note-math

flat-metric_(tag) (ref-2, vol)

flat metric := 存在坐标使得 $\forall 𝑥, 𝑔 (𝑥) = 𝜂$ 标准 metric

$𝕊^{2}$ 的继承自 $ℝ^{3}$ 的子流形 metric 不是 #link(<flat-metric>)[], 而是 #link(<quadratic-manifold-is-constant-sectional-curvature>)[constant-sectional-curvature].

何时存在 flat metric?

选取一个坐标 $𝑦$ , with metric $𝑔$

假设变换到 $𝑥$ with metric $𝜂$ , 则

$𝑔 = \frac{\partial 𝑥}{\partial 𝑦}^{⊺} \cdot 𝜂 \cdot \frac{\partial 𝑥}{\partial 𝑦}$
$𝑔^{- 1} = \frac{\partial 𝑦}{\partial 𝑥} \cdot 𝜂 \cdot \frac{\partial 𝑦}{\partial 𝑥}^{⊺}$
$\frac{\partial 𝑥}{\partial 𝑦} \cdot 𝑔^{- 1} \cdot \frac{\partial 𝑥}{\partial 𝑦}^{⊺} = 𝜂$

#link(<connection-transformations>)[联络的变换]

Γ (𝑦) = \frac{\partial 𝑦}{\partial 𝑥} \cdot Γ (𝑥) \cdot \frac{\partial 𝑥}{\partial 𝑦} + \frac{\partial 𝑦}{\partial 𝑥} \cdot \frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦})

对于 flat metric $Γ (𝑥) = 0$ , 从而

Γ (𝑦) = \frac{\partial 𝑦}{\partial 𝑥} \cdot \frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦})

等价地

\frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦}) = \frac{\partial 𝑥}{\partial 𝑦} \cdot Γ (𝑦)

这个性质, 加上 $𝑝$ 的初值条件, 使得可以用 PDE 恢复 flat metric, i.e. 用于证明

\frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦} \cdot 𝑔^{- 1} \cdot \frac{\partial 𝑥}{\partial 𝑦}^{⊺}) = 0

Proof

product-rule 展开以上的微分

\begin{matrix} = & (\frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦})) \cdot 𝑔^{- 1} \cdot {(\frac{\partial 𝑥}{\partial 𝑦})}^{⊺} \\ + \frac{\partial 𝑥}{\partial 𝑦} \cdot (\frac{\partial}{\partial 𝑦} (𝑔^{- 1})) \cdot {(\frac{\partial 𝑥}{\partial 𝑦})}^{⊺} \\ + \frac{\partial 𝑥}{\partial 𝑦} \cdot 𝑔^{- 1} \cdot (\frac{\partial}{\partial 𝑦} {(\frac{\partial 𝑥}{\partial 𝑦})}^{⊺}) & then use \frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦}) = \frac{\partial 𝑥}{\partial 𝑦} \cdot Γ (𝑦) \\ = & \frac{\partial 𝑥}{\partial 𝑦} \cdot (𝑔^{- 1} (Γ,) + \partial 𝑔^{- 1} + 𝑔^{- 1} (, Γ)) \cdot {(\frac{\partial 𝑥}{\partial 𝑦})}^{⊺} & see #link(<differenial-of-metric-inverse-vs-connection>)[link] \\ = & 0 \end{matrix}

关于 $\frac{\partial 𝑥}{\partial 𝑦}$ 的线性 PDE

\frac{\partial}{\partial 𝑦} (\frac{\partial 𝑥}{\partial 𝑦}) = \frac{\partial 𝑥}{\partial 𝑦} \cdot Γ (𝑦)

是可解的 <==> 满足 #link(<linear-PDE-integrable-condition>)[]

[\nabla_{𝑖}, \nabla_{𝑖^{'}}] = 0

where $\nabla = \partial + Γ$ is #link(<geodesic-derivative>)[]

[\nabla_{𝑣}, \nabla_{𝑣^{'}}] - \nabla_{[𝑣, 𝑣^{'}]} = 0

如果解出了 $\frac{\partial 𝑥}{\partial 𝑦} (𝑦)$ , 用初值条件再次积分得到 $𝑥 (𝑦)$ , 得到从坐标 $𝑦$ 到 flat-metric 坐标 $𝑥$ 的转换函数

