{
  "index": "2011-A-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "\\[ \\{g_1=e,g_2,\\dots,g_k\\} \\subsetneqq G \\]\nbe a (not necessarily minimal) set of distinct generators of $G$. A special\ndie, which randomly selects one of the elements $g_1,g_2,...,g_k$ with equal\nprobability, is rolled $m$ times and the selected elements are multiplied\nto produce an element $g \\in G$.  Prove that there exists a real number\n$b \\in (0,1)$ such that\n\n\\[ \\lim_{m\\to\\infty} \\frac{1}{b^{2m}} \\sum_{x\\in G} \\left(\\mathrm{Prob}(g=x)\n    - \\frac{1}{n}\\right)^2 \\]\nis positive and finite.",
  "solution": "Choose some ordering $h_1,\\dots, h_n$ of the elements of $G$ with $h_1 = e$.\nDefine an $n \\times n$ matrix $M$\nby settting $M_{ij} = 1/k$ if $h_j = h_i g$ for some $g \\in \\{g_1,\\dots,g_k\\}$ and $M_{ij} = 0$ otherwise.\nLet $v$ denote the column vector $(1,0,\\dots,0)$. The probability\nthat the product of $m$ random elements of $\\{g_1,\\dots,g_k\\}$\nequals $h_i$ can then be interpreted as the $i$-th component of the vector\n$M^m v$.\n\nLet $\\hat{G}$ denote the dual group of $G$, i.e., the group of complex-valued characters of $G$.\nLet $\\hat{e} \\in \\hat{G}$ denote the trivial character.\nFor each $\\chi \\in \\hat{G}$, the vector $v_\\chi = (\\chi(h_i))_{i=1}^n$ is an eigenvector of $M$\nwith eigenvalue $\\lambda_\\chi = (\\chi(g_1) + \\cdots + \\chi(g_k))/k$.\nIn particular, $v_{\\hat{e}}$ is the all-ones vector and $\\lambda_{\\hat{e}} = 1$.\nPut\n\\[\nb = \\max\\{|\\lambda_\\chi|: \\chi \\in \\hat{G} - \\{\\hat{e}\\}\\};\n\\]\nwe show that $b \\in (0,1)$ as follows.\nFirst suppose $b=0$; then\n\\[\n1 = \\sum_{\\chi \\in \\hat{G}} \\lambda_\\chi\n= \\frac{1}{k} \\sum_{i=1}^k \\sum_{\\chi \\in \\hat{G}} \\chi(g_i) = \\frac{n}{k}\n\\]\nbecause $\\sum_{\\chi \\in \\hat(G)} \\chi(g_i)$ equals $n$ for $i=1$ and $0$ otherwise.\nHowever, this contradicts the hypothesis that $\\{g_1, \\dots, g_k\\}$ is not all of $G$.\nHence $b > 0$. Next suppose $b=1$, and choose $\\chi \\in \\hat{G} - \\{\\hat{e}\\}$ with $|\\lambda_\\chi| = 1$.\nSince each of\n$\\chi(g_1), \\dots, \\chi(g_k)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.\nSince $\\chi(g_1) = \\chi(e) = 1$, $\\chi$ must map each of $g_1,\\dots, g_k$ to 1, but this is impossible because\n$\\chi$ is a nontrivial character and $g_1,\\dots,g_k$ form a set of generators of $G$.\nThis contradiction yields $b < 1$.\n\nSince $v = \\frac{1}{n} \\sum_{\\chi \\in \\hat{G}} v_\\chi$\nand $M v_\\chi = \\lambda_\\chi v_\\chi$, we have\n\\[\nM^m v - \\frac{1}{n} v_{\\hat{e}}\n= \\frac{1}{n} \\sum_{\\chi \\in \\hat{G} - \\{\\hat{e}\\}} \\lambda_\\chi^m v_\\chi.\n\\]\nSince the vectors $v_\\chi$ are pairwise orthogonal,\nthe limit we are interested in can be written as\n\\[\n\\lim_{m \\to \\infty} \\frac{1}{b^{2m}} (M^m v - \\frac{1}{n} v_{\\hat{e}}) \\cdot (M^m v - \\frac{1}{n} v_{\\hat{e}}).\n\\]\nand then rewritten as\n\\[\n\\lim_{m \\to \\infty} \\frac{1}{b^{2m}} \\sum_{\\chi \\in \\hat{G} - \\{\\hat{e}\\}} |\\lambda_\\chi|^{2m}\n= \\#\\{\\chi \\in \\hat{G}: |\\lambda_\\chi| = b\\}.\n\\]\nBy construction, this last quantity is nonzero and finite.\n\n\n\\textbf{Remark.}\nIt is easy to see that the result fails if we do not assume $g_1 = e$: take $G = \\ZZ/2\\ZZ$,\n$n=1$, and $g_1 = 1$.\n\n\\textbf{Remark.}\nHarm Derksen points out that a similar argument applies even if $G$ is not assumed to be abelian,\nprovided that the operator $g_1 + \\cdots + g_k$ in the group algebra $\\ZZ[G]$ is \\emph{normal},\ni.e., it commutes with the operator $g_1^{-1} + \\cdots + g_k^{-1}$.\nThis includes the cases where the set $\\{g_1,\\dots,g_k\\}$ is closed under taking inverses\nand where it is a union of conjugacy classes (which in turn includes the case of $G$ abelian).\n\n\\textbf{Remark}.\nThe matrix $M$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $m$ such that $M^m$ has positive entries. For any such matrix,\nthe Perron-Frobenius theorem implies that\nthe sequence of vectors $M^m v$ converges to a limit $w$, and there exists $b \\in [0,1)$\nsuch that\n\\[\n\\limsup_{m \\to \\infty} \\frac{1}{b^{2m}} \\sum_{i=1}^n ((M^m v - w)_i)^2\n\\]\nis nonzero and finite. (The intended interpretation in case $b=0$ is that $M^m v = w$ for all large $m$.)\nHowever, the limit need not exist in general.",
  "vars": [
    "g",
    "g_1",
    "g_2",
    "g_k",
    "g_i",
    "g_j",
    "h_i",
    "h_1",
    "h_n",
    "x",
    "m",
    "i",
    "j",
    "w",
    "\\\\chi",
    "\\\\lambda_\\\\chi"
  ],
  "params": [
    "G",
    "k",
    "n",
    "b",
    "e",
    "M",
    "v",
    "\\\\lambda_\\\\hat{e}"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "g": "element",
        "g_1": "firstgen",
        "g_2": "secondgen",
        "g_k": "lastgen",
        "g_i": "ithgen",
        "g_j": "jthgen",
        "h_i": "ithgroup",
        "h_1": "firstgroup",
        "h_n": "lastgroup",
        "x": "someelem",
        "m": "rollcount",
        "i": "indexi",
        "j": "indexj",
        "w": "limitvec",
        "\\\\chi": "charfunc",
        "\\\\lambda_\\\\chi": "eigenval",
        "G": "groupg",
        "k": "gencount",
        "n": "gsize",
        "b": "spectrad",
        "e": "identity",
        "M": "probmat",
        "v": "startvec",
        "\\\\lambda_\\\\hat{e}": "identeval"
      },
      "question": "\\[ \\{firstgen=identity,secondgen,\\dots,lastgen\\} \\subsetneqq groupg \\]\nbe a (not necessarily minimal) set of distinct generators of $groupg$. A special\ndie, which randomly selects one of the elements $firstgen,secondgen,\\dots,lastgen$ with equal\nprobability, is rolled $rollcount$ times and the selected elements are multiplied\nto produce an element $element \\in groupg$.  Prove that there exists a real number\n$spectrad \\in (0,1)$ such that\n\n\\[ \\lim_{rollcount\\to\\infty} \\frac{1}{spectrad^{2\\,rollcount}} \\sum_{someelem\\in groupg} \\left(\\mathrm{Prob}(element=someelem)\n    - \\frac{1}{gsize}\\right)^2 \\]\nis positive and finite.",
