note-math

flat-metric_(tag)

flat metric := a metric such that there exist coordinates $𝑦$ where $\forall 𝑥, 𝑔 (𝑥) = 𝜂$ standard metric

The submanifold metric of $𝕊^{2}$ inherited from $ℝ^{3}$ is not #link(<flat-metric>)[], but rather #link(<quadratic-manifold-is-constant-sectional-curvature>)[constant-sectional-curvature].

When does a flat metric exist?

Choose a coordinate $𝑦$ , with metric $𝑔$

Assume transformation to $𝑥$ with metric $𝜂$ , then

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

#link(<connection-transformations>)[Connection transformations]

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})

For flat metric $Γ (𝑥) = 0$ , thus

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

Equivalently

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

This property, combined with the initial condition of $𝑝$ , allows for the recovery of the flat metric using PDE, i.e., for proving

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

Proof

product-rule expansion of the above differential

\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}

linear PDE for $\frac{\partial 𝑥}{\partial 𝑦}$

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

is solvable <==> satisfies #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

If $\frac{\partial 𝑥}{\partial 𝑦} (𝑦)$ is solved, integrating again with the initial condition yields $𝑥 (𝑦)$ , which gives the conversion function from coordinate $𝑦$ to the flat-metric coordinate $𝑥$

In flat-metric coordinates $Γ = 0$ , so the geodesic ODE is $\ddot{𝑥} = 0$ , so flat-metric coordinates will be geodesic coordinates

When not exist flat metric coordinate, choose #link(<Einstein-metric>)[] as minimal #link(<scalar-curvature>)[]

Now, do not assume flat metric

curvature-of-metric_(tag)

Curvature ( $𝑅$ from "Riemann")

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

$[\nabla_{𝑖}, \nabla_{𝑖^{'}}]$ is a tensor (even though $\nabla$ is not)

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

In coordinates $𝑅_{𝑖 𝑖 𝑗^{'} 𝑗^{'}} = 𝑔_{𝑗^{'} 𝑘} {(𝑅_{𝑖 𝑖^{'}})}_{𝑗}^{𝑘}$

flat-metric-iff-curvature-0_(tag) flat-metric <==> curvature is zero

curvature-determine-metric-locally_(tag)

"flat-metric <==> curvature zero" can be generalized to curvature determines local metric

If two metrics $𝑔, 𝑔^{'}$ and their curvatures are related by a local diffeomorphism between points $𝑝, 𝑝^{'}$ , and the differential is an isometry between the tangent spaces $𝑝, 𝑞$ , then the local diffeomorphism is a local isometry of $𝑔, 𝑔^{'}$ .

curvature-in-geodesic-coordinate_(tag)

At the origin $𝑝$ of geodesic coordinates, by calculation, through

#link(<metric-connection>)[] definition of $Γ$ and definition of curvature $𝑅$
$Γ (𝑝) = 0$

we have

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^{'}})

==> If in geodesic coordinates, the second order derivative of the Taylor expansion of the metric is also zero $𝑔 (𝑝 + 𝑣) = 𝜂 + 𝑜 (𝑣^{2}),$ then the curvature is also zero $𝑅 = 0$ , which leads to a flat-metric, and thus the higher order derivatives are also zero $𝑜 (𝑣^{2}) = 0$

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 Definition of curvature $𝑅$ in geodesic coordinates, expressed using $Γ$ or $𝑔$

algebraic-curvature-tensor_(tag) The algebraic curvature tensor is defined to satisfy the above symmetries (ref-6, lect 8)

curvature-product_(tag)

Mimicking #link(<curvature-in-geodesic-coordinate>)[the definition of curvature in geodesic coordinates], for second-order symmetric tensors $𝑇, 𝑆 \in ⨀^{2} 𝑉^{⊺}$ , define the 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^{'}}

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

At the origin of geodesic coordinates, the curvature is (formally)

𝑅 = \frac{1}{2} (\partial^{2} ⧀ 𝑔)

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

$𝑓$ maps $𝑇 \in ⨀^{2} (⋀^{2} 𝑉^{⊺})$ to itself and $𝑓^{2} = 𝑓$ , i.e. wiki:Projection_(linear_algebra), so $⨀^{2} (⋀^{2} 𝑉^{⊺}) = ker 𝑓 \oplus im 𝑓$
$ker 𝑓 = curvature$
For alternating tensors $𝑇, 𝑆 \in ⋀^{2} 𝑉^{⊺}$ , $𝑓 (𝑇 ⊙ 𝑆) = 𝑇 \land 𝑆$ , so $im (𝑓) = ⋀^{4} 𝑉^{⊺}$

dimension-of-algebraic-curvature_(tag) Using $\frac{domain}{kernel} ≃ image$ , we have the dimension of the algebraic curvature tensor space