在 flat-metric 坐标 $Γ = 0$ 所以测地线 ODE 是 $\ddot{𝑥} = 0$ ,所以 flat-metric 坐标将是测地线坐标

不存在 flate metric 坐标时, 则选取 #link(<Einstein-metric>)[] 作为最小 #link(<scalar-curvature>)[纯量曲率]

现在不假设 flat metric

curvature-of-metric_(tag)

曲率 ( $𝑅$ from "Riemann")

$𝑅_{𝑖 𝑖^{'}} = [\nabla_{𝑖}, \nabla_{𝑖^{'}}]$
${(𝑅_{𝑖 𝑖^{'}})}_{𝑗}^{𝑗^{'}} = {[\nabla_{𝑖}, \nabla_{𝑖^{'}}]}_{𝑗}^{𝑗^{'}}$

$[\nabla_{𝑖}, \nabla_{𝑖^{'}}]$ 是张量 (尽管 $\nabla$ 不是)

name-overload: 曲率 := 曲率的 #link(<metric-dual>)[] $𝑅 ≔ 𝑔 [\nabla_{𝑖}, \nabla_{𝑖^{'}}]$

在坐标中 $𝑅_{𝑖 𝑖 𝑗^{'} 𝑗^{'}} = 𝑔_{𝑗^{'} 𝑘} {(𝑅_{𝑖 𝑖^{'}})}_{𝑗}^{𝑘}$

flat-metric-iff-curvature-0_(tag) flat-metric <==> 曲率是零

curvature-determine-metric-locally_(tag)

"flat-metric <==> 曲率零" 可以推广到曲率决定局部 metric

如果两个 metric $𝑔, 𝑔^{'}$ 及其曲率通过 $𝑝, 𝑝^{'}$ 点之间局部微分同胚联系起来, 且微分是 $𝑝, 𝑞$ 切空间之间的 isometry, 则局部微分同胚是 $𝑔, 𝑔^{'}$ 的局部 isometry

curvature-in-geodesic-coordinate_(tag)

在测地线坐标的原点 $𝑝$ , 通过计算, 通过

#link(<metric-connection>)[] $Γ$ 的定义和曲率 $𝑅$ 定义
$Γ (𝑝) = 0$

有

𝑅_{𝑖_{1} 𝑖_{2} 𝑗_{1} 𝑗_{2}} = \frac{1}{2} \sum_{𝑖, 𝑗 \in 2!} sign (𝑖) sign (𝑗) \partial_{𝑖_{1} 𝑗_{1}} 𝑔_{𝑖_{2} 𝑗_{2}}

𝑅_{𝑖 𝑖^{'} 𝑗 𝑗^{'}} = \frac{1}{2} (\partial_{𝑖 𝑗} 𝑔_{𝑖^{'} 𝑗^{'}} + \partial_{𝑖^{'} 𝑗^{'}} 𝑔_{𝑖 𝑗} - \partial_{𝑖 𝑗^{'}} 𝑔_{𝑖^{'} 𝑗} - \partial_{𝑖^{'} 𝑗} 𝑔_{𝑖 𝑗^{'}})

==> 如果在测地线坐标, metric 的 Taylor 展开二阶微分也是零 $𝑔 (𝑝 + 𝑣) = 𝜂 + 𝑜 (𝑣^{2}),$ 则曲率也是零 $𝑅 = 0$ , 从而导致 flat-metric, 从而高阶微分也是零 $𝑜 (𝑣^{2}) = 0$

symmetry-of-curvature_(tag)