      "solution": "Choose some ordering $firstgroup,\\dots, lastgroup$ of the elements of $groupg$ with $firstgroup = identity$.\nDefine an $gsize \\times gsize$ matrix $probmat$\nby settting $probmat_{indexi indexj} = 1/gencount$ if $jthgroup = ithgroup element$ for some $element \\in \\{firstgen,\\dots,lastgen\\}$ and $probmat_{indexi indexj} = 0$ otherwise.\nLet $startvec$ denote the column vector $(1,0,\\dots,0)$. The probability\nthat the product of $rollcount$ random elements of $\\{firstgen,\\dots,lastgen\\}$\nequals $ithgroup$ can then be interpreted as the $indexi$-th component of the vector\n$probmat^{rollcount} startvec$.\n\nLet $\\hat{groupg}$ denote the dual group of $groupg$, i.e., the group of complex-valued characters of $groupg$.\nLet $\\hat{identity} \\in \\hat{groupg}$ denote the trivial character.\nFor each $charfunc \\in \\hat{groupg}$, the vector $startvec_{charfunc} = (charfunc(ithgroup))_{indexi=1}^{gsize}$ is an eigenvector of $probmat$\nwith eigenvalue $eigenval = (charfunc(firstgen) + \\cdots + charfunc(lastgen))/gencount$.\nIn particular, $startvec_{\\hat{identity}}$ is the all-ones vector and $identeval = 1$.\nPut\n\\[\nspectrad = \\max\\{ |eigenval| : charfunc \\in \\hat{groupg} - \\{\\hat{identity}\\} \\};\n\\]\nwe show that $spectrad \\in (0,1)$ as follows.\nFirst suppose $spectrad=0$; then\n\\[\n1 = \\sum_{charfunc \\in \\hat{groupg}} eigenval\n= \\frac{1}{gencount} \\sum_{indexi=1}^{gencount} \\sum_{charfunc \\in \\hat{groupg}} charfunc(ithgen) = \\frac{gsize}{gencount}\n\\]\nbecause $\\sum_{charfunc \\in \\hat{groupg}} charfunc(ithgen)$ equals $gsize$ for $indexi=1$ and $0$ otherwise.\nHowever, this contradicts the hypothesis that $\\{firstgen, \\dots, lastgen\\}$ is not all of $groupg$.\nHence $spectrad > 0$. Next suppose $spectrad=1$, and choose $charfunc \\in \\hat{groupg} - \\{\\hat{identity}\\}$ with $|eigenval| = 1$.\nSince each of\n$charfunc(firstgen), \\dots, charfunc(lastgen)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.\nSince $charfunc(firstgen) = charfunc(identity) = 1$, $charfunc$ must map each of $firstgen,\\dots, lastgen$ to 1, but this is impossible because\n$charfunc$ is a nontrivial character and $firstgen,\\dots,lastgen$ form a set of generators of $groupg$.\nThis contradiction yields $spectrad < 1$.\n\nSince $startvec = \\frac{1}{gsize} \\sum_{charfunc \\in \\hat{groupg}} startvec_{charfunc}$\nand $probmat startvec_{charfunc} = eigenval startvec_{charfunc}$, we have\n\\[\nprobmat^{rollcount} startvec - \\frac{1}{gsize} startvec_{\\hat{identity}}\n= \\frac{1}{gsize} \\sum_{charfunc \\in \\hat{groupg} - \\{\\hat{identity}\\}} eigenval^{rollcount} startvec_{charfunc}.\n\\]\nSince the vectors $startvec_{charfunc}$ are pairwise orthogonal,\nthe limit we are interested in can be written as\n\\[\n\\lim_{rollcount \\to \\infty} \\frac{1}{spectrad^{2\\,rollcount}} \\left( probmat^{rollcount} startvec - \\frac{1}{gsize} startvec_{\\hat{identity}} \\right) \\cdot \\left( probmat^{rollcount} startvec - \\frac{1}{gsize} startvec_{\\hat{identity}} \\right)\n\\]\nand then rewritten as\n\\[\n\\lim_{rollcount \\to \\infty} \\frac{1}{spectrad^{2\\,rollcount}} \\sum_{charfunc \\in \\hat{groupg} - \\{\\hat{identity}\\}} |eigenval|^{2\\,rollcount}\n= \\#\\{ charfunc \\in \\hat{groupg} : |eigenval| = spectrad \\}.\n\\]\nBy construction, this last quantity is nonzero and finite.\n\n\\textbf{Remark.}\nIt is easy to see that the result fails if we do not assume $firstgen = identity$: take $groupg = \\ZZ/2\\ZZ$,\n$gsize=1$, and $firstgen = 1$.\n\n\\textbf{Remark.}\nHarm Derksen points out that a similar argument applies even if $groupg$ is not assumed to be abelian,\nprovided that the operator $firstgen + \\cdots + lastgen$ in the group algebra $\\ZZ[groupg]$ is \\emph{normal},\ni.e., it commutes with the operator $firstgen^{-1} + \\cdots + lastgen^{-1}$.\nThis includes the cases where the set $\\{firstgen,\\dots,lastgen\\}$ is closed under taking inverses\nand where it is a union of conjugacy classes (which in turn includes the case of $groupg$ abelian).\n\n\\textbf{Remark}.\nThe matrix $probmat$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $rollcount$ such that $probmat^{rollcount}$ has positive entries. For any such matrix,\nthe Perron-Frobenius theorem implies that\nthe sequence of vectors $probmat^{rollcount} startvec$ converges to a limit $limitvec$, and there exists $spectrad \\in [0,1)$\nsuch that\n\\[\n\\limsup_{rollcount \\to \\infty} \\frac{1}{spectrad^{2\\,rollcount}} \\sum_{indexi=1}^{gsize} ((probmat^{rollcount} startvec - limitvec)_{indexi})^2\n\\]\nis nonzero and finite. (The intended interpretation in case $spectrad=0$ is that $probmat^{rollcount} startvec = limitvec$ for all large $rollcount$.)\nHowever, the limit need not exist in general."
    },
    "descriptive_long_confusing": {
      "map": {
        "g": "watermelon",
        "g_1": "tangerine",
        "g_2": "blueberry",
        "g_k": "porcupine",
        "g_i": "marsupial",
        "g_j": "anachrony",
        "h_i": "caterpillar",
        "h_1": "salamander",
        "h_n": "platypus",
        "x": "orangutan",
        "m": "dandelion",
        "i": "armadillo",
        "j": "butterfly",
        "w": "rhinoceros",
        "\\chi": "megafauna",
        "\\lambda_\\chi": "tarantula",
        "G": "lemniscate",
        "k": "honeycomb",
        "n": "bluewhale",
        "b": "chrysalis",
        "M": "hibiscus",
        "v": "mongoose",
        "\\lambda_\\hat{e}": "aardvarks"
      },
      "question": "\\[ \\{tangerine=e,blueberry,\\dots,porcupine\\} \\subsetneqq lemniscate \\]\nbe a (not necessarily minimal) set of distinct generators of $lemniscate$. A special\ndie, which randomly selects one of the elements $tangerine,blueberry,...,porcupine$ with equal\nprobability, is rolled $dandelion$ times and the selected elements are multiplied\nto produce an element $watermelon \\in lemniscate$.  Prove that there exists a real number\n$chrysalis \\in (0,1)$ such that\n\n\\[ \\lim_{dandelion\\to\\infty} \\frac{1}{chrysalis^{2dandelion}} \\sum_{orangutan\\in lemniscate} \\left(\\mathrm{Prob}(watermelon=orangutan)\n    - \\frac{1}{bluewhale}\\right)^2 \\]\nis positive and finite.",
