{
  "index": "1960-A-5",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Consider a polynomial \\( f(x) \\) with real coefficients having the property \\( f(g(x))=g(f(x)) \\) for every polynomial \\( g(x) \\) with real coefficients. Determine and prove the nature of \\( f(x) \\).",
  "solution": "Solution. Consider a constant function \\( g \\), say \\( g(x)=a \\). Then \\( f(g(x))= \\) \\( g(f(x)) \\) becomes \\( f(a)=a \\). Since this is true for all real \\( a, f \\) is the identity function, i.e., \\( f(x)=x \\).",
  "vars": [
    "x",
    "g"
  ],
  "params": [
    "f",
    "a"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvariable",
        "g": "genericpoly",
        "f": "fixedpoly",
        "a": "fixedreal"
      },
      "question": "5. Consider a polynomial \\( fixedpoly(inputvariable) \\) with real coefficients having the property \\( fixedpoly(genericpoly(inputvariable))=genericpoly(fixedpoly(inputvariable)) \\) for every polynomial \\( genericpoly(inputvariable) \\) with real coefficients. Determine and prove the nature of \\( fixedpoly(inputvariable) \\).",
      "solution": "Solution. Consider a constant function \\( genericpoly \\), say \\( genericpoly(inputvariable)=fixedreal \\). Then \\( fixedpoly(genericpoly(inputvariable))= \\) \\( genericpoly(fixedpoly(inputvariable)) \\) becomes \\( fixedpoly(fixedreal)=fixedreal \\). Since this is true for all real \\( fixedreal, fixedpoly \\) is the identity function, i.e., \\( fixedpoly(inputvariable)=inputvariable \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "marblecup",
        "g": "tangentleaf",
        "f": "planetveil",
        "a": "coppertwig"
      },
      "question": "5. Consider a polynomial \\( planetveil(marblecup) \\) with real coefficients having the property \\( planetveil(tangentleaf(marblecup))=tangentleaf(planetveil(marblecup)) \\) for every polynomial \\( tangentleaf(marblecup) \\) with real coefficients. Determine and prove the nature of \\( planetveil(marblecup) \\).",
      "solution": "Solution. Consider a constant function \\( tangentleaf \\), say \\( tangentleaf(marblecup)=coppertwig \\). Then \\( planetveil(tangentleaf(marblecup))= \\) \\( tangentleaf(planetveil(marblecup)) \\) becomes \\( planetveil(coppertwig)=coppertwig \\). Since this is true for all real \\( coppertwig, planetveil \\) is the identity function, i.e., \\( planetveil(marblecup)=marblecup \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "g": "staticnumber",
        "f": "randommap",
        "a": "variablequantity"
      },
      "question": "<<<\n5. Consider a polynomial \\( randommap(fixedvalue) \\) with real coefficients having the property \\( randommap(staticnumber(fixedvalue))=staticnumber(randommap(fixedvalue)) \\) for every polynomial \\( staticnumber(fixedvalue) \\) with real coefficients. Determine and prove the nature of \\( randommap(fixedvalue) \\).\n>>>",
      "solution": "<<<\nSolution. Consider a constant function \\( staticnumber \\), say \\( staticnumber(fixedvalue)=variablequantity \\). Then \\( randommap(staticnumber(fixedvalue))=staticnumber(randommap(fixedvalue)) \\) becomes \\( randommap(variablequantity)=variablequantity \\). Since this is true for all real \\( variablequantity, randommap \\) is the identity function, i.e., \\( randommap(fixedvalue)=fixedvalue \\).\n>>>"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "g": "hjgrksla",
