ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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A two-dimensional Jordan-block extension does not split

Example

Fix $h \in C$ and let $V = C e_{1} \oplus C e_{2}$ . Define $N e_{1} = 0$ , $N e_{2} = e_{1}$ and $T = h I_{V} + N$ . Give $V$ the $C [t]$ -module action $f (t) \cdot v = f (T) v$ , where $f (T) = \sum_{j = 0}^{m} a_{j} T^{j}$ for $f (t) = \sum_{j = 0}^{m} a_{j} t^{j}$ . Give $K = C e_{1}$ and $C_{h} = C$ the scalar actions $f (t) \cdot a e_{1} = f (h) a e_{1}$ and $f (t) \cdot b = f (h) b$ , respectively. With $i : K \to V$ the inclusion and $q : V \to C_{h}$ given by $q (a e_{1} + b e_{2}) = b$ , the sequence $0 \to K \overset{i}{\to} V \overset{q}{\to} C_{h} \to 0$ is exact and does not split as a sequence of $C [t]$ -modules. Equivalently, $K$ has no $T$ -invariant vector-space complement in $V$ .

Facts & Assumptions

Given: The displayed vector spaces, operators, polynomial actions and linear maps. Exactness means that the maps are module homomorphisms, $i$ is injective, $q$ is surjective and $im i = \ker q$ . Splitting means that a module homomorphism $s : C_{h} \to V$ satisfies $q \circ s = I_{C_{h}}$ .

Verification

1.1givenalgebra

Polynomial evaluation defines a unital module action: $1 (T) = I_{V}$ , evaluation is complex-linear, and $(f g) (T) = f (T) g (T)$ because $T^{j} T^{k} = T^{j + k}$ and both polynomial sums are finite. Scalar evaluation at $h$ gives the same properties on $K$ and $C_{h}$ . Since $T e_{1} = h e_{1}$ , one has $f (T) e_{1} = f (h) e_{1}$ , so $K$ is invariant and $i$ is a module homomorphism.

2.1givenstep 1.1algebra

For $v = a e_{1} + b e_{2}$ one has $T v = (h a + b) e_{1} + h b e_{2}$ , hence $q (T v) = h q (v)$ . Iteration and linearity give $q (f (T) v) = f (h) q (v)$ , so $q$ is a module homomorphism. Inclusion is injective, $q (e_{2}) = 1$ proves surjectivity, and $q (v) = 0$ holds exactly when $b = 0$ , that is, when $v \in K = im i$ . Thus the sequence is exact.

3.1givenstep 2.1algebra

Every vector-space complement $U$ to $K$ is a line spanned by $u = a e_{1} + b e_{2}$ with $b \neq 0$ . If $U$ were $T$ -invariant, it would be invariant under $N = T - h I_{V}$ . But $N u = b e_{1}$ is a nonzero element of $K$ , whereas $N u \in U$ would force $N u \in K \cap U = 0$ . Hence there is no $T$ -invariant complement. Conversely, a $T$ -invariant complement is invariant under every polynomial in $T$ , and $q ∣_{U}$ would be a bijective module map; its inverse would split the sequence.

4.1givenstep 2.1step 3.1algebra∎

If a splitting $s$ existed, then $s (1) = a e_{1} + e_{2}$ because $q (s (1)) = 1$ . Module-linearity would imply $T s (1) = s (t \cdot 1) = h s (1)$ , while direct computation gives $(T - h I_{V}) s (1) = e_{1} \neq 0$ . This contradiction proves nonsplitting, consistently with the complement obstruction.

Remarks

This generated finite-dimensional example isolates an elementary obstruction from an actual invariant Jordan block. Constructing a module for a vertex algebra such as the triplet $W (p)$ additionally requires state fields, lower truncation, the vacuum axiom and the full module Jacobi identity, together with any claimed grading and exact-sequence identifications. None of those additional constructions follows from this example.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.