      "solution": "Choose some ordering $salamander,\\dots, platypus$ of the elements of $lemniscate$ with $salamander = e$.\nDefine an $bluewhale \\times bluewhale$ matrix $hibiscus$\nby settting $hibiscus_{armadillobutterfly} = 1/honeycomb$ if $butterfly$-th element equals $armadillo$-th element multiplied by some $watermelon \\in \\{tangerine,\\dots,porcupine\\}$ and $hibiscus_{armadillobutterfly} = 0$ otherwise.\nLet $mongoose$ denote the column vector $(1,0,\\dots,0)$. The probability\nthat the product of $dandelion$ random elements of $\\{tangerine,\\dots,porcupine\\}$\nequals $caterpillar$ can then be interpreted as the $armadillo$-th component of the vector\n$hibiscus^{dandelion} mongoose$.\n\nLet $\\hat{lemniscate}$ denote the dual group of $lemniscate$, i.e., the group of complex-valued characters of $lemniscate$.\nLet $\\hat{e} \\in \\hat{lemniscate}$ denote the trivial character.\nFor each $megafauna \\in \\hat{lemniscate}$, the vector $mongoose_{megafauna} = (megafauna(caterpillar))_{armadillo=1}^{bluewhale}$ is an eigenvector of $hibiscus$\nwith eigenvalue $tarantula = (megafauna(tangerine) + \\cdots + megafauna(porcupine))/honeycomb$.\nIn particular, $mongoose_{\\hat{e}}$ is the all-ones vector and $aardvarks = 1$.\nPut\n\\[\nchrysalis = \\max\\{|tarantula|: megafauna \\in \\hat{lemniscate} - \\{\\hat{e}\\}\\};\n\\]\nwe show that $chrysalis \\in (0,1)$ as follows.\nFirst suppose $chrysalis=0$; then\n\\[\n1 = \\sum_{megafauna \\in \\hat{lemniscate}} tarantula\n= \\frac{1}{honeycomb} \\sum_{armadillo=1}^{honeycomb} \\sum_{megafauna \\in \\hat{lemniscate}} megafauna(marsupial) = \\frac{bluewhale}{honeycomb}\n\\]\nbecause $\\sum_{megafauna \\in \\hat(lemniscate)} megafauna(marsupial)$ equals $bluewhale$ for $armadillo=1$ and $0$ otherwise.\nHowever, this contradicts the hypothesis that $\\{tangerine, \\dots, porcupine\\}$ is not all of $lemniscate$.\nHence $chrysalis > 0$. Next suppose $chrysalis=1$, and choose $megafauna \\in \\hat{lemniscate} - \\{\\hat{e}\\}$ with $|tarantula| = 1$.\nSince each of\n$megafauna(tangerine), \\dots, megafauna(porcupine)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.\nSince $megafauna(tangerine) = megafauna(e) = 1$, $megafauna$ must map each of $tangerine,\\dots, porcupine$ to 1, but this is impossible because\n$megafauna$ is a nontrivial character and $tangerine,\\dots,porcupine$ form a set of generators of $lemniscate$.\nThis contradiction yields $chrysalis < 1$.\n\nSince $mongoose = \\frac{1}{bluewhale} \\sum_{megafauna \\in \\hat{lemniscate}} mongoose_{megafauna}$\nand $hibiscus mongoose_{megafauna} = tarantula mongoose_{megafauna}$, we have\n\\[\nhibiscus^{dandelion} mongoose - \\frac{1}{bluewhale} mongoose_{\\hat{e}}\n= \\frac{1}{bluewhale} \\sum_{megafauna \\in \\hat{lemniscate} - \\{\\hat{e}\\}} tarantula^{dandelion} mongoose_{megafauna}.\n\\]\nSince the vectors $mongoose_{megafauna}$ are pairwise orthogonal,\nthe limit we are interested in can be written as\n\\[\n\\lim_{dandelion \\to \\infty} \\frac{1}{chrysalis^{2dandelion}} (hibiscus^{dandelion} mongoose - \\frac{1}{bluewhale} mongoose_{\\hat{e}}) \\cdot (hibiscus^{dandelion} mongoose - \\frac{1}{bluewhale} mongoose_{\\hat{e}}).\n\\]\nand then rewritten as\n\\[\n\\lim_{dandelion \\to \\infty} \\frac{1}{chrysalis^{2dandelion}} \\sum_{megafauna \\in \\hat{lemniscate} - \\{\\hat{e}\\}} |tarantula|^{2dandelion}\n= \\#\\{megafauna \\in \\hat{lemniscate}: |tarantula| = chrysalis\\}.\n\\]\nBy construction, this last quantity is nonzero and finite.\n\n\\textbf{Remark.}\nIt is easy to see that the result fails if we do not assume $tangerine = e$: take $lemniscate = \\ZZ/2\\ZZ$,\n$bluewhale=1$, and $tangerine = 1$.\n\n\\textbf{Remark.}\nHarm Derksen points out that a similar argument applies even if $lemniscate$ is not assumed to be abelian,\nprovided that the operator $tangerine + \\cdots + porcupine$ in the group algebra $\\ZZ[lemniscate]$ is \\emph{normal},\ni.e., it commutes with the operator $tangerine^{-1} + \\cdots + porcupine^{-1}$.\nThis includes the cases where the set $\\{tangerine,\\dots,porcupine\\}$ is closed under taking inverses\nand where it is a union of conjugacy classes (which in turn includes the case of $lemniscate$ abelian).\n\n\\textbf{Remark}.\nThe matrix $hibiscus$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $dandelion$ such that $hibiscus^{dandelion}$ has positive entries. For any such matrix,\nthe Perron-Frobenius theorem implies that\nthe sequence of vectors $hibiscus^{dandelion} mongoose$ converges to a limit $rhinoceros$, and there exists $chrysalis \\in [0,1)$\nsuch that\n\\[\n\\limsup_{dandelion \\to \\infty} \\frac{1}{chrysalis^{2dandelion}} \\sum_{armadillo=1}^{bluewhale} ((hibiscus^{dandelion} mongoose - rhinoceros)_{armadillo})^2\n\\]\nis nonzero and finite. (The intended interpretation in case $chrysalis=0$ is that $hibiscus^{dandelion} mongoose = rhinoceros$ for all large $dandelion$.)\nHowever, the limit need not exist in general."
    },
    "descriptive_long_misleading": {
      "map": {
        "g": "staticelement",
        "g_1": "absorbingone",
        "g_2": "absorbingtwo",
        "g_k": "absorbinglast",
        "g_i": "absorbingindi",
        "g_j": "absorbingindj",
        "h_i": "disorderedindi",
        "h_1": "disorderedone",
        "h_n": "disorderedn",
        "x": "specificelement",
        "m": "stabilitycount",
        "i": "unitindex",
        "j": "wholeindex",
        "w": "zeropointv",
        "\\chi": "scalarfield",
        "\\lambda_\\chi": "nullvalue",
        "G": "nongroup",
        "k": "unbounded",
        "n": "fractional",
        "b": "exceedone",
        "M": "scalarunit",
        "v": "covectorian",
        "\\lambda_\\hat{e}": "zeroeigen"
      },
      "question": "\\[ \\{absorbingone=e,absorbingtwo,\\dots,absorbinglast\\} \\subsetneqq nongroup \\]\nbe a (not necessarily minimal) set of distinct generators of $nongroup$. A special\ndie, which randomly selects one of the elements absorbingone,absorbingtwo,...,absorbinglast with equal\nprobability, is rolled $stabilitycount$ times and the selected elements are multiplied\nto produce an element $staticelement \\in nongroup$.  Prove that there exists a real number\n$exceedone \\in (0,1)$ such that\n\n\\[ \\lim_{stabilitycount\\to\\infty} \\frac{1}{exceedone^{2stabilitycount}} \\sum_{specificelement\\in nongroup} \\left(\\mathrm{Prob}(staticelement=specificelement)\n    - \\frac{1}{fractional}\\right)^2 \\]\nis positive and finite.",
