{
  "source_sha256": "8adb9681e2cd5e4bdcb292162a32796b00ae7b757a5a900fbd580d0bc939e34e",
  "review_policy": {
    "schema": 1,
    "cutoff_date": "2015-07-28",
    "basis": "journal_received",
    "evidence": "Oxford Academic publisher indexed article DOI 10.1093/imrn/rnw136: Received 28 July 2015; revision 3 March 2016; accepted 27 May 2016; online 5 August 2016. Search-accessible publisher record; direct HTML/PDF downloads failed.",
    "status": "recorded",
    "rule": "external-results-available-by-cutoff"
  },
  "nodes": {
    "toptwoth": {
      "outcome": "confirmed_defect",
      "origin": "curated_source_check",
      "source_quote": "\\begin{thm}\\label{toptwoth}\nFor $n\\geq 3$ fixed, let $G=\\mathrm{SL}_n(\\mathbb R)$, and let $\\mu$ be Haar measure on $\\mathrm{SL}_n(\\mathbb R)$.  For $g\\in G$, denote by $a_1(g),a_2(g),\\dots,a_n(g)$ the nonzero entries of the diagonal matrix $a_g$ in the $KAK$ decomposition of $g$ in (\\ref{KAK}) with $a_1(g)\\geq a_2(g)\\geq\\cdots\\geq a_n(g)>0$.  Fix $\\eta>4$. Then\n$$0<\\frac{\\lim_{X\\rightarrow\\infty} \\, \\mu(\\{g\\in G\\, | \\, ||g||\\leq X, a_1(g)/a_2(g)\\geq\\eta^2\\})}{\\lim_{X\\rightarrow\\infty} \\, \\mu(\\{g\\in G\\, | \\, ||g||\\leq X\\})}<\\frac{1}{\\eta^{2(n^2-3n+2)}}.$$\n\\end{thm}",
      "summary": "The n=3 limiting tail is 2 eta^(-4)-eta^(-8), larger than the proposed eta^(-4) for every eta>4. The normalized interpretation of the displayed divergent-limit ratio is refuted too.",
      "statement_status": "refuted",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#false-numerical-bound",
      "statement_argument": "Cartan rescaling and dominated convergence give density 8 exp(-3y) sinh y; its exact tail at eta=5 is 1249/390625 > 1/625. See the report and the exact checker."
    },
    "twoops": {
      "outcome": "confirmed_defect",
      "origin": "curated_source_check",
      "source_quote": "\\begin{lemma}\n\\label{twoops}\nLet $A, B \\in \\GLn(\\mathbb Z)$ be such that $\\lambda_1(A), \\lambda_1(B), \\lambda_1(A^{-1}), \\lambda_1(B^{-1}) \\geq \\lambda \\gg 1.$ Assume, further, that all the inner products $\\langle u_1(A), v_1(A)\\rangle, \\langle u_1(A), v_1(B)\\rangle, \\langle u_1(A),\\\\ v_1(B), \\langle(u_1(B), v_1(B)\\rangle$ are bigger than $\\epsilon$ in absolute value. Let $w(A, B)$ be a reduced word of length $k$ in $A, B, A^{-1}, B^{-1}.$ Then \\begin{equation}\n\\| w(A, B) u_1(A)\\| \\geq (\\epsilon \\lambda)^k.\n\\end{equation}\n\\end{lemma}",
      "summary": "The fixed input vector cannot satisfy the expansion estimate for the reduced word A inverse; the supplied inverse-singular-vector identity is wrong.",
      "statement_status": "refuted",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#false-product-lemma",
      "statement_argument": "The exact integral determinant-one matrices A=[[13,8],[8,5]], B=[[13,-8],[-8,5]] satisfy epsilon=1/3 and all four norm hypotheses with lambda=9+4 sqrt(5). Yet ||A inverse u1(A)||=9-4 sqrt(5)<lambda/3."
    },
    "freelen": {
      "outcome": "confirmed_defect",
      "origin": "independent_arbitration",
      "source_quote": "\\begin{fact}\n\\label{freelen}\nLet $w(x, y)$ be a random product of $x, y, x^{-1}, y^{-1},$ of length $n.$ Then, with probability tending to one as $n$ goes to infinity, the reduced length of $w$ is at least $3\\ell(w)/4.$\n\\end{fact}",
      "summary": "Accepted Gemini arbitration refutes the 3n/4 high-probability lower bound for uniform independent four-letter input.",
      "statement_status": "refuted",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#false-reduced-length-estimate",
      "statement_argument": "The return-probability bound gives E L_n <= n/2+O(1), and hence limsup P(L_n>=3n/4)<=2/3. The correct almost-sure speed is 1/2 by bounded martingale differences and finitely many returns."
