Induced subdivisions in graphs of large girth

Peiru Kuang; Yan Wang

arxiv: 2605.17218 · v2 · pith:3UMPKLYInew · submitted 2026-05-17 · 🧮 math.CO

Induced subdivisions in graphs of large girth

Peiru Kuang , Yan Wang This is my paper

Pith reviewed 2026-05-19 23:35 UTC · model grok-4.3

classification 🧮 math.CO MSC 05C35

keywords induced subdivisionsgraph girthminimum degreeMader's theoremcomplete graph subdivisionsextremal graph theory

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The pith

Graphs with minimum degree at least k and girth above a fixed constant contain an induced subdivision of K_{k+1}.

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

The paper establishes that there exists an absolute constant g0 such that for every integer k at least 3, any graph with minimum degree at least k and girth at least g0 contains an induced subdivision of the complete graph on k+1 vertices. This provides a strong affirmative answer to a question originally posed by Shi and later by Kuhn and Osthus about induced subdivisions in high-girth graphs. A sympathetic reader would care because the result shows that girth can be increased to force these specific induced structures to appear uniformly, no matter how large the minimum degree becomes. The argument proceeds by reducing the problem to an auxiliary statement about induced subdivisions in bounded-degree high-girth graphs with high average degree.

Core claim

There exists an absolute constant g0 such that for every integer k greater than or equal to 3, every graph G with minimum degree delta(G) at least k and girth g(G) at least g0 contains an induced subdivision of K_{k+1}. The proof relies on an induced variant of Mader's theorem asserting that for every fixed s, eta, and D, every graph J with maximum degree at most D, average degree greater than s minus 2 plus eta, and sufficiently large girth contains an induced subdivision of K_s.

What carries the argument

The induced variant of Mader's theorem, which guarantees an induced subdivision of a complete graph in any graph of bounded maximum degree, average degree above a linear threshold, and large girth.

If this is right

Once girth exceeds the fixed g0, minimum degree k alone suffices to guarantee an induced subdivision of K_{k+1}.
The girth threshold g0 works simultaneously for every minimum degree k at least 3.
The result strengthens classical subdivision theorems by requiring the subdivision to be induced rather than merely topological.
High-girth graphs cannot avoid induced subdivisions of complete graphs when their minimum degree is large.

Where Pith is reading between the lines

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

The same reduction technique might extend to induced subdivisions of other fixed graphs beyond complete graphs.
Explicit upper bounds on g0 could be pursued by refining the dependence on the parameters in the induced Mader variant.
The statement suggests that large girth acts as a uniform forcing condition for induced dense substructures across all degree regimes.

Load-bearing premise

The argument requires that an induced version of Mader's theorem holds for graphs with bounded maximum degree, average degree exceeding s minus 2 plus eta, and sufficiently large girth.

What would settle it

A graph with minimum degree k, girth larger than the claimed g0, and no induced subdivision of K_{k+1} for some k at least 3 would refute the central claim.

Figures

Figures reproduced from arXiv: 2605.17218 by Peiru Kuang, Yan Wang.

**Figure 1.** Figure 1: Local construction of the path Peλ and the corresponding edge eλ in the auxiliary graph H∗ . Claim 4.4. Let y ∈ S ∗ , z ∈ NJ (y) and let λ, λ′ ∈ Lz(y) be distinct. Then NF (eλ) ∩ NF (eλ′) = {y}. Suppose that w ∈ NF (eλ) ∩ NF (eλ′) \ {y}. For each ξ ∈ {λ, λ′}, there exist a vertex aξ ∈ V (Peξ ) and an aξ-w path Qξ of length at most 2ℓ + 1 whose internal vertices lie in Tw. Let u be the last common vertex of… view at source ↗

read the original abstract

In this paper, we prove that there exists an absolute constant $g_0$ such that, for every integer $k\ge 3$, every graph $G$ with $\delta(G)\ge k$ and $g(G)\ge g_0$ contains an induced subdivision of $K_{k+1}$. This fully resolves a problem raised by K\"{u}hn and Osthus (originally attributed to Shi), and improves a recent result of Gir\~{a}o and Hunter. Our proof uses some ideas from Gir\~{a}o and Hunter. Another main ingredient in our proof is an induced variant of Mader's theorem: for every fixed $s,\eta,D$, every graph $J$ with $\Delta(J)\le D$, $d(J)>s-2+\eta$ and sufficiently large girth contains an induced subdivision of $K_s$.

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.

Desk Editor's Note private letter to a colleague

Answers the Kühn-Osthus question on induced subdivisions but the absolute girth constant looks vulnerable to parameter dependence in the induced Mader lemma.

