{"id":"b0fedfee-9317-4369-8981-ca9c1a9a65d3","arxiv_id":"1908.09544","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":5.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"A corrected proof of the Logarithmic Law for intrinsic algebraic entropy, with a counterexample to the previously relied-upon Lemma 3.11(b).","lead":"This note corrects a flawed proof of the Logarithmic Law for intrinsic algebraic entropy of Abelian group endomorphisms and supplies a valid replacement proof. It also gives a concrete counterexample showing that one of the lemmas used in the original proof is false.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The proof depends on [DGSV, Prop. 5.6] as stated, but the right Bernoulli shift on ⊕_N Z shows T_m(φ,F) need never be φ-inert; unless the cited proposition has unstated hypotheses, the construction of H fails.","rationale":"The central claim is the corrected proof of the Logarithmic Law. The most load-bearing step is the construction of an inert subgroup H inside a finitely generated trajectory: without it, Prop. 3.16(b) and the index computation cannot be applied. The note obtains H by first applying [DGSV, Prop. 5.6] to make T_m(φ,F) inert, then applying Lemma 1(a). The reader flagged Lemma 1(a) as suspect, but that lemma is actually easy to prove and is not the soft spot. The soft spot is Prop. 5.6 as quoted: in the torsion-free Bernoulli shift example, every T_m(β,F) fails to be β-inert because the quotient T_{m+1}/T_m is infinite cyclic. Since the entropy of this example is finite, the note's restriction to ~ent(φ)<∞ does not protect it. This does not disprove the Logarithmic Law, and the proof may be repairable by choosing a different inert subgroup or by stating the true hypotheses of Prop. 5.6, but as written the finite-generation case has a genuine gap. Therefore the verdict should remain conditional: the paper should be accepted only after the Prop. 5.6 dependency is clarified or the finite-generation proof is replaced.","tokens_in":3656,"tokens_out":30505,"duration_ms":317008,"concrete_test":"Check the cited [DGSV, Prop. 5.6] against the right Bernoulli shift β on G=⊕_N Z with F=Z e_0. Compute T_m(β,F)=⊕_{i<m}Z e_i; then (T_m+βT_m)/T_m ≅ Z, so no T_m is β-inert. If Prop. 5.6 really implies otherwise, this is a counterexample. If Prop. 5.6 has additional hypotheses, verify whether they hold in the note's setting and add the missing verification; otherwise replace the proof of the finite-generation case with an argument that does not require T_m(φ,F) to be inert.","verdict_should_be":"CONDITIONAL","load_bearing_attack":"Section 1's finite-generation case is built on the assertion that, for G=T(φ,F) with F finitely generated, [DGSV, Prop. 5.6] gives m with T_m(φ,F) φ-inert. As quoted, this assertion is false. Let G=⊕_{i∈N}Z, let β be the right shift β(x_0,x_1,...)=(0,x_0,x_1,...), and let F=Z e_0. Then G=T(β,F). For every m, T_m(β,F)=⊕_{i<m} Z e_i, so (T_m+βT_m)/T_m ≅ Z e_m, which is infinite; hence T_m is not β-inert. The standing assumption ~ent(φ)<∞ does not exclude this example: the intrinsic entropy of this shift is finite (in fact 0, since all β-inert subgroups here are commensurable with β-invariant subgroups). Thus the proof cannot construct the required inert subgroup H=T_k(φ,T_m(φ,F)) in this case. If the original Prop. 5.6 carries extra hypotheses, the note must state and verify them; if not, the finite-case argument is invalid. The reader's concern about Lemma 1(a) is less acute: part (a) is elementary and true, because H'/H is finite and T_n(φ,H')=T_{n+k-1}(φ,H), so the finite correction does not affect the limit.