《视觉 SLAM 十四讲》V2 第 4 讲 李群与李代数 【什么样的相机位姿 最符合 当前观测数据】

P71

文章目录

    • [4.1 李群与李代数基础](#4.1 李群与李代数基础)
      • [4.1.3 李代数的定义](#4.1.3 李代数的定义)
      • [4.1.4 李代数 so(3)](#4.1.4 李代数 so(3))
      • [4.1.5 李代数 se(3)](#4.1.5 李代数 se(3))
    • [4.2 指数与对数映射](#4.2 指数与对数映射)
    • [4.3 李代数求导与扰动模型](#4.3 李代数求导与扰动模型)
      • [4.3.2 SO(3)上的李代数求导](#4.3.2 SO(3)上的李代数求导)
      • [4.3.3 李代数求导](#4.3.3 李代数求导)
      • [4.3.4 扰动模型(左乘)【更简单 的导数计算模型】](#4.3.4 扰动模型(左乘)【更简单 的导数计算模型】)
      • [4.3.5 SE(3)上的李代数求导](#4.3.5 SE(3)上的李代数求导)
    • [4.4 Sophus应用 【Code】](#4.4 Sophus应用 【Code】)
      • [4.4.2 评估轨迹的误差 【Code】](#4.4.2 评估轨迹的误差 【Code】)
    • [4.5 相似变换群 与 李代数](#4.5 相似变换群 与 李代数)
    • 习题
      • 题1
      • 题2
      • 题4
      • [√ 题5](#√ 题5)
      • [√ 题6](#√ 题6)
        • [6.2 SE(3)伴随性质](#6.2 SE(3)伴随性质)
      • [√ 题7](#√ 题7)
      • [√ 题8](#√ 题8)
    • LaTex

什么样的相机位姿 最符合 当前观测数据。

求解最优的 R , t \bm{R, t} R,t, 使得误差最小化。
李群-李代数间转换 旋转矩阵_正交且行列式为1 位姿估计变成无约束的优化问题

4.1 李群与李代数基础

: 只有一个(良好的)运算的集合。



封结幺逆 、 丰俭由你


李群: 具有连续(光滑)性质的群。



在 t = 0 附近,旋转矩阵可以由 e x p ( ϕ 0 ∧ t ) exp(\phi_0^{\land}t) exp(ϕ0∧t)计算得到

4.1.3 李代数的定义

李代数 描述了李群的局部性质

  • 单位元 附近的正切空间。

g = ( R 3 , R , × ) \mathfrak{g}=(\mathbb{R}^3, \mathbb{R}, \times) g=(R3,R,×)构成了一个李代数

4.1.4 李代数 so(3)

s o ( 3 ) \mathfrak{so}(3) so(3): 一个由三维向量组成的集合 ,每个向量对应一个反对称矩阵,可以用于表达旋转矩阵的导数
s o ( 3 ) = { ϕ ∈ R 3 , Φ = ϕ ∧ ∈ R 3 × 3 } \mathfrak{so}(3)=\{\bm{\phi}\in\mathbb{R}^3,\bm{\Phi}=\bm{\phi}^{\land}\in\mathbb{R}^{3\times3}\} so(3)={ϕ∈R3,Φ=ϕ∧∈R3×3}

4.1.5 李代数 se(3)

李群 S E ( 3 ) SE(3) SE(3) 对应的李代数 s e ( 3 ) \mathfrak{se}(3) se(3)



4.2 指数与对数映射

4.2.1 SO(3)上的指数映射

s o ( 3 ) \mathfrak{so}(3) so(3): 旋转向量 组成的空间
指数映射: 罗德里格斯公式

R = e x p ( ϕ ∧ ) = e x p ( θ a ∧ ) = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ \bm{R}=exp(\phi^{\land})=exp(\theta\bm{a}^{\land})=cos\theta\bm{I} + (1-cos\theta)\bm{a}\bm{a}^T+sin\theta\bm{a}^{\land} R=exp(ϕ∧)=exp(θa∧)=cosθI+(1−cosθ)aaT+sinθa∧

对数映射: 李群 S O ( 3 ) SO(3) SO(3) 中的元素 ------> 李代数 s o ( 3 ) \mathfrak{so}(3) so(3)


