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Mixed Hodge Structures And Strictness: Examples

1 · Prerequisites

2 · Summary

This companion to mixed-hodge-structures-and-strictness works the smallest example that distinguishes a mixed Hodge structure from a direct sum of pure pieces. A mixed Hodge extension without a rational splitting takes H=Qe0⊕Qe1 with weight filtration concentrated in weights −2 and 0 and Hodge filtration F0=C(e0+ie1); the graded pieces are the Tate structures Q(1) and Q(0), both checks of the opposition axiom are carried out in the displayed basis, and the resulting short exact sequence 0→Q(1)→H→Q(0)→0 is shown to admit no rational splitting: a rational section sends the generator of Q(0) to e0+qe1 with q∈Q, while preservation of F0 forces q=i. The example is used by the rest of the track as the standard warning that the Deligne splitting is a decomposition over C only.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

A mixed Hodge extension without a rational splitting

Example

A rational mixed Hodge structure can be an extension of two pure structures and still admit no splitting over Q. On H=Qe0⊕Qe1 put W−3=0, W−2=W−1=Qe1, W0=H, and F1=0, F0=C(e0+ie1), F−1=HC. This is a mixed Hodge structure sitting in a short exact sequence 0→Q(1)→H→Q(0)→0 which has no splitting in the category of rational mixed Hodge structures.

Facts & Assumptions

Given: The rational vector space H=Qe0⊕Qe1 (Rational mixed Hodge structures), the increasing filtration W−3=0, W−2=W−1=Qe1, W0=H and the decreasing filtration F1HC=0, F0HC=C(e0+ie1), F−1HC=HC, with i∈C satisfying i2=−1 and conjugation σ(z)=zˉ (The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i).

[F1]

A rational mixed Hodge structure is the datum of a finite increasing W and a finite decreasing F such that on every Gr⁡nWHC the induced filtrations are n-opposed, i.e. split into the pieces Hnp,q with Hnp,q‾=Hnq,p; a morphism preserves W and F (Rational mixed Hodge structures).

[F2]

The Tate structure Q(r) is the pure structure of weight −2r with Q(r)C=C concentrated in type (−r,−r); in particular Q(1) has weight −2 and type (−1,−1), while Q(0) has weight 0 and type (0,0) (Rational mixed Hodge structures).

[F3]

On a graded piece Gr⁡nWHC the induced filtration F is the image of F∩Wn and Fˉ=σ(F); the summands are Hnp,q=FpGr⁡nW∩FˉqGr⁡nW (Rational mixed Hodge structures).

Verification

technique · direct
1.1givenF1F2F3algebra

The weight filtration is finite with Gr⁡−2W=Qe1, Gr⁡−1W=0 and Gr⁡0W=H/Qe1≅Qe0. On Gr⁡−2W the induced F has F0=0 and F−1=Gr⁡−2W, since F0∩W−2=C(e0+ie1)∩Ce1=0; the conjugate filtration satisfies Fˉ−1=Gr⁡−2W and Fˉ0=0 for the same reason, so the pieces are H−1,−1=Gr⁡−2W and nothing else, and H−1,−1 is spanned by the image of e1, which is rational. On Gr⁡0W the image of F0∩W0=C(e0+ie1) is all of Gr⁡0W because e0+ie1↦e0≠0, and F1=0; the conjugate Fˉ0=σ(F0)=C(e0−ie1) likewise has image all of Gr⁡0W, so the pieces are H0,0=Gr⁡0W and nothing else. Both graded pieces are therefore n-opposed, with the rational structure of e1 and e0 respectively, so (H,W,F) is a rational mixed Hodge structure; its graded pieces are Q(1) and Q(0) in the convention [F2].

2.1givenF1F2step 1.1algebra

Let ι ⁣:Q(1)→H be the rational inclusion 1↦e1, whose complexification is C→HC and let π ⁣:H→Q(0) be the Q-linear map e0↦1, e1↦0. Both preserve the filtrations: ι sends the weight-(−2) structure into W−2H=Qe1 and F−1Q(1)C=C into F−1HC=HC, while F0Q(1)C=0; and π satisfies π(W−1H)=0=W−1Q(0) and π(W0H)=Q(0)=W0Q(0), π(F0HC)⊆C=F0Q(0)C and π(F1HC)=0. Hence 0→Q(1)→ιH→πQ(0)→0 is a short exact sequence of rational mixed Hodge structures, with ker⁡π=Qe1.

3.1givenF1step 2.1algebra∎

Every Q-linear section s of π satisfies s(1)=e0+qe1 with q∈Q, because π(e0+qe1)=1 and π(e1)=0. If s were a morphism of mixed Hodge structures, it would satisfy sC(F0Q(0)C)⊆F0HC=C(e0+ie1); since F0Q(0)C=C and sC(1)=e0+qe1, there would be λ∈C with e0+qe1=λ(e0+ie1). Comparing the coefficients of e0 and e1 in the basis (e0,e1) gives λ=1 and q=i, which is impossible for q∈Q. Hence the extension does not split in the category of rational mixed Hodge structures. The obstruction is visible directly in the (0,0)-direction: F0HC∩Fˉ0HC=C(e0+ie1)∩C(e0−ie1)=0, while a rational section must send 1 into a rational vector of F0, which is then also fixed by conjugation and hence would have to lie in this zero intersection.

Source notes

Deligne II gives the theory, the Tate object and the morphism condition; the two-dimensional witness with F0=C(e0+ie1) was not located verbatim in the fetched texts, so the statement and verification are recorded as generated, with the definitional background cited. The closest statement-level analogue is the one-dimensional example on the punctured line, El Zein, Example 1.3. No choice principle is used: the only quantifiers run over elements of the displayed two-dimensional space.

Sources