note-math

为了处理 $ℝ^{1, 𝑛}$ 的幂级数, 需要先处理值域 $ℝ^{1, 𝑛}$ 的级数

暂时只处理类时未来的情况

sum-preserve-time-future_(tag) let $𝑣, 𝑤$ 类时未来, then $𝑣 + 𝑤$ 类时未来

${⟨ 𝑣 + 𝑤 ⟩}^{2} = {⟨ 𝑣 ⟩}^{2} + {⟨ 𝑤 ⟩}^{2} + 2 ⟨ 𝑣, 𝑤 ⟩$

$⟨ 𝑣, 𝑤 ⟩ > 0$

$| 𝑣 + 𝑤 | \geq | 𝑣 | + | 𝑤 |$ (#link(<quadratic-form-inequality-Minkowski>)[三角不等式])

$| 𝑣 + 𝑤 | > | 𝑣 |, | 𝑤 |$

let $𝑥_{𝑛}$ 类时未来 or $0$

then $𝑥_{1} + \dots + 𝑥_{𝑛}$ 类时未来 or $0$

$| 𝑥_{1} + \dots + 𝑥_{𝑛} | \geq | 𝑥_{1} | + \dots + | 𝑥_{𝑛} |$

$| 𝑥_{1} + \dots + 𝑥_{𝑛} | \geq | 𝑥_{1} + \dots + 𝑥_{𝑛 - 1} |$ , or $𝑛 ⇝ | 𝑥_{1} + \dots + 𝑥_{𝑛} | \geq 0$ 单调递增 increasing-length-of-time-future-series_(tag)

let $𝑣, 𝑤$ time future, not co-linear. $span (𝑣, 𝑤) ≃ ℝ^{1, 1}$ cf. #link(<signature-of-2d-subspace-of-spacetime>)[]

通过 $SO (1, 𝑛)$ 变换, 可以假设测地线球的中心是 $(0, \dots, 0, 1)$

sum-preserve-angle-range_(tag)

let $𝑣, 𝑤$ 的方向 $\frac{𝑣}{| 𝑣 |}, \frac{𝑤}{| 𝑤 |}$ 包含在 $ℍ 𝕪$ 的测地线球 $exph ([- 𝑅, 𝑅] i_{split})$

==> $𝑣 + 𝑤$ 的方向 $\frac{𝑣 + 𝑤}{| 𝑣 + 𝑤 |}$ 在测地线球 $exph ([- 𝑅, 𝑅] i_{split})$

Proof

Question 有更直接的证明吗?

let

$\begin{matrix} 𝑣 = | 𝑣 | exph (ϕ i_{split}) \\ 𝑤 = | 𝑤 | exph (𝜓 i_{split}) \end{matrix}$

let $| ϕ |, | 𝜓 | \leq 𝑅$

双曲线的测地线长度 $𝑟$ 是双射. 双曲线映射到到空间轴是单射. 复合后是 $sinh 𝑟$ 保持单调的双射

\begin{matrix} argh (\frac{𝑣 + 𝑤}{| 𝑣 + 𝑤 |}) & \to & im \frac{𝑣 + 𝑤}{| 𝑣 + 𝑤 |} & (im of ℂ_{split}) \\ = & \frac{| 𝑣 | sinh ϕ + | 𝑤 | sinh 𝜓}{{(| 𝑣 |^{2} + | 𝑤 |^{2} + 2 | 𝑣 | | 𝑤 | cosh (𝜓 - ϕ))}^{\frac{1}{2}}} & cosh \geq 1 \\ \leq & \frac{| 𝑣 | sinh 𝑅 + | 𝑤 | sinh 𝑅}{{(| 𝑣 |^{2} + | 𝑤 |^{2} + 2 | 𝑣 | | 𝑤 | cosh (𝑅 - 𝑅))}^{\frac{1}{2}}} \\ = & sinh 𝑅 \\ | argh (\frac{𝑣 + 𝑤}{| 𝑣 + 𝑤 |}) | & \leq & 𝑅 \end{matrix}

let $𝑥_{𝑛}$ 方向 $\frac{𝑥_{𝑛}}{| 𝑥_{𝑛} |}$ in ${ℍ 𝕪}^{𝑛} (time, future)$ 测地线球 $𝔹$

==> $𝑥_{1} + \dots + 𝑥_{𝑛} \in 𝔹$

Proof use $span (𝑥_{1} + \dots + 𝑥_{𝑛 - 1}, 𝑥_{𝑛})$ signature $1, 1$ , 嵌入 $ℍ 𝕪$ , 归纳

