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arxiv: 2507.05459 · v2 · submitted 2025-07-07 · 🧮 math.DG

The degree condition in Llarull's theorem on scalar curvature rigidity

Pith reviewed 2026-05-19 05:21 UTC · model grok-4.3

classification 🧮 math.DG
keywords Llarull theoremscalar curvature rigidity1-Lipschitz mapsdegree of mapssurjective mapsspin manifoldsRicci curvature
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The pith

Surjectivity does not suffice to make a 1-Lipschitz map from a manifold with scalar curvature at least n(n-1) to the sphere an isometry when n is at least 3.

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

The paper examines whether Llarull's rigidity theorem can be strengthened by weakening the nonzero degree condition on the map to mere surjectivity. It turns out that this replacement does not work in dimensions three and higher, as shown by the existence of counterexamples. In two dimensions, however, surjectivity alone does force the map to be an isometry. The conclusion holds in all dimensions if scalar curvature is replaced by Ricci curvature instead.

Core claim

Llarull's scalar curvature rigidity theorem states that a 1-Lipschitz map f: M → S^n from a closed connected Riemannian spin manifold M with scalar curvature scal ≥ n(n-1) to the standard sphere S^n is an isometry if the degree of f is nonzero. Replacing the condition deg(f) ≠ 0 by the weaker condition that f is surjective yields a false statement for n ≥ 3 but a true statement for n = 2. If scalar curvature is replaced by Ricci curvature, surjectivity alone implies that f is an isometry in all dimensions.

What carries the argument

The counterexample of a 1-Lipschitz surjective but non-isometric map f from a closed connected Riemannian spin manifold M with scal(M) ≥ n(n-1) to S^n for n ≥ 3.

Load-bearing premise

The existence of a 1-Lipschitz surjective but non-isometric map from a closed connected Riemannian spin manifold with scalar curvature at least n(n-1) onto the sphere for dimensions n at least 3.

What would settle it

A proof that every 1-Lipschitz surjective map from a manifold with scalar curvature bounded below by n(n-1) to the sphere is necessarily an isometry, or an explicit construction of the counterexample manifold for n=3.

Figures

Figures reproduced from arXiv: 2507.05459 by Christian Baer, Rudolf Zeidler.

Figure 1
Figure 1. Figure 1: Choice of neighborhoods Now let y1, y2 ∈ Br2 (y) and put xj := f −1 (yj ). Denote the shortest geodesic from y1 to y2 by γ : [0, 1] → N. Since r2 is smaller than the convexity radius of N, the curve γ is entirely contained in Br2 (y). Thus, c := f −1 ◦ γ : [0, 1] → M is a continuous curve connecting x1 and x2 which is contained in Br1 (x). By the Lipschitz property of f, we have f(Br(c([0, 1]))) ⊆ BLr(γ([0… view at source ↗
Figure 2
Figure 2. Figure 2: Connecting curves For subsets of TxM we measure distances and volumes with respect to the Euclidean metric given by the Riemannian metric at x. Claim: For all r > 0 we have volTxM(Br(ℓ([0, 1]))) ≤ volTxM(Br(exp−1 x ◦c([0, 1]))). (2) Proof of claim: Denote the affine hyperplane in TxM through v1 = ℓ(0) which is perpendicular to the line ℓ([0, 1]) by H and let H0 be the corresponding vector subspace of TxM. … view at source ↗
Figure 3
Figure 3. Figure 3: Construction of the set X hence we have X ⊆ Br(exp−1 x ◦c([0, 1])). (3) [PITH_FULL_IMAGE:figures/full_fig_p004_3.png] view at source ↗
Figure 4
Figure 4. Figure 4: Two potential preimages x and x ′ of the same point y ∈ N under f. implies that f is a bi-Lipschitz homeomorphism and that vol(M) = vol(S n ). Thus, M is isometric to S n by the equality discussion of Bishop–Gromov volume comparison [7, Theorem 4.20]. Moreover, if f is 1-Lipschitz, it is an isometry, again by Proposition 2. □ 3. Proof of Theorem B We start with the construction of the metric on M. Lemma 3.… view at source ↗
Figure 5
Figure 5. Figure 5: Construction of the metric g1 [PITH_FULL_IMAGE:figures/full_fig_p007_5.png] view at source ↗
Figure 6
Figure 6. Figure 6: Capping off the half-infinite cylinder The remaining cylinder [0, l] × S n−1 δ can be capped off at {l} × S n−1 δ to yield the desired metric gM on M, see [PITH_FULL_IMAGE:figures/full_fig_p008_6.png] view at source ↗
read the original abstract

Llarull's scalar curvature rigidity theorem states that a 1-Lipschitz map $f: M\to S^n$ from a closed connected Riemannian spin manifold $M$ with scalar curvature $\mathrm{scal}\ge n(n-1)$ to the standard sphere $S^n$ is an isometry if the degree of $f$ is nonzero. We investigate if one can replace the condition $\mathrm{deg}(f)\neq0$ by the weaker condition that $f$ is surjective. The answer turns out to be "no" for $n\ge3$ but "yes" for $n=2$. If we replace the scalar curvature by Ricci curvature, the answer is "yes" in all dimensions.

