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arxiv: 2604.15649 · v1 · submitted 2026-04-17 · 🧮 math.CO

Signless Laplacian index conditions for trebly chorded cycles in graphs with given order

Jin Cai , Bo Zhou This is my paper

Pith reviewed 2026-05-10 08:32 UTC · model grok-4.3

classification 🧮 math.CO MSC 05C5005C38
keywords signless Laplacian indexspectral radiustrebly chorded cyclechorded cyclesextremal graph theorycycle subgraphseigenvalue conditions
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The pith

If the signless Laplacian index of an n-vertex graph meets or exceeds a threshold, it contains a trebly chorded cycle unless it is one of two exceptions.

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

The paper proves a sufficient condition based on the signless Laplacian spectral radius that forces the presence of a trebly chorded cycle in any graph of order n at least 6. A trebly chorded cycle is a cycle equipped with three chords that all meet at a single common vertex. The bound holds except for two explicitly identified graphs that achieve the index without the cycle. Readers care because the result translates an algebraic invariant into a guaranteed structural feature, offering a computable test for complex cycle subgraphs without enumerating substructures directly. It builds on extremal graph theory by replacing edge-count or degree conditions with an eigenvalue threshold.

Core claim

For every graph G of order n with n ≥ 6, if the signless Laplacian index q(G) is at least a specific value depending on n, then G contains a trebly chorded cycle unless G is isomorphic to one of two specified exceptional graphs.

What carries the argument

The signless Laplacian index, the largest eigenvalue of the signless Laplacian matrix Q = D + A, acts as the central invariant; a lower bound on this eigenvalue is shown to force the existence of a cycle plus three chords sharing one vertex.

If this is right

  • Graphs whose signless Laplacian index meets the bound must contain a cycle with three chords from one vertex, implying they cannot be too sparse in their cycle space.
  • The two exceptional graphs achieve the highest possible index without the desired cycle and thereby certify that the bound is tight.
  • The condition supplies a practical algebraic criterion for detecting the cycle structure via matrix eigenvalue computation.
  • The result extends prior degree-based or edge-based conditions on chorded cycles to this trebly case using spectral methods.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same spectral approach could be tested for related structures such as cycles with more than three chords or with chords in different configurations.
  • Computational verification of the bound for small n would directly confirm whether additional exceptions exist beyond the two identified.
  • The threshold might connect to other spectral invariants like the adjacency spectral radius in the study of cycle-rich graphs.

Load-bearing premise

The chosen threshold is the minimal value that works for all n-vertex graphs except the two exceptions, and the case analysis identifies precisely those exceptions without overlooking others.

What would settle it

Any graph of order n whose signless Laplacian index meets or exceeds the stated threshold yet contains no trebly chorded cycle and differs from the two exceptions.

Figures

Figures reproduced from arXiv: 2604.15649 by Bo Zhou, Jin Cai.

Figure 1
Figure 1. Figure 1: The graphs U1, . . . , U12. Let U1, . . . , U12 be the graphs displayed in [PITH_FULL_IMAGE:figures/full_fig_p004_1.png] view at source ↗
read the original abstract

It is proved that for a graph of order $n$, where $n\ge 6$, if the signless Laplacian index is larger than or equal to certain value depending on $n$, then the graph contains a trebly chorded cycle, where the chords incident to a common vertex, unless it is one of two specified graphs.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

0 major / 3 minor

Summary. The manuscript proves that for a graph G of order n (n ≥ 6), if the signless Laplacian spectral radius q(G) is at least a certain n-dependent threshold, then G contains a trebly chorded cycle (a cycle with three chords incident to one common vertex) unless G is isomorphic to one of two specified exceptional graphs on n vertices.

