Exercise PLAN-105

Functional equivariance is insufficient even for entire maps on finite unitary irreps. For explicitly linear endomorphisms, positive finite dimension and algebraic closure supply an eigenvalue; the shifted linear Schur lemma then gives the scalar commutant, with zero-space and exterior-field/dimension limits explicit.

1. The nonlinear complex sign-irrep example refutes the functional-only claim; for an explicitly linear endomorphism on nonzero finite-dimensional V over algebraically closed F, the characteristic polynomial supplies an eigenvalue, while V=0 is handled separately.
2. The shifted map h=f-lambda I is F-linear and equivariant because linear action maps commute with scalars.
3. The nonzero eigenvector lies in ker(h), so h is not invertible; a nonlinear function would not justify this endomorphism/eigenvalue argument.
4. The shared linear kernel/image Schur lemma forces h=0 and f=lambda I; finite/closedness supplies the eigenvalue rather than an invisible convention.
5. Inside End_F(V), all commuting X and all scalar maps give exactly F*I, with uniqueness only on nonzero V and genuine outside-hypothesis field/dimension boundaries; this is not a classification of all equivariant functions or all topological variants.

The literal claim is false without F-linearity even under the stated field and dimension conditions. Use G={+1,-1}, V=C and phi(epsilon)z=epsilon*z. This nonzero one-dimensional complex representation is irreducible and unitary; the finite discrete group is compact and acts continuously. The entire map f(z)=z^3 satisfies f(epsilon*z)=epsilon*f(z) for each group element. It is not a scalar multiple of the identity: f(1)=1 would force lambda=1, but f(2)=8!=2. It is also noninjective, since1 and omega=(-1+i*sqrt(3))/2 are distinct and both cube to1. This is the same concrete functional-equivariance failure used in104, not a different promise. Holomorphicity or continuity does not make a function C-linear. A nonlinear function has no endomorphism matrix or characteristic polynomial in End_F(V), so the old eigenvalue proof cannot even start on that carrier. Likewise, a commutant taken among all functions is not the linear-operator commutant below.

Now impose the conditional hypotheses actually stated in the revised question: f is F-linear, V is finite-dimensional over algebraically closed F, and the representation is algebraically irreducible. If V=0, its only linear endomorphism is0 and lambda*I is0 for every lambda; the conclusion holds directly and there is no nonzero eigenvector. If irreducibility by definition excludes the zero space, this branch is outside that convention. In the nonzero case n=dim_F(V)>0. In any basis f has an n-by-n matrix, and chi_f(t)=det(tI-f) is a polynomial over F of degree n. Algebraic closure gives a root lambda in F. Since det(lambda I-f)=0, finite-dimensional linear algebra gives a nonzero v with (f-lambda I)v=0. Finite positive dimension supplies the nonconstant characteristic polynomial and the matrix kernel criterion; algebraic closure supplies its root. Neither follows for an arbitrary nonlinear function.

Set h=f-lambda I. It is F-linear because both f and I are. For every g, F-linearity of phi(g) and scalar commutation give h phi(g)=f phi(g)-lambda phi(g)=phi(g)f-phi(g)lambda I=phi(g)h. Thus h is a linear intertwiner, not just a function satisfying an unexplained formal subtraction. The eigenvector v is a nonzero kernel vector, so h cannot be injective or a representation isomorphism.

The linear kernel/image lemma now applies. Explicitly, ker(h) and im(h) are linear subspaces by linearity, and the displayed intertwining relation makes them G-invariant. Since ker(h) contains a nonzero vector, irreducibility forces ker(h)=V. Hence h=0 and f=lambda I. Equivalently, the zero-or-isomorphism lemma says h is zero or a linear isomorphism and the latter is excluded. This is precisely the104 subproof used after a new eigenvalue/shift step;104 by itself does not imply every endomorphism is scalar over every field.

For the corollary define End_G(V) INSIDE End_F(V). Any X in that set is linear and equivariant under the same finite-dimensional/algebraically closed hypotheses, so the preceding argument gives X=lambda I. Conversely, for every lambda in F and every g, phi(g)(lambda v)=lambda phi(g)v, so lambda I commutes and belongs to End_G(V). Thus the sets are equal. If V is nonzero, (lambda-mu)I=0 implies lambda=mu by evaluation at a nonzero vector, so the scalar is unique and F*I is a one-dimensional operator space. At V0 the set equality is still true but all scalars give the same zero operator; no one-dimensionality or unique-scalar claim is made there.

The root argument actually needs only that this particular linear f have an eigenvalue: if f-lambda I has a nonzero kernel, algebraic irreducibility already forces it to vanish. Finite dimension and algebraic closure are general sufficient hypotheses guaranteeing that property, not necessary conditions for each individual operator to be scalar. No averaging over G, finite-group premise, characteristic-zero restriction, inner product or unitarity enters the proof.

There are genuine boundaries to deleting those sufficient hypotheses wholesale. Over R take G=C_4 acting on R^2 by powers of J=[[0,-1],[1,0]]. A nonzero invariant real line would give Jv=a v with real a, and then -v=J^2v=a^2v, impossible. The representation is irreducible and orthogonal, yet J commutes with every action matrix and is not a real scalar. This shows why algebraic closure cannot be dropped from a universal scalar statement merely because the maps are linear.

For an algebraic infinite-dimensional boundary over C, let K=C(t), V=K as a C-vector space, and G=K^times act by multiplication. Powers1,t,t^2,... are C-linearly independent, so V is infinite-dimensional. Every nonzero invariant C-linear subspace W contains some w!=0; for any v!=0, multiplication by the group element v/w sends w to v, so W=V. Thus this is an algebraically irreducible representation. Multiplication T(v)=t*v is C-linear and commutes with G, but is not lambda I for any complex lambda. Indeed T-lambda I is invertible by multiplication with1/(t-lambda), so T has no complex eigenvalue. This is a genuine irreducible counterexample outside finite dimension, not a shift on a reducible trivial representation. No compactness, Hilbert-space boundedness or continuity hypothesis is asserted for this exterior example, and it does not refute differently hypothesized topological/unitary Schur theorems. The finite-group nonlinear examples already meet the continuity/finite-dimensional conditions of the actual literal defect.

1. Refutes the literal function-only claim by a genuine nonlinear continuous/holomorphic finite-unitary-irrep example; then under the QUESTION'S explicit F-linearity/finite-dimensional/algebraically-closed conditions produces an eigenvalue only for V!=0 and treats V=0 separately if admitted. (1 weight)
2. Shows h=f-lambda I is an F-linear intertwiner by actual scalar/linear commutation, not merely a formally subtracted nonlinear function. (1 weight)
3. Uses a nonzero eigenvector to give nonzero kernel, applies the linear invariant kernel/image or zero-or-isomorphism lemma correctly, and concludes h=0 and f=lambda I with no missing morphism premise. (1 weight)
4. Defines the commutant inside End_F(V), proves both inclusions and the correct uniqueness/zero-space boundary; identifies the weaker eigenvalue condition and genuine field/dimension boundary examples without claiming all infinite/topological variants fail. Distinguishes105's extra eigenvalue target from shared104 subproof. (1 weight)

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