把旋转角固定在 ± π ±\pi ±π 之间,则李群和李代数 元素 一一对应。


罗德里格斯公式推导

e x p ( ϕ ∧ ) = e x p ( θ a ∧ ) = ∑ n = 0 ∞ 1 n ! ( θ a ∧ ) n n = 0 , 1 , 2 , 3 , . . . 原式 = I + θ a ∧ + 1 2 ! θ 2 a ∧ a ∧ + 1 3 ! θ 3 a ∧ a ∧ a ∧ + 1 4 ! θ 4 ( a ∧ ) 4 + ⋅ ⋅ ⋅ 将 a ∧ a ∧ = a a T − I ; a ∧ a ∧ a ∧ = ( a ∧ ) 3 = − a ∧ 代入 原式 = a a T − a ∧ a ∧ + θ a ∧ + 1 2 ! θ 2 a ∧ a ∧ − 1 3 ! θ 3 a ∧ − 1 4 ! θ 4 ( a ∧ ) 2 + ⋅ ⋅ ⋅ = a a T + ( θ − 1 3 ! θ 3 + 1 5 ! θ 5 − ⋅ ⋅ ⋅ ) a ∧ − ( 1 − 1 2 ! θ 2 + 1 4 ! θ 4 − ⋅ ⋅ ⋅ ) a ∧ a ∧ = a ∧ a ∧ + I + s i n θ a ∧ − c o s θ a ∧ a ∧ = ( 1 − c o s θ ) a ∧ a ∧ + I + s i n θ a ∧ = ( 1 − c o s θ ) ( a a T − I ) + I + s i n θ a ∧ = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ \begin{align*}exp(\bm{\phi}^{\land}) &=exp(\theta\bm{a}^{\land})=\sum\limits_{n=0}^{\infty}\frac{1}{n!} (\theta\bm{a}^{\land})^n \\ & n = 0, 1, 2, 3, ... \\ 原式 & = \bm{I} + \theta\bm{a}^{\land} + \frac{1}{2!}\theta^2\bm{a}^{\land}\bm{a}^{\land} + \frac{1}{3!}\theta^3\bm{a}^{\land}\bm{a}^{\land}\bm{a}^{\land} +\frac{1}{4!}\theta^4(\bm{a}^{\land})^4+···\\ & 将\bm{a}^{\land}\bm{a}^{\land} = \bm{a}\bm{a}^T - \bm{I}; \bm{a}^{\land}\bm{a}^{\land}\bm{a}^{\land} = (\bm{a}^{\land})^3 =-\bm{a}^{\land} 代入 \\ 原式& = \bm{a}\bm{a}^T - \bm{a}^{\land}\bm{a}^{\land} + \theta\bm{a}^{\land} + \frac{1}{2!}\theta^2\bm{a}^{\land}\bm{a}^{\land}-\frac{1}{3!}\theta^3\bm{a}^{\land}-\frac{1}{4!}\theta^4(\bm{a}^{\land})^2+···\\ & = \bm{a}\bm{a}^T + (\theta - \frac{1}{3!}\theta^3+\frac{1}{5!}\theta^5-···)\bm{a}^{\land}-(1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-···)\bm{a}^{\land}\bm{a}^{\land}\\ & = \bm{a}^{\land}\bm{a}^{\land} + \bm{I}+sin\theta \bm{a}^{\land}-cos\theta\bm{a}^{\land}\bm{a}^{\land}\\ &=(1-cos\theta)\bm{a}^{\land}\bm{a}^{\land}+\bm{I} + sin\theta\bm{a}^{\land}\\ &= (1-cos\theta)(\bm{a}\bm{a}^T-\bm{I})+\bm{I} + sin\theta\bm{a}^{\land}\\ & = cos\theta\bm{I}+ (1-cos\theta)\bm{a}\bm{a}^T + sin\theta\bm{a}^{\land} \end{align*} exp(ϕ∧)原式原式=exp(θa∧)=n=0∑∞n!1(θa∧)nn=0,1,2,3,...=I+θa∧+2!1θ2a∧a∧+3!1θ3a∧a∧a∧+4!1θ4(a∧)4+⋅⋅⋅将a∧a∧=aaT−I;a∧a∧a∧=(a∧)3=−a∧代入=aaT−a∧a∧+θa∧+2!1θ2a∧a∧−3!1θ3a∧−4!1θ4(a∧)2+⋅⋅⋅=aaT+(θ−3!1θ3+5!1θ5−⋅⋅⋅)a∧−(1−2!1θ2+4!1θ4−⋅⋅⋅)a∧a∧=a∧a∧+I+sinθa∧−cosθa∧a∧=(1−cosθ)a∧a∧+I+sinθa∧=(1−cosθ)(aaT−I)+I+sinθa∧=cosθI+(1−cosθ)aaT+sinθa∧

e x p ( θ a ∧ ) = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ exp(\theta\bm{a}^{\land})=cos\theta\bm{I} + (1-cos\theta)\bm{a}\bm{a}^T+sin\theta\bm{a}^{\land} exp(θa∧)=cosθI+(1−cosθ)aaT+sinθa∧


bash 复制代码
$\bm{a}^{\land}$

a ∧ \bm{a}^{\land} a∧


4.2.2 SE(3) 上的指数映射


推导:
∑ n = 0 ∞ 1 ( n + 1 ) ! ( ϕ ∧ ) n = ∑ n = 0 ∞ 1 ( n + 1 ) ! ( θ a ∧ ) n = I + 1 2 ! θ a ∧ + 1 3 ! θ 2 ( a ∧ ) 2 + 1 4 ! θ 3 ( a ∧ ) 3 + 1 5 ! θ 4 ( a ∧ ) 4 + ⋅ ⋅ ⋅ = 1 θ ( 1 2 ! θ 2 − 1 4 ! θ 4 + ⋅ ⋅ ⋅ ) a ∧ + 1 θ ( 1 3 ! θ 3 − 1 5 ! θ 5 + ⋅ ⋅ ⋅ ) ( a ∧ ) 2 + I = 1 θ ( 1 − c o s θ ) a ∧ + 1 θ ( θ − s i n θ ) ( a a T − I ) + I = s i n θ θ I + ( 1 − s i n θ θ ) a a T + 1 − c o s θ θ a ∧ = d e f J \begin{align*} \sum\limits_{n=0}^{\infty}\frac{1}{(n+1)!}(\bm{\phi}^{\land})^n &= \sum\limits_{n=0}^{\infty}\frac{1}{(n+1)!}(\theta\bm{a}^{\land})^n\\ & = \bm{I} + \frac{1}{2!}\theta\bm{a}^{\land} + \frac{1}{3!}\theta^2(\bm{a}^{\land})^2 + \frac{1}{4!}\theta^3(\bm{a}^{\land})^3 + \frac{1}{5!}\theta^4(\bm{a}^{\land})^4+···\\ & = \frac{1}{\theta}( \frac{1}{2!}\theta^2- \frac{1}{4!}\theta^4+···)\bm{a}^{\land