      "solution": "Choose some ordering $disorderedone,\\dots, disorderedn$ of the elements of $nongroup$ with $disorderedone = e$.\nDefine an $fractional \\times fractional$ matrix $scalarunit$\nby settting $scalarunit_{unitindex wholeindex} = 1/unbounded$ if $h_{wholeindex} = h_{unitindex} staticelement$ for some $staticelement \\in \\{absorbingone,\\dots,absorbinglast\\}$ and $scalarunit_{unitindex wholeindex} = 0$ otherwise.\nLet $covectorian$ denote the column vector $(1,0,\\dots,0)$. The probability\nthat the product of $stabilitycount$ random elements of $\\{absorbingone,\\dots,absorbinglast\\}$\nequals $disorderedindi$ can then be interpreted as the $unitindex$-th component of the vector\n$scalarunit^{stabilitycount} covectorian$.\n\nLet $\\hat{nongroup}$ denote the dual group of $nongroup$, i.e., the group of complex-valued characters of $nongroup$.\nLet $\\hat{e} \\in \\hat{nongroup}$ denote the trivial character.\nFor each $scalarfield \\in \\hat{nongroup}$, the vector $covectorian_{scalarfield} = (scalarfield(disorderedindi))_{unitindex=1}^{fractional}$ is an eigenvector of $scalarunit$\nwith eigenvalue $nullvalue = (scalarfield(absorbingone) + \\cdots + scalarfield(absorbinglast))/unbounded$.\nIn particular, $covectorian_{\\hat{e}}$ is the all-ones vector and $zeroeigen = 1$.\nPut\n\\[\nexceedone = \\max\\{|nullvalue|: scalarfield \\in \\hat{nongroup} - \\{\\hat{e}\\}\\};\n\\]\nwe show that $exceedone \\in (0,1)$ as follows.\nFirst suppose $exceedone=0$; then\n\\[\n1 = \\sum_{scalarfield \\in \\hat{nongroup}} nullvalue\n= \\frac{1}{unbounded} \\sum_{unitindex=1}^{unbounded} \\sum_{scalarfield \\in \\hat{nongroup}} scalarfield(absorbingindi) = \\frac{fractional}{unbounded}\n\\]\nbecause $\\sum_{scalarfield \\in \\hat{nongroup}} scalarfield(absorbingindi)$ equals $fractional$ for $unitindex=1$ and $0$ otherwise.\nHowever, this contradicts the hypothesis that $\\{absorbingone, \\dots, absorbinglast\\}$ is not all of $nongroup$.\nHence $exceedone > 0$. Next suppose $exceedone=1$, and choose $scalarfield \\in \\hat{nongroup} - \\{\\hat{e}\\}$ with $|nullvalue| = 1$.\nSince each of\n$scalarfield(absorbingone), \\dots, scalarfield(absorbinglast)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.\nSince $scalarfield(absorbingone) = scalarfield(e) = 1$, $scalarfield$ must map each of $absorbingone,\\dots, absorbinglast$ to $1$, but this is impossible because\n$scalarfield$ is a nontrivial character and $absorbingone,\\dots,absorbinglast$ form a set of generators of $nongroup$.\nThis contradiction yields $exceedone < 1$.\n\nSince $covectorian = \\frac{1}{fractional} \\sum_{scalarfield \\in \\hat{nongroup}} covectorian_{scalarfield}$\nand $scalarunit covectorian_{scalarfield} = nullvalue covectorian_{scalarfield}$, we have\n\\[\nscalarunit^{stabilitycount} covectorian - \\frac{1}{fractional} covectorian_{\\hat{e}}\n= \\frac{1}{fractional} \\sum_{scalarfield \\in \\hat{nongroup} - \\{\\hat{e}\\}} nullvalue^{stabilitycount} covectorian_{scalarfield}.\n\\]\nSince the vectors $covectorian_{scalarfield}$ are pairwise orthogonal,\nthe limit we are interested in can be written as\n\\[\n\\lim_{stabilitycount \\to \\infty} \\frac{1}{exceedone^{2stabilitycount}} \\left(scalarunit^{stabilitycount} covectorian - \\frac{1}{fractional} covectorian_{\\hat{e}}\\right) \\cdot \\left(scalarunit^{stabilitycount} covectorian - \\frac{1}{fractional} covectorian_{\\hat{e}}\\right).\n\\]\nand then rewritten as\n\\[\n\\lim_{stabilitycount \\to \\infty} \\frac{1}{exceedone^{2stabilitycount}} \\sum_{scalarfield \\in \\hat{nongroup} - \\{\\hat{e}\\}} |nullvalue|^{2stabilitycount}\n= \\#\\{scalarfield \\in \\hat{nongroup}: |nullvalue| = exceedone\\}.\n\\]\nBy construction, this last quantity is nonzero and finite.\n\n\n\\textbf{Remark.}\nIt is easy to see that the result fails if we do not assume absorbingone = e: take $nongroup = \\ZZ/2\\ZZ$,\n$fractional=1$, and absorbingone = 1.\n\n\\textbf{Remark.}\nHarm Derksen points out that a similar argument applies even if $nongroup$ is not assumed to be abelian,\nprovided that the operator absorbingone + \\cdots + absorbinglast in the group algebra $\\ZZ[nongroup]$ is \\emph{normal},\ni.e., it commutes with the operator absorbingone^{-1} + \\cdots + absorbinglast^{-1}.\nThis includes the cases where the set $\\{absorbingone,\\dots,absorbinglast\\}$ is closed under taking inverses\nand where it is a union of conjugacy classes (which in turn includes the case of $nongroup$ abelian).\n\n\\textbf{Remark}.\nThe matrix $scalarunit$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $stabilitycount$ such that $scalarunit^{stabilitycount}$ has positive entries. For any such matrix,\nthe Perron-Frobenius theorem implies that\nthe sequence of vectors $scalarunit^{stabilitycount} covectorian$ converges to a limit $zeropointv$, and there exists $exceedone \\in [0,1)$\nsuch that\n\\[\n\\limsup_{stabilitycount \\to \\infty} \\frac{1}{exceedone^{2stabilitycount}} \\sum_{unitindex=1}^{fractional} ((scalarunit^{stabilitycount} covectorian - zeropointv)_{unitindex})^2\n\\]\nis nonzero and finite. (The intended interpretation in case $exceedone=0$ is that $scalarunit^{stabilitycount} covectorian = zeropointv$ for all large $stabilitycount$.)\nHowever, the limit need not exist in general."
    },
    "garbled_string": {
      "map": {
        "g": "qzxwvtnp",
        "g_1": "hjgrksla",
        "g_2": "mfldkseo",
        "g_k": "pldrasqe",
        "g_i": "xvcnlmzr",
        "g_j": "tbuyemqk",
        "h_i": "ksjdferu",
        "h_1": "pvbhsqma",
        "h_n": "ztqwerop",
        "x": "lajdmsnt",
        "m": "cbordfix",
        "w": "ktmslqpo",
        "\\chi": "uxbkzyqn",
        "\\lambda_\\chi": "fgdqrpou",
        "G": "rtemasuv",
        "k": "azqnklyt",
        "n": "sdfghjkl",
        "b": "weprlmsa",
        "M": "cvbnpqte",
        "v": "ouijzxcv",
        "\\lambda_\\hat{e}": "opytkjhg"
      },
      "question": "\\[ \\{hjgrksla=e,mfldkseo,\\dots,pldrasqe\\} \\subsetneqq rtemasuv \\]\nbe a (not necessarily minimal) set of distinct generators of $rtemasuv$. A special\ndie, which randomly selects one of the elements $hjgrksla,mfldkseo,...,pldrasqe$ with equal\nprobability, is rolled $cbordfix$ times and the selected elements are multiplied\nto produce an element $qzxwvtnp \\in rtemasuv$.  Prove that there exists a real number\n$\\weprlmsa \\in (0,1)$ such that\n\n\\[ \\lim_{cbordfix\\to\\infty} \\frac{1}{\\weprlmsa^{2cbordfix}} \\sum_{lajdmsnt\\in rtemasuv} \\left(\\mathrm{Prob}(qzxwvtnp=lajdmsnt)\n    - \\frac{1}{sdfghjkl}\\right)^2 \\]\nis positive and finite.",
