/2)\mathbf 1_E$ lies in the intersection of the two base polytopes. If one could always write $(1/2)\mathbf 1_E$ as the average of four common-base vertices, one would get the exact double-cover statement immediately. I do not see a mechanism forcing such a $4$-vertex decomposition. General Carathéodory bounds are far too weak. # What Seems Most Promising The strongly base orderable proof suggests the right intermediate object. Fix complementary bases $A,B$ and a bijection $\varphi : A \to B$. The coloring always singles out a nonempty affine subspace of $\{0,1\}^r$ consisting of the parity-compatible transversals $T_I$. So a possible route to the full conjecture is: > Find a structural reason that, for some good choice of $A,B,\varphi$, at least three of those parity-compatible transversals are bases. Strongly base orderable matroids are the extreme case where all of them are bases. For a general matroid, I do not currently know how to force even one parity-compatible transversal to be a basis from a fixed bijection, let alone three. # Bottom Line I could not prove the full conjecture. I did prove the following self-contained special case: > If $M$ is strongly base orderable and $E$ is the disjoint union of two bases, then for every coloring of $E$ into $r$ pairs there exist two rainbow bases covering $E$. That is the strongest proof-level progress I found in this turn.