note-math

cf. wiki:Symmetric_space wiki:Simple_Lie_group

symmetric-space-locally_(tag) := $\nabla 𝑅 = 0$

Example quadratic manifold, simple-Lie-group and related symmetric-space

constant-sectional-curvature ==> symmetric-space

simple-Lie-group := Lie-algebra & Lie-bracket cannot decompose

Killing form := $𝑔 (𝑋, 𝑌) ≔ \pm tr [𝑋, [𝑌,]] = \pm tr (ad 𝑋 ad 𝑌)$ for $𝟙$ 处的切空间的 $𝑋, 𝑌$

然后定义 $𝟙$ 处的 metric 通过 action 生成的其它地方的 metric, 而且是 bi-invariant 的 i.e. 群作用的两种形式都给出相同的 metric

这样的定义使得群作用是 Killing-form 的 isometry

不是 simple-Lie-group 也可以定义 Killing-form

Question Killing-form 的定义的动机?

simple-Lie-group <==> Killing-form 非退化

simple-Lie-group and its symmetric-space 的 Killing-form 是 Einstein-metric

Proof of simple-Lie-group 的情况

$𝑔 ([𝑋, 𝑋^{'}], 𝑋^{″}) + 𝑔 (𝑋^{'}, [𝑋, 𝑋^{″}]) = 0$ for Lie algebra $𝑋, 𝑋^{'}, 𝑋^{″}$

Proof

Lie-algebra ==> δ-isometry ==> for δ-group-action $𝑋$ , $\partial_{𝑋} 𝑔 = 0$

因为 Killing-form 是群作用生成的 metric 所以 Lie-derivative 是零 $𝐿_{𝑋} 𝑔 = 0$

对于 $𝑋, 𝑋^{'}, 𝑋^{″}$ 生成的场

\begin{matrix} 0 & = & 𝐿_{𝑋} (𝑔 (𝑋^{'}, 𝑋^{″})) \\ = & (\partial_{𝑋} 𝑔) (𝑋^{'}, 𝑋^{″}) \\ 𝑠 - (𝑔 (𝐿_{𝑋} 𝑋^{'}, 𝑋^{″}) + 𝑔 (𝑋^{'}, 𝐿_{𝑋} 𝑋^{″})) \\ = & 𝑔 ([𝑋, 𝑋^{'}], 𝑋^{″}) + 𝑔 (𝑋^{'}, [𝑋, 𝑋^{″}]) \end{matrix}

geodesic-derivative $\nabla = \frac{1}{2} [,]$ . Proof see below
curvature $[\nabla_{𝑖}, \nabla_{𝑖^{'}}] 𝑗 = - \frac{1}{4} [[𝑖, 𝑖^{'}], 𝑗]$
$\nabla 𝑅 = 0$ . hence symmetric-space-locally
curvature $𝑔 ([\nabla_{𝑖}, \nabla_{𝑖^{'}}] 𝑗, 𝑗^{'}) = - \frac{1}{4} 𝑔 ([𝑖, 𝑖^{'}], [𝑗, 𝑗^{'}])$
sectional-curvature for orthonormal $- \frac{1}{4} | [𝑋, 𝑌] |^{2}$
Ricci-curvature $Ric (𝑋, 𝑌) = \frac{1}{4} tr [𝑋, [𝑌,]] = \pm \frac{1}{4} 𝑔$ . hence Einstein-metric
scalar-curvature $scal = \pm \frac{1}{4} dim$

Prop $\nabla_{𝑋} 𝑌 = \frac{1}{2} [𝑋, 𝑌]$ at $𝟙$ , 同理对 $𝑋, 𝑌$ 生成的场 (bi-invariant)

Proof

Prop $\nabla_{𝑋} 𝑋 = 0$

这给出 $\nabla_{𝑋} 𝑌 + \nabla_{𝑌} 𝑋 = 0$

with $\nabla_{𝑋} 𝑌 - \nabla_{𝑌} 𝑋 = [𝑋, 𝑌]$ , 这给出 $\nabla = \frac{1}{2} [,]$

Proof of $\nabla_{𝑋} 𝑋 = 0$

need $𝑔 (𝑌, \nabla_{𝑋} 𝑋) = 0$

由于群作用生成 $𝑔, 𝑋^{'}, 𝑋^{″}$ , $𝑔 (𝑋^{'}, 𝑋^{″}) (at 𝑝) \equiv 𝑔 (𝑋^{'}, 𝑋^{″}) (at 𝟙)$ 常值 ==> $\partial 𝑔 (𝑋^{'}, 𝑋^{″}) = 0$