$𝑅_{𝑖 (12) 𝑗 (12)} =$

$- 𝑅_{𝑖 (21) 𝑗 (12)}$
$- 𝑅_{(𝑖 (12) 𝑗 (21)}$
$𝑅_{𝑗 (12) 𝑖 (12)}$

$\sum_{cyclic (123)} 𝑅_{𝑖 (123) 𝑗} = 0$

$𝑅_{𝑖 𝑖^{'} 𝑗 𝑗^{'}} =$

$- 𝑅_{𝑖^{'} 𝑖 𝑗 𝑗^{'}}$
$- 𝑅_{𝑖 𝑖^{'} 𝑗^{'} 𝑗}$
$𝑅_{𝑗 𝑗^{'} 𝑖 𝑖^{'}}$

$𝑅_{𝑖 𝑖^{'} 𝑖^{″} 𝑗} + 𝑅_{𝑖^{'} 𝑖^{″} 𝑖 𝑗} + 𝑅_{𝑖^{″} 𝑖 𝑖^{'} 𝑗} = 0$

==> $𝑅 \in ⨀^{2} (⋀^{2} 𝑉^{⊺})$

Proof 在测地线坐标, 用 $Γ$ or $𝑔$ 表示的曲率 $𝑅$ 的定义

algebraic-curvature-tensor_(tag) 代数曲率张量定义为满足上述对称性 (ref-6, lect 8)

curvature-product_(tag)

模仿 #link(<curvature-in-geodesic-coordinate>)[曲率在测地线坐标的定义], 对于二阶对称张量 $𝑇, 𝑆 \in ⨀^{2} 𝑉^{⊺}$ 定义 curvature-product

{(𝑇 ⧀ 𝑆)}_{𝑖_{1} 𝑖_{2} 𝑗_{1} 𝑗_{2}} = \sum_{𝑖, 𝑗 \in 2!} sign (𝑖) sign (𝑗) 𝑇_{𝑖_{1} 𝑗_{1}} 𝑆_{𝑖_{2} 𝑗_{2}}

{(𝑇 ⧀ 𝑆)}_{𝑖 𝑖^{'} 𝑗 𝑗^{'}} = 𝑇_{𝑖 𝑗} 𝑆_{𝑖^{'} 𝑗^{'}} + 𝑇_{𝑖^{'} 𝑗^{'}} 𝑆_{𝑖 𝑗} - 𝑇_{𝑖 𝑗^{'}} 𝑆_{𝑖^{'} 𝑗} - 𝑇_{𝑖^{'} 𝑗} 𝑆_{𝑖 𝑗^{'}}

$𝑇 ⧀ 𝑆$ 满足 #link(<symmetry-of-curvature>)[], 从而 $𝑇 ⧀ 𝑆 \in curvature$ , or $⧀ : {(⨀^{2} 𝑉^{⊺})}^{2} \to curvature$

在测地线坐标的原点, 曲率是 (formally)

𝑅 = \frac{1}{2} (\partial^{2} ⧀ 𝑔)

Def $𝑓 (𝑅_{𝑖 (1234)}) ≔ \sum_{cyclic (123)} 𝑅_{𝑖 (123) 𝑖 (4)}$

$𝑓$ 将 $𝑇 \in ⨀^{2} (⋀^{2} 𝑉^{⊺})$ 映射到自身且 $𝑓^{2} = 𝑓$ , i.e. wiki:Projection_(linear_algebra), so $⨀^{2} (⋀^{2} 𝑉^{⊺}) = ker 𝑓 \oplus im 𝑓$
$ker 𝑓 = curvature$
对于交错张量 $𝑇, 𝑆 \in ⋀^{2} 𝑉^{⊺}$ , $𝑓 (𝑇 ⊙ 𝑆) = 𝑇 \land 𝑆$ , so $im (𝑓) = ⋀^{4} 𝑉^{⊺}$

dimension-of-algebraic-curvature_(tag) 使用 $\frac{domain}{kernel} ≃ image$ , 有代数曲率张量空间的维数

\begin{matrix} dim (curvature) & = & dim (⨀^{2} (\overset{2}{⋀} 𝑉^{⊺})) - dim (\overset{4}{⋀} 𝑉^{⊺}) \\ = & \frac{1}{12} 𝑛^{2} (𝑛^{2} - 1) \end{matrix}

where $𝑛 = dim 𝑉$

metric 是一种张量 $𝑔 \in ⨀^{2} 𝑉^{⊺}$

映射 $𝑔 ⧀ : ⨀^{2} 𝑉^{⊺} \to curvature$

adjoint-of-curvature-product_(tag) ${(𝑔 ⧀)}^{†} : curvature \to ⨀^{2} 𝑉^{⊺}$ :=

对于 $𝑇 \in ⨀^{2} 𝑉^{⊺}$ 和 $𝑆 \in curvature$ 和 $curvature, ⨀^{2} 𝑉^{⊺}$ 的 #link(<tensor-induced-metric>)[]