      "solution": "Choose some ordering $pvbhsqma,\\dots, ztqwerop$ of the elements of $rtemasuv$ with $pvbhsqma = e$.\nDefine an $sdfghjkl \\times sdfghjkl$ matrix $cvbnpqte$\nby settting $cvbnpqte_{ij} = 1/azqnklyt$ if $h_j = ksjdferu qzxwvtnp$ for some $qzxwvtnp \\in \\{hjgrksla,\\dots,pldrasqe\\}$ and $cvbnpqte_{ij} = 0$ otherwise.\nLet $ouijzxcv$ denote the column vector $(1,0,\\dots,0)$. The probability\nthat the product of $cbordfix$ random elements of $\\{hjgrksla,\\dots,pldrasqe\\}$\nequals $ksjdferu$ can then be interpreted as the $i$-th component of the vector\n$cvbnpqte^{cbordfix} ouijzxcv$.\n\nLet $\\hat{rtemasuv}$ denote the dual group of $rtemasuv$, i.e., the group of complex-valued characters of $rtemasuv$.\nLet $\\hat{e} \\in \\hat{rtemasuv}$ denote the trivial character.\nFor each $uxbkzyqn \\in \\hat{rtemasuv}$, the vector $v_{uxbkzyqn} = (uxbkzyqn(h_i))_{i=1}^{sdfghjkl}$ is an eigenvector of $cvbnpqte$\nwith eigenvalue $fgdqrpou = (uxbkzyqn(hjgrksla) + \\cdots + uxbkzyqn(pldrasqe))/azqnklyt$.\nIn particular, $v_{\\hat{e}}$ is the all-ones vector and $opytkjhg = 1$.\nPut\n\\[\nweprlmsa = \\max\\{|fgdqrpou|: uxbkzyqn \\in \\hat{rtemasuv} - \\{\\hat{e}\\}\\};\n\\]\nwe show that $\\weprlmsa \\in (0,1)$ as follows.\nFirst suppose $\\weprlmsa=0$; then\n\\[\n1 = \\sum_{uxbkzyqn \\in \\hat{rtemasuv}} fgdqrpou\n= \\frac{1}{azqnklyt} \\sum_{i=1}^{azqnklyt} \\sum_{uxbkzyqn \\in \\hat{rtemasuv}} uxbkzyqn(g_i) = \\frac{sdfghjkl}{azqnklyt}\n\\]\nbecause $\\sum_{uxbkzyqn \\in \\hat(rtemasuv)} uxbkzyqn(g_i)$ equals $sdfghjkl$ for $i=1$ and $0$ otherwise.\nHowever, this contradicts the hypothesis that $\\{hjgrksla, \\dots, pldrasqe\\}$ is not all of $rtemasuv$.\nHence $\\weprlmsa > 0$. Next suppose $\\weprlmsa=1$, and choose $uxbkzyqn \\in \\hat{rtemasuv} - \\{\\hat{e}\\}$ with $|fgdqrpou| = 1$.\nSince each of\n$uxbkzyqn(hjgrksla), \\dots, uxbkzyqn(pldrasqe)$ is a complex number of norm 1, the triangle inequality forces them all to be equal.\nSince $uxbkzyqn(hjgrksla) = uxbkzyqn(e) = 1$, $uxbkzyqn$ must map each of $hjgrksla,\\dots, pldrasqe$ to 1, but this is impossible because\n$uxbkzyqn$ is a nontrivial character and $hjgrksla,\\dots,pldrasqe$ form a set of generators of $rtemasuv$.\nThis contradiction yields $\\weprlmsa < 1$.\n\nSince $ouijzxcv = \\frac{1}{sdfghjkl} \\sum_{uxbkzyqn \\in \\hat{rtemasuv}} v_{uxbkzyqn}$\nand $cvbnpqte v_{uxbkzyqn} = fgdqrpou v_{uxbkzyqn}$, we have\n\\[\ncvbnpqte^{cbordfix} ouijzxcv - \\frac{1}{sdfghjkl} v_{\\hat{e}}\n= \\frac{1}{sdfghjkl} \\sum_{uxbkzyqn \\in \\hat{rtemasuv} - \\{\\hat{e}\\}} fgdqrpou^{cbordfix} v_{uxbkzyqn}.\n\\]\nSince the vectors $v_{uxbkzyqn}$ are pairwise orthogonal,\nthe limit we are interested in can be written as\n\\[\n\\lim_{cbordfix \\to \\infty} \\frac{1}{\\weprlmsa^{2cbordfix}} (cvbnpqte^{cbordfix} ouijzxcv - \\frac{1}{sdfghjkl} v_{\\hat{e}}) \\cdot (cvbnpqte^{cbordfix} ouijzxcv - \\frac{1}{sdfghjkl} v_{\\hat{e}}).\n\\]\nand then rewritten as\n\\[\n\\lim_{cbordfix \\to \\infty} \\frac{1}{\\weprlmsa^{2cbordfix}} \\sum_{uxbkzyqn \\in \\hat{rtemasuv} - \\{\\hat{e}\\}} |fgdqrpou|^{2cbordfix}\n= \\#\\{uxbkzyqn \\in \\hat{rtemasuv}: |fgdqrpou| = \\weprlmsa\\}.\n\\]\nBy construction, this last quantity is nonzero and finite.\n\n\n\\textbf{Remark.}\nIt is easy to see that the result fails if we do not assume hjgrksla = e: take $rtemasuv = \\ZZ/2\\ZZ$,\n$sdfghjkl=1$, and hjgrksla = 1.\n\n\\textbf{Remark.}\nHarm Derksen points out that a similar argument applies even if $rtemasuv$ is not assumed to be abelian,\nprovided that the operator hjgrksla + \\cdots + pldrasqe in the group algebra $\\ZZ[rtemasuv]$ is \\emph{normal},\ni.e., it commutes with the operator hjgrksla^{-1} + \\cdots + pldrasqe^{-1}.\nThis includes the cases where the set $\\{hjgrksla,\\dots,pldrasqe\\}$ is closed under taking inverses\nand where it is a union of conjugacy classes (which in turn includes the case of $rtemasuv$ abelian).\n\n\\textbf{Remark}.\nThe matrix $cvbnpqte$ used above has nonnegative entries with row sums equal to 1 (i.e., it corresponds to a Markov chain), and there exists a positive integer $cbordfix$ such that $cvbnpqte^{cbordfix}$ has positive entries. For any such matrix,\nthe Perron-Frobenius theorem implies that\nthe sequence of vectors $cvbnpqte^{cbordfix} ouijzxcv$ converges to a limit $ktmslqpo$, and there exists $\\weprlmsa \\in [0,1)$\nsuch that\n\\[\n\\limsup_{cbordfix \\to \\infty} \\frac{1}{\\weprlmsa^{2cbordfix}} \\sum_{i=1}^{sdfghjkl} ((cvbnpqte^{cbordfix} ouijzxcv - ktmslqpo)_i)^2\n\\]\nis nonzero and finite. (The intended interpretation in case $\\weprlmsa=0$ is that $cvbnpqte^{cbordfix} ouijzxcv = ktmslqpo$ for all large $cbordfix$.)\nHowever, the limit need not exist in general."
    },
    "kernel_variant": {
      "question": "Let $G$ be a finite (not necessarily abelian) group of order $N$.  \nFix pairwise distinct elements  \n\\[\nB=\\{\\,b_{1},\\ldots ,b_{s-1},\\;b_{s}=e\\,\\}\\subset G\n\\]\nthat generate $G$, and choose positive rational numbers $p_{1},\\ldots ,p_{s}$ with  \n\\[\n\\sum_{r=1}^{s}p_{r}=1\n\\quad\\text{and}\\quad\n(p_{1},\\ldots ,p_{s})\\neq\\Bigl(\\frac1s,\\ldots ,\\frac1s\\Bigr).\n\\]\n\nPut  \n\\[\n\\mu=\\sum_{r=1}^{s}p_{r}\\,\\delta_{b_{r}},\\qquad\nT=\\sum_{r=1}^{s}p_{r}\\,b_{r}\\ \\in\\ \\mathbf C[G],\n\\]\nwhere $\\delta_{g}$ denotes the point mass at $g$.  \nAssume that $T$ is \\emph{normal} in $\\mathbf C[G]$, i.e.\\ $T\\,T^{*}=T^{*}T$ with respect to the involution $g\\mapsto g^{-1}$ extended linearly.\n\nLet $X_{1},X_{2},\\ldots$ be i.i.d.\\ $G$–valued random variables with law $\\mu$ and form the ordered products  \n\\[\nZ_{\\ell}=X_{1}X_{2}\\cdots X_{\\ell},\\qquad\\ell\\ge 1.\n\\]\nFor $g\\in G$ write  \n\\[\n\\pi_{\\ell}(g)=\\Pr\\bigl\\{Z_{\\ell}=g\\bigr\\},\\qquad\n\\Delta_{\\ell}(g)=\\pi_{\\ell}(g)-\\frac1N,\\qquad\n\\|\\Delta_{\\ell}\\|_{2}^{2}=\\sum_{g\\in G}\\Delta_{\\ell}(g)^{2}.\n\\]\n\n1.  Prove that there exists a constant $c$ with $0<c<1$ such that the limit  \n\\[\nL=\\lim_{\\ell\\to\\infty}\\frac{\\|\\Delta_{\\ell}\\|_{2}^{2}}{c^{\\,2\\ell}}\n\\tag{$\\dagger$}\n\\]\nexists, is finite and strictly positive.