$0 = \partial_{𝑋} (𝑔 (𝑋^{'}, 𝑋^{″})) = 𝑔 (\nabla_{𝑋} 𝑋^{'}, 𝑋^{″}) + 𝑔 (𝑋^{'}, \nabla_{𝑋} 𝑋^{″})$

need $𝑔 (\nabla_{𝑋} 𝑌, 𝑋) = 0$

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

$𝑔 ([𝑋, 𝑌], 𝑋) = - 𝑔 (𝑌, [𝑋, 𝑋]) = 0$

need $𝑔 (\nabla_{𝑌} 𝑋, 𝑋) = 0$

by $0 = \partial_{𝑌} (𝑔 (𝑋, 𝑋))$

Question any more intuitive proof?

Prop for simple-Lie-group

Lie algebra 生成的 bi-invariant vector field 的积分曲线都是 Killing-form 测地线, 因为

测地线可以写为 $\nabla_{\dot{𝑥}} \dot{𝑥} = 0$
假设 $\dot{𝑥}$ 是 $𝑋$ 的积分曲线 $\dot{𝑥} (𝑡) = 𝑋 (𝑥 (𝑡))$
$\nabla_{𝑋} 𝑋 = 0$

二次型流形. $ℝ^{𝑝, 𝑞}$ 的对称群 $SO (𝑝, 𝑞)$

orbit type $| 𝑥 |^{2} = 1$ or $ℚ^{𝑝, 𝑞} (1)$
- induced metric signature $(𝑝 - 1, 𝑞)$ (normal vector $| 𝑥 |^{2} = 1 > 0$ )
- isotropy-group $SO (𝑝 - 1, 𝑞)$
- quotient $\frac{SO (𝑝, 𝑞)}{SO (𝑝 - 1, 𝑞)} = ℚ^{𝑝, 𝑞} (1)$
- isometry of $ℚ^{𝑝, 𝑞} (1)$ is $SO (𝑝, 𝑞)$ (isometry 假设保持方向)
orbit type $| 𝑥 |^{2} = - 1$ or $ℚ^{𝑝, 𝑞} (- 1)$
- induced metric signature $(𝑝, 𝑞 - 1)$ (normal vector $| 𝑥 |^{2} = - 1 < 0$ )
- isotropy-group $SO (𝑝, 𝑞 - 1)$
- quotient $\frac{SO (𝑝, 𝑞)}{SO (𝑝, 𝑞 - 1)} = ℚ^{𝑝, 𝑞} (- 1)$
- isometry of $ℚ^{𝑝, 𝑞} (- 1)$ is $SO (𝑝, 𝑞)$

$ℚ^{𝑝, 𝑞} (\pm 1) = ℚ^{𝑞, 𝑝} (\mp 1)$

Example

$(0, 𝑛)$ spatial manifold 有 $ℚ^{1, 𝑛} (1) = {ℍ 𝕪}^{𝑛} = \frac{SO (1, 𝑛)}{SO (𝑛)}, ℚ^{0, 𝑛 + 1} (- 1) = 𝕊^{𝑛} = \frac{SO (𝑛 + 1)}{SO (𝑛)}$
$(1, 𝑛 - 1)$ 时空二次型流形有 $ℚ^{2, 𝑛 - 1} (1) = \frac{SO (2, 𝑛 - 1)}{SO (1, 𝑛 - 1)}$ 和单叶双曲面 $ℚ^{1, 𝑛} (- 1) = \frac{SO (1, 𝑛)}{SO (1, 𝑛 - 1)}$

二次型流形的例子有这种性质

simple-Lie-group $𝐺$ , simple-Lie-group isotropy $𝐻$ , orbit $\frac{𝐺}{𝐻}$

Lie-algebra 有正交分解 $g = h \oplus \frac{g}{h}$ , 不是 Lie bracket 分解

$h$ 是 $𝐻$ 的 Lie-algebra, $\frac{g}{h} = h^{⟂}$ 是正交补

$exp$ 给出 $𝐺, 𝐻, \frac{𝐺}{𝐻}$ 的坐标

$g$ 的 Killing-form 导出 $h$ 的 Killing-form 和 $\frac{g}{h}$ 的 Einstein metric

$[h, h] \subset h$
$[h, \frac{g}{h}] \subset \frac{g}{h}$
$[\frac{g}{h}, \frac{g}{h}] \subset h$