        "f": "mndplkqe",
        "a": "rsvchmzt"
      },
      "question": "5. Consider a polynomial \\( mndplkqe(qzxwvtnp) \\) with real coefficients having the property \\( mndplkqe(hjgrksla(qzxwvtnp))=hjgrksla(mndplkqe(qzxwvtnp)) \\) for every polynomial \\( hjgrksla(qzxwvtnp) \\) with real coefficients. Determine and prove the nature of \\( mndplkqe(qzxwvtnp) \\).",
      "solution": "Solution. Consider a constant function \\( hjgrksla \\), say \\( hjgrksla(qzxwvtnp)=rsvchmzt \\). Then \\( mndplkqe(hjgrksla(qzxwvtnp))= \\) \\( hjgrksla(mndplkqe(qzxwvtnp)) \\) becomes \\( mndplkqe(rsvchmzt)=rsvchmzt \\). Since this is true for all real \\( rsvchmzt, mndplkqe \\) is the identity function, i.e., \\( mndplkqe(qzxwvtnp)=qzxwvtnp \\)."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 1$ and let $\\mathbf P^{n}=\\mathbf P^{n}(\\mathbf C)$ be complex projective $n$–space.  \nA dominant rational self–map  \n\n\\[\nF:\\mathbf P^{n}\\dashrightarrow\\mathbf P^{n}\n\\]\n\nis given, in homogeneous coordinates, by an $(n+1)$–tuple of homogeneous polynomials of the same degree $d\\ge 1$  \n\n\\[\nH_{0},\\dots ,H_{n}\\in\\mathbf C[X_{0},\\dots ,X_{n}],\\qquad \\deg H_{i}=d\\;(0\\le i\\le n),\n\\tag{$\\star$}\n\\]\n\nsuch that the set of common zeros of $H_{0},\\dots ,H_{n}$ in $\\mathbf P^{n}$ is empty\n(equivalently, the ideal $(H_{0},\\dots ,H_{n})$ is the unit ideal in the homogeneous coordinate ring).  \nCondition $(\\star)$ guarantees that $F$ is everywhere defined, hence a morphism,\n\n\\[\nF([X_{0}:\\dots :X_{n}])=[\\,H_{0}(X):\\dots :H_{n}(X)\\,].\n\\tag{1}\n\\]\n\nAssume moreover that $F$ centralises the entire projective linear group:\n\n\\[\nF\\circ G = G\\circ F\\quad\\text{in }\\operatorname{Rat}(\\mathbf P^{n},\\mathbf P^{n})\n\\quad\\text{for every }G\\in\\operatorname{PGL}_{n+1}(\\mathbf C).\n\\tag{2}\n\\]\n\nDetermine all such morphisms $F$ and give a complete proof.",
      "solution": "Throughout put $V=\\mathbf C^{\\,n+1}$ (column vectors) and identify $\\mathbf P^{n}$ with $\\mathbf P(V)$.\n\nStep 0.  Non–vanishing of $H$ on $V\\setminus\\{0\\}$.  \nDefine  \n\n\\[\nH(v):=(H_{0}(v),\\dots ,H_{n}(v)),\\qquad v\\in V.\n\\]\n\nBecause of $(\\star)$ the base locus of $H_{0},\\dots ,H_{n}$ in $\\mathbf P^{n}$ is empty, so none of the points of $\\mathbf P^{n}$ is annihilated by all the $H_{i}$.  \nConsequently\n\n\\[\nH(v)\\neq 0\\quad\\text{for every }v\\neq 0.\n\\tag{3}\n\\]\n\nIndeed, suppose $H(v)=0$ for some non–zero $v$.  For any $A\\in\\operatorname{GL}(V)$ let $G=[A]$.  Since $H(Av)$ is just $H$ evaluated at another non–zero vector, $H(Av)$ may still vanish; we therefore employ centrality: from $(2)$ we get  \n\n\\[\nH(Av)=0\\;\\Longrightarrow\\;F\\bigl([Av]\\bigr)=G\\bigl(F([v])\\bigr)=[0:\\dots :0],\n\\]\n\ncontradicting the fact that $F$ is a morphism.  Hence $H(v)\\neq 0$ for all $v\\neq 0$, and, by homogeneity,\n\n\\[\nH(\\lambda v)=\\lambda^{d}H(v)\\qquad(\\lambda\\in\\mathbf C,v\\in V).\n\\tag{4}\n\\]\n\nThus $F([v])=[H(v)]$ is well defined.\n\nStep 1.  Equivariance up to a scalar.  \nFix $A\\in\\operatorname{GL}(V)$ and write $G=[A]\\in\\operatorname{PGL}(V)$.  From $(2)$ we obtain\n\n\\[\nH(Av)=\\sigma_{A}(v)\\,A\\,H(v)\\qquad(v\\ne 0),\n\\tag{5}\n\\]\n\nwhere $\\sigma_{A}$ is a homogeneous rational function of degree $0$, hence a rational function on $\\mathbf P^{n}$.\n\nStep 2.  $\\sigma_{A}$ is constant.  \nChoose a non–zero vector $v_{0}\\in V$.  Put  \n\n\\[\n\\mu(A):=\\sigma_{A}(v_{0}).\n\\]\n\nUsing (5) successively for $A,B\\in\\operatorname{GL}(V)$ gives  \n\n\\[\n\\sigma_{AB}(v)=\\sigma_{A}(Bv)\\,\\sigma_{B}(v)\\qquad(v\\neq 0),\n\\tag{6}\n\\]\n\nwhence  \n\n\\[\n\\mu(AB)=\\sigma_{A}(Bv_{0})\\,\\mu(B).\n\\tag{7}\n\\]\n\nSolving (7) for $\\sigma_{A}(Bv_{0})$ and observing that the $\\operatorname{PGL}$–orbit of $[v_{0}]$ is the whole $\\mathbf P^{n}$, one finds that the rational map  \n\n\\[\n\\tau_{A}:=\\sigma_{A}/\\mu(A)\\colon\\mathbf P^{n}\\dashrightarrow\\mathbf C^{\\times}\n\\]\n\nis $1$ on a Zariski–dense subset, hence identically $1$.  Therefore  \n\n\\[\n\\sigma_{A}(v)=\\mu(A)\\qquad(v\\neq 0).\n\\tag{8}\n\\]\n\nStep 3.  $\\mu$ is a character of $\\operatorname{GL}(V)$.  \nWith (8) equation (5) becomes\n\n\\[\nH(Av)=\\mu(A)\\,A\\,H(v).\n\\tag{9}\n\\]\n\nApplying (9) twice yields $\\mu(AB)=\\mu(A)\\mu(B)$, so  \n\n\\[\n\\mu:\\operatorname{GL}_{n+1}(\\mathbf C)\\longrightarrow\\mathbf C^{\\times}\n\\]\n\nis a group homomorphism.  All algebraic characters of $\\operatorname{GL}_{n+1}(\\mathbf C)$ are integral powers of the determinant; hence  \n\n\\[\n\\mu(A)=\\det(A)^{k}\\qquad\\text{for some }k\\in\\mathbf Z.\n\\tag{10}\n\\]\n\nStep 4.  Relating $k$ and $d$.  \nInsert $A=c\\cdot\\operatorname{Id}_{V}$ ($c\\in\\mathbf C^{\\times}$) into (9):\n\n\\[\nc^{d}H(v)=H(cv)=\\mu(c\\,\\operatorname{Id})\\,c\\,H(v)=c^{1+(n+1)k}H(v),\n\\]\n\nso  \n\n\\[\nd=1+(n+1)k\\quad\\Longrightarrow\\quad k=\\frac{d-1}{n+1}\\in\\mathbf Z.\n\\tag{11}\n\\]\n\nStep 5.  Representation–theoretic obstruction for $d>1$.  \nRewrite (9) using (10):\n\n\\[\nH(Av)=\\det(A)^{k}\\,A\\,H(v).\n\\tag{12}\n\\]\n\nRegard $H$ as a $\\operatorname{GL}(V)$–equivariant linear map\n\n\\[\nh:\\operatorname{Sym}^{d}(V)\\longrightarrow V\\otimes\\det^{k}.\n\\tag{13}\n\\]\n\nThe $\\operatorname{GL}(V)$–representations on the two sides are irreducible with highest weights  \n\n\\[\n\\operatorname{Sym}^{d}(V):\\;(d,0,\\dots ,0),\\qquad  \nV\\otimes\\det^{k}:\\;(1+k,k,\\dots ,k).\n\\]\n\nEquality of highest weights forces $d=1+k$ and $k=0$.  Substituting in (11) gives $d=1$.  Therefore $d>1$ is impossible.\n\nStep 6.  The linear case $d=1$.  \nWith $d=1$ and $k=0$, (12) takes the simpler form\n\n\\[\nH(Av)=A\\,H(v)\\qquad(A\\in\\operatorname{GL}(V)).\n\\tag{14}\n\\]\n\nThus $H$ is $\\operatorname{GL}(V)$–equivariant for the standard representation, hence linear:\n\n\\[\nH(v)=L\\,v\\qquad(L\\in\\operatorname{End}(V)).\n\\tag{15}\n\\]\n\nEquation (14) implies $LA=AL$ for all $A\\in\\operatorname{GL}(V)$; therefore $L=c\\,\\operatorname{Id}_{V}$ with $c\\in\\mathbf C^{\\times}$.  Multiplying every $H_{i}$ by $c^{-1}$ does not change the projective map, so  \n\n\\[\nF=\\operatorname{Id}_{\\mathbf P^{n}}.\n\\]\n\nStep 7.  The one–dimensional case $n=1$ (verification).  \nWhen $n=1$ the above argument still applies.  