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"This note is an erratum to the paper by Dikranjan, Giordano Bruno, Salce, and Virili on intrinsic algebraic entropy. The authors identify a false statement in the original Lemma 3.11(b), provide an explicit counterexample involving the right Bernoulli shift on a direct sum of copies of Z(2), and then propose a new proof of the Logarithmic Law, ~ent(φ^k)=k·~ent(φ), for endomorphisms of Abelian groups. The new proof first treats the case G=T(φ,F) with F finitely generated, using a cited result [DGSV, Prop. 5.6] to obtain an inert subgroup, and then extends to arbitrary G by a direct-limit argument and upper continuity.","tokens_in":4001,"tokens_out":14481,"duration_ms":143568,"significance":"If the proposed proof were valid, the note would be a valuable correction: the counterexample to Lemma 1(b) is explicit, self-contained, and correctly computed, with the two entropy limits (2 log 2 and log 2) genuinely disagreeing. The note is also transparent about the source of the original error. However, the new proof relies on a citation to [DGSV, Prop. 5.6] that is not justified in the setting used and is in fact false as applied. This is a load-bearing gap, because the finite-generation case is the core of the argument. The Logarithmic Law itself is likely true, and the gap appears repairable, but the manuscript as submitted does not provide a complete correct proof.","major_comments":[{"comment":"The proof invokes [DGSV, Prop. 5.6] to assert the existence of m∈N+ such that T_m(φ,F) is φ-inert. This assertion is false in the exact setting where it is used. Let G=⊕_{i∈N}Z, let β be the right shift β(x_0,x_1,...)=(0,x_0,x_1,...), and let F=Z e_0. Then G=T(β,F), and ~ent(β)=0 because every β-inert subgroup H satisfies βH⊆H and hence has zero trajectory entropy. Yet for every m, T_m(β,F)=⊕_{i<m}Z e_i and (T_m+βT_m)/T_m ≅ Z e_m, which is infinite, so T_m is not β-inert. Thus the construction of H fails in this case, and the finite-generation part of the proof is invalid. The authors must either state and prove the precise version of Prop. 5.6 they need, with all hypotheses, or replace this step with a different argument.","section":"Section 1, finite-generation case"}],"minor_comments":[{"comment":"The proof cites [DGSV, Lem. 3.11(a)] without proof. Since part (b) of that lemma is false, the note should explicitly verify part (a) or cite an independent proof. This is not a substantive obstacle—part (a) follows because H'/H is finite and T_n(φ,H')=T_{n+k-1}(φ,H)—but the independence should be stated for the reader.","section":"Section 1, use of Lemma 1(a)"},{"comment":"The displayed formula for ~ent(φ^k) as a supremum over the subgroups T(φ,F) needs justification. Upper continuity gives a supremum over T(φ^k,F), not directly over T(φ,F). The equality can be recovered using T(φ,F)=T(φ^k,T_k(φ,F)) and the inclusion T(φ^k,F)⊆T(φ,F), but the text should spell this out.","section":"General case, Eq. (III)"},{"comment":"The notation Z(2) for the cyclic group of order 2 is nonstandard; Z/2Z or F_2 would be clearer.","section":"Example 2"}],"recommendation":"major_revision","confidential_remarks":"The paper is an erratum by an author of the original work, which is appropriate, and the counterexample is useful. The main concern is the unverified and apparently false use of [DGSV, Prop. 5.6]; this is repairable but requires substantive rewriting of the finite-generation case. I therefore recommend major revision rather than rejection."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The useful part of this note is the counterexample to DGSV Lemma 3.11(b). The right Bernoulli shift on ⊕N Z(2) with H a single coordinate does show ~ent(β², H′) ≠ ~ent(β², H), and the calculation is straightforward and correct. That alone is worth knowing and is likely new. The identity Tn(φ^k, H) = T_{kn−k+1}(φ, H) is also neat and could be useful elsewhere.