Significance. If correct, the result adds a sharp spectral Turán-type theorem for a specific chorded cycle structure to the literature on signless Laplacian conditions in extremal graph theory. The explicit identification of exactly two exceptions indicates the bound is tight, which strengthens the contribution when the case analysis is exhaustive.

minor comments (3)
  1. [Abstract] Abstract: the phrase 'certain value depending on n' should be replaced by an explicit formula or reference to the main theorem statement so that the threshold is immediately clear to readers.
  2. [Introduction] The definition of a trebly chorded cycle (three chords sharing a vertex) should be stated formally in the introduction or preliminaries section with a small diagram for clarity, as the term is not entirely standard.
  3. [Main result / Proof] Ensure that the two exceptional graphs are explicitly constructed and their signless Laplacian indices computed in a dedicated subsection or lemma so that the sharpness claim can be verified directly.

Circularity Check

0 steps flagged

No significant circularity detected

full rationale

The paper establishes an extremal theorem: for n ≥ 6, q(G) ≥ f(n) implies G contains a trebly chorded cycle unless G is one of two explicit exceptions. The threshold f(n) is the signless Laplacian index of those exceptions, identified via case analysis on maximal graphs without the structure. This is a standard direct proof using eigenvalue bounds and structural arguments; no step reduces by construction to a fitted parameter, self-definition, or load-bearing self-citation chain. The derivation is self-contained against external graph-theoretic benchmarks.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The result rests on the standard definitions of the signless Laplacian matrix and the trebly chorded cycle together with the axioms of finite simple undirected graphs.

axioms (1)
  • domain assumption Graphs are finite, simple, and undirected.
    Implicit in all statements about order n and the signless Laplacian index.

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Reference graph

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37 extracted references · 37 canonical work pages

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    Then g′ 1(x) = 4x3 −(3n+ 45)x 2 + (32n+ 116)x−80n+ 24, g′′ 1(x) = 12x2 −(6n+ 90)x+ 32n+ 116

    Proof ofq(G 1)< q(K + 1,1,n−2): Note thatQ(G 1) has an equitable quotient matrixB 1 = n−2 4n−6 0 1 6 0 1 1 0 7 0 0 4 0 4 with CP g1(x) =x 4 −(n+ 15)x 3 + (16n+ 58)x 2 −(80n−24)x+ 120n−272. Then g′ 1(x) = 4x3 −(3n+ 45)x 2 + (32n+ 116)x−80n+ 24, g′′ 1(x) = 12x2 −(6n+ 90)x+ 32n+ 116. Forx≥n+ 1, asn≥10, we have g′′ 1(x)≥g ′′ 1(n+ 1) = 6n 2 −40n+ 38>0, g′ 1(x)...

    work page

  23. [23]

    19 Then g′ 2(x) = 5x4 −(4n+ 80)x 3 + (63n+ 399)x 2 −(320n+ 520)x+ 520n−452, g′′ 2(x) = 20x3 −(12n+ 240)x 2 + (126n+ 798)x−320n−520, g(3) 2 (x) = 60x2 −(24n+ 480)x+ 126n+ 798

    Proof ofq(G 2)< q(K + 1,1,n−2): Note thatQ(G 2) has an equitable quotient matrixB 2 = n−2 3 4n−9 0 1 5 0 0 0 1 0 6 0 1 1 0 0 7 0 0 0 4 0 4 ! with CP g2(x) =x 5 −(n+ 20)x 4 + (21n+ 133)x 3 −(160n+ 260)x 2 + (520n−452)x−600n+ 1480. 19 Then g′ 2(x) = 5x4 −(4n+ 80)x 3 + (63n+ 399)x 2 −(320n+ 520)x+ 520n−452, g′′ 2(x) = 20x3 −(12n+ 240)x 2 + (126n+ 798)x−320n−...

    work page

  24. [24]

    Then g′ 3(x) = 5x4 −(4n+ 64)x 3 + (51n+ 225)x 2 −(196n+ 84)x+ 214n−348, g′′ 3(x) = 20x3 −(12n+ 192)x 2 + (102n+ 450)x−196n−84, g(3) 3 (x) = 60x2 −(24n+ 384)x+ 102n+ 450