Alternatively, $\\operatorname{PGL}_{2}(\\mathbf C)$ acts triply transitively on $\\mathbf P^{1}$; a morphism commuting with every projective automorphism fixes three distinct points and must therefore be the identity.\n\nConclusion.  \nFor every $n\\ge 1$ the only dominant rational self–map of $\\mathbf P^{n}(\\mathbf C)$ that commutes with every projective automorphism is\n\n\\[\nF=\\operatorname{Id}_{\\mathbf P^{n}}.\\qquad\\Box\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.519017",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-Dimensional & Projective Setting  \n • We moved from ℂⁿ to projective space ℙⁿ, introducing homogeneous coordinates, rational maps, and dominance issues.  \n • The automorphism group escalates from affine translations to the full algebraic group PGL_{n+1}(ℂ).\n\n2. Additional Structural Constraints  \n • Commutation is required with every element of an algebraic group of dimension (n+1)² – 1, not merely with all entire maps.\n\n3. Deeper Theory Needed  \n • Characters of GL_{n+1}, homogeneous lifting, and projective equivalence.  \n • Representation theory of GL_{n+1}: highest weights, symmetric powers, Schur’s lemma, vanishing of certain Hom-spaces.  \n • Algebraic group arguments to identify μ(A)=det(A)^{k}.  \n\n4. Multi-Step Proof  \n • Establishing a lifted functional equation with a scalar factor.  \n • Showing the scalar factor is a character and determining its exponent.  \n • Translating the problem into one about equivariant homomorphisms between representations and ruling them out for d>1.  \n • Final linear-algebra centralizer argument.\n\n5. Significantly Harder  \n • The original problem collapses immediately by plugging in constants.  \n • Here constants are unavailable; one must deploy algebraic-group and representation-theoretic machinery, execute weight computations, and control dominant rational maps on ℙⁿ.  \n • Each step is non-trivial and relies on advanced graduate-level material, making the new variant far more challenging than both the original and the previous kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 1$ and let $\\mathbf P^{n}=\\mathbf P^{n}(\\mathbf C)$ be complex projective $n$–space.  \nA dominant rational self–map  \n\n\\[\nF:\\mathbf P^{n}\\dashrightarrow\\mathbf P^{n}\n\\]\n\nis given, in homogeneous coordinates, by an $(n+1)$–tuple of homogeneous polynomials of the same degree $d\\ge 1$  \n\n\\[\nH_{0},\\dots ,H_{n}\\in\\mathbf C[X_{0},\\dots ,X_{n}],\\qquad \\deg H_{i}=d\\;(0\\le i\\le n),\n\\tag{$\\star$}\n\\]\n\nsuch that the set of common zeros of $H_{0},\\dots ,H_{n}$ in $\\mathbf P^{n}$ is empty\n(equivalently, the ideal $(H_{0},\\dots ,H_{n})$ is the unit ideal in the homogeneous coordinate ring).  \nCondition $(\\star)$ guarantees that $F$ is everywhere defined, hence a morphism,\n\n\\[\nF([X_{0}:\\dots :X_{n}])=[\\,H_{0}(X):\\dots :H_{n}(X)\\,].\n\\tag{1}\n\\]\n\nAssume moreover that $F$ centralises the entire projective linear group:\n\n\\[\nF\\circ G = G\\circ F\\quad\\text{in }\\operatorname{Rat}(\\mathbf P^{n},\\mathbf P^{n})\n\\quad\\text{for every }G\\in\\operatorname{PGL}_{n+1}(\\mathbf C).\n\\tag{2}\n\\]\n\nDetermine all such morphisms $F$ and give a complete proof.",
