{
  "index": "2009-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Say that a polynomial with real coefficients in two variables, $x,y$, is \\emph{balanced} if\nthe average value of the polynomial on each circle centered at the origin is $0$.\nThe balanced polynomials of degree at most $2009$ form a vector space $V$ over $\\mathbb{R}$.\nFind the dimension of $V$.",
  "solution": "Any polynomial $P(x,y)$ of degree at most $2009$ can be written uniquely\nas a sum $\\sum_{i=0}^{2009} P_i(x,y)$ in which $P_i(x,y)$ is a homogeneous\npolynomial of degree $i$.\nFor $r>0$, let $C_r$ be the path $(r\\cos \\theta, r\\sin \\theta)$\nfor $0 \\leq \\theta \\leq 2\\pi$. Put $\\lambda(P_i) = \\oint_{C_1} P_i$; then\nfor $r>0$,\n\\[\n\\oint_{C_r} P = \\sum_{i=0}^{2009} r^i \\lambda(P_i).\n\\]\nFor fixed $P$, the right side is a polynomial in $r$, which vanishes for\nall $r>0$ if and only if its coefficients vanish.\nIn other words,\n$P$ is balanced\nif and only if $\\lambda(P_i) = 0$ for $i=0,\\dots,2009$.\n\nFor $i$ odd, we have $P_i(-x,-y) = -P_i(x,y)$.\nHence $\\lambda(P_i) = 0$, e.g.,\nbecause the contributions to the integral from\n$\\theta$ and $\\theta + \\pi$ cancel.\n\nFor $i$ even, $\\lambda(P_i)$ is a linear function of the coefficients of\n$P_i$. This function is not identically zero, e.g., because for $P_i =\n(x^2 + y^2)^{i/2}$, the integrand is always positive and so\n$\\lambda(P_i) > 0$. The kernel of $\\lambda$ on the space of homogeneous\npolynomials of degree $i$ is thus a subspace of codimension 1.\n\nIt follows that the dimension of $V$ is\n\\[\n(1 + \\cdots + 2010) - 1005 = (2011 - 1) \\times 1005 = 2020050.\n\\]",
  "vars": [
    "x",
    "y",
    "r",
    "P",
    "P_i",
    "C_r",
    "i",
    "\\\\theta",
    "V",
    "\\\\lambda"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoordinate",
        "y": "ycoordinate",
        "r": "radiusvar",
        "P": "polyfunct",
        "P_i": "polyindcom",
        "C_r": "circlecurve",
        "i": "indexvar",
        "\\theta": "angletheta",
        "V": "vectorspace",
        "\\lambda": "lambdaoper"
      },
      "question": "Say that a polynomial with real coefficients in two variables, $xcoordinate,ycoordinate$, is \\emph{balanced} if\nthe average value of the polynomial on each circle centered at the origin is $0$.\nThe balanced polynomials of degree at most $2009$ form a vector space $vectorspace$ over $\\mathbb{R}$.\nFind the dimension of $vectorspace$.",
      "solution": "Any polynomial $polyfunct(xcoordinate,ycoordinate)$ of degree at most $2009$ can be written uniquely\nas a sum $\\sum_{indexvar=0}^{2009} polyindcom(xcoordinate,ycoordinate)$ in which $polyindcom(xcoordinate,ycoordinate)$ is a homogeneous\npolynomial of degree $indexvar$.\nFor $radiusvar>0$, let $circlecurve$ be the path $(radiusvar\\cos angletheta, radiusvar\\sin angletheta)$\nfor $0 \\leq angletheta \\leq 2\\pi$. Put $lambdaoper(polyindcom) = \\oint_{C_1} polyindcom$; then\nfor $radiusvar>0$,\n\\[\n\\oint_{circlecurve} polyfunct = \\sum_{indexvar=0}^{2009} radiusvar^{indexvar} lambdaoper(polyindcom).\n\\]\nFor fixed $polyfunct$, the right side is a polynomial in $radiusvar$, which vanishes for\nall $radiusvar>0$ if and only if its coefficients vanish.\nIn other words,\n$polyfunct$ is balanced\nif and only if $lambdaoper(polyindcom) = 0$ for $indexvar=0,\\dots,2009$.\n\nFor $indexvar$ odd, we have $polyindcom(-xcoordinate,-ycoordinate) = -polyindcom(xcoordinate,ycoordinate)$.