    Proof ofq(G 3)< q(K + 1,1,n−2): Note thatQ(G 3) has an equitable quotient matrixB 3 = n−2 2 2n−6 0 1 5 2 0 0 1 2 4 0 1 1 0 0 7 0 0 0 2 0 2 ! with CP g3(x) =x 5 −(n+ 16)x 4 + (17n+ 75)x 3 −(98n+ 48)x 2 + (214n−288)x−132n+ 288. Then g′ 3(x) = 5x4 −(4n+ 64)x 3 + (51n+ 225)x 2 −(196n+ 84)x+ 214n−348, g′′ 3(x) = 20x3 −(12n+ 192)x 2 + (102n+ 450)x−196n−84, g(3)...

    work page

  25. [25]

    20 Forx≥n+ 1, asn≥7, we haveg 4(x)≥g 4(n+ 1)>0

    Proof ofq(G 4)< q(K + 1,1,n−2): Note thatQ(G 4) has an equitable quotient matrixB 4 =   n−2 1 2 2n−7 0 1 1 0 0 0 0 1 0 5 2 0 0 1 0 2 4 0 1 1 0 0 0 7 0 0 0 0 2 0 2   with CP g4(x) = (x−1)g 3(x). 20 Forx≥n+ 1, asn≥7, we haveg 4(x)≥g 4(n+ 1)>0. From Lemmas 2.3 and 2.8, we have q(G4)< n+ 1< n+ 2− 4 n+2 < q(K + 1,1,n−2)

    work page

  26. [26]

    Then g′ 5(x) = 5x4 −(4n+ 64)x 3 + (51n+ 219)x 2 −(192n+ 56)x+ 200n−356, g′′ 5(x) = 20x3 −(12n+ 192)x 2 + (102n+ 438)x−192n−56, g(3) 5 (x) = 60∗x 2 −(24n+ 384)x+ 102n+ 438

    Proof ofq(G 5)< q(K + 1,1,n−2): Note thatQ(G 5) has an equitable quotient matrixB 5 = n−2 1 4n−7 0 1 1 0 0 0 1 0 6 0 1 1 0 0 7 0 0 0 4 0 4 ! with CP g5(x) =x 5 −(n+ 16)x 4 + (17n+ 73)x 3 −(96n+ 28)x 2 + (200n−356)x−120n+ 392. Then g′ 5(x) = 5x4 −(4n+ 64)x 3 + (51n+ 219)x 2 −(192n+ 56)x+ 200n−356, g′′ 5(x) = 20x3 −(12n+ 192)x 2 + (102n+ 438)x−192n−56, g(3)...

    work page

  27. [27]

    Proof ofq(G 6)< q(K + 1,1,n−2): Note thatQ(G 6) has an equitable quotient matrixB 6 =   n−2 3 2 2n−9 0 1 5 0 0 0 0 1 0 4 2 0 1 1 0 2 5 0 0 1 0 0 0 7 0 0 0 2 0 0 2   with CP g6(x) =x 6 −(n+ 21)x 5 + (22n+ 155)x 4 −(183n+ 417)x 3 + (704n−114)x 2 + (1920−1202n)x+ 660n−1572. Then g′ 6(x) = 6x5 −(5n+ 105)x 4 + (88n+ 620)x 3 −(549n+ 1251)x 2 + (1408n−228)x−...

    work page 1920

  28. [28]

    Letf(x) = g7(x) x−6

    Proof ofq(G 7)< q(K + 1,1,n−2): Note thatQ(G 7) has an equitable quotient matrixB 7 = n−2 3 3n−8 0 1 6 0 0 1 1 0 5 0 0 1 0 0 7 0 0 3 0 0 3 ! with CP g7(x) = (x−6) x4 −(n+ 13)x 3 + (14n+ 40)x 2 + (42−60n)x+ 75n−180 . Letf(x) = g7(x) x−6 . Then f ′(x) = 4x3 −(3n+ 39)x 2 + (28n+ 80)x−60n+ 42, f ′′(x) = 12x2 −(6n+ 78)x+ 28n+ 80. Forx≥n+ 1, asn≥8, we have f ′′...