      "solution": "Throughout put $V=\\mathbf C^{\\,n+1}$ (column vectors) and identify $\\mathbf P^{n}$ with $\\mathbf P(V)$.\n\nStep 0.  Non–vanishing of $H$ on $V\\setminus\\{0\\}$.  \nDefine  \n\n\\[\nH(v):=(H_{0}(v),\\dots ,H_{n}(v)),\\qquad v\\in V.\n\\]\n\nBecause of $(\\star)$ the base locus of $H_{0},\\dots ,H_{n}$ in $\\mathbf P^{n}$ is empty, so none of the points of $\\mathbf P^{n}$ is annihilated by all the $H_{i}$.  \nConsequently\n\n\\[\nH(v)\\neq 0\\quad\\text{for every }v\\neq 0.\n\\tag{3}\n\\]\n\nIndeed, suppose $H(v)=0$ for some non–zero $v$.  For any $A\\in\\operatorname{GL}(V)$ let $G=[A]$.  Since $H(Av)$ is just $H$ evaluated at another non–zero vector, $H(Av)$ may still vanish; we therefore employ centrality: from $(2)$ we get  \n\n\\[\nH(Av)=0\\;\\Longrightarrow\\;F\\bigl([Av]\\bigr)=G\\bigl(F([v])\\bigr)=[0:\\dots :0],\n\\]\n\ncontradicting the fact that $F$ is a morphism.  Hence $H(v)\\neq 0$ for all $v\\neq 0$, and, by homogeneity,\n\n\\[\nH(\\lambda v)=\\lambda^{d}H(v)\\qquad(\\lambda\\in\\mathbf C,v\\in V).\n\\tag{4}\n\\]\n\nThus $F([v])=[H(v)]$ is well defined.\n\nStep 1.  Equivariance up to a scalar.  \nFix $A\\in\\operatorname{GL}(V)$ and write $G=[A]\\in\\operatorname{PGL}(V)$.  From $(2)$ we obtain\n\n\\[\nH(Av)=\\sigma_{A}(v)\\,A\\,H(v)\\qquad(v\\ne 0),\n\\tag{5}\n\\]\n\nwhere $\\sigma_{A}$ is a homogeneous rational function of degree $0$, hence a rational function on $\\mathbf P^{n}$.\n\nStep 2.  $\\sigma_{A}$ is constant.  \nChoose a non–zero vector $v_{0}\\in V$.  Put  \n\n\\[\n\\mu(A):=\\sigma_{A}(v_{0}).\n\\]\n\nUsing (5) successively for $A,B\\in\\operatorname{GL}(V)$ gives  \n\n\\[\n\\sigma_{AB}(v)=\\sigma_{A}(Bv)\\,\\sigma_{B}(v)\\qquad(v\\neq 0),\n\\tag{6}\n\\]\n\nwhence  \n\n\\[\n\\mu(AB)=\\sigma_{A}(Bv_{0})\\,\\mu(B).\n\\tag{7}\n\\]\n\nSolving (7) for $\\sigma_{A}(Bv_{0})$ and observing that the $\\operatorname{PGL}$–orbit of $[v_{0}]$ is the whole $\\mathbf P^{n}$, one finds that the rational map  \n\n\\[\n\\tau_{A}:=\\sigma_{A}/\\mu(A)\\colon\\mathbf P^{n}\\dashrightarrow\\mathbf C^{\\times}\n\\]\n\nis $1$ on a Zariski–dense subset, hence identically $1$.  Therefore  \n\n\\[\n\\sigma_{A}(v)=\\mu(A)\\qquad(v\\neq 0).\n\\tag{8}\n\\]\n\nStep 3.  $\\mu$ is a character of $\\operatorname{GL}(V)$.  \nWith (8) equation (5) becomes\n\n\\[\nH(Av)=\\mu(A)\\,A\\,H(v).\n\\tag{9}\n\\]\n\nApplying (9) twice yields $\\mu(AB)=\\mu(A)\\mu(B)$, so  \n\n\\[\n\\mu:\\operatorname{GL}_{n+1}(\\mathbf C)\\longrightarrow\\mathbf C^{\\times}\n\\]\n\nis a group homomorphism.  All algebraic characters of $\\operatorname{GL}_{n+1}(\\mathbf C)$ are integral powers of the determinant; hence  \n\n\\[\n\\mu(A)=\\det(A)^{k}\\qquad\\text{for some }k\\in\\mathbf Z.\n\\tag{10}\n\\]\n\nStep 4.  Relating $k$ and $d$.  \nInsert $A=c\\cdot\\operatorname{Id}_{V}$ ($c\\in\\mathbf C^{\\times}$) into (9):\n\n\\[\nc^{d}H(v)=H(cv)=\\mu(c\\,\\operatorname{Id})\\,c\\,H(v)=c^{1+(n+1)k}H(v),\n\\]\n\nso  \n\n\\[\nd=1+(n+1)k\\quad\\Longrightarrow\\quad k=\\frac{d-1}{n+1}\\in\\mathbf Z.\n\\tag{11}\n\\]\n\nStep 5.  Representation–theoretic obstruction for $d>1$.  \nRewrite (9) using (10):\n\n\\[\nH(Av)=\\det(A)^{k}\\,A\\,H(v).\n\\tag{12}\n\\]\n\nRegard $H$ as a $\\operatorname{GL}(V)$–equivariant linear map\n\n\\[\nh:\\operatorname{Sym}^{d}(V)\\longrightarrow V\\otimes\\det^{k}.