Exercise PLAN-106

The unrestricted algebraic abelian-irrep claim fails on C(t). In nonzero finite complex dimension, commuting action operators are scalar by eigenvalue-based Schur, so an invariant line is the whole irrep; real and zero/infinite-dimensional boundaries remain explicit.

1. Abelianness makes every action operator commute with the action, but without finite dimension this does not imply scalarity; the C(t)^times multiplication representation is a genuine irreducible counterexample.
2. In nonzero finite-dimensional complex V, the linear operator phi(h) has an eigenvalue, and the invariant-kernel Schur argument gives phi(h)=lambda_h I for every h.
3. Consequently the line through any nonzero v is G-invariant, and the nonzero scalar coefficients form a multiplicative character.
4. Irreducibility forces that line to be V, so dim V=1; the zero-carrier case has dimension0 and is not hidden inside this positive theorem.
5. The real C4 quarter-turn representation is irreducible of dimension2, showing a field boundary; the algebraic infinite-dimensional and continuous/unitary settings remain distinct from the stated positive theorem.

The unrestricted assertion is false as an algebraic theorem. Put K=C(t), the field of rational functions in an indeterminate t, G=K^times, and V=K as a COMPLEX vector space. G is abelian. Multiplication phi(g)v=gv is complex-linear, invertible with inverse multiplication by g^{-1}, and respects multiplication and the identity, so it defines a representation. The powers 1,t,t^2,... are complex-linearly independent: a finite relation among them would make a nonzero complex polynomial in the indeterminate vanish. Thus V has infinite complex dimension. If a nonzero invariant complex-linear subspace W contains w!=0, then for every v!=0, the group element g=v/w sends w to v. Hence W=V, proving actual algebraic irreducibility. This is not a reducible infinite-dimensional trivial representation.

Now assume the revised positive hypotheses, in particular 0<dim_C V<infinity. Fix h in G. Abelianness gives phi(g)phi(h)=phi(gh)=phi(hg)=phi(h)phi(g), so the LINEAR operator phi(h) commutes with every action map. Scalar Schur applies in the stated finite-dimensional complex setting. Explicitly, positive finite dimension makes det(sI-phi(h)) a polynomial of positive degree; C being algebraically closed gives a root lambda_h and a nonzero eigenvector. The linear map phi(h)-lambda_h I has a nonzero kernel, which commutation makes G-invariant. Irreducibility forces that kernel to be all of V, so phi(h)=lambda_h I. This is the eigenvalue/kernel reason for scalarity, not an averaging or simultaneous-diagonalization assertion with hidden hypotheses.

For any nonzero v, the line Cv is invariant under EVERY h, because phi(h)v=lambda_h v. Algebraic irreducibility now gives V=Cv, proving dim_C V=1. Since each phi(h) is invertible and V!=0, lambda_h!=0. Applying the representation identities to a nonzero vector yields lambda_{gh}=lambda_g lambda_h, lambda_e=1 and lambda_{g^{-1}}=lambda_g^{-1}; thus h->lambda_h is a multiplicative character into C^times. Conversely such a character acts on C by invertible scalar multiplication, respects the group laws, and is irreducible because a nonzero one-dimensional complex vector space has only zero and itself as linear subspaces. No assertion of continuity of an arbitrary algebraic character is implicit.

The real boundary can be made finite and exact. Let G=C_4 act on R^2 by powers of J=[[0,-1],[1,0]]. Then J^4=I and J is orthogonal. A proper nonzero invariant real subspace would be a line Rv, so Jv=a v for a real number a; applying J twice gives -v=a^2 v, impossible for v!=0. The real representation is therefore irreducible and two-dimensional. J commutes with the action but has no real eigenvalue and is not a real scalar. A quarter-turn, not an identity or half-turn rotation, is the needed witness. It does not contradict the complex theorem: over C the same matrix has eigenlines with eigenvalues i and -i.

If V=0 is admitted by a purely invariant-subspace definition, its unique action is vacuously irreducible by that predicate but dim V=0, not 1. The positive statement expressly requires V!=0. There is no nonzero vector, eigenvector or invariant-line argument on the zero carrier; if one's irrep convention already excludes zero, this case simply lies outside that convention.

Finiteness or compactness of G is unnecessary for the positive FINITE-DIMENSIONAL proof. It uses no group averaging, inner product, unitarity or topology and no group-order/characteristic-zero division. The same eigenvalue argument works over any algebraically closed field with a finite positive-dimensional carrier. Algebraic closure can be replaced, for this particular representation, by existence of an eigenvalue in the base field for every action map; it cannot simply be omitted uniformly. The C(t) example explains the dimension boundary: multiplication by t has no complex eigenvalue, since for each lambda in C its difference from lambda I is invertible by multiplication with 1/(t-lambda). No compact, continuous, Hilbert-space or bounded-operator structure is asserted for that exterior example. It does not refute topological/unitary Schur theorems with their own hypotheses. Plan105 supplies scalar Schur for linear endomorphisms; the extra step here is applying it to abelian GROUP ACTION operators and classifying their invariant lines. Shared source/counterexample mechanisms are dependencies, not proof of whole-question equivalence.

1. Explicitly refutes the unrestricted algebraic assertion by a genuine infinite-dimensional complex irreducible action of an abelian group, proving group action/linearity, infinite dimension and irreducibility rather than assuming them. (1 weight)
2. For the revised nonzero finite-dimensional complex theorem, uses abelianness to place each phi(h) in the linear commutant and justifies scalar Schur with the positive-dimensional/eigenvalue/algebraic-closure hypotheses actually present. (1 weight)
3. Shows every nonzero vector spans a G-invariant line, uses irreducibility to prove dimension one, and verifies the nonzero scalars form a multiplicative character and the converse one-dimensional action is irreducible. (1 weight)
4. Provides the precise real C4 quarter-turn irreducibility counterexample and separates the zero carrier, algebraic infinite-dimensional boundary and continuous/unitary variants. Does not add finite G, averaging or unitarity premises or claim a plan105 alias, a gate/resource theorem or admission. (1 weight)

No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.

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