\nHence $lambdaoper(polyindcom) = 0$, e.g.,\nbecause the contributions to the integral from\n$angletheta$ and $angletheta + \\pi$ cancel.\n\nFor $indexvar$ even, $lambdaoper(polyindcom)$ is a linear function of the coefficients of\npolyindcom. This function is not identically zero, e.g., because for polyindcom =\n$(xcoordinate^2 + ycoordinate^2)^{indexvar/2}$, the integrand is always positive and so\n$lambdaoper(polyindcom) > 0$. The kernel of $lambdaoper$ on the space of homogeneous\npolynomials of degree $indexvar$ is thus a subspace of codimension 1.\n\nIt follows that the dimension of $vectorspace$ is\n\\[\n(1 + \\cdots + 2010) - 1005 = (2011 - 1) \\times 1005 = 2020050.\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sunflower",
        "y": "blueberry",
        "r": "horseshoe",
        "P": "raincloud",
        "P_i": "raincloudi",
        "C_r": "gallopwing",
        "i": "tangerine",
        "\\\\theta": "driftwood",
        "V": "starlight",
        "\\\\lambda": "waterfall"
      },
      "question": "Say that a polynomial with real coefficients in two variables, $sunflower,blueberry$, is \\emph{balanced} if\nthe average value of the polynomial on each circle centered at the origin is $0$.\nThe balanced polynomials of degree at most $2009$ form a vector space $starlight$ over $\\mathbb{R}$.\nFind the dimension of $starlight$.",
      "solution": "Any polynomial $raincloud(sunflower,blueberry)$ of degree at most $2009$ can be written uniquely\nas a sum $\\sum_{tangerine=0}^{2009} raincloudi(sunflower,blueberry)$ in which $raincloudi(sunflower,blueberry)$ is a homogeneous\npolynomial of degree $tangerine$.\nFor $horseshoe>0$, let $gallopwing$ be the path $(horseshoe\\cos driftwood, horseshoe\\sin driftwood)$\nfor $0 \\leq driftwood \\leq 2\\pi$. Put $waterfall(raincloudi) = \\oint_{C_1} raincloudi$; then\nfor $horseshoe>0$,\n\\[\n\\oint_{gallopwing} raincloud = \\sum_{tangerine=0}^{2009} horseshoe^{tangerine} waterfall(raincloudi).\n\\]\nFor fixed $raincloud$, the right side is a polynomial in $horseshoe$, which vanishes for\nall $horseshoe>0$ if and only if its coefficients vanish.\nIn other words,\n$raincloud$ is balanced\nif and only if $waterfall(raincloudi) = 0$ for $tangerine=0,\\dots,2009$.\n\nFor $tangerine$ odd, we have $raincloudi(-sunflower,-blueberry) = -raincloudi(sunflower,blueberry)$.\nHence $waterfall(raincloudi) = 0$, e.g.,\nbecause the contributions to the integral from\ndriftwood and driftwood + $\\pi$ cancel.\n\nFor $tangerine$ even, $waterfall(raincloudi)$ is a linear function of the coefficients of\nraincloudi. This function is not identically zero, e.g., because for raincloudi =\n$(sunflower^2 + blueberry^2)^{tangerine/2}$, the integrand is always positive and so\n$waterfall(raincloudi) > 0$. The kernel of $waterfall$ on the space of homogeneous\npolynomials of degree $tangerine$ is thus a subspace of codimension 1.\n\nIt follows that the dimension of $starlight$ is\n\\[\n(1 + \\cdots + 2010) - 1005 = (2011 - 1) \\times 1005 = 2020050.\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "r": "nonradius",
        "P": "unbalanced",
        "P_i": "unbalancedpart",
        "C_r": "linesegment",
        "i": "wholenumber",
        "\\theta": "nodirection",
        "V": "scalarspace",
        "\\lambda": "zerofunction"
      },