\n\\tag{13}\n\\]\n\nThe $\\operatorname{GL}(V)$–representations on the two sides are irreducible with highest weights  \n\n\\[\n\\operatorname{Sym}^{d}(V):\\;(d,0,\\dots ,0),\\qquad  \nV\\otimes\\det^{k}:\\;(1+k,k,\\dots ,k).\n\\]\n\nEquality of highest weights forces $d=1+k$ and $k=0$.  Substituting in (11) gives $d=1$.  Therefore $d>1$ is impossible.\n\nStep 6.  The linear case $d=1$.  \nWith $d=1$ and $k=0$, (12) takes the simpler form\n\n\\[\nH(Av)=A\\,H(v)\\qquad(A\\in\\operatorname{GL}(V)).\n\\tag{14}\n\\]\n\nThus $H$ is $\\operatorname{GL}(V)$–equivariant for the standard representation, hence linear:\n\n\\[\nH(v)=L\\,v\\qquad(L\\in\\operatorname{End}(V)).\n\\tag{15}\n\\]\n\nEquation (14) implies $LA=AL$ for all $A\\in\\operatorname{GL}(V)$; therefore $L=c\\,\\operatorname{Id}_{V}$ with $c\\in\\mathbf C^{\\times}$.  Multiplying every $H_{i}$ by $c^{-1}$ does not change the projective map, so  \n\n\\[\nF=\\operatorname{Id}_{\\mathbf P^{n}}.\n\\]\n\nStep 7.  The one–dimensional case $n=1$ (verification).  \nWhen $n=1$ the above argument still applies.  Alternatively, $\\operatorname{PGL}_{2}(\\mathbf C)$ acts triply transitively on $\\mathbf P^{1}$; a morphism commuting with every projective automorphism fixes three distinct points and must therefore be the identity.\n\nConclusion.  \nFor every $n\\ge 1$ the only dominant rational self–map of $\\mathbf P^{n}(\\mathbf C)$ that commutes with every projective automorphism is\n\n\\[\nF=\\operatorname{Id}_{\\mathbf P^{n}}.\\qquad\\Box\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.434467",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-Dimensional & Projective Setting  \n • We moved from ℂⁿ to projective space ℙⁿ, introducing homogeneous coordinates, rational maps, and dominance issues.  \n • The automorphism group escalates from affine translations to the full algebraic group PGL_{n+1}(ℂ).\n\n2. Additional Structural Constraints  \n • Commutation is required with every element of an algebraic group of dimension (n+1)² – 1, not merely with all entire maps.\n\n3. Deeper Theory Needed  \n • Characters of GL_{n+1}, homogeneous lifting, and projective equivalence.  \n • Representation theory of GL_{n+1}: highest weights, symmetric powers, Schur’s lemma, vanishing of certain Hom-spaces.  \n • Algebraic group arguments to identify μ(A)=det(A)^{k}.  \n\n4. Multi-Step Proof  \n • Establishing a lifted functional equation with a scalar factor.  \n • Showing the scalar factor is a character and determining its exponent.  \n • Translating the problem into one about equivariant homomorphisms between representations and ruling them out for d>1.  \n • Final linear-algebra centralizer argument.\n\n5. Significantly Harder  \n • The original problem collapses immediately by plugging in constants.  \n • Here constants are unavailable; one must deploy algebraic-group and representation-theoretic machinery, execute weight computations, and control dominant rational maps on ℙⁿ.  \n • Each step is non-trivial and relies on advanced graduate-level material, making the new variant far more challenging than both the original and the previous kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}