    },
    "freesln": {
      "outcome": "confirmed_defect",
      "origin": "independent_arbitration",
      "source_quote": "\\begin{prop}\\label{freesln}\nLet $G=\\mathrm{SL}_n(\\ZZ)$, and let $\\Gamma(g)$ and $\\mu_X'$ be as above.  Then we have\n\\begin{equation*}\n\\lim_{X\\rightarrow\\infty}\\mu'_X(\\{g=(g_1,g_2)\\in G^2\\; |\\; \\Gamma(g) \\mbox{\\emph{ is free}}\\})=1\n\\end{equation*}\n\\end{prop}",
      "summary": "Accepted Gemini arbitration finds a defective proof: impossible odd-dimensional integral bound and missing angular/ping-pong justification. It leaves the generic-freeness statement unresolved.",
      "statement_status": "unresolved",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#generic-freeness-and-the-main-theorem"
    },
    "middletwo": {
      "outcome": "confirmed_defect",
      "origin": "curated_source_check",
      "source_quote": "\\begin{lemma}\\label{middletwo}\nLet $G=\\mathrm{SL}_n(\\ZZ)$, and let $\\mu'_X$ be as above.  For $g\\in G$, denote by $a_1(g),\\dots, a_n(g)$ the nonzero entries of the diagonal matrix $a_g$ in the $KAK$ decomposition of $g$ in (\\ref{KAK}) with $a_1(g)\\geq a_2(g)\\geq \\cdots\\geq a_n(g)>0$.  Fix $\\eta>4$. Then, if $n$ is even, we have \n$$\\lim_{X\\rightarrow\\infty} \\, \\mu'_X\\left(\\left\\{g\\in G\\, \\Big| \\, \\frac{a_k(g)}{a_{k+1}(g)}\\geq\\eta^2 \\mbox{\\emph{ and }} \\frac{a_k(g^{-1})}{a_{k+1}(g^{-1})}\\geq\\eta^2\\right\\}\\right)=1,$$\nwhere $k=n/2$.  If $n$ is odd, we have that \n$$\\lim_{X\\rightarrow\\infty} \\, \\mu'_X\\left(\\left\\{g\\in G\\, \\Big| \\, \\frac{a_k(g)}{a_{k+1}(g)}, \\frac{a_{k+1}(g)}{a_{k+2}(g)}\\geq\\eta^2, \\mbox{\\emph{ and }} \\frac{a_k(g^{-1})}{a_{k+1}(g^{-1})},  \\frac{a_{k+1}(g^{-1})}{a_{k+2}(g^{-1})}\\geq\\eta^2\\right\\}\\right)=1,$$\nwhere $k=(n-1)/2$.\n\\end{lemma}",
      "summary": "The odd-dimensional proof contains an impossible vanishing bound for a positive integral over an expanding region. The corrected volume/gap argument remains a proposed replacement; the lemma itself is not refuted here.",
      "statement_status": "unresolved",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#generic-freeness-and-the-main-theorem"
    },
    "prox": {
      "outcome": "confirmed_defect",
      "origin": "curated_source_check",
      "source_quote": "\\begin{lemma}\\label{prox}\nLet $G=\\mathrm{SL}_n(\\ZZ)$, where $n>2$, and let $\\mu'_X$ be as before.  For $g\\in G$, let $g=k_ga_gk_g'$ be a Cartan decomposition of $g$ and let $k=n/2$ if $n$ is even, and $k=(n-1)/2$ if $n$ is odd.  Denote by $H_g$ the projective hyperplane in $\\mathbb P(\\bigwedge^k(\\mathbb R^n))$ spanned by $\\{k_{g}'^{-1}(\\mathbf{v})\\;|\\; \\mathbf v\\in B\\backslash e_1\\wedge e_2\\wedge\\cdots\\wedge e_k\\}$ where $B$ is as in Lemma~\\ref{epscontractextpower}, and let $v_g=k_{g}(e_1\\wedge e_2\\wedge\\cdots\\wedge e_k)$.  Then, as $\\epsilon'$ tends to $0$, we have\n$$\\lim_{X\\rightarrow\\infty}\\, \\mu'_X(\\{g\\in G\\, | \\, d(v_g,H_g)\\leq 2\\epsilon'\\})\\ll \\epsilon'.$$\n\\end{lemma}",
      "summary": "Positive-dimensional compact-group orbits alone do not prove the printed linear incidence-measure estimate. The actual sector-counting hypotheses and all inverse-generator incidences need checking.",
      "statement_status": "unresolved",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#generic-freeness-and-the-main-theorem"
    },
    "thinsln": {
      "outcome": "unresolved",
      "origin": "curated_source_check",
      "source_quote": "\\begin{thm}\\label{thinsln}\nLet $G=\\mathrm{SL}_n(\\ZZ)$ where $n\\geq 2$, and let $B_X'(G)$ and $\\mu_X'$ be as above.  Then we have\n\\begin{equation*}\n\\lim_{X\\rightarrow\\infty}\\mu'_X(\\{(g_1,g_2)\\in (B_X'(G))^2\\; |\\; \\langle g_1,g_2\\rangle \\mbox{\\emph{ is thin}}\\})=1\n\\end{equation*}\n\\end{thm}",
      "summary": "Several local statements and the supplied generic-freeness proof are defective. No counterexample to main generic thinness is established; the replacement proof chain remains incomplete.",
      "statement_status": "unresolved",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#generic-freeness-and-the-main-theorem"
    },
    "thm:line:734": {
      "outcome": "unresolved",
      "origin": "curated_source_check",
      "source_quote": "\\begin{thm}\nLet $A, B$ be picked uniformly at random from the symmetrized ball of radius $R.$ Then, the Lyapunov exponent of the group generated by $A, B$ goes to infinity with $R,$ with high probability.\n\\end{thm}",
      "summary": "The supplied Lyapunov proof uses two refuted inputs and lacks a checked uniform replacement cone argument. This does not establish falsity of the theorem.",
      "statement_status": "unresolved",
      "explanation_url": "/papers/human-fuchs-rivin-generic-thinness/panel/#generic-freeness-and-the-main-theorem"
    }
  }
}