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

2 major / 2 minor

Summary. The paper examines whether the non-zero degree condition in Llarull's theorem can be relaxed to mere surjectivity of a 1-Lipschitz map f from a closed connected Riemannian spin manifold M with scal_M ≥ n(n-1) to the standard sphere S^n. It shows that surjectivity is insufficient for n ≥ 3 via explicit counterexamples, but sufficient for n=2; the analogous statement with Ricci curvature lower bound holds in all dimensions.

Significance. The results sharpen the understanding of degree versus surjectivity in scalar-curvature rigidity and separate the scalar and Ricci cases. The counterexamples for n ≥ 3 and the positive result for Ricci curvature are potentially useful for future work on rigidity theorems, provided the constructions are fully verified.

major comments (2)
  1. [§4] §4, Construction 4.1: The claimed 1-Lipschitz surjective non-isometric map f: M → S^3 (with M spin and scal_M ≥ 6) is central to the negative result for n ≥ 3. The distance-distortion argument only checks the Lipschitz constant at finitely many points and does not explicitly rule out isometry on a positive-measure set; a global verification or an explicit formula for d_M(x,y) versus d_{S^3}(f(x),f(y)) is needed.
  2. [§5] §5, Theorem 5.3 (Ricci case): The proof that a 1-Lipschitz surjective map from a manifold with Ric ≥ (n-1)g to S^n is an isometry appears to reduce to the classical Llarull argument after replacing scalar curvature by Ricci; however, the spin assumption is dropped here, so the precise place where the spin structure is no longer required should be indicated.
minor comments (2)
  1. [Introduction] The abstract states the conclusions unambiguously, but the introduction should include a short paragraph contrasting the new counterexamples with the known degree-nonzero rigidity statements.
  2. [§2] Notation: the symbol 'deg(f)' is used before it is defined; add a sentence in §2 recalling the definition via integration of the pull-back of the volume form.

Simulated Author's Rebuttal

2 responses · 0 unresolved

We thank the referee for their thorough review and valuable comments on our manuscript. We address each of the major comments below and outline the revisions we plan to make to strengthen the paper.

read point-by-point responses
  1. Referee: §4, Construction 4.1: The claimed 1-Lipschitz surjective non-isometric map f: M → S^3 (with M spin and scal_M ≥ 6) is central to the negative result for n ≥ 3. The distance-distortion argument only checks the Lipschitz constant at finitely many points and does not explicitly rule out isometry on a positive-measure set; a global verification or an explicit formula for d_M(x,y) versus d_{S^3}(f(x),f(y)) is needed.

    Authors: We appreciate the referee's observation regarding the verification of the Lipschitz property in Construction 4.1. The construction involves a specific metric on M obtained by modifying the standard metric in a controlled way, allowing us to compute distances explicitly along geodesics. While the local checks at sample points suffice due to the smoothness and the way the map is defined, we acknowledge that a more global argument would enhance clarity. In the revised version, we will include an explicit formula for the Riemannian metric on M and a direct comparison showing that d_M(x,y) ≤ d_{S^3}(f(x),f(y)) with strict inequality for some pairs, thereby confirming it is not an isometry on a positive measure set. revision: yes

  2. Referee: §5, Theorem 5.3 (Ricci case): The proof that a 1-Lipschitz surjective map from a manifold with Ric ≥ (n-1)g to S^n is an isometry appears to reduce to the classical Llarull argument after replacing scalar curvature by Ricci; however, the spin assumption is dropped here, so the precise place where the spin structure is no longer required should be indicated.

    Authors: We agree that clarification is needed regarding the spin assumption in the Ricci case. The proof of Theorem 5.3 does not rely on spinorial methods; instead, it uses the Ricci lower bound to derive a contradiction via the maximum principle applied to the function dist(f(x), p) or similar, which holds without spin structure. We will add a sentence in the proof indicating that the spin condition from the scalar case is not invoked here, making the result valid for non-spin manifolds as well. revision: yes

Circularity Check

0 steps flagged

No circularity: derivation relies on independent counterexample construction and direct analysis

full rationale

The paper extends Llarull's theorem by distinguishing nonzero degree from surjectivity. The negative result for n≥3 rests on an explicit construction of a closed connected Riemannian spin manifold M with scal ≥ n(n-1) admitting a 1-Lipschitz surjective non-isometric map to S^n; this construction is verified directly against the curvature bound, spin condition, Lipschitz constant, surjectivity, and failure of isometry without reducing to any fitted parameter or self-definition inside the paper. Positive results for n=2 and the Ricci-curvature case are obtained by separate analytic arguments. No load-bearing step invokes a self-citation chain, uniqueness theorem from the authors' prior work, or an ansatz smuggled via citation; the cited Llarull theorem is treated as external input. The derivation chain is therefore self-contained against external benchmarks.

Axiom & Free-Parameter Ledger

0 free parameters · 2 axioms · 0 invented entities

The paper rests on standard background assumptions of Riemannian spin geometry and the original Llarull setup; no new free parameters or invented entities appear in the abstract.

axioms (2)
  • domain assumption M is a closed connected Riemannian spin manifold with scal(M) ≥ n(n-1)
    This is the curvature hypothesis taken directly from Llarull's theorem.
  • domain assumption f is a 1-Lipschitz map from M to S^n
    Standard Lipschitz condition in the rigidity statement.

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17 extracted references · 17 canonical work pages · 1 internal anchor

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