      "question": "Say that a polynomial with real coefficients in two variables, $verticalaxis,horizontalaxis$, is \\emph{balanced} if\nthe average value of the polynomial on each circle centered at the origin is $0$.\nThe balanced polynomials of degree at most $2009$ form a vector space $scalarspace$ over $\\mathbb{R}$.\nFind the dimension of $scalarspace$.",
      "solution": "Any polynomial $unbalanced(verticalaxis,horizontalaxis)$ of degree at most $2009$ can be written uniquely\nas a sum $\\sum_{wholenumber=0}^{2009} unbalancedpart(verticalaxis,horizontalaxis)$ in which $unbalancedpart(verticalaxis,horizontalaxis)$ is a homogeneous\npolynomial of degree $wholenumber$.\nFor $nonradius>0$, let $linesegment$ be the path $(nonradius\\cos nodirection, nonradius\\sin nodirection)$\nfor $0 \\leq nodirection \\leq 2\\pi$. Put $zerofunction(unbalancedpart) = \\oint_{C_1} unbalancedpart$; then\nfor $nonradius>0$,\n\\[\n\\oint_{linesegment} unbalanced = \\sum_{wholenumber=0}^{2009} nonradius^{wholenumber} zerofunction(unbalancedpart).\n\\]\nFor fixed $unbalanced$, the right side is a polynomial in $nonradius$, which vanishes for\nall $nonradius>0$ if and only if its coefficients vanish.\nIn other words,\n$unbalanced$ is balanced\nif and only if $zerofunction(unbalancedpart) = 0$ for $wholenumber=0,\\dots,2009$.\n\nFor $wholenumber$ odd, we have $unbalancedpart(-verticalaxis,-horizontalaxis) = -unbalancedpart(verticalaxis,horizontalaxis)$.\nHence $zerofunction(unbalancedpart) = 0$, e.g.,\nbecause the contributions to the integral from\n$nodirection$ and $nodirection + \\pi$ cancel.\n\nFor $wholenumber$ even, $zerofunction(unbalancedpart)$ is a linear function of the coefficients of\n$unbalancedpart$. This function is not identically zero, e.g., because for $unbalancedpart =\n(verticalaxis^2 + horizontalaxis^2)^{wholenumber/2}$, the integrand is always positive and so\n$zerofunction(unbalancedpart) > 0$. The kernel of $zerofunction$ on the space of homogeneous\npolynomials of degree $wholenumber$ is thus a subspace of codimension 1.\n\nIt follows that the dimension of $scalarspace$ is\n\\[\n(1 + \\cdots + 2010) - 1005 = (2011 - 1) \\times 1005 = 2020050.\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "sjkdfhgn",
        "y": "qlazmwer",
        "r": "hvnckpqe",
        "P": "zxbqwepo",
        "P_i": "lkjweoru",
        "C_r": "vmskdjfa",
        "i": "nopqrsuv",
        "\\\\theta": "abcdlmno",
        "V": "wertyuii",
        "\\\\lambda": "zxcvmnbq"
      },
      "question": "Say that a polynomial with real coefficients in two variables, $sjkdfhgn,qlazmwer$, is \\emph{balanced} if\nthe average value of the polynomial on each circle centered at the origin is $0$.\nThe balanced polynomials of degree at most $2009$ form a vector space $wertyuii$ over $\\mathbb{R}$.\nFind the dimension of $wertyuii$.",
      "solution": "Any polynomial $zxbqwepo(sjkdfhgn,qlazmwer)$ of degree at most $2009$ can be written uniquely\nas a sum $\\sum_{nopqrsuv=0}^{2009} lkjweoru(sjkdfhgn,qlazmwer)$ in which $lkjweoru(sjkdfhgn,qlazmwer)$ is a homogeneous\npolynomial of degree $nopqrsuv$.\nFor $hvnckpqe>0$, let $vmskdjfa$ be the path $(hvnckpqe\\cos abcdlmno, hvnckpqe\\sin abcdlmno)$\nfor $0 \\leq abcdlmno \\leq 2\\pi$. Put $zxcvmnbq(lkjweoru) = \\oint_{vmskdjfa_{1}} lkjweoru$; then\nfor $hvnckpqe>0$,\n\\[\n\\oint_{vmskdjfa} zxbqwepo = \\sum_{nopqrsuv=0}^{2009} hvnckpqe^{nopqrsuv} zxcvmnbq(lkjweoru).\n\\]\nFor fixed $zxbqwepo$, the right side is a polynomial in $hvnckpqe$, which vanishes for\nall $hvnckpqe>0$ if and only if its coefficients vanish.