\n\n2.  Let $\\widehat G$ be a complete set of inequivalent irreducible representations of $G$, let $d_{\\rho}$ be the degree of $\\rho$, and set  \n\\[\nT_{\\rho}= \\sum_{r=1}^{s} p_{r}\\,\\rho(b_{r})\\in M_{d_{\\rho}}(\\mathbf C).\n\\]\nShow that  \n\\[\nc=\\max_{\\rho\\in \\widehat G\\setminus\\{1\\}}\\|T_{\\rho}\\|_{\\operatorname{op}},\n\\tag{$\\ddagger$}\n\\]\nand that the limit in $(\\dagger)$ is  \n\\[\nL=\\frac1N\\sum_{\\substack{\\rho\\in \\widehat G\\setminus\\{1\\}\\\\\\|T_{\\rho}\\|_{\\operatorname{op}}=c}}\nd_{\\rho}\\,m_{\\rho},\n\\tag{$\\sharp$}\n\\]\nwhere  \n\\[\nm_{\\rho}= \\dim\\Bigl\\{v\\in\\mathbf C^{d_{\\rho}}\\;:\\;\n\\text{there exists }\\lambda\\text{ with }|\\lambda|=c\\text{ and }T_{\\rho}v=\\lambda v\\Bigr\\}.\n\\]\nIn particular, $L$ is a positive rational number.\n\n(Examples in which the normality hypothesis holds automatically are:  \n$\\bullet$ $\\{b_{1},\\ldots ,b_{s-1}\\}$ is closed under taking inverses;  \n$\\bullet$ $\\{b_{1},\\ldots ,b_{s-1}\\}$ is a union of conjugacy classes.  \nNeither of these additional conditions is required.)\n\n--------------------------------------------------------------------",
      "solution": "We keep the six–step structure of the previous proof, but give complete arguments for the crucial inequalities $0<c<1$.\n\n\\medskip\n\\textbf{Step 0.  Fourier set–up.}  \nLet $\\lambda_{G}:G\\!\\longrightarrow\\!U(N)$ denote the left–regular representation.  \nUnder the Fourier transform  \n\\[\n\\mathcal F:\\mathbf C[G]\\ \\longrightarrow\\ \n\\bigoplus_{\\rho\\in\\widehat G}\\! \\operatorname{End}(\\mathbf C^{d_{\\rho}}),\n\\qquad\n\\lambda_{G}(x)\\ \\mapsto\\ \n\\bigoplus_{\\rho\\in\\widehat G} \\bigl(I_{d_{\\rho}}\\otimes \\rho(x)\\bigr),\n\\]\nthe element $T$ becomes the block–diagonal operator  \n\\[\n\\mathcal F(T)=\\bigoplus_{\\rho\\in\\widehat G}\\bigl(I_{d_{\\rho}}\\otimes T_{\\rho}\\bigr),\n\\qquad\nT_{\\rho}=\\sum_{r=1}^{s}p_{r}\\,\\rho(b_{r}).\n\\]\nSince $T$ is normal, each block $T_{\\rho}$ is a normal matrix, hence unitarily diagonalizable and  \n\\[\n\\|T_{\\rho}\\|_{\\operatorname{op}}=\\max\\{\\,|\\lambda| : \\lambda\\text{ eigenvalue of }T_{\\rho}\\}.\n\\]\n\n\\medskip\n\\textbf{Step 1.  The $\\boldsymbol\\ell$–step distribution in the group algebra.}  \nLet $\\nu_{\\ell}=\\mu^{\\!*\\,\\ell}\\in\\mathbf C[G]$ be the $\\ell$–fold convolution of $\\mu$.  Then  \n\\[\n\\pi_{\\ell}(g)=\\nu_{\\ell}(g),\\qquad \n\\Delta_{\\ell}= \\nu_{\\ell}-\\frac1N\\,\\mathbf 1_{G},\n\\]\nwhere $\\mathbf 1_{G}=\\sum_{g\\in G}\\delta_{g}$.  \nInside the regular representation,\n\\[\n\\lambda_{G}(\\nu_{\\ell})=T^{\\ell},\n\\qquad\n\\lambda_{G}(\\mathbf 1_{G}/N)=P,\n\\tag{1}\n\\]\nwith $P$ the rank–one projection onto the constant functions.\n\n\\medskip\n\\textbf{Step 2.  $L^{2}$–norm in spectral form.}  \nBecause $\\lambda_{G}$ is unitary,\n\\[\n\\|\\Delta_{\\ell}\\|_{2}^{2}\n=\\frac1N\\operatorname{tr}\\bigl[(T^{\\ell}-P)(T^{\\ell}-P)^{*}\\bigr]\n=\\frac1N\\operatorname{tr}\\bigl(T^{\\ell}T^{*\\,\\ell}-P\\bigr),\n\\tag{2}\n\\]\nthe cross–terms vanish since $TP=PT=P$.\n\n\\medskip\n\\textbf{Step 3.  Block decomposition with multiplicity.}  \nApply $\\mathcal F$ to \\eqref{2}.  Using $\\lambda_{G}\\cong\\bigoplus_{\\rho} d_{\\rho}\\,\\rho$ we obtain\n\\[\n\\|\\Delta_{\\ell}\\|_{2}^{2}\n=\\frac1N \\sum_{\\rho\\in\\widehat G\\setminus\\{1\\}} d_{\\rho}\\;\n\\operatorname{tr}\\bigl(T_{\\rho}^{\\ell}T_{\\rho}^{*\\,\\ell}\\bigr).\n\\tag{3}\n\\]\nDiagonalising $T_{\\rho}=U_{\\rho}\\operatorname{Diag}\n\\bigl(\\lambda_{\\rho,1},\\ldots ,\\lambda_{\\rho,d_{\\rho}}\\bigr)U_{\\rho}^{*}$ gives  \n\\[\n\\operatorname{tr}\\bigl(T_{\\rho}^{\\ell}T_{\\rho}^{*\\,\\ell}\\bigr)\n=\\sum_{j=1}^{d_{\\rho}}|\\lambda_{\\rho,j}|^{2\\ell}.\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 4.  The dominant spectral radius: proof that $0<c<1$.}  \n\n\\emph{Definition.}  \n\\[\nc:=\\max_{\\rho\\neq 1}\\|T_{\\rho}\\|_{\\operatorname{op}}\n       =\\max_{\\rho,j}|\\lambda_{\\rho,j}|.\n\\tag{5}\n\\]\n\n\\emph{Upper bound $c<1$.}  \nFix $\\rho\\neq 1$ and a unit vector $v$ with $\\|T_{\\rho}\\|_{\\operatorname{op}}=\\|T_{\\rho}v\\|$.  \nBecause each $\\rho(b_{r})$ is unitary and $\\sum_{r}p_{r}=1$,\n\\[\n\\|T_{\\rho}v\\|=\\Bigl\\|\\sum_{r=1}^{s}p_{r}\\,\\rho(b_{r})v\\Bigr\\|\n\\le \\sum_{r=1}^{s}p_{r}\\|\\rho(b_{r})v\\|=1.\n\\]\nEquality can occur only if \\emph{all} vectors $\\rho(b_{r})v$ are parallel and point in the same direction; otherwise the strict triangle inequality would apply.  \nConsequently there exist scalars $\\omega_{r}\\in\\mathbf C$ with $|\\omega_{r}|=1$ such that  \n\\[\n\\rho(b_{r})v=\\omega_{r}v\\quad(\\forall r),\n\\qquad\\text{and}\\qquad \n\\sum_{r=1}^{s}p_{r}\\omega_{r}=1.\n\\tag{6}\n\\]\nSince $b_{s}=e$ one has $\\omega_{s}=1$, and the equality on the right of \\eqref{6} forces \\emph{all} $\\omega_{r}=1$.  \nThus $v$ is a common eigenvector with eigenvalue $1$ for every $\\rho(b_{r})$.  \nBecause the $b_{r}$ generate $G$, this means $\\rho(g)v=v$ for every $g\\in G$; hence $\\langle v\\rangle$ is a $G$–invariant subspace.  By irreducibility $\\rho$ must be one–dimensional.  But then each $\\rho(b_{r})=1$, whence $\\rho$ is the trivial character, contradicting $\\rho\\neq 1$.  \nTherefore $\\|T_{\\rho}\\|_{\\operatorname{op}}<1$ for all $\\rho\\neq 1$, so $c<1$.\n\n\\emph{Lower bound $c>0$.}  \nAssume for contradiction that $\\|T_{\\rho}\\|_{\\operatorname{op}}=0$ for every $\\rho\\neq 1$.  \nThen $T_{\\rho}=0$ for all $\\rho\\neq 1$, so the Fourier transform of $T$ consists of a single non–zero block at the trivial representation, equal to $\\sum_{r}p_{r}=1$.  \nFourier inversion therefore gives $T=(1/N)\\,\\mathbf 1_{G}$, i.e. all Fourier coefficients of $T$ are equal.  \nBut in the basis $\\{\\delta_{g}\\}_{g\\in G}$ we have $T=\\sum_{r=1}^{s}p_{r}\\,b_{r}$, whose coefficient at $e$ equals $p_{s}$, whereas the coefficient at $b_{1}$ equals $p_{1}\\neq p_{s}$ (because the vector $(p_{1},\\ldots ,p_{s})$ is not constant).  This is impossible, whence $c>0$.\n\n\\medskip\n\\textbf{Step 5.  Identification of the limit.}  \nSet\n\\[\nS_{\\ell}:=\\frac{\\|\\Delta_{\\ell}\\|_{2}^{2}}{c^{\\,2\\ell}}.