\n\nThe problem is in the new proof of the Logarithmic Law. The finite-generation case asserts that, when G = T(φ, F) with F finitely generated, DGSV Prop. 5.6 gives an m with T_m(φ, F) φ-inert. That assertion is false as stated. Take G = ⊕_N Z, β the right shift, and F = Z e₀. Then T(β, F) = G, but T_m(β, F) = ⊕_{i<m} Z e_i, and (T_m + βT_m)/T_m ≅ Z e_m, infinite, so T_m is never β-inert. The intrinsic entropy of this β is finite (indeed 0), so the standing assumption ~ent(φ) < ∞ does not exclude it. Unless DGSV Prop. 5.6 carries hypotheses the note fails to state and verify, the construction of H collapses, and with it the first part of the proof. This is not a minor hole; it is the central step.\n\nThe reader's worry about Lemma 1(a) is less serious: that part is elementary and appears true, as the stress-test note says. So the paper does not need to defend (a) in the way the reader suggested.\n\nNet: the counterexample is a real contribution, and the authors are honest about the original flaw. But the paper's stated goal—to give a correct proof of the Logarithmic Law—is not achieved, because the proof leans on a false quoted result. The authors should either find a valid way to construct the inert subgroup H or present the counterexample as a separate erratum without claiming a full proof.\n\nWho is this for? Algebraists working on intrinsic entropy will want to know the original lemma is false. A serious referee should look at this, mainly to check whether the proof can be repaired. I'd send it out, but with the expectation of major revision.","headline":"The counterexample to the original lemma is valid and valuable, but the replacement proof of the Logarithmic Law relies on a false claim lifted from DGSV, so the correction does not close the gap.","tokens_in":4454,"tokens_out":2859,"would_cite":false,"duration_ms":26230,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["20K30","20K27","20K15","22B05","16D10","37A35","11R06"],"pacs":[],"model":"deepseek-v4-flash","headline":"The Logarithmic Law for intrinsic algebraic entropy is true, and this note replaces the flawed original proof with a correct one.","keywords":["intrinsic algebraic entropy","Logarithmic Law","endomorphisms","Abelian groups","algebraic dynamics","errata corrige","counterexample","Bernoulli shift"],"falsifier":"A concrete way to test the central claim is to compute $\\widetilde{\\mathrm{ent}}(\\varphi^2)$ and $2\\,\\widetilde{\\mathrm{ent}}(\\varphi)$ for an explicit family of endomorphisms where the supremum over inert subgroups can be examined, for instance the right Bernoulli shift on the direct sum of countably many copies of $\\mathbb{Z}(2)$; the note itself shows that one inert subgroup gives $\\widetilde{\\mathrm{ent}}(\\beta^2,H)=\\log(2)$ while $H'=T_2(\\beta,H)$ gives $2\\log(2)$, so what must be checked is whether the global suprema still coincide. Alternatively, a direct failure of Lemma 1(a), witnessed by a $\\varphi$-inert subgroup $H$ with $\\widetilde{\\mathrm{ent}}(\\varphi,H)\\neq \\widetilde{\\mathrm{ent}}(\\varphi,T_k(\\varphi,H))$, would break the new proof even if the Logarithmic Law itself might still be true.","tokens_in":3496,"feed_emoji":"📈","tokens_out":11919,"duration_ms":95041,"temperature":0.7,"pith_summary":"The paper is an erratum that repairs the proof of the Logarithmic Law for the intrinsic algebraic entropy $\\widetilde{\\mathrm{ent}}(\\varphi)$ of an endomorphism $\\varphi$ of an Abelian group. The law states that iterating the endomorphism $k$ times multiplies the entropy by exactly $k$: $\\widetilde{\\mathrm{ent}}(\\varphi^k)=k\\,\\widetilde{\\mathrm{ent}}(\\varphi)$. The authors show that the original proof relied on a lemma whose second part is false, and they provide an explicit counterexample to that lemma. They then prove the law through a different route, using trajectory