\nIn other words,\nzxbqwepo is balanced\nif and only if $zxcvmnbq(lkjweoru) = 0$ for $nopqrsuv=0,\\dots,2009$.\n\nFor $nopqrsuv$ odd, we have $lkjweoru(-sjkdfhgn,-qlazmwer) = -lkjweoru(sjkdfhgn,qlazmwer)$.\nHence $zxcvmnbq(lkjweoru) = 0$, e.g.,\nbecause the contributions to the integral from\n$abcdlmno$ and $abcdlmno + \\pi$ cancel.\n\nFor $nopqrsuv$ even, $zxcvmnbq(lkjweoru)$ is a linear function of the coefficients of\nlkjweoru. This function is not identically zero, e.g., because for $lkjweoru =\n(sjkdfhgn^2 + qlazmwer^2)^{nopqrsuv/2}$, the integrand is always positive and so\n$zxcvmnbq(lkjweoru) > 0$. The kernel of $zxcvmnbq$ on the space of homogeneous\npolynomials of degree $nopqrsuv$ is thus a subspace of codimension 1.\n\nIt follows that the dimension of $wertyuii$ is\n\\[\n(1 + \\cdots + 2010) - 1005 = (2011 - 1) \\times 1005 = 2020050.\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\nP(x_{1},x_{2},x_{3},x_{4})\\in\\mathbb{R}[x_{1},x_{2},x_{3},x_{4}]\n\\]\nbe a real polynomial in four variables.  \nFor every radius $r>0$ denote the (Euclidean) $3$–sphere of radius $r$ by  \n\\[\n\\Sigma_{r}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{4}\\;:\\;\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=r^{2}\\bigr\\},\n\\qquad dS=\\hbox{ surface measure on }\\Sigma_{r}.\n\\]\n\nThe polynomial $P$ is called \\emph{$2$–spherically balanced} if, for every $r>0$, its spherical moments of orders $0,1,2$ vanish, i.e.  \n\\[\n\\tag{M0}\\int_{\\Sigma_{r}} P\\,dS = 0,\n\\qquad\n\\tag{M1}\\int_{\\Sigma_{r}} x_{j}\\,P\\,dS = 0\\;(j=1,2,3,4),\n\\]\n\\[\n\\tag{M2}\\int_{\\Sigma_{r}}\\Bigl(x_{j}x_{k}-\\tfrac{r^{2}}{4}\\delta_{jk}\\Bigr)P\\,dS = 0\n\\quad(1\\le j\\le k\\le 4).\n\\]\n\nLet $U$ be the vector space of all $2$–spherically balanced real polynomials of total degree $\\le 2022$.  \nDetermine $\\dim U$.\n\n──────────────────────────────────────────────────",
      "solution": "Throughout put  \n\\[\nr=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}},\\qquad\n\\omega=\\frac{x}{r}\\in S^{3}.\n\\]\n\n1. Harmonic decomposition.  \nLet $\\mathcal{H}_{\\ell}$ be the space of spherical harmonics of degree $\\ell$ on $S^{3}$.  \nEvery homogeneous polynomial $Q$ of (total) degree $d$ admits the unique expansion  \n\\[\nQ(r,\\omega)=\\sum_{\\substack{0\\le \\ell\\le d\\\\ d-\\ell\\text{ even}}}\nr^{\\,d-\\ell}\\,H_{\\ell}(\\omega),\\qquad H_{\\ell}\\in\\mathcal{H}_{\\ell}. \\tag{1}\n\\]\nFor $n=4$ it is classical that  \n\\[\n\\dim\\mathcal{H}_{\\ell}=(\\ell+1)^{2},\\qquad\\ell\\ge 0. \\tag{2}\n\\]\n\n2. Spherical moment operators.  \nLet  \n\\[\nV:=\\mathbb{R}^{4},\\qquad W:=\\operatorname{Sym}^{2}_{0}(V)\n=\\Bigl\\{A\\in\\operatorname{Sym}^{2}(V)\\ :\\ \\operatorname{tr}A=0\\Bigr\\}.\n\\]\nDefine $\\mathrm{SO}(4)$–equivariant maps\n\\[\n\\begin{aligned}\nM_{0}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow\\mathbb{R},\n& M_{0}(f)&=\\int_{S^{3}}f(\\omega)\\,d\\omega,\\\\[2pt]\nM_{1}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow V,\n& M_{1}(f)&=\\int_{S^{3}}\\omega\\,f(\\omega)\\,d\\omega,\\\\[2pt]\nM_{2}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow W,\n& M_{2}(f)&=\\int_{S^{3}}\\bigl(\\omega\\omega^{\\mathrm{T}}-\\tfrac{1}{4}I_{4}\\bigr)f(\\omega)\\,d\\omega,\n\\end{aligned}\\tag{3}\n\\]\nwhere $d\\omega$ is the normalised surface measure ($\\int_{S^{3}}d\\omega=1$).