𝑔 (𝑔 ⧀ 𝑇, 𝑆) = 𝑔 (𝑇, {(𝑔 ⧀)}^{†} 𝑆)

对每个 $𝑆 \in curvature$ , 线性函数

(𝑇 ⇝ 𝑔 (𝑔 ⧀ 𝑇, 𝑆)) \in {(⨀^{2} 𝑉^{⊺})}^{⊺}

有 $⨀^{2} 𝑉^{⊺}$ 空间的 metric-adjoint :=

{(𝑔 ⧀)}^{†} (𝑆) \in ⨀^{2} 𝑉^{⊺}

线性函数 $𝑇 ⇝ 𝑔 (𝑔 ⧀ 𝑇, 𝑆)$ 可以用 $⨀^{2} 𝑉^{⊺}$ 空间的 metric 表示

𝑇 ⇝ 𝑔 (𝑇, {(𝑔 ⧀)}^{†} 𝑆)

在坐标中 ${({(𝑔 ⧀)}^{†} 𝑆)}_{𝑖 𝑗} = 4 𝑔^{𝑖^{'} 𝑗^{'}} 𝑆_{𝑖 𝑖^{'} 𝑗 𝑗^{'}}$

$𝑔 ⧀$ 是单射, ${(𝑔 ⧀)}^{†}$ 是满射. Proof 使用复合映射的前置逆和后置逆, 构造方式参考 #link(<curvature-decomposition>)[] 的计算

$dim ({im (𝑔 ⧀)}^{†}) = dim (im (𝑔 ⧀)) = dim ⨀^{2} 𝑉^{⊺}$

metric-adjoint ${(𝑔 ⧀)}^{†}$ ==> ${ker (𝑔 ⧀)}^{†} ⟂ im (𝑔 ⧀) \subset curvature$

线性映射 ${(𝑔 ⧀)}^{†}$ ==> $dim {ker (𝑔 ⧀)}^{†} + dim {im (𝑔 ⧀)}^{†} = curvature$

==> ${ker (𝑔 ⧀)}^{†} ⟂ im (𝑔 ⧀) = curvature$

映射 $𝑔 \cdot : ℝ \to ⨀^{2} 𝑉^{⊺}$

metric-adjoint ${(𝑔 \cdot)}^{†} : ⨀^{2} 𝑉^{⊺} \to ℝ$ :=

for $𝑟 \in ℝ$ and $𝑇 \in ⨀^{2} 𝑉^{⊺}$

𝑔 (𝑔 \cdot 𝑟, 𝑇) = 𝑔 (𝑟, {(𝑔 \cdot)}^{†} 𝑇) = 𝑟 \cdot ({(𝑔 \cdot)}^{†} 𝑇)

$𝑔 (𝑔 \cdot 𝑟, 𝑇) = 𝑟 \cdot 𝑔 (𝑔, 𝑇)$ so ${(𝑔 \cdot)}^{†} 𝑇 = 𝑔 (𝑔, 𝑇)$

在坐标中 $𝑔 (𝑔, 𝑇) = 𝑔^{𝑖 𝑗} 𝑇_{𝑖 𝑗}$

$𝑔 \cdot$ 是单射, ${(𝑔 \cdot)}^{†}$ 是满射

$dim ({im (𝑔 \cdot)}^{†}) = dim (im (𝑔 \cdot))$

${ker (𝑔 \cdot)}^{†} ⟂ im (𝑔 \cdot) = ⨀^{2} 𝑉^{⊺} = {im (𝑔 ⧀)}^{†}$

curvature-decomposition-space_(tag)

curvature = {ker (𝑔 ⧀)}^{†} ⟂ {ker (𝑔 \cdot)}^{†} ⟂ im (𝑔 \cdot)

