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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Isolated complex meridian disks inside the product of the dihedral cone

Statement

Let $I = {a > 0}$ and $H_{k} = {b + k a = 0}$ be as in The infinite dihedral Tits cone is an open half-plane with its vertex. Identify the product complement of Complexified complements in an open convex cone with $M = (I + i I) ∖ ⋃_{k} H_{k}^{C}$ . For every integer $k$ there is a closed disk $D_{k} \subset I + i I$ meeting $H_{k}^{C}$ only at its centre and avoiding every $H_{ℓ}^{C}$ for $ℓ \neq k$ . Its positively oriented boundary $γ_{k}$ lies in $M$ , and $L_{k} (γ_{k} (θ)) = ε e^{i θ}$ , where $L_{k} = b + k a$ and $ε > 0$ .

Facts & Assumptions

[F1]

Given: The cone and root hyperplanes of The infinite dihedral Tits cone is an open half-plane with its vertex, with the product topology of Complexified complements in an open convex cone. Goal: Construct disks without replacing $I + i I$ by a larger tube domain.

Proof

technique · direct

1.1F1algebra

Extend $a, b$ complex linearly. Choose the centre $z_{k}$ with $a (z_{k}) = 1 + i$ and $b (z_{k}) = - k (1 + i)$ . Both its real and imaginary parts belong to $I$ . Choose a real vector $v$ with $a (v) = 0$ and $b (v) = 1$ , and set $D_{k} = {z_{k} + ζ v : ∣ ζ ∣ \leq ε}$ for $0 < ε < 1$ . All its points have real and imaginary $a$ -coordinate equal to $1$ , so the whole disk lies in $I + i I$ .

2.1step 1.1algebra∎

On this disk $L_{ℓ} = (ℓ - k) (1 + i) + ζ$ . For $ℓ \neq k$ the first summand has modulus at least $\sqrt{2} > ε$ , so $L_{ℓ}$ never vanishes. For $ℓ = k$ it vanishes exactly at $ζ = 0$ . The boundary parametrisation $ζ = ε e^{i θ}$ has the required properties.

Depends on

Used by · 2

Dependency tree · 2 levels

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