\n\\]\nCombining \\eqref{3}–\\eqref{4},\n\\[\nS_{\\ell}\n=\\frac1N\\sum_{\\rho\\neq 1} d_{\\rho}\\sum_{j=1}^{d_{\\rho}}\n\\Bigl(\\frac{|\\lambda_{\\rho,j}|}{c}\\Bigr)^{2\\ell}.\n\\tag{7}\n\\]\nEach ratio $|\\lambda_{\\rho,j}|/c$ lies in $[0,1]$ and equals $1$ iff $|\\lambda_{\\rho,j}|=c$.  \nBy dominated convergence,\n\\[\n\\lim_{\\ell\\to\\infty}S_{\\ell}\n=\\frac1N \\sum_{\\substack{\\rho\\neq 1\\\\|\\lambda_{\\rho,j}|=c}}\nd_{\\rho}\n=\\frac1N\\sum_{\\substack{\\rho\\neq 1\\\\\\|T_{\\rho}\\|_{\\operatorname{op}}=c}}\nd_{\\rho}\\,m_{\\rho},\n\\tag{8}\n\\]\nbecause $m_{\\rho}$ counts the number of eigenvalues of $T_{\\rho}$ (counted with multiplicity) whose modulus equals $c$.  Formula $(\\sharp)$ follows.\n\n\\medskip\n\\textbf{Step 6.  Positivity, finiteness and rationality.}  \nAt least one block attains the maximum $c$, so the sum in \\eqref{8} is non–zero; finiteness is automatic.  As each $d_{\\rho}$ and $m_{\\rho}$ is an integer, $L\\in\\mathbf Q_{>0}$.\n\n\\medskip\nThe conclusions $(\\dagger)$, $(\\ddagger)$ and $(\\sharp)$ are therefore completely established.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.820046",
        "was_fixed": false,
        "difficulty_analysis": "1. From abelian to non-abelian:  characters were one-dimensional in the original problem; here full matrix-valued representation theory and the regular representation are required.\n\n2. Normal but non-central generator:  The element T need not lie in the center of C[G]; we only assume it is normal.  Thus T_ρ is no longer a scalar but an arbitrary normal matrix whose whole eigen-spectrum matters.  Handling its operator norm, diagonalization, and multiplicities is significantly subtler.\n\n3. Eigenvalue multiplicities:  The limit involves the total multiplicity m_ρ of eigenvalues of maximal modulus, a new layer absent from the original statement.\n\n4. New algebraic obstacles:  Proving 0 < c < 1 now needs an argument combining convexity of spectra for normal matrices, simultaneous diagonalization, and the generating property of B.\n\n5. Broader technical toolkit:  One must invoke the spectral theorem for normal matrices, the decomposition of the regular representation, trace identities, and basic algebraic number theory to show L is rational.\n\nBecause of these additional structural, representational, and spectral complexities, solving the enhanced variant demands substantially deeper insight and a more sophisticated blend of techniques than the original kernel problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $G$ be a finite (not necessarily abelian) group of order $N$.  \nFix pairwise distinct elements  \n\\[\nB=\\{\\,b_{1},\\ldots ,b_{s-1},\\;b_{s}=e\\,\\}\\subset G\n\\]\nthat generate $G$, and choose positive rational numbers $p_{1},\\ldots ,p_{s}$ with  \n\\[\n\\sum_{r=1}^{s}p_{r}=1\n\\quad\\text{and}\\quad\n(p_{1},\\ldots ,p_{s})\\neq\\Bigl(\\frac1s,\\ldots ,\\frac1s\\Bigr).\n\\]\n\nPut  \n\\[\n\\mu=\\sum_{r=1}^{s}p_{r}\\,\\delta_{b_{r}},\\qquad\nT=\\sum_{r=1}^{s}p_{r}\\,b_{r}\\ \\in\\ \\mathbf C[G],\n\\]\nwhere $\\delta_{g}$ denotes the point mass at $g$.  \nAssume that $T$ is \\emph{normal} in $\\mathbf C[G]$, i.e.\\ $T\\,T^{*}=T^{*}T$ with respect to the involution $g\\mapsto g^{-1}$ extended linearly.\n\nLet $X_{1},X_{2},\\ldots$ be i.i.d.\\ $G$–valued random variables with law $\\mu$ and form the ordered products  \n\\[\nZ_{\\ell}=X_{1}X_{2}\\cdots X_{\\ell},\\qquad\\ell\\ge 1.\n\\]\nFor $g\\in G$ write  \n\\[\n\\pi_{\\ell}(g)=\\Pr\\bigl\\{Z_{\\ell}=g\\bigr\\},\\qquad\n\\Delta_{\\ell}(g)=\\pi_{\\ell}(g)-\\frac1N,\\qquad\n\\|\\Delta_{\\ell}\\|_{2}^{2}=\\sum_{g\\in G}\\Delta_{\\ell}(g)^{2}.\n\\]\n\n1.  Prove that there exists a constant $c$ with $0<c<1$ such that the limit  \n\\[\nL=\\lim_{\\ell\\to\\infty}\\frac{\\|\\Delta_{\\ell}\\|_{2}^{2}}{c^{\\,2\\ell}}\n\\tag{$\\dagger$}\n\\]\nexists, is finite and strictly positive.\n\n2.  Let $\\widehat G$ be a complete set of inequivalent irreducible representations of $G$, let $d_{\\rho}$ be the degree of $\\rho$, and set  \n\\[\nT_{\\rho}= \\sum_{r=1}^{s} p_{r}\\,\\rho(b_{r})\\in M_{d_{\\rho}}(\\mathbf C).\n\\]\nShow that  \n\\[\nc=\\max_{\\rho\\in \\widehat G\\setminus\\{1\\}}\\|T_{\\rho}\\|_{\\operatorname{op}},\n\\tag{$\\ddagger$}\n\\]\nand that the limit in $(\\dagger)$ is  \n\\[\nL=\\frac1N\\sum_{\\substack{\\rho\\in \\widehat G\\setminus\\{1\\}\\\\\\|T_{\\rho}\\|_{\\operatorname{op}}=c}}\nd_{\\rho}\\,m_{\\rho},\n\\tag{$\\sharp$}\n\\]\nwhere  \n\\[\nm_{\\rho}= \\dim\\Bigl\\{v\\in\\mathbf C^{d_{\\rho}}\\;:\\;\n\\text{there exists }\\lambda\\text{ with }|\\lambda|=c\\text{ and }T_{\\rho}v=\\lambda v\\Bigr\\}.\n\\]\nIn particular, $L$ is a positive rational number.\n\n(Examples in which the normality hypothesis holds automatically are:  \n$\\bullet$ $\\{b_{1},\\ldots ,b_{s-1}\\}$ is closed under taking inverses;  \n$\\bullet$ $\\{b_{1},\\ldots ,b_{s-1}\\}$ is a union of conjugacy classes.  \nNeither of these additional conditions is required.)\n\n--------------------------------------------------------------------",
      "solution": "We keep the six–step structure of the previous proof, but give complete arguments for the crucial inequalities $0<c<1$.\n\n\\medskip\n\\textbf{Step 0.  Fourier set–up.}  \nLet $\\lambda_{G}:G\\!\\longrightarrow\\!U(N)$ denote the left–regular representation.  \nUnder the Fourier transform  \n\\[\n\\mathcal F:\\mathbf C[G]\\ \\longrightarrow\\ \n\\bigoplus_{\\rho\\in\\widehat G}\\! \\operatorname{End}(\\mathbf C^{d_{\\rho}}),\n\\qquad\n\\lambda_{G}(x)\\ \\mapsto\\ \n\\bigoplus_{\\rho\\in\\widehat G} \\bigl(I_{d_{\\rho}}\\otimes \\rho(x)\\bigr),\n\\]\nthe element $T$ becomes the block–diagonal operator  \n\\[\n\\mathcal F(T)=\\bigoplus_{\\rho\\in\\widehat G}\\bigl(I_{d_{\\rho}}\\otimes T_{\\rho}\\bigr),\n\\qquad\nT_{\\rho}=\\sum_{r=1}^{s}p_{r}\\,\\rho(b_{r}).\n\\]\nSince $T$ is normal, each block $T_{\\rho}$ is a normal matrix, hence unitarily diagonalizable and  \n\\[\n\\|T_{\\rho}\\|_{\\operatorname{op}}=\\max\\{\\,|\\lambda| : \\lambda\\text{ eigenvalue of }T_{\\rho}\\}.\n\\]\n\n\\medskip\n\\textbf{Step 1.  The $\\boldsymbol\\ell$–step distribution in the group algebra.}  \nLet $\\nu_{\\ell}=\\mu^{\\!*\\,\\ell}\\in\\mathbf C[G]$ be the $\\ell$–fold convolution of $\\mu$.  