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Simultaneous generating sets for flags

T0 review · 1 major / 5 minor · reviewed 2026-08-07 · deepseek-v4-flash

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Pith's one-line read For any three complete flags in $\mathbb{R}^d$, a common generating set of size at most $\lfloor 5d/3\rfloor$ always exists, and some flag triples need that many.

desk verdict The main theorem is likely right, but the central claim about ordering non-crossing cycles in Section 5.3 is unproved and load-bearing for the upper bound. read the letter →

arxiv 2502.09530 v1 pith:POZZHYP6 submitted 2025-02-13 math.CO math.AG

Federico Glaudo , Noah Kravitz , Chayim Lowen This is my paper · ORCID
classification math.COmath.AG MSC 05C7015A0314M1505B35
keywords simultaneousgeneratingsetscompleteflagsBruhatdecompositionedgecovernumberintersectiondimensionsflagvarietiestransversalitymatroid
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

A complete flag in $\mathbb{R}^d$ is a nested chain of subspaces, one of each dimension. The paper asks how many vectors are needed to generate three flags at once—that is, to supply a vector inside each layer that is not in the previous layer. The answer is exactly $\lfloor 5d/3\rfloor$ in the worst case, and the same kind of exact formula settles the question for any number $m$ of flags. This extends the classical two-flag result, where $d$ vectors suffice, and it shows the worst case is larger than the naive generic-flags guess of $\lceil 3d/2\rceil$. For odd $m$, the full formula is $md/2 + \lfloor 2d/3\rfloor - d/2$.

What carries the argument

The central object is the tripartite graph $G$ with vertices $U_1,\ldots,U_d,V_1,\ldots,V_d,W_1,\ldots,W_d$ and one perfect matching between each pair of flag families; edges mark the layers that pair according to the permutations associated to two flags. A second graph $\widetilde G$ contains an edge whenever two layers are compatible, meaning some single vector is new for both of them, and the number of compatible pairs needed to cover all vertices is exactly the edge-cover number of $\widetilde G$. The proof partitions the cycles of $G$ into groups $A$, $B$, $C$ by how efficiently they can be covered, then runs a lattice walk through the grid of intersection dimensions $\dim(U^i\cap V^j\cap W^k)$. The cost of each vertex, the increment of this dimension when that vertex is hopped on, is either $0$ or $1$, and the total cost is always $d$; the walk is chosen so that the costs of $B$-cycles are bounded by one third of their size and every $C$-triangle has cost $0$, which yields the inequality $|A|\ge 2|C|$ needed to reach the $5d/3$ bound.

What would settle it

Compute the minimum simultaneous generating set for all triples of complete flags in $\mathbb{R}^6$, for example by exhaustive search over the finitely many possible intersection-dimension grids that are known to be representable for three flags; any instance whose optimal generating set has size 11 would disprove the asserted sharp bound $\mu(3,6)=10$. A smaller falsifier for the proof mechanism would be a triple of flags whose auxiliary graph contains an odd cycle interlacing a triangle, so that no lattice path can bound the cycle cost by one third of its size.

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Extended reading notes

Core claim

The main discovery is an exact formula for $\mu(m,d)$, the largest possible value of the smallest simultaneous generating set for $m$ complete flags in $\mathbb{R}^d$. For $m=1$ or $d=1$ it is $d$; for even $m\ge2$ and $d\ge2$ it is $md/2$; for odd $m\ge3$ and $d\ge2$ it is $md/2 + \lfloor 2d/3\rfloor - d/2$. The heart is the triple case: $\mu(3,d)=\lfloor 5d/3\rfloor$, with the bound achieved by explicit equality cases. The proof encodes the three flags in an auxiliary tripartite graph whose vertices are the layers and whose edges record which pairs of layers admit a common new vector, then covers the graph with compatible pairs at a carefully controlled rate.