    work page

  29. [29]

    Then g′ 8(x) = 5x4 −(4n+ 68)x 3 + (54n+ 264)x 2 −(224n+ 168)x+ 276n−384, g′′ 8(x) = 20x3 −(12n+ 204)x 2 + (108n+ 528)x−224n−168, g(3) 8 (x) = 60x2 −(24n+ 408)x+ 108n+ 528

    Proof ofq(G 8)< q(K + 1,1,n−2): Note thatQ(G 8) has an equitable quotient matrixB 8 = n−2 1 3n−6 0 1 2 0 0 1 1 0 6 0 1 1 0 0 7 0 0 1 3 0 4 ! with CP g8(x) =x 5 −(n+ 17)x 4 + (18n+ 88)x 3 −(112n+ 84)x 2 + (276n−384)x−216n+ 624. Then g′ 8(x) = 5x4 −(4n+ 68)x 3 + (54n+ 264)x 2 −(224n+ 168)x+ 276n−384, g′′ 8(x) = 20x3 −(12n+ 204)x 2 + (108n+ 528)x−224n−168, g...

    work page

  30. [30]

    Proof ofq(G 9)< q(K + 1,1,n−2): Note thatQ(G 9) has an equitable quotient matrixB 9 =   n−2 2 1 1n−6 0 1 5 1 0 0 1 1 2 4 1 0 0 1 0 1 3 0 1 1 0 0 0 7 0 0 2 0 1 0 3   with CP g9(x) =x 6 −(n+ 20)x 5 + (21n+ 140)x 4 −(167n+ 358)x 3 + (620n−91)x 2 + (1612−1051n)x+ 618n−1524. Then g′ 9(x) = 6x5 −(5n+ 100)x 4 + (84n+ 560)x 3 −(501n+ 1074)x 2 + (1240n−182)x −...

    work page

  31. [31]

    Supposen≥11

    Proof ofq(G 10)< q(K + 1,1,n−2): Forn= 7,q(G 10) = 8.2351<8.7355 =q(K + 1,1,n−2). Supposen≥11. Note thatQ(G 10) has an equitable quotient matrixB 10 =   n−2 2 1 1 1n−7 0 1 5 1 0 0 0 1 1 2 5 1 1 0 0 1 0 1 2 0 0 0 1 0 1 0 3 0 1 1 0 0 0 0 7 0 0 2 0 0 1 0 3   with CP g10(x) =x 7 −(n+ 23)x 6 + (24n+ 197)x 5 −(227n+ 731)x 4 + (1073n+ 739)x 3 + (2307−2643n)x...

    work page

  32. [32]

    Suppose thatn≥12

    Proof ofq(G 11)< q(K + 1,1,n−2): Forn= 8,q(G 11) = 9.0478<9.7324 =q(K + 1,1,n−2). Suppose thatn≥12. Note thatQ(G 11) has an equitable quotient matrixB 11 =   n−2 2 1 2 1n−8 0 1 5 1 0 0 0 1 1 2 6 2 1 0 0 1 0 1 2 0 0 0 1 0 1 0 3 0 1 1 0 0 0 0 7 0 0 2 0 0 1 0 3   with CP g11(x) =x 7 −(n+ 24)x 6 + (25n+ 214)x 5 −(245n+ 824)x 4 + (1192n+ 841)x 3 + (2952−29...

    work page

  33. [33]

    Leth 1(x) =g 12(x, s)−g 12(x, s+ 4)