\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 is a tensor $𝑔 \in ⨀^{2} 𝑉^{⊺}$

Mapping $𝑔 ⧀ : ⨀^{2} 𝑉^{⊺} \to curvature$

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

For $𝑇 \in ⨀^{2} 𝑉^{⊺}$ and $𝑆 \in curvature$ and the #link(<tensor-induced-metric>)[] of $curvature, ⨀^{2} 𝑉^{⊺}$

𝑔 (𝑔 ⧀ 𝑇, 𝑆) = 𝑔 (𝑇, {(𝑔 ⧀)}^{†} 𝑆)

For each $𝑆 \in curvature$ , the linear function

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

has the metric-adjoint of the $⨀^{2} 𝑉^{⊺}$ space :=

{(𝑔 ⧀)}^{†} (𝑆) \in ⨀^{2} 𝑉^{⊺}

The linear function $𝑇 ⇝ 𝑔 (𝑔 ⧀ 𝑇, 𝑆)$ can be represented by the metric of the $⨀^{2} 𝑉^{⊺}$ space

𝑇 ⇝ 𝑔 (𝑇, {(𝑔 ⧀)}^{†} 𝑆)

In coordinates ${({(𝑔 ⧀)}^{†} 𝑆)}_{𝑖 𝑗} = 4 𝑔^{𝑖^{'} 𝑗^{'}} 𝑆_{𝑖 𝑖^{'} 𝑗 𝑗^{'}}$

$𝑔 ⧀$ is injective, and ${(𝑔 ⧀)}^{†}$ is surjective. Proof uses the pre-inverse and post-inverse of composite mappings. The construction method refers to the calculation in #link(<curvature-decomposition>)[]

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

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

Linear mapping ${(𝑔 ⧀)}^{†}$ ==> $dim {ker (𝑔 ⧀)}^{†} + dim {im (𝑔 ⧀)}^{†} = curvature$

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

Mapping $𝑔 \cdot : ℝ \to ⨀^{2} 𝑉^{⊺}$

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 ${(𝑔 \cdot)}^{†} 𝑇 = 𝑔 (𝑔, 𝑇)$

In coordinates $𝑔 (𝑔, 𝑇) = 𝑔^{𝑖 𝑗} 𝑇_{𝑖 𝑗}$

$𝑔 \cdot$ is injective, ${(𝑔 \cdot)}^{†}$ is surjective

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

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

curvature-decomposition-space_(tag)

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

Orthogonal decomposition into tensor subspace, and cannot be decomposed further, i.e., irreducible

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

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) $Ric ≔ \frac{1}{4} {(𝑔 ⧀)}^{†} 𝑅$

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

In coordinates $scal = 𝑔^{𝑖 𝑗} 𝑔^{𝑖^{'} 𝑗^{'}} 𝑅_{𝑖 𝑖^{'} 𝑗 𝑗^{'}}$
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")

Similarly, the orthogonal decomposition $⨀^{2} 𝑉^{⊺} = {ker (𝑔 \cdot)}^{†} \oplus im (𝑔 \cdot)$ is $(𝑆 - \frac{1}{𝑛} \cdot 𝑔 \cdot {(𝑔 \cdot)}^{†} 𝑆) + \frac{1}{𝑛} \cdot 𝑔 \cdot {(𝑔 \cdot)}^{†} 𝑆$

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

Curvature Orthogonal Decomposition to Tensor Subspace

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

\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 $𝑔 ⧀ 𝑔 \sim 𝑔_{11} 𝑔_{12} - 𝑔_{12}^{2} = det (𝑔)$
$Ric = \frac{1}{2} scal 𝑔$

$𝑛 = 3 ⟹ dim (curvature) = 6$

$𝑔 ⧀ : ⨀^{2} 𝑉^{⊺} \to curvature$ is a bijection
$𝑊 = 0$
$𝑅$ is completely determined by $Ric = \frac{1}{4} {(𝑔 ⧀)}^{†} 𝑅$

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

Curvature also appears in the Taylor expansion of metric geodesic coordinates

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

and satisfies

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

Note that this is an equality of sums $\sum_{𝑗 𝑗^{'}}$ , not an equality of coefficients

"trace" ${(𝑔 ⧀)}^{†}$ also appears in Taylor-expansion of metric volume-form $| 𝑔 | = | det (𝑔_{𝑖 𝑗}) |^{\frac{1}{2}}$ , related to $det (𝟙 + 𝐴) = 𝟙 + tr (𝐴) + 𝑜 (𝐴)$ 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" ${(𝑔 \cdot)}^{†}$ appears again in the volume of geodesic ball (for spacetime manifold should use 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 $\frac{1}{𝜆} scal$

When representing spacetime metric with signature $(3, 1)$ , it is obtained by multiplying the $(1, 3)$ signature by $𝜆 = - 1$