Load-bearing premise

The most fragile structural premise is that after greedily removing crossing odd cycles, all remaining odd cycles are pairwise non-crossing and can be strictly ordered by height, with every middle-length cycle lying entirely between two consecutive non-compatible triangles in all three coordinates, so that the special lattice walk forces the desired cost bounds.

Editorial extensions

If this is right

  • Every triple of complete flags in $\mathbb{R}^d$ admits a common generating set of size at most $\lfloor 5d/3\rfloor$, and some triples require that many, so the bound is sharp.
  • For any even number $m$ of flags, the worst case is exactly $md/2$; for any odd $m\ge5$, the formula is obtained by combining the three-flag bound with the even case.
  • The statement and proof hold over any infinite field, and over every finite field with at least $m-1$ elements; for sufficiently large fields the same numbers hold in all dimensions.
  • Equality in the three-flag bound occurs only in rigid configurations: the auxiliary graph consists of $d/3$ non-compatible, non-crossing triangles plus even cycles, each with a specific alternating crossing behavior.
  • Generic triples of flags need only $\lceil 3d/2\rceil$ vectors in the worst generic case, but the true worst case is larger; transversality does not maximize the generating number.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The graph-theoretic reduction suggests an algorithm for triples: $\mu(U,V,W)$ is the edge-cover number of $\widetilde G$, so once the intersection dimensions are known, the exact value can be computed in polynomial time via matching algorithms.
  • The sharp gap between generic and worst-case sizes raises a natural average-case question: for random triples of flags, how does the minimal simultaneous generating set concentrate, and does it track the generic bound rather than the worst case?
  • The proof's reliance on sandwiching the leftover cycles between consecutive non-compatible triangles indicates that the same $5d/3$ bound may fail for more general flag-like objects where layer intersections behave non-linearly, exactly the obstruction identified in the matroid discussion.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

1 major / 5 minor

Summary. This paper studies the simultaneous generating number µ(m,d): the largest, over all m-tuples of complete flags in R^d, of the minimum size of a set of vectors that contains a new vector for every layer of every flag. The main theorem determines µ(m,d) exactly: µ(2,d)=d, µ(m,d)=md/2 for even m≥2, and µ(m,d)=md/2+⌊2d/3⌋−d/2 for odd m≥3, with the three-flag case µ(3,d)=⌊5d/3⌋ as the heart. The upper bound is proved by reducing to µ(3,d) and then, for three flags, encoding the layer-incidence data in two tripartite graphs G and eG. The vertices are covered by families of compatible sets at rates 1/2, 5/9, and 2/3 on three groups of cycles (A,B,C), and the critical inequality (3) is derived from a lattice path in the dimension grid, with costs controlled by lemmas about G-edges and non-compatible triangles. The lower bound uses generic (transverse) flags and a direct-sum superadditivity argument.

Significance. If the main proof is completed, this is a worthwhile exact result. It extends the classical Bruhat-decomposition fact for two flags and exhibits a genuinely non-generic worst case (5d/3 rather than the naive 3d/2). The proof method is interesting: the graph-theoretic encoding via permutations, the compatibility-pair criterion in Lemma 5.1, and the dimension-cost lattice path are natural and elegant, and the lower-bound construction is simple and robust. The paper is self-contained and does not rely on numerical fitting or external conjectures; the main unproved step is a single structural lemma about non-crossing cycles.