quadratic-form-inequality-Minkowski-another_(tag)

let $𝑣, 𝑤$ 的方向 in $ℍ 𝕪$ 测地线球 $exph ([- 𝑅, 𝑅] i_{split})$

==> $⟨ 𝑣, 𝑤 ⟩ \leq | 𝑣 | | 𝑤 | cosh (2 𝑅)$

Proof use $⟨ 𝑣, 𝑤 ⟩ = | 𝑣 | | 𝑤 | cosh (𝜓 - ϕ)$ cf. #link(<hyperbolic-cosine-formula>)[]

let $𝑥_{𝑛}$ time future, 方向 $\frac{𝑥_{𝑛}}{| 𝑥_{𝑛} |}$ 在 ${ℍ 𝕪}^{𝑛} (time, future)$ 测地线球 $𝔹$ with 半径 $𝑅$

\begin{matrix} | 𝑥_{1} + \dots + 𝑥_{𝑛} |^{2} & = & | 𝑥_{1} |^{2} + \dots + | 𝑥_{𝑛} |^{2} + 2 \sum_{1 \leq 𝑖 < 𝑗 \leq 𝑛} ⟨ 𝑥_{𝑖}, 𝑥_{𝑗} ⟩ \\ \leq & | 𝑥_{1} |^{2} + \dots + | 𝑥_{𝑛} |^{2} + 2 cosh (2 𝑅) \sum_{1 \leq 𝑖 < 𝑗 \leq 𝑛} | 𝑥_{𝑖} | | 𝑥_{𝑗} | \end{matrix}

==> use $cosh \geq 1$

\begin{matrix} | 𝑥_{1} + \dots + 𝑥_{𝑛} |^{2} & \leq & cosh (2 𝑅) {(| 𝑥_{1} | + \dots + | 𝑥_{𝑛} |)}^{2} \end{matrix}

absolute-convergence-Minkowski-distance_(tag) $\sum_{𝑖 = 1}^{\infty} | 𝑥_{𝑖} | < \infty$ ==> #link(<increasing-length-of-time-future-series>)[单调递增] 有界 $lim_{𝑛 \to \infty} | 𝑥_{1} + \dots + 𝑥_{𝑛} | < \infty$ 存在极限

Minkowski-power-series_(tag)

let $\sum_{𝑖 = 1}^{\infty} | 𝑥_{𝑖} | < \infty$

距离 $| 𝑦 | = lim_{𝑛 \to \infty} | 𝑥_{1} + \dots + 𝑥_{𝑛} |$ 极限存在 (上一个定理)
方向 $\frac{𝑦}{| 𝑦 |} = lim_{𝑛 \to \infty} \frac{𝑥_{1} + \dots + 𝑥_{𝑛}}{| 𝑥_{1} + \dots + 𝑥_{𝑛} |} \in 𝔹$ , 极限存在待证明

have property

$𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛})$ time future
$lim_{𝑛 \to \infty} | 𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛}) | = 0$

称为 Minkowski 级数收敛 $\sum_{𝑖 = 1}^{\infty} 𝑥_{𝑖} = 𝑦$

Proof of convergence in direction space

方向 $\frac{𝑥_{1} + \dots + 𝑥_{𝑛}}{| 𝑥_{1} + \dots + 𝑥_{𝑛} |}$ 收敛

Question 有无更直接的证明

use #link(<isom-top-hyperbolic-Euclidean>)[]. $ℝ^{1, 𝑛}$ 距离限制在 ${ℍ 𝕪}^{𝑛} (time, future)$ 等价于作为 Riemman manifold 的测地线距离, ${ℍ 𝕪}^{𝑛}$ 两个元素相减是 $ℝ^{1, 𝑛}$ 类空

we prove $\frac{𝑥_{1} + \dots + 𝑥_{𝑛}}{| 𝑥_{1} + \dots + 𝑥_{𝑛} |}$ Cauchy in ${ℍ 𝕪}^{𝑛} (time, future)$

$lim_{𝑛 \to \infty} | 𝑥_{1} + \dots + 𝑥_{𝑛} | < \infty$ ==> $| 𝑥_{1} + \dots + 𝑥_{𝑛} |$ 是 Cauchy 序列

