CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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A complex-linear map need not preserve a chosen real form

Statement refuted

Every complex-linear operator on a complex vector space preserves every chosen real form; equivalently, a complex-linear operator always descends to the fixed real form of a given conjugation.

Facts & Assumptions

Given: The conjugation $σ_{0}$ on $C^{2}$ , its fixed real form $V = R^{2}$ , and the displayed operator $T$ .

[L1]

The fixed real form of a conjugation is the real subspace of its fixed points (The fixed real form of a conjugation).

[L2]

A complex-linear operator comes from a real operator on the fixed real form exactly when it commutes with the chosen conjugation (A complex-linear operator comes from a real operator exactly when it commutes with the chosen conjugation).

Counterexample

Take $W = C^{2}$ with the coordinatewise conjugation $σ_{0} (z_{1}, z_{2}) = ({\overline{z}}_{1}, {\overline{z}}_{2})$ , whose fixed real form is $R^{2}$ . The complex-linear operator

$T (z_{1}, z_{2}) = (i z_{1}, z_{2})$

does not commute with $σ_{0}$ : at $w = (1, 0)$ one has $T σ_{0} (w) = (i, 0)$ while $σ_{0} T (w) = (- i, 0)$ . Consequently $T$ does not come from a real operator on $R^{2}$ , and $T$ does not even carry $R^{2}$ into itself, since $T (1, 0) = (i, 0) \notin R^{2}$ .

Proof technique: direct.

1.1

The map $σ_{0}$ is a conjugation, and by [L1] its fixed real form is ${(z_{1}, z_{2}) : {\overline{z}}_{1} = z_{1}, {\overline{z}}_{2} = z_{2}} = R^{2}$ .

L1algebra

1.2

The map $T$ is complex-linear: $T (λ z_{1}, λ z_{2}) = (i λ z_{1}, λ z_{2}) = λ T (z_{1}, z_{2})$ .

algebra

1.3

The two maps do not commute: $T σ_{0} (1, 0) = T (1, 0) = (i, 0)$ , while $σ_{0} T (1, 0) = σ_{0} (i, 0) = (- i, 0)$ .

algebra

2.1

By [L2], $T$ does not come from any real operator on $R^{2}$ ; concretely $T (1, 0) = (i, 0) \notin R^{2}$ , so $T$ does not even preserve the chosen real form as a set.

L2step 1.3

3.1

Steps 1.2, 1.3 and 2.1 refute the claimed universality: a complex-linear map can fail to preserve a chosen real form.

step 1.2step 1.3step 2.1∎

Depends on

Used by

FALSE: every complex-linear operator descends to every chosen real form False statement

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Mikhail Troshkin, Real-complex linear algebra and abelian varieties (standard reference, not scraped)