major comments (1)
  1. [§5.5] The proof of Lemma 5.3(2) rests on an unproved structural claim in §5.3: 'If two cycles are non-crossing, then one cycle must lie completely below the other; thus any collection of pairwise non-crossing cycles is totally ordered by height.' This is used in §5.5 to assert that every B-cycle lies entirely in one chunk between consecutive C-triangles (or below/above the extremal ones), i.e., that all of its U-, V-, and W-indices lie strictly between the corresponding indices of the surrounding triangles. That containment is what makes the cost argument work: a U- or V-vertex of S is hopped before its G-neighbor in W, so Lemma 5.4 gives cost 0 and hence cost(S) ≤ |S|/3. Without a proof of the ordering/chunk-containment lemma, the manuscript does not rule out a B-cycle that interlaces a C-triangle in one coordinate while having no individual edge crossing; in such a case the cost-0 argument could fail and cost(S) could be as large as |S|/2, breaking inequality (3) and the bound in Theorem 4.5. Please provide a precise definition of 'height'/'completely below' and a formal proof that non-crossing cycles are totally ordered and that this order gives the simultaneous coordinate-wise separation asserted here.
minor comments (5)
  1. [§5.5] The lattice-path construction in the proof of Lemma 5.3 assumes M ≥ 1, since it refers to U_{i_1}, V_{j_1}, W_{k_1} and 'hopping over the first triangle'. The case where C is empty should be handled separately; for example, hopping all U-vertices, then all V-vertices, then all W-vertices works.
  2. [§3] In the proof of Proposition 3.2, in the odd d case the range '1 ≤ i ≤ d/2' should read '1 ≤ i ≤ ⌊d/2⌋'.
  3. [§5.7] The equality-case characterization is asserted ('A close examination ... reveals', 'A separate clever argument then shows') without proof. If these are intended as theorems, they need proofs; otherwise the passage should be explicitly labeled as conjectural or heuristic.
  4. [§6.2] Propositions 6.1 and 6.2 are stated with only a citation to a table and no proof. Since they are peripheral to the main theorem, a proof sketch or a more precise reference would strengthen the presentation.
  5. [§5.1] The terms 'crossing', 'height', and 'completely below' are introduced informally for edges and cycles. I recommend giving formal definitions (for instance, coordinatewise comparison of vertex indices) so that the ordering and chunk-containment claims can be checked.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the main bound is derived from explicit constructions and linear algebra, not from assuming its conclusion.

full rationale

The paper's derivation chain is self-contained in the sense relevant to circularity. Theorem 1.1 is reduced to Theorem 4.5 by elementary superadditivity and subadditivity arguments (Corollaries 4.2 and 4.4) together with the two-flag theorem (Theorem 2.1), which is proved directly in the paper by a dimension-counting argument. The core upper bound Theorem 4.5 is proved from explicit combinatorial data: the graph G whose edges are compatible pairs, Lemma 5.1's characterization of compatible edges in terms of G, Lemma 5.2's edge-cover counting, and Lemma 5.3's lattice-path cost bound. No parameter is fitted to the target quantity; no equality-case claim from Section 5.7 is used; and the reference list contains no works by the present authors. The only fragile point is the unproved structural assertion in Section 5.3 that the remaining odd cycles are totally ordered by height and that each B-cycle lies entirely in one chunk between consecutive C-triangles; if false, Lemma 5.3(2) and the final 5d/3 bound would fail. That is a potential gap or correctness risk, not a circular step: the assertion is not assumed as the theorem, nor is it obtained by renaming an input. Accordingly the circularity score is 0.

Assumptions & free parameters 0 free parameters · 4 assumptions · 0 invented entities

The central claim rests on standard linear algebra and the Bruhat decomposition; the only nonstandard structural assumption is the unproved height-ordering of non-crossing cycles. No free parameters are fitted and no new external entities are introduced.

assumptions (4)
  • standard math Bruhat decomposition of GL_d(R): any pair of complete flags admits a unique permutation sigma and a compatible basis, as stated in Theorem 2.1.
    Used throughout Section 5 to define the graph G. The paper gives a self-contained proof of the needed combinatorial version, so this is standard background rather than an ad hoc input.
  • domain assumption Infinite-field linear algebra: a subspace cannot be covered by a finite union of proper subspaces.
    Used in Lemma 5.1 to pass from 'not contained in either of two proper subspaces' to the existence of a vector in the difference of two layers. This is valid over R and all infinite fields, as noted in Section 6.2.
  • domain assumption Transverse m-tuples of flags form a dense open subset of Fl_d(R)^m and satisfy the dimension formulas of Definition 3.1.
    Used in Proposition 3.2 to obtain lower bounds for generic flags, which feed the lower-bound half of Theorem 1.1. Standard for complete flags over an infinite field.
  • ad hoc to paper After the greedy pairing of crossing odd cycles, the remaining odd cycles in groups B and C are pairwise non-crossing, totally ordered by height, and every B-cycle lies entirely between two consecutive C-triangles in all three coordinates.
    Asserted in Section 5.3 and used in Section 5.5 to define the chunks of the lattice path and to force U- and V-vertices in B to have cost 0. This is the most fragile unstated structural premise.