Then  \n\\[\n\\pi_{\\ell}(g)=\\nu_{\\ell}(g),\\qquad \n\\Delta_{\\ell}= \\nu_{\\ell}-\\frac1N\\,\\mathbf 1_{G},\n\\]\nwhere $\\mathbf 1_{G}=\\sum_{g\\in G}\\delta_{g}$.  \nInside the regular representation,\n\\[\n\\lambda_{G}(\\nu_{\\ell})=T^{\\ell},\n\\qquad\n\\lambda_{G}(\\mathbf 1_{G}/N)=P,\n\\tag{1}\n\\]\nwith $P$ the rank–one projection onto the constant functions.\n\n\\medskip\n\\textbf{Step 2.  $L^{2}$–norm in spectral form.}  \nBecause $\\lambda_{G}$ is unitary,\n\\[\n\\|\\Delta_{\\ell}\\|_{2}^{2}\n=\\frac1N\\operatorname{tr}\\bigl[(T^{\\ell}-P)(T^{\\ell}-P)^{*}\\bigr]\n=\\frac1N\\operatorname{tr}\\bigl(T^{\\ell}T^{*\\,\\ell}-P\\bigr),\n\\tag{2}\n\\]\nthe cross–terms vanish since $TP=PT=P$.\n\n\\medskip\n\\textbf{Step 3.  Block decomposition with multiplicity.}  \nApply $\\mathcal F$ to \\eqref{2}.  Using $\\lambda_{G}\\cong\\bigoplus_{\\rho} d_{\\rho}\\,\\rho$ we obtain\n\\[\n\\|\\Delta_{\\ell}\\|_{2}^{2}\n=\\frac1N \\sum_{\\rho\\in\\widehat G\\setminus\\{1\\}} d_{\\rho}\\;\n\\operatorname{tr}\\bigl(T_{\\rho}^{\\ell}T_{\\rho}^{*\\,\\ell}\\bigr).\n\\tag{3}\n\\]\nDiagonalising $T_{\\rho}=U_{\\rho}\\operatorname{Diag}\n\\bigl(\\lambda_{\\rho,1},\\ldots ,\\lambda_{\\rho,d_{\\rho}}\\bigr)U_{\\rho}^{*}$ gives  \n\\[\n\\operatorname{tr}\\bigl(T_{\\rho}^{\\ell}T_{\\rho}^{*\\,\\ell}\\bigr)\n=\\sum_{j=1}^{d_{\\rho}}|\\lambda_{\\rho,j}|^{2\\ell}.\n\\tag{4}\n\\]\n\n\\medskip\n\\textbf{Step 4.  The dominant spectral radius: proof that $0<c<1$.}  \n\n\\emph{Definition.}  \n\\[\nc:=\\max_{\\rho\\neq 1}\\|T_{\\rho}\\|_{\\operatorname{op}}\n       =\\max_{\\rho,j}|\\lambda_{\\rho,j}|.\n\\tag{5}\n\\]\n\n\\emph{Upper bound $c<1$.}  \nFix $\\rho\\neq 1$ and a unit vector $v$ with $\\|T_{\\rho}\\|_{\\operatorname{op}}=\\|T_{\\rho}v\\|$.  \nBecause each $\\rho(b_{r})$ is unitary and $\\sum_{r}p_{r}=1$,\n\\[\n\\|T_{\\rho}v\\|=\\Bigl\\|\\sum_{r=1}^{s}p_{r}\\,\\rho(b_{r})v\\Bigr\\|\n\\le \\sum_{r=1}^{s}p_{r}\\|\\rho(b_{r})v\\|=1.\n\\]\nEquality can occur only if \\emph{all} vectors $\\rho(b_{r})v$ are parallel and point in the same direction; otherwise the strict triangle inequality would apply.  \nConsequently there exist scalars $\\omega_{r}\\in\\mathbf C$ with $|\\omega_{r}|=1$ such that  \n\\[\n\\rho(b_{r})v=\\omega_{r}v\\quad(\\forall r),\n\\qquad\\text{and}\\qquad \n\\sum_{r=1}^{s}p_{r}\\omega_{r}=1.\n\\tag{6}\n\\]\nSince $b_{s}=e$ one has $\\omega_{s}=1$, and the equality on the right of \\eqref{6} forces \\emph{all} $\\omega_{r}=1$.  \nThus $v$ is a common eigenvector with eigenvalue $1$ for every $\\rho(b_{r})$.  \nBecause the $b_{r}$ generate $G$, this means $\\rho(g)v=v$ for every $g\\in G$; hence $\\langle v\\rangle$ is a $G$–invariant subspace.  By irreducibility $\\rho$ must be one–dimensional.  But then each $\\rho(b_{r})=1$, whence $\\rho$ is the trivial character, contradicting $\\rho\\neq 1$.  \nTherefore $\\|T_{\\rho}\\|_{\\operatorname{op}}<1$ for all $\\rho\\neq 1$, so $c<1$.\n\n\\emph{Lower bound $c>0$.}  \nAssume for contradiction that $\\|T_{\\rho}\\|_{\\operatorname{op}}=0$ for every $\\rho\\neq 1$.  \nThen $T_{\\rho}=0$ for all $\\rho\\neq 1$, so the Fourier transform of $T$ consists of a single non–zero block at the trivial representation, equal to $\\sum_{r}p_{r}=1$.  \nFourier inversion therefore gives $T=(1/N)\\,\\mathbf 1_{G}$, i.e. all Fourier coefficients of $T$ are equal.  \nBut in the basis $\\{\\delta_{g}\\}_{g\\in G}$ we have $T=\\sum_{r=1}^{s}p_{r}\\,b_{r}$, whose coefficient at $e$ equals $p_{s}$, whereas the coefficient at $b_{1}$ equals $p_{1}\\neq p_{s}$ (because the vector $(p_{1},\\ldots ,p_{s})$ is not constant).  This is impossible, whence $c>0$.\n\n\\medskip\n\\textbf{Step 5.  Identification of the limit.}  \nSet\n\\[\nS_{\\ell}:=\\frac{\\|\\Delta_{\\ell}\\|_{2}^{2}}{c^{\\,2\\ell}}.\n\\]\nCombining \\eqref{3}–\\eqref{4},\n\\[\nS_{\\ell}\n=\\frac1N\\sum_{\\rho\\neq 1} d_{\\rho}\\sum_{j=1}^{d_{\\rho}}\n\\Bigl(\\frac{|\\lambda_{\\rho,j}|}{c}\\Bigr)^{2\\ell}.\n\\tag{7}\n\\]\nEach ratio $|\\lambda_{\\rho,j}|/c$ lies in $[0,1]$ and equals $1$ iff $|\\lambda_{\\rho,j}|=c$.  \nBy dominated convergence,\n\\[\n\\lim_{\\ell\\to\\infty}S_{\\ell}\n=\\frac1N \\sum_{\\substack{\\rho\\neq 1\\\\|\\lambda_{\\rho,j}|=c}}\nd_{\\rho}\n=\\frac1N\\sum_{\\substack{\\rho\\neq 1\\\\\\|T_{\\rho}\\|_{\\operatorname{op}}=c}}\nd_{\\rho}\\,m_{\\rho},\n\\tag{8}\n\\]\nbecause $m_{\\rho}$ counts the number of eigenvalues of $T_{\\rho}$ (counted with multiplicity) whose modulus equals $c$.  Formula $(\\sharp)$ follows.\n\n\\medskip\n\\textbf{Step 6.  Positivity, finiteness and rationality.}  \nAt least one block attains the maximum $c$, so the sum in \\eqref{8} is non–zero; finiteness is automatic.  As each $d_{\\rho}$ and $m_{\\rho}$ is an integer, $L\\in\\mathbf Q_{>0}$.\n\n\\medskip\nThe conclusions $(\\dagger)$, $(\\ddagger)$ and $(\\sharp)$ are therefore completely established.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.627493",
        "was_fixed": false,
        "difficulty_analysis": "1. From abelian to non-abelian:  characters were one-dimensional in the original problem; here full matrix-valued representation theory and the regular representation are required.\n\n2. Normal but non-central generator:  The element T need not lie in the center of C[G]; we only assume it is normal.  Thus T_ρ is no longer a scalar but an arbitrary normal matrix whose whole eigen-spectrum matters.  Handling its operator norm, diagonalization, and multiplicities is significantly subtler.\n\n3. Eigenvalue multiplicities:  The limit involves the total multiplicity m_ρ of eigenvalues of maximal modulus, a new layer absent from the original statement.\n\n4. New algebraic obstacles:  Proving 0 < c < 1 now needs an argument combining convexity of spectra for normal matrices, simultaneous diagonalization, and the generating property of B.\n\n5. Broader technical toolkit:  One must invoke the spectral theorem for normal matrices, the decomposition of the regular representation, trace identities, and basic algebraic number theory to show L is rational.\n\nBecause of these additional structural, representational, and spectral complexities, solving the enhanced variant demands substantially deeper insight and a more sophisticated blend of techniques than the original kernel problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}