equalities, a reduction to finitely generated subgroups, and upper-continuity. The point that matters is that a basic property one expects from any entropy function is confirmed for the intrinsic algebraic entropy despite the flaw in its original demonstration.","feed_headline":"Intrinsic algebraic entropy follows Logarithmic Law after all","feed_subtitle":"The original proof leaned on a false lemma; a counterexample and a new argument close the gap.","key_machinery":"The machinery is the trajectory calculus of the intrinsic algebraic entropy. For a subgroup $H$, the $n$-th partial $\\varphi$-trajectory is $T_n(\\varphi,H)=H+\\varphi(H)+\\cdots+\\varphi^{n-1}(H)$, and the entropy with respect to $H$ is the logarithmic growth rate of $\\lvert T_n(\\varphi,H)/H\\rvert$ as $n\\to\\infty$; the intrinsic entropy $\\widetilde{\\mathrm{ent}}(\\varphi)$ is the supremum of these growth rates over all $\\varphi$-inert subgroups $H$. The central identity in the corrected proof is $T_n(\\varphi^k,H)=T_{kn-k+1}(\\varphi,H)$, which reindexes the $k$-fold iteration as a single trajectory and produces the factor $k$ in the growth rate. Around this identity, the argument uses Lemma 1(a) to find a finitely generated $H$ whose trajectory is all of $G$, Lemma 2.7(b) to transfer inertness from $\\varphi$ to $\\varphi^k$, Proposition 3.16(b) to identify the global entropy with the entropy on $H$, and Lemma 3.14 (upper-continuity) to extend from finitely generated trajectories to arbitrary $G$. The counterexample to Lemma 1(b) is a right Bernoulli shift on a direct sum of copies of $\\mathbb{Z}(2)$, where two different inert subgroups give different partial entropies for $\\beta^2$.","core_discovery":"The central claim is that the Logarithmic Law, $\\widetilde{\\mathrm{ent}}(\\varphi^k)=k\\,\\widetilde{\\mathrm{ent}}(\\varphi)$, holds for every endomorphism $\\varphi$ of an Abelian group and every positive integer $k$, even though the original argument for it was invalid. The note exhibits a right Bernoulli shift on the direct sum of countably many copies of $\\mathbb{Z}(2)$ for which the equality asserted in Lemma 1(b) of the original paper fails, thereby showing that the old proof of the inequality $\\widetilde{\\mathrm{ent}}(\\varphi^k)\\le k\\,\\widetilde{\\mathrm{ent}}(\\varphi)$ cannot stand. A new proof reconstructs the missing inequality without the false lemma: for a finitely generated subgroup $F$ with $G=T(\\varphi,F)$, the subgroup $H=T_{m+k-1}(\\varphi,F)$ is both $\\varphi$-inert and $\\varphi^k$-inert, the partial trajectories satisfy $T_n(\\varphi^k,H)=T_{kn-k+1}(\\varphi,H)$, and the general case follows by taking suprema over finitely generated $\\varphi$-invariant subgroups. The theorem originally stated in the paper being corrected is therefore true, with a proof that no longer depends on Lemma 1(b).","pith_inferences":["The new proof leans on Lemma 1(a) without re-verifying it; a separate proof of that part, independent of the flawed Lemma 1(b), would make the corrected argument fully self-contained.","The Bernoulli-shift example suggests a general caution: for a fixed endomorphism, different inert subgroups can yield very different partial entropies, so only the supremum over all inert subgroups is a reliable invariant.","One testable extension is to carry the trajectory-reindexing argument to other entropy-like invariants defined by suprema over inert subgroups in settings that admit a direct-limit structure, where the same $k$-factor formula might appear."],"forward_implications":["For any endomorphism of an Abelian group, the entropy of the $k$-th iterate is exactly $k$ times the entropy of the endomorphism, so the invariant measures trajectory growth at the expected logarithmic rate.","The corrected proof does not disturb the other intrinsic