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Cite this review

Pith. "Pith review of Simultaneous generating sets for flags." pith.science (2026). https://pith.science/paper/POZZHYP6

@misc{pith2026250209530,
  author       = {Pith},
  title        = {Pith review of: Simultaneous generating sets for flags},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/POZZHYP6}},
  note         = {Machine review of arXiv:2502.09530}
}
abstract

We prove that any triple of complete flags in $\mathbb R^d$ admits a common generating set of size $\lfloor 5d/3\rfloor$ and that this bound is sharp. This result extends the classical linear-algebraic fact -- a consequence of the Bruhat decomposition of $\text{GL}_d(\mathbb R)$ -- that any pair of complete flags in $\mathbb R^d$ admits a common generating set of size $d$. We also deduce an analogue for $m$-tuples of flags with $m>3$.

Figures

Figures reproduced from arXiv: 2502.09530 by the authors.

Figure 1
Figure 1. The graphs G (left) and Ge (right) for a triple of generic flags in R 3 . The graph G consists of a triangle (in bold) and a 6-cycle; these cycles cross. The resulting edges present in Ge but not in G are drawn dashed. (V j \ V j−1 ) ∩ U i ̸= ∅. The edges of the form (V j , Wk ) and (Wk , Ui ) are determined analogously. • We include the edge (U i , V j ) in Ge if and only if (U i \ U i−1 ) ∩ (V j \ V j−1 ) ̸= ∅, i.… view at source ↗
Figure 2
Figure 2. An illustration of Lemma 5.1. The left part of the diagram shows a pair of crossing edges in G, with the corresponding additional edge (drawn dotted) in Ge. The right part of the diagram shows a pair of non-crossing edges in G, for which there is no additional edge in Ge. Proof of Lemma 5.1. Recall that j ′ is minimal in [d] such that (U i \ U i−1 ) ∩ V j ′ ̸= ∅. It follows that if j < j′ , then (U i \ U i−1 ) ∩ (V … view at source ↗
Figure 3
Figure 3. The edge cover (shaded) used in the third part of Lemma 5.2. 5.3. Grouping cycles by efficiency. Recall that G is a disjoint union of cycles with lengths divisible by 3. We divide these cycles into three groups A, B, C according to how efficiently we can cover them with compatible sets, as follows. First, put all of the even cycles in A, thereby removing them from consideration. Do the same with each triangle in G t… view at source ↗
Figures from the paper (3 more)
Figure 4
Figure 4. Figure 4: A schematic of the types of cycles appearing in the groups A, B, C. In A, the isolated triangle represents a compatible triple, and the two odd cycles at the bottom are crossing (with the dotted edges representing edges present in Ge but not G) [PITH_FULL_IMAGE:figure…
Figure 5
Figure 5. Figure 5: Schematic “snapshots” (from left to right) of four stages of the hopping procedure from the proof of Lemma 5.3. The triangles in C are drawn in bold. In each snapshot the already-hopped-on vertices are those in the shaded region. (i) We have just hopped on the vertices…
Figure 6
Figure 6. Figure 6: The graph G for several equality cases. The large graph on the left is a particularly symmetric example which contains a 12-cycle. The four smaller graphs on the right are randomly-chosen examples. This hope is dashed when one examines the hundreds of different equalit…

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Forward citations

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Reference graph

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