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QC24 Projective Chart Dimension

Statement

For every integer D≥1, the standard patches Uj of QC24 Complex Projective Space and Its Standard Patches are smooth charts with coordinate space CD−1=R2D−2. If D≥2, each fixed patch is proper, so a single standard coordinate patch does not cover all of CPD−1.

Facts & Assumptions

Given: An integer D≥1, the quotient ray space CPD−1, standard patches Uj, and candidate coordinate maps ϕj and inverses from QC24 Complex Projective Space and Its Standard Patches.

[F1]

Uj={[z]:zj≠0} and ϕj([z])=(zi/zj)i≠j; the candidate inverse uses the representative with jth coordinate 1 (QC24 Complex Projective Space and Its Standard Patches).

Proof

Given: As in Facts & Assumptions.

1.1F1algebra

If [z]∈Uj, multiplying z by any nonzero complex scalar leaves every ratio zi/zj unchanged, so ϕj is well-defined on rays. The representative z/zj has jth coordinate 1 and all other coordinates equal to those ratios, so the stated inverse sends ϕj([z]) back to [z].

2.1F1step 1.1algebra

Conversely, applying ϕj to the inverse representative with jth coordinate 1 returns each entry of w; hence the two maps are inverse bijections. Let π be the quotient map and Oj={z:zj≠0}=π−1(Uj). Since Oj is open, Uj is open. The restriction πj:Oj→Uj is a quotient map: for B⊆Uj, if πj−1(B) is open in Oj, it is open in CD∖{0}, so B is open in the quotient and hence in Uj; the reverse implication follows from continuity of π. The composite ϕj∘πj consists of continuous coordinate ratios on Oj, so ϕj is continuous. The inverse is the quotient map composed with the continuous representative map w↦ej+∑i≠jwiei, hence is continuous.

3.1F1step 2.1algebra

Let π:CD∖{0}→CPD−1 be the quotient map. Then π−1(Uj)={z:zj≠0} is Euclidean open, so Uj is open by the quotient topology. On Uj∩Uk with j≠k, write w=ϕj([z]). Then wk≠0, the transition coordinates satisfy wj′=1/wk and wi′=wi/wk for i≠j,k. Conversely, wk=1/wj′ and wi=wi′/wj′ for i≠j,k. These rational maps are smooth as real maps on their open coordinate domains, so the charts are compatible and form a smooth atlas.

4.1F1algebrachoose∎

The patches cover every ray because a nonzero vector has at least one nonzero coordinate. If D≥2, choose k≠j; the ray of the basis vector ek has jth coordinate zero and is not in Uj, so each fixed standard patch is proper. Finally, CD−1 has D−1 complex, or 2D−2 real, coordinates.

Depends on

Used by

Cited to discharge well-definedness by QC24 Complex Projective Space and Its Standard Patches.

Dependency tree · two levels

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Sources