    Proof ofq(G 12) =q(G 12(n, s))< q(G 13)< q(K + 1,1,n−2) wheres≥3 andn≥s+ 3: Note thatQ(G 12(n, s)) has an equitable quotient matrixB 12 = n−2s1n−s−3 0 1 3 1 0 1 1s s+1 0 0 1 0 0 7 0 0s0 0s ! with CP g12(x, s) =x 5 −(n+ 2s+ 9)x 4 + (s2 + (2n+ 15)s+ 10n+ 11)x 3 −((n+ 6)s 2 + (17n+ 2)s + 27n−39)x 2 + ((7n−14)s 2 + (32n−44)s+ 18n−54)x+ 6s 3 −(6n−2)s 2 −(12n−3...

    work page

  34. [34]

    Ifn= 9, thenq(G) = 10.3723<10.7381 =q(K + 1,1,n−2)

    Proof ofq(K 1 ∨ n−1 4 K4)< q(K + 1,1,n−2): LetG=K 1 ∨ n−1 4 K4. Ifn= 9, thenq(G) = 10.3723<10.7381 =q(K + 1,1,n−2). Suppose thatn≥13. Note thatQ(G) has an equitable quotient matrixB 14 = ( n−1n−1 1 7 ) with CP g14(x) =x 2 −(n+ 6)x+ 6n−6. Forx≥n+ 1, we haveg 14(x)≥g 14(n+ 1) =n−11>0. From Lemmas 2.3 and 2.8, we have q(G)< n+ 1< n+ 2− 4 n+2 < q(K + 1,1,n−2)

    work page

  35. [35]

    Ifn= 10, thenq(G) = 11.0666<11.7474 =q(K + 1,1,n−2)

    Proof ofq(K 1 ∨(K 1 ∪ n−2 4 K4))< q(K + 1,1,n−2): LetG=K 1 ∨(K 1 ∪ n−2 4 K4). Ifn= 10, thenq(G) = 11.0666<11.7474 =q(K + 1,1,n−2). Suppose that n≥14. Note thatQ(G) has an equitable quotient matrixB 15 = n−1 1n−2 1 1 0 1 0 7 with CP g15(x) =x 3 −(n+ 7)x 2 + 7nx−6n+ 12. Then g′ 15(x) = 3x2 −(2n+ 14)x+ 7n. Forx≥n+ 1, asn≥14, we have g′ 15(x)≥g ′ 15(n+ 1) =n ...

    work page

  36. [36]

    Forn= 7,q(G) = 8.7016<8.7355 =q(K + 1,1,n−2)

    Proof ofq(K 1 ∨ K1,1 ∪ n−3 4 K4 )< q(K + 1,1,n−2): LetG=K 1 ∨ K1,1 ∪ n−3 4 K4 . Forn= 7,q(G) = 8.7016<8.7355 =q(K + 1,1,n−2). Suppose that n≥11. Note thatQ(G) has an equitable quotient matrixB 16 = n−1 2n−3 1 3 0 1 0 7 with CP g16(x) =x 3 −(n+ 9)x 2 + (9n+ 12)x−18n+ 26. We have g′ 16(x) = 3x2 −(2n+ 18)x+ 9n+ 12. Forx≥n+ 1, we have g′ 16(x)≥g ′ 16(n+ 1) =n...

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  37. [37]

    Ifs= 2, thenQ(G) has an equitable quotient matrixB 17 = n−1 3n−4 1 5 0 1 0 7 with CP g17(x) =x 3 −(n+ 11)x 2 + (11n+ 24)x−30n+ 36

    Proof ofq(K 1 ∨(K + 1,s ∪ n−s−2 4 K4))< q(K + 1,1,n−2) where 2≤s≤n−6: LetG=K 1 ∨(K + 1,s ∪ n−s−2 4 K4). Ifs= 2, thenQ(G) has an equitable quotient matrixB 17 = n−1 3n−4 1 5 0 1 0 7 with CP g17(x) =x 3 −(n+ 11)x 2 + (11n+ 24)x−30n+ 36. 26 Then g′ 17(x) = 3x2 −(2n+ 22)x+ 11n+ 24 Forx≥n+ 1, asn≥8, we have g′ 17(x)≥g ′ 17(n+ 1) =n 2 −7n+ 5>0. Then, forx≥n+ 2−...

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