正交分解为子张量空间, 且不可再这样分解 i.e. irreducible

curvature-decomposition_(tag) forall $𝑇 \in curvature$ , exists $𝑆 \in ⨀^{2} 𝑉^{⊺}$ , 正交分解 $𝑇 = 𝑈 + 𝑔 ⧀ 𝑆 \in {ker (𝑔 ⧀)}^{†} \oplus im (𝑔 ⧀)$

Proof if it's true then

$\begin{matrix} \frac{1}{4} {(𝑔 ⧀)}^{†} 𝑇 & = & {(𝑔 ⧀)}^{†} (𝑔 ⧀) 𝑆 \\ = & (𝑛 - 2) 𝑆 + 𝑔 \cdot 𝑔 (𝑔, 𝑆) \end{matrix}$

$\begin{matrix} \frac{1}{4} {(𝑔 \cdot)}^{†} {(𝑔 ⧀)}^{†} 𝑇 & = & (𝑛 - 2) \cdot 𝑔 (𝑔, 𝑆) + 𝑔 (𝑔, 𝑔) \cdot 𝑔 (𝑔, 𝑆) \\ = & 2 (𝑛 - 1) \cdot 𝑔 (𝑔, 𝑆) \end{matrix}$

$𝑆 = \frac{1}{4 (𝑛 - 2)} ({(𝑔 ⧀)}^{†} 𝑇 - \frac{1}{2 (𝑛 - 1)} 𝑔 \cdot {(𝑔 \cdot)}^{†} {(𝑔 ⧀)}^{†} 𝑇)$

Ricci-curvature_(tag) $Ric ≔ \frac{1}{4} {(𝑔 ⧀)}^{†} 𝑅$

在坐标中 $Ric (𝑖 𝑗) = 𝑔^{𝑖^{'} 𝑗^{'}} 𝑅_{𝑖 𝑖^{'} 𝑗 𝑗^{'}}$
scalar-curvature_(tag) $scal ≔ \frac{1}{4} {(𝑔 \cdot)}^{†} {(𝑔 ⧀)}^{†} 𝑅$

在坐标中 $scal = 𝑔^{𝑖 𝑗} 𝑔^{𝑖^{'} 𝑗^{'}} 𝑅_{𝑖 𝑖^{'} 𝑗 𝑗^{'}}$
conformal-curvature_(tag) $𝑊 ≔ 𝑅 - 𝑔 ⧀ \frac{1}{𝑛 - 2} (Ric - \frac{1}{2 (𝑛 - 1)} \cdot 𝑔 \cdot scal) \in {ker (𝑔 ⧀)}^{†}$ (named so because if it vanish then the metric conformally flat when $𝑛 \geq 4$ ) ( $𝑊$ from "Weyl")

类似地, 正交分解 $⨀^{2} 𝑉^{⊺} = {ker (𝑔 \cdot)}^{†} \oplus im (𝑔 \cdot)$ 是 $(𝑆 - \frac{1}{𝑛} \cdot 𝑔 \cdot {(𝑔 \cdot)}^{†} 𝑆) + \frac{1}{𝑛} \cdot 𝑔 \cdot {(𝑔 \cdot)}^{†} 𝑆$

trace-free Ricci-curvature $tr-free-Ric ≔ Ric - \frac{1}{𝑛} \cdot 𝑔 \cdot scal \in {ker (𝑔 \cdot)}^{†}$

曲率正交子张量空间分解

$curvature = {ker (𝑔 ⧀)}^{†} \oplus {ker (𝑔 \cdot)}^{†} \oplus im (𝑔 \cdot)$

\begin{matrix} 𝑅 = & 𝑊 \\ + & \frac{1}{𝑛 - 2} \cdot 𝑔 ⧀ tr-free-Ric \\ + & \frac{1}{2 𝑛 (𝑛 - 1)} \cdot scal \cdot 𝑔 ⧀ 𝑔 \end{matrix}

quadratic-form

\begin{matrix} | 𝑅 |^{2} & = & | 𝑊 |^{2} + \frac{4}{𝑛 - 2} | tr-free-Ric |^{2} + \frac{2}{𝑛 (𝑛 - 1)} {scal}^{2} \\ = & | 𝑊 |^{2} + \frac{4}{𝑛 - 2} | Ric |^{2} - \frac{2}{𝑛 (𝑛 - 1)} {scal}^{2} \end{matrix}

