ScalingStacks

Verified tagged author-source HTML · 1904.03696v1 · cited publication edition alignment unverified.

00M2

Lemma 5.5. Assume that ϕ\phi is a Fubini-Study metric. For any ϵ>0\epsilon>0, there exists 𝜹∈ℝd+1\boldsymbol{\delta}\in\mathbb{R}^{d+1} with |𝜹|≤ϵ|\boldsymbol{\delta}|\leq\epsilon such that

∀n∈ℕ,dist⁡(∥⋅∥n​ϕ,∥⋅∥n​ϕ​(𝜹))≤n​ϵ.\forall n\in\mathbb{N},\quad\dist(\lVert\mathord{\cdot}\rVert_{n\phi},\lVert\mathord{\cdot}\rVert_{n\phi(\boldsymbol{\delta})})\leq n\epsilon.
00M3

Proof. Choose an arbitrary 𝜹\boldsymbol{\delta} with |𝜹|≤ϵ|\boldsymbol{\delta}|\leq\epsilon. Then

dist⁡(∥⋅∥ϕ,∥⋅∥ϕ⁡(𝜹))≤ϵ.\dist(\lVert\mathord{\cdot}\rVert_{\phi},\lVert\mathord{\cdot}\rVert_{\phi(\boldsymbol{\delta})})\leq\epsilon.

By Proposition 3.12, we have

dist⁡(FS⁡(∥⋅∥ϕ),FS⁡(∥⋅∥ϕ⁡(𝜹)))=dist⁡(FS⁡(∥⋅∥ϕ),ϕ⁡(𝜹))≤ϵ,\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\phi(\boldsymbol{\delta})}))=\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\phi}),\phi(\boldsymbol{\delta}))\leq\epsilon,

by the assumption and Proposition 3.11,

FS⁡(∥⋅∥ϕ)=ϕ,\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\phi})=\phi,

so the conlusion holds. ∎

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