read the letter

The paper claims there is a single absolute g0 such that any graph with minimum degree k and girth at least g0 contains an induced subdivision of K_{k+1}, for every k at least 3. This directly targets the open problem and packages the answer as strong. The main new piece is the induced variant of Mader's theorem: for fixed s, η, D, a graph with maximum degree at most D, average degree above s-2+η, and large enough girth has an induced K_s subdivision. If that lemma holds with reasonable bounds, it is a useful technical addition for this area of graph theory. The authors apply it to reach the main result without obvious circularity or heavy self-citation. That part earns credit for being a concrete advance on the stated question. The soft spot is exactly the uniformity issue raised in the stress-test note. The lemma is stated only for fixed s, η, D, so the required girth size can depend on those values. When the main argument sets s to k+1 and likely needs D at least on the order of k to handle the minimum-degree condition, the girth threshold for the lemma may grow with k. If it does, no fixed g0 can serve for all k simultaneously, and the central claim would not hold as written. The abstract gives no indication of how the dependence is capped, so the full proof needs to show explicit control over the parameters or a different arrangement that keeps the girth bound independent of k. This paper is for people working on subdivisions, girth conditions, and extremal graph theory. A reader who wants to see whether the open question is settled would get value from the details, but only after checking that the girth bound really stays uniform. I would send it to peer review so the argument can be examined closely; the question is sharp enough and the method is reasonable even if revisions are required to fix the bound.

Referee Report

1 major / 1 minor

Summary. The paper proves that there exists an absolute constant g_0 such that for every integer k ≥ 3, every graph G with minimum degree δ(G) ≥ k and girth g(G) ≥ g_0 contains an induced subdivision of K_{k+1}. The proof relies on a new induced variant of Mader's theorem: for every fixed s, η, D, every graph J with Δ(J) ≤ D, average degree d(J) > s-2 + η, and sufficiently large girth contains an induced subdivision of K_s. This is presented as answering a problem of Kühn and Osthus (attributed to Shi) in a strong form.

Significance. If the central claim holds, the result gives a uniform girth threshold independent of k for the existence of induced K_{k+1}-subdivisions in graphs of minimum degree k, which is a substantial strengthening of known results on subdivisions in high-girth graphs. The induced Mader theorem is a potentially reusable tool for studying induced subgraphs in sparse graphs with large girth.

major comments (1)

[Abstract] Abstract: The main theorem asserts a single absolute g_0 independent of k. The induced Mader ingredient is stated only for fixed s, η, D with 'sufficiently large girth' (whose dependence on the fixed parameters is not quantified). To derive the main result, the argument must set s = k+1 and select D, η to accommodate δ(G) ≥ k (likely via some core or auxiliary graph whose parameters grow with k). If the girth threshold implicit in the proof of the induced Mader statement grows with s or D, the resulting g_0 would depend on k, contradicting the claimed uniformity. The manuscript must explicitly confirm that the girth bound remains bounded independently of k or supply the dependence function.

minor comments (1)

[Abstract] The abstract refers to the problem 'originally attributed to Shi' without a citation; a reference should be added for completeness.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their careful reading of the manuscript and for identifying this important point regarding the uniformity of the girth threshold. We address the comment in detail below.

read point-by-point responses

Referee: [Abstract] Abstract: The main theorem asserts a single absolute g_0 independent of k. The induced Mader ingredient is stated only for fixed s, η, D with 'sufficiently large girth' (whose dependence on the fixed parameters is not quantified). To derive the main result, the argument must set s = k+1 and select D, η to accommodate δ(G) ≥ k (likely via some core or auxiliary graph whose parameters grow with k). If the girth threshold implicit in the proof of the induced Mader statement grows with s or D, the resulting g_0 would depend on k, contradicting the claimed uniformity. The manuscript must explicitly confirm that the girth bound remains bounded independently of k or supply the dependence function.

Authors: We agree that the dependence of the girth threshold on the parameters must be made fully explicit to avoid any ambiguity. In the proof of the induced Mader theorem, the required girth is determined by a probabilistic argument that depends only on η and D (specifically, on ensuring the existence of a suitable expander-like subgraph after deleting low-degree vertices and short cycles); the target clique size s enters only in the final embedding step and does not affect the girth lower bound. Consequently, once η and D are fixed independently of k (which is possible by working in a suitable auxiliary graph derived from the k-core of G), the same absolute girth g_0 suffices for every s = k+1. We will revise the manuscript by (i) adding an explicit sentence to the statement of the induced Mader theorem quantifying the girth bound as a function of η and D only, and (ii) inserting a short paragraph in the proof of the main theorem that records the parameter choices and confirms that g_0 can be chosen independently of k. revision: yes

Circularity Check

0 steps flagged

No significant circularity; derivation self-contained

full rationale

The paper states a theorem asserting an absolute constant g0 independent of k, with the proof relying on a stated induced variant of Mader's theorem for fixed s, η, D and sufficiently large girth. No equations or steps in the abstract reduce a claimed result to a fitted parameter, self-definition, or load-bearing self-citation by construction. The ingredient is presented as an internal component of the argument rather than an external assumption that loops back to the target claim. The derivation chain therefore remains non-circular and self-contained as a mathematical proof.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The central claim rests on standard graph-theoretic definitions plus the unproven (in the abstract) induced variant of Mader's theorem.

axioms (1)

standard math Standard definitions of girth, minimum degree, induced subdivision, and Mader's theorem in graph theory.
Invoked throughout the abstract as background.

pith-pipeline@v0.9.0 · 5643 in / 1190 out tokens · 31698 ms · 2026-05-19T23:35:39.186918+00:00 · methodology

Induced subdivisions in graphs of large girth

Core claim

What carries the argument

If this is right

Where Pith is reading between the lines

Load-bearing premise

What would settle it

discussion (0)