\n\nThe restrictions $M_{i}|_{\\mathcal{H}_{\\ell}}$ are intertwiners of irreducible\n$\\mathrm{SO}(4)$–representations; hence they are either $0$ or isomorphisms onto their images.\n\n3. Irreducible representation data.  \nWrite $\\mathrm{SO}(4)\\cong(\\mathrm{SU}(2)_{+}\\times\\mathrm{SU}(2)_{-})/\\{\\pm(1,1)\\}$.\nIrreducible real representations are indexed by pairs of non–negative integers\n\\[\n(j_{+},j_{-}),\\qquad\\dim(j_{+},j_{-})=(2j_{+}+1)(2j_{-}+1).\n\\]\nThe relevant ones are (see, e.g. Weyl’s character formula):\n\n•$\\;$ Standard representation $V$ : type $(\\tfrac12,\\tfrac12)$, dimension $4$.\n\n•$\\;$ Trace–free quadratic forms $W$ : type $(1,1)$, dimension $9$.\n\n•$\\;$ Spherical harmonics $\\mathcal{H}_{\\ell}$ : type $\\bigl(\\tfrac{\\ell}{2},\\tfrac{\\ell}{2}\\bigr)$, dimension $(\\ell+1)^{2}$  \n(the well–known fact that $\\mathcal{H}_{\\ell}$ is irreducible).\n\nTherefore  \n\n\\[\n\\operatorname{Hom}_{\\mathrm{SO}(4)}\\!\\bigl(\\mathcal{H}_{\\ell},V\\bigr)\n\\neq 0\\Longleftrightarrow \\tfrac{\\ell}{2}=\\tfrac12\\Longleftrightarrow\\ell=1,\n\\]\n\\[\n\\operatorname{Hom}_{\\mathrm{SO}(4)}\\!\\bigl(\\mathcal{H}_{\\ell},W\\bigr)\n\\neq 0\\Longleftrightarrow \\tfrac{\\ell}{2}=1\\Longleftrightarrow\\ell=2.\n\\]\n\nConsequences for the moment maps:\n\n\\[\n\\begin{aligned}\n&M_{0}|_{\\mathcal{H}_{0}}\\ \\text{is non–zero (constant functions), whereas }M_{0}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\ge 1);\\\\\n&M_{1}|_{\\mathcal{H}_{1}}:\\mathcal{H}_{1}\\xrightarrow{\\;\\cong\\;}V\\ \\text{is an isomorphism, while }M_{1}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\neq 1);\\\\\n&M_{2}|_{\\mathcal{H}_{2}}:\\mathcal{H}_{2}\\xrightarrow{\\;\\cong\\;}W\\ \\text{is an isomorphism, while }M_{2}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\neq 2).\n\\end{aligned}\\tag{4}\n\\]\n\n4. Independence of different radial powers.  \nLet $H_{\\ell}\\in\\mathcal{H}_{\\ell}$ be fixed.  \nOn the sphere $\\Sigma_{r}$ the measure satisfies $dS=r^{3}d\\omega$, so\n\\[\n\\int_{\\Sigma_{r}}r^{\\,d-\\ell}H_{\\ell}(\\omega)\\,dS=r^{\\,d-\\ell+3}\\int_{S^{3}}H_{\\ell}(\\omega)\\,d\\omega.\n\\]\nIf a finite linear combination of such terms vanishes for every $r>0$, then\nthe coefficients of the distinct powers of $r$ must vanish separately.\nHence the moment conditions act independently on every summand\n$r^{\\,d-\\ell}H_{\\ell}$ in (1).\n\n5. Which $\\ell$ survive?  \nBy (4) each of the three moment conditions annihilates exactly one\nspherical–harmonic isotype:\n\n\\[\n\\ell=0\\ \\text{ruled out by }(M0),\\qquad\n\\ell=1\\ \\text{ruled out by }(M1),\\qquad\n\\ell=2\\ \\text{ruled out by }(M2).\n\\]\n\nConversely all $\\ell\\notin\\{0,1,2\\}$ automatically satisfy $(M0)$–$(M2)$.\nThus a homogeneous piece $r^{\\,d-\\ell}H_{\\ell}$ is $2$–spherically balanced\niff $\\ell\\notin\\{0,1,2\\}$.\n\n6. Parity restriction from (1).  \nBecause $d-\\ell$ is even,\n\n\\[\n\\begin{cases}\nd\\ \\text{even} (\\ge 2): &\\ell\\ \\text{even},\\ \\ell\\neq 0,2;\\\\\nd=0: &\\ell=0\\ (\\text{excluded});\\\\\nd\\ \\text{odd}: &\\ell\\ \\text{odd},\\ \\ell\\neq 1.\n\\end{cases}\\tag{5}\n\\]\n\n7. Codimension in every degree.  \nWith $\\dim\\mathcal{H}_{0}=1,\\dim\\mathcal{H}_{1}=4,\\dim\\mathcal{H}_{2}=9$ we obtain\n\n\\[\n\\dim(\\text{deleted part in degree }d)=\n\\begin{cases}\n1,& d=0,\\\\[2pt]\n1+9=10,& d\\ \\text{even},\\ d\\ge 2,\\\\[2pt]\n4,& d\\ \\text{odd}.\n\\end{cases}\\tag{6}\n\\]\n\n8. Total codimension up to degree $2022$.  \nEven degrees: $d=0,2,\\dots,2022$ – exactly $1012$ of them.\n\n\\[\n1+10\\cdot 1011 = 10\\,111. \\tag{7}\n\\]\n\nOdd degrees: $d=1,3,\\dots,2021$ – exactly $1011$ of them.\n\n\\[\n4\\cdot 1011 = 4\\,044. \\tag{8}\n\\]\n\nHence  \n\n\\[\n\\text{total codimension}=10\\,111+4\\,044=14\\,155. \\tag{9}\n\\]\n\n9. Ambient dimension.  \nAll real polynomials in four variables of total degree $\\le n$ form a space of dimension $\\binom{n+4}{4}$.  \nFor $n=2022$,\n\\[\n\\binom{2026}{4}=\\frac{2026\\cdot 2025\\cdot 2024\\cdot 2023}{24}=699\\,938\\,073\\,450. \\tag{10}\n\\]\n\n10. Dimension of $U$.  \n\\[\n\\dim U=\\binom{2026}{4}-14\\,155\n=699\\,938\\,073\\,450-14\\,155\n=699\\,938\\,059\\,295. \\tag{11}\n\\]\n\n\\[\n\\boxed{\\dim U = 699\\,938\\,059\\,295}\n\\]\n\n──────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.815622",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:\n  The original problem worked in two variables (integration over\n  circles); here we are in \\(\\mathbb R^4\\) and integrate over\n  3-spheres, which forces the use of higher-dimensional harmonic\n  analysis.\n\n• Additional constraints:\n  Instead of annihilating only the mean value (order 0 moment),\n  the polynomial must annihilate all order 0, 1 and 2 spherical\n  moments—nine independent conditions for each admissible degree.\n\n• More sophisticated structures:\n  The solution requires the full spherical-harmonic decomposition in\n  four dimensions, the explicit dimension formula\n  \\(\\dim\\mathcal H_{\\ell}=4\\ell+1\\), and careful parity considerations\n  of the indices \\((d,\\ell)\\).\n\n• Deeper theory and more steps:\n  One must combine\n  – orthogonality of spherical harmonics,\n  – representation of homogeneous polynomials as radial powers times\n    harmonics,\n  – parity restrictions,\n  – enumeration of dimensions in each degree,\n  – summations producing very large integers.\n\n  Each of these ingredients goes beyond the elementary integral\n  cancellation used in the original 2-variable problem,\n  and they interact non-trivially.\n\nFor these reasons the enhanced variant is\nsignificantly harder than both the original problem\nand the previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nP(x_{1},x_{2},x_{3},x_{4})\\in\\mathbb{R}[x_{1},x_{2},x_{3},x_{4}]\n\\]\nbe a real polynomial in four variables.  \nFor every radius $r>0$ denote the (Euclidean) $3$–sphere of radius $r$ by  \n\\[\n\\Sigma_{r}\\;=\\;\\bigl\\{x\\in\\mathbb{R}^{4}\\;:\\;\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=r^{2}\\bigr\\},\n\\qquad dS=\\hbox{ surface measure on }\\Sigma_{r}.\n\\]\n\nThe polynomial $P$ is called \\emph{$2$–spherically balanced} if, for every $r>0$, its spherical moments of orders $0,1,2$ vanish, i.e.  \n\\[\n\\tag{M0}\\int_{\\Sigma_{r}} P\\,dS = 0,\n\\qquad\n\\tag{M1}\\int_{\\Sigma_{r}} x_{j}\\,P\\,dS = 0\\;(j=1,2,3,4),\n\\]\n\\[\n\\tag{M2}\\int_{\\Sigma_{r}}\\Bigl(x_{j}x_{k}-\\tfrac{r^{2}}{4}\\delta_{jk}\\Bigr)P\\,dS = 0\n\\quad(1\\le j\\le k\\le 4).\n\\]\n\nLet $U$ be the vector space of all $2$–spherically balanced real polynomials of total degree $\\le 2022$.  \nDetermine $\\dim U$.\n\n──────────────────────────────────────────────────",