entropy properties—additivity, upper-continuity, and the Intrinsic Yuzvinski Formula—because their proofs do not rely on the false Lemma 1(b).","The explicit counterexample to Lemma 1(b) warns that passing a subgroup to its trajectory closure can change the entropy contribution, so any future proof using that lemma must check the equality independently.","The identity $T_n(\\varphi^k,H)=T_{kn-k+1}(\\varphi,H)$ gives a direct computational shortcut for the entropy of iterates whenever a finitely generated subgroup generates the whole trajectory.","The reduction to finitely generated $\\varphi$-invariant subgroups via upper-continuity means intrinsic entropy can be computed from the directed system of finitely generated trajectory subgroups."],"supporting_citations":[{"why":"Supplies an m such that T_m(phi,F) is phi-inert for a finitely generated F, used to build the subgroup H.","marker":"[DGSV, Prop. 5.6]"},{"why":"Shows that a phi-inert subgroup is also phi^k-inert, so the constructed H is admissible for both endomorphisms.","marker":"[DGSV, Lem. 2.7(b)]"},{"why":"Gives equality between global intrinsic entropy and the entropy on H when G is the full phi-trajectory of H.","marker":"[DGSV, Prop. 3.16(b)]"},{"why":"Upper-continuity lets the proof pass from finitely generated trajectory subgroups to all of G by taking suprema.","marker":"[DGSV, Lem. 3.14]"},{"why":"The statement whose part (b) is shown false by Example 2, while part (a) remains load-bearing in the new proof.","marker":"[DGSV, Lem. 3.11]"}],"fun_headline_variants":["Logarithmic Law holds for intrinsic algebraic entropy","Intrinsic algebraic entropy: Logarithmic Law proven","New proof secures Logarithmic Law for intrinsic algebraic entropy","Logarithmic Law for intrinsic algebraic entropy proven after flaw"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The corrected proof assumes that Lemma 1(a) from the original paper remains valid even though the neighboring part 1(b) is false; if Lemma 1(a) also fails, the construction of the subgroup $H$ whose trajectories equalize $\\varphi$ and $\\varphi^k$ would not go through and the proof would have to be repaired again.","fun_headline_variants_meta":{"raw":{"variants":["Logarithmic Law holds for intrinsic algebraic entropy","Intrinsic algebraic entropy: Logarithmic Law proven","New proof secures Logarithmic Law for intrinsic algebraic entropy","Logarithmic Law for intrinsic algebraic entropy proven after flaw"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000415,"raw_usage":{"total_tokens":2117,"prompt_tokens":895,"completion_tokens":1222,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":511,"completion_tokens_details":{"reasoning_tokens":1159}},"tokens_in":511,"tokens_out":1222,"duration_ms":8836,"temperature":1.0,"reasoning_tokens":1159,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T11:07:36.029028+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"A concrete way to test the central claim is to compute $\\widetilde{\\mathrm{ent}}(\\varphi^2)$ and $2\\,\\widetilde{\\mathrm{ent}}(\\varphi)$ for an explicit family of endomorphisms where the supremum over inert subgroups can be examined, for instance the right Bernoulli shift on the direct sum of countably many copies of $\\mathbb{Z}(2)$; the note itself shows that one inert subgroup gives $\\widetilde{\\mathrm{ent}}(\\beta^2,H)=\\log(2)$ while $H'=T_2(\\beta,H)$ gives $2\\log(2)$, so what must be checked is whether the global suprema still coincide. Alternatively, a direct failure of Lemma 1(a), witnessed by a $\\varphi$-inert subgroup $H$ with $\\widetilde{\\mathrm{ent}}(\\varphi,H)\\neq \\widetilde{\\mathrm{ent}}(\\varphi,T_k(\\varphi,H))$, would break the new proof even if the Logarithmic Law itself might still be true.","supporting_citations":[],"review_version":1}