==> all $𝜀 > 0$ , exist $𝑛 \in ℕ$ , all $𝑚 \in ℕ$
$| 𝑥_{1} + \dots + 𝑥_{𝑛 + 𝑚} | - | 𝑥_{1} + \dots + 𝑥_{𝑛} | < 𝜀$

let

$𝑣 = 𝑥_{1} + \dots + 𝑥_{𝑛}, 𝑤 = 𝑥_{1} + \dots + 𝑥_{𝑛 + 𝑚}$ . use ${ℍ 𝕪}^{𝑛}$ 两点相减是 $ℝ^{1, 𝑛}$ 类空

\begin{matrix} 0 & \leq & - {⟨ \frac{𝑣}{| 𝑣 |} - \frac{𝑤}{| 𝑤 |} ⟩}^{2} \\ = & - 2 (1 - ⟨ \frac{𝑣}{| 𝑣 |}, \frac{𝑤}{| 𝑤 |} ⟩) \\ = & 2 (\frac{⟨ 𝑣, 𝑤 ⟩ - | 𝑣 | | 𝑤 |}{| 𝑣 | | 𝑤 |}) \end{matrix}

use $\frac{1}{| 𝑣 | | 𝑤 |} = \frac{1}{| 𝑥_{1} + \dots + 𝑥_{𝑛 + 𝑚} | | 𝑥_{1} + \dots + 𝑥_{𝑛} |} \leq \frac{1}{| 𝑥_{1} |^{2}}$ (or other) 不影响极限 $\underset{𝑛 \to \infty}{\to} 0$

use #link(<sum-preserve-time-future>)[]. $𝑤 - 𝑣$ 类时, $| 𝑤 | > | 𝑣 |$ . use #link(<triangel-inequality-Minkowski>)[三角不等式] $| 𝑤 | \geq | 𝑤 - 𝑣 | + | 𝑣 |$ , use Cauchy $| 𝑤 | - | 𝑣 | < 𝜀$

==> $| 𝑤 - 𝑣 | \leq | 𝑤 | - | 𝑣 | < 𝜀$

==> $| 𝑤 - 𝑣 |^{2} \leq {(| 𝑤 | - | 𝑣 |)}^{2} < 𝜀^{2}$

==> $0 \leq {(| 𝑤 | - | 𝑣 |)}^{2} - | 𝑤 - 𝑣 |^{2} < 𝜀^{2}$

==> $2 (⟨ 𝑣, 𝑤 ⟩ - | 𝑣 | | 𝑤 |) < 𝜀^{2}$

==> $\frac{𝑥_{1} + \dots + 𝑥_{𝑛}}{| 𝑥_{1} + \dots + 𝑥_{𝑛} |}$ Cauchy in ${ℍ 𝕪}^{𝑛} (time, future)$

内积收敛

let $𝑦 = | 𝑦 | exph ϕ$
let $𝑧 = | 𝑧 | exph 𝜓$ time future
for $𝑦_{𝑛} \to 𝑦$ use 双曲余弦

\begin{matrix} ⟨ 𝑦, 𝑧 ⟩ & = & | 𝑦 | | 𝑧 | cosh (| ϕ - 𝜓 |) \\ = & lim_{𝑛 \to \infty} | 𝑦_{𝑛} | | 𝑧 | cosh (| ϕ_{𝑛} - 𝜓 |) & distance | 𝑦_{𝑛} | \land direction ϕ_{𝑛} \\ = & lim_{𝑛 \to \infty} ⟨ 𝑦_{𝑛}, 𝑧 ⟩ \end{matrix}

$𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛})$ 类时

\begin{matrix} {⟨ 𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛}) ⟩}^{2} & = & {⟨ 𝑦 - 𝑦_{𝑛} ⟩}^{2} \\ = & 𝑦^{2} + 𝑦_{𝑛}^{2} - 2 ⟨ 𝑦, 𝑦_{𝑛} ⟩ \\ = & lim_{𝑚 \to \infty} (𝑦_{𝑛 + 𝑚}^{2} + 𝑦_{𝑛}^{2} - 2 ⟨ 𝑦_{𝑛 + 𝑚}, 𝑦_{𝑛} ⟩) \\ = & lim_{𝑚 \to \infty} {⟨ 𝑦_{𝑛 + 𝑚} - 𝑦_{𝑛} ⟩}^{2} \end{matrix}

where ${⟨ 𝑦_{𝑛 + 𝑚} - 𝑦_{𝑛} ⟩}^{2} = {⟨ 𝑥_{𝑛 + 1} + \dots + 𝑥_{𝑛 + 𝑚} ⟩}^{2} \geq 0$

$𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛})$ 未来

\begin{matrix} ⟨ 𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛}), 𝑥_{1} + \dots + 𝑥_{𝑛} ⟩ & = & ⟨ 𝑦, 𝑦_{𝑛} ⟩ - ⟨ 𝑦_{𝑛}, 𝑦_{𝑛} ⟩ \\ = & (lim_{𝑚 \to \infty} ⟨ 𝑦_{𝑛 + 𝑚}, 𝑦_{𝑛} ⟩) - ⟨ 𝑦_{𝑛}, 𝑦_{𝑛} ⟩ \\ = & lim_{𝑚 \to \infty} ⟨ 𝑦_{𝑛 + 𝑚} - 𝑦_{𝑛}, 𝑦_{𝑛} ⟩ \end{matrix}

where $⟨ 𝑦_{𝑛 + 𝑚} - 𝑦_{𝑛}, 𝑦_{𝑛} ⟩ = ⟨ 𝑥_{𝑛 + 1} + \dots + 𝑥_{𝑛 + 𝑚}, 𝑥_{1} + \dots + 𝑥_{𝑛} ⟩ \geq 0$

\begin{matrix} lim_{𝑛 \to \infty} {⟨ 𝑦 - (𝑥_{1} + \dots + 𝑥_{𝑛}) ⟩}^{2} & = & lim_{𝑛 \to \infty} (𝑦^{2} + 𝑦_{𝑛}^{2} - 2 ⟨ 𝑦, 𝑦_{𝑛} ⟩) \\ = & 0 \end{matrix}

处理幂级数

let $𝐴_{𝑛} \in Lin (⊙^{𝑛} ℝ^{1, 3} \to ℝ^{1, 3})$

Minkowski-analytic_(tag) Minkowski 解析

$\sum_{𝑛 = 1}^{\infty} 𝐴_{𝑛} (𝑣^{𝑛})$ (零阶不影响)

let 测地线坐标 $argh : {ℍ 𝕪}^{𝑛} \to ℝ^{𝑛}$

目标的 $𝐴_{𝑛}$ property

$𝑣$ time future ==> $𝐴_{𝑛} (𝑣^{𝑛})$ time future
定义范数给出绝对收敛

use 几何级数控制 like #link(<linear-map-induced-norm>)[the case of Euclidean]

\begin{matrix} | 𝐴_{𝑛} | (𝑅) & ≔ & \sup_{\begin{matrix} 𝑣 \in {ℍ 𝕪}^{𝑛} \\ | argh (𝑣) | \leq 𝑅 \end{matrix}} | 𝐴_{𝑛} (𝑣^{𝑛}) | = \sup_{| argh (𝑣) | \leq 𝑅} \frac{| 𝐴_{𝑛} (𝑣^{𝑛}) |}{| 𝑣 |^{𝑛}} \\ 𝜌 (𝑅) & ≔ & \underset{𝑛 \in ℕ}{lim sup} {| 𝐴_{𝑛} | {(𝑅)}^{\frac{1}{𝑛}}}^{- 1} \end{matrix}

==> $| argh (𝑣) | < 𝑅$ and $| 𝑣 | < 𝜌 (𝑅)$ ==> $\sum | 𝐴_{𝑛} (𝑣^{𝑛}) | \leq \sum {(\frac{| 𝑣 |}{𝜌 (𝑅)})}^{𝑛} = \frac{1}{1 - \frac{| 𝑣 |}{𝜌 (𝑅)}} - 1 < \infty$

Question 类似 Euclidean 的情况, 收敛半径内蕴含 Minkowski 连续性 i.e. 距离和方向的连续, 而且有绝对一致收敛性质

Example Question

$exph$ is $ℂ_{split} ≃ ℝ^{1, 1}$ Minkowski 解析
类似复分析, $ℂ_{split}$ 解析 ==> $ℝ^{1, 1}$ Minkowski 解析
类似于 Euclidean 的情况, PDE, 特征函数, 特殊函数可能会给出更多 Minkowski-analytic 的例子

Question let $𝑓 : ℝ^{1, 𝑑} \to ℝ^{1, 𝑑}$ Euclidean 解析 $ℝ^{1 + 𝑑}$ , $𝑑 𝑓 \in SO (1, 𝑑)$ ==> $𝑓$ Minkowski 解析?

Question 对于 Minkowski 解析, 考虑对应到 Euclidean 的 #link(<analytic-continuation>)[], #link(<power-series-space>)[], #link(<analytic-space>)[]

多了 $| argh (𝑣) | \leq 𝑅$ , 参数 $𝑅$

级数 or 函数三角不等式可能需要加上参数 $cosh (2 𝑅)$ 修正的版本

解析函数空间的网的逼近时还需要 $𝑅 \to \infty$ 作为整个方向空间的极限