      "solution": "Throughout put  \n\\[\nr=\\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}},\\qquad\n\\omega=\\frac{x}{r}\\in S^{3}.\n\\]\n\n1. Harmonic decomposition.  \nLet $\\mathcal{H}_{\\ell}$ be the space of spherical harmonics of degree $\\ell$ on $S^{3}$.  \nEvery homogeneous polynomial $Q$ of (total) degree $d$ admits the unique expansion  \n\\[\nQ(r,\\omega)=\\sum_{\\substack{0\\le \\ell\\le d\\\\ d-\\ell\\text{ even}}}\nr^{\\,d-\\ell}\\,H_{\\ell}(\\omega),\\qquad H_{\\ell}\\in\\mathcal{H}_{\\ell}. \\tag{1}\n\\]\nFor $n=4$ it is classical that  \n\\[\n\\dim\\mathcal{H}_{\\ell}=(\\ell+1)^{2},\\qquad\\ell\\ge 0. \\tag{2}\n\\]\n\n2. Spherical moment operators.  \nLet  \n\\[\nV:=\\mathbb{R}^{4},\\qquad W:=\\operatorname{Sym}^{2}_{0}(V)\n=\\Bigl\\{A\\in\\operatorname{Sym}^{2}(V)\\ :\\ \\operatorname{tr}A=0\\Bigr\\}.\n\\]\nDefine $\\mathrm{SO}(4)$–equivariant maps\n\\[\n\\begin{aligned}\nM_{0}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow\\mathbb{R},\n& M_{0}(f)&=\\int_{S^{3}}f(\\omega)\\,d\\omega,\\\\[2pt]\nM_{1}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow V,\n& M_{1}(f)&=\\int_{S^{3}}\\omega\\,f(\\omega)\\,d\\omega,\\\\[2pt]\nM_{2}&:\\ \\mathcal{C}^{\\infty}(S^{3})\\longrightarrow W,\n& M_{2}(f)&=\\int_{S^{3}}\\bigl(\\omega\\omega^{\\mathrm{T}}-\\tfrac{1}{4}I_{4}\\bigr)f(\\omega)\\,d\\omega,\n\\end{aligned}\\tag{3}\n\\]\nwhere $d\\omega$ is the normalised surface measure ($\\int_{S^{3}}d\\omega=1$).\n\nThe restrictions $M_{i}|_{\\mathcal{H}_{\\ell}}$ are intertwiners of irreducible\n$\\mathrm{SO}(4)$–representations; hence they are either $0$ or isomorphisms onto their images.\n\n3. Irreducible representation data.  \nWrite $\\mathrm{SO}(4)\\cong(\\mathrm{SU}(2)_{+}\\times\\mathrm{SU}(2)_{-})/\\{\\pm(1,1)\\}$.\nIrreducible real representations are indexed by pairs of non–negative integers\n\\[\n(j_{+},j_{-}),\\qquad\\dim(j_{+},j_{-})=(2j_{+}+1)(2j_{-}+1).\n\\]\nThe relevant ones are (see, e.g. Weyl’s character formula):\n\n•$\\;$ Standard representation $V$ : type $(\\tfrac12,\\tfrac12)$, dimension $4$.\n\n•$\\;$ Trace–free quadratic forms $W$ : type $(1,1)$, dimension $9$.\n\n•$\\;$ Spherical harmonics $\\mathcal{H}_{\\ell}$ : type $\\bigl(\\tfrac{\\ell}{2},\\tfrac{\\ell}{2}\\bigr)$, dimension $(\\ell+1)^{2}$  \n(the well–known fact that $\\mathcal{H}_{\\ell}$ is irreducible).\n\nTherefore  \n\n\\[\n\\operatorname{Hom}_{\\mathrm{SO}(4)}\\!\\bigl(\\mathcal{H}_{\\ell},V\\bigr)\n\\neq 0\\Longleftrightarrow \\tfrac{\\ell}{2}=\\tfrac12\\Longleftrightarrow\\ell=1,\n\\]\n\\[\n\\operatorname{Hom}_{\\mathrm{SO}(4)}\\!\\bigl(\\mathcal{H}_{\\ell},W\\bigr)\n\\neq 0\\Longleftrightarrow \\tfrac{\\ell}{2}=1\\Longleftrightarrow\\ell=2.\n\\]\n\nConsequences for the moment maps:\n\n\\[\n\\begin{aligned}\n&M_{0}|_{\\mathcal{H}_{0}}\\ \\text{is non–zero (constant functions), whereas }M_{0}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\ge 1);\\\\\n&M_{1}|_{\\mathcal{H}_{1}}:\\mathcal{H}_{1}\\xrightarrow{\\;\\cong\\;}V\\ \\text{is an isomorphism, while }M_{1}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\neq 1);\\\\\n&M_{2}|_{\\mathcal{H}_{2}}:\\mathcal{H}_{2}\\xrightarrow{\\;\\cong\\;}W\\ \\text{is an isomorphism, while }M_{2}|_{\\mathcal{H}_{\\ell}}=0\\;(\\ell\\neq 2).\n\\end{aligned}\\tag{4}\n\\]\n\n4. Independence of different radial powers.  \nLet $H_{\\ell}\\in\\mathcal{H}_{\\ell}$ be fixed.  \nOn the sphere $\\Sigma_{r}$ the measure satisfies $dS=r^{3}d\\omega$, so\n\\[\n\\int_{\\Sigma_{r}}r^{\\,d-\\ell}H_{\\ell}(\\omega)\\,dS=r^{\\,d-\\ell+3}\\int_{S^{3}}H_{\\ell}(\\omega)\\,d\\omega.\n\\]\nIf a finite linear combination of such terms vanishes for every $r>0$, then\nthe coefficients of the distinct powers of $r$ must vanish separately.\nHence the moment conditions act independently on every summand\n$r^{\\,d-\\ell}H_{\\ell}$ in (1).\n\n5. Which $\\ell$ survive?  \nBy (4) each of the three moment conditions annihilates exactly one\nspherical–harmonic isotype:\n\n\\[\n\\ell=0\\ \\text{ruled out by }(M0),\\qquad\n\\ell=1\\ \\text{ruled out by }(M1),\\qquad\n\\ell=2\\ \\text{ruled out by }(M2).\n\\]\n\nConversely all $\\ell\\notin\\{0,1,2\\}$ automatically satisfy $(M0)$–$(M2)$.\nThus a homogeneous piece $r^{\\,d-\\ell}H_{\\ell}$ is $2$–spherically balanced\niff $\\ell\\notin\\{0,1,2\\}$.\n\n6. Parity restriction from (1).  \nBecause $d-\\ell$ is even,\n\n\\[\n\\begin{cases}\nd\\ \\text{even} (\\ge 2): &\\ell\\ \\text{even},\\ \\ell\\neq 0,2;\\\\\nd=0: &\\ell=0\\ (\\text{excluded});\\\\\nd\\ \\text{odd}: &\\ell\\ \\text{odd},\\ \\ell\\neq 1.\n\\end{cases}\\tag{5}\n\\]\n\n7. Codimension in every degree.  \nWith $\\dim\\mathcal{H}_{0}=1,\\dim\\mathcal{H}_{1}=4,\\dim\\mathcal{H}_{2}=9$ we obtain\n\n\\[\n\\dim(\\text{deleted part in degree }d)=\n\\begin{cases}\n1,& d=0,\\\\[2pt]\n1+9=10,& d\\ \\text{even},\\ d\\ge 2,\\\\[2pt]\n4,& d\\ \\text{odd}.\n\\end{cases}\\tag{6}\n\\]\n\n8. Total codimension up to degree $2022$.  \nEven degrees: $d=0,2,\\dots,2022$ – exactly $1012$ of them.\n\n\\[\n1+10\\cdot 1011 = 10\\,111. \\tag{7}\n\\]\n\nOdd degrees: $d=1,3,\\dots,2021$ – exactly $1011$ of them.\n\n\\[\n4\\cdot 1011 = 4\\,044. \\tag{8}\n\\]\n\nHence  \n\n\\[\n\\text{total codimension}=10\\,111+4\\,044=14\\,155. \\tag{9}\n\\]\n\n9. Ambient dimension.  \nAll real polynomials in four variables of total degree $\\le n$ form a space of dimension $\\binom{n+4}{4}$.  \nFor $n=2022$,\n\\[\n\\binom{2026}{4}=\\frac{2026\\cdot 2025\\cdot 2024\\cdot 2023}{24}=699\\,938\\,073\\,450. \\tag{10}\n\\]\n\n10. Dimension of $U$.  \n\\[\n\\dim U=\\binom{2026}{4}-14\\,155\n=699\\,938\\,073\\,450-14\\,155\n=699\\,938\\,059\\,295. \\tag{11}\n\\]\n\n\\[\n\\boxed{\\dim U = 699\\,938\\,059\\,295}\n\\]\n\n──────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.624266",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension:\n  The original problem worked in two variables (integration over\n  circles); here we are in \\(\\mathbb R^4\\) and integrate over\n  3-spheres, which forces the use of higher-dimensional harmonic\n  analysis.\n\n• Additional constraints:\n  Instead of annihilating only the mean value (order 0 moment),\n  the polynomial must annihilate all order 0, 1 and 2 spherical\n  moments—nine independent conditions for each admissible degree.\n\n• More sophisticated structures:\n  The solution requires the full spherical-harmonic decomposition in\n  four dimensions, the explicit dimension formula\n  \\(\\dim\\mathcal H_{\\ell}=4\\ell+1\\), and careful parity considerations\n  of the indices \\((d,\\ell)\\).\n\n• Deeper theory and more steps:\n  One must combine\n  – orthogonality of spherical harmonics,\n  – representation of homogeneous polynomials as radial powers times\n    harmonics,\n  – parity restrictions,\n  – enumeration of dimensions in each degree,\n  – summations producing very large integers.\n\n  Each of these ingredients goes beyond the elementary integral\n  cancellation used in the original 2-variable problem,\n  and they interact non-trivially.\n\nFor these reasons the enhanced variant is\nsignificantly harder than both the original problem\nand the previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}