curvature-low-dimension_(tag) low dimension curvature

$𝑛 = 0, 1 ⟹ dim (curvature) = 0$
$𝑛 = 2 ⟹ dim (curvature) = 1$ span by $𝑔 ⧀ 𝑔$

$𝑅 = \frac{1}{2} scal 𝑔 ⧀ 𝑔$ , only type $𝑅_{1212} \neq 0$ , and $𝑔 ⧀ 𝑔 \sim 𝑔_{11} 𝑔_{12} - 𝑔_{12}^{2} = det (𝑔)$
$Ric = \frac{1}{2} scal 𝑔$

$𝑛 = 3 ⟹ dim (curvature) = 6$

$𝑔 ⧀ : ⨀^{2} 𝑉^{⊺} \to curvature$ 是双射
$𝑊 = 0$
$𝑅$ 完全决定于 $Ric = \frac{1}{4} {(𝑔 ⧀)}^{†} 𝑅$

let $\frac{\partial 𝑔_{𝑖 𝑖^{'}}}{\partial 𝑥^{𝑗} \partial 𝑥^{𝑗^{'}}} = 𝑔_{𝑖 𝑖^{'}, 𝑗 𝑗^{'}}$

metric 测地线坐标展开也有曲率的出现

\begin{matrix} 𝑔_{𝑖 𝑖^{'}} & = & 𝜂_{𝑖 𝑖^{'}} + (\partial_{𝑗 𝑗^{'}} 𝑔_{𝑖 𝑖^{'}}) 𝑣^{𝑗} 𝑣^{𝑗^{'}} + 𝑜 (𝑣^{2}) \end{matrix}

且满足

(\partial_{𝑗 𝑗^{'}} 𝑔_{𝑖 𝑖^{'}}) 𝑣^{𝑗} 𝑣^{𝑗^{'}} = - \frac{1}{3} 𝑅_{𝑖 𝑗 𝑖^{'} 𝑗^{'}} 𝑤^{𝑗} 𝑤^{𝑗^{'}}

注意这是求和 $\sum_{𝑗 𝑗^{'}}$ 的等式, 而不是系数的等式

"trace" ${(𝑔 ⧀)}^{†}$ 也出现在 Taylor-expansion of metric volume-form $| 𝑔 | = | det (𝑔_{𝑖 𝑗}) |^{\frac{1}{2}}$ , 相关于 $det (𝟙 + 𝐴) = 𝟙 + tr (𝐴) + 𝑜 (𝐴)$ and ${(1 + 𝑥)}^{\frac{1}{2}} = 1 + \frac{1}{2} 𝑥 + 𝑜 (𝑥)$
$| 𝑔 | = 1 - \frac{1}{6} Ric (𝑣, 𝑣) + 𝑜 (𝑣^{2})$
"trace" ${(𝑔 \cdot)}^{†}$ 再次出现在 volume of geodesic ball (for spacetime manifold 应该用 multi radius?)
$\frac{Vol (geodesic-ball (𝑟))}{Vol (ball (𝑟))} = 1 - \frac{1}{6 (𝑛 + 2)} scal 𝑟^{2} + 𝑜 (𝑟^{2})$

$\nabla_{𝑋} 𝑌 - \nabla_{𝑌} 𝑋 = [𝑋, 𝑌]$

if scale matric $𝑔 ⇝ 𝜆 𝑔$

geodesic-derivative $\nabla$
curvature $[\nabla_{𝑖}, \nabla_{𝑖^{'}}]$
curvature metric-dual $𝜆 𝑔 [\nabla_{𝑖}, \nabla_{𝑖^{'}}]$
Ricci-curvature $\frac{1}{𝜆} Ric$
#link(<sectional_curvature>)[] $\frac{1}{𝜆} 𝐾$
scalar-curvature $\frac{1}{𝜆} scal$

用 signature $(3, 1)$ 来表示时空 metric 时, 就是对 $(1, 3)$ signature 乘以 $𝜆 = - 1$