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

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Different conjugations on $C^{2}$ can have different fixed real forms

Example

On $W = C^{2}$ define the coordinatewise conjugation

$σ_{0} (z_{1}, z_{2}) = ({\overline{z}}_{1}, {\overline{z}}_{2})$

and the transposed conjugation

$σ^{'} (z_{1}, z_{2}) = ({\overline{z}}_{2}, {\overline{z}}_{1}) .$

Their fixed real forms are $W^{σ_{0}} = R^{2}$ and $W^{σ^{'}} = {(w, \overline{w}) : w \in C}$ , a different real two-plane of $(C^{2})_{R}$ . One complex vector space therefore carries two different real forms, attached to two different choices of conjugation.

Facts & Assumptions

Given: The complex vector space $W = C^{2}$ and the two displayed maps $σ_{0}, σ^{'}$ .

[L1]

The fixed points of a conjugation form a real subspace whose complexification recovers the ambient complex space (The fixed points of a conjugation form a real vector space whose complexification recovers the ambient complex space).

[L2]

Real forms of a complex vector space correspond exactly to conjugations (Real forms of a complex vector space correspond exactly to conjugations).

Verification

technique · direct

1.1

The map $σ_{0}$ is a conjugation: it is additive, $σ_{0} (λ z) = (\overline{λ z_{1}}, \overline{λ z_{2}}) = \overline{λ} σ_{0} (z)$ , and applying it twice is the identity.

algebra

1.2

The map $σ^{'}$ is also a conjugation: additivity is clear, $σ^{'} (λ z) = (\overline{λ z_{2}}, \overline{λ z_{1}}) = \overline{λ} σ^{'} (z)$ , and $σ^{'} σ^{'} (z) = (z_{1}, z_{2})$ .

algebra

2.1

The fixed set of $σ_{0}$ is ${(z_{1}, z_{2}) : {\overline{z}}_{1} = z_{1}, {\overline{z}}_{2} = z_{2}} = R^{2}$ , the real coordinate plane inside $(C^{2})_{R}$ .

step 1.1algebra

2.2

The fixed set of $σ^{'}$ is ${(z_{1}, z_{2}) : z_{1} = {\overline{z}}_{2}} = {(w, \overline{w}) : w \in C} = {(a + b i, a - b i) : a, b \in R}$ , a real two-plane.

step 1.2algebra

3.1

The two fixed sets are different real subspaces: $(1, 0)$ is fixed by $σ_{0}$ but $σ^{'} (1, 0) = (0, 1) \neq (1, 0)$ . By [L1] each is a real form whose complexification recovers $C^{2}$ , and by [L2] the distinct conjugations give distinct real forms.

step 2.1step 2.2L1L2

4.1

Steps 1.1 through 3.1 exhibit two different conjugations and two different fixed real forms on one complex vector space.

step 2.1step 2.2step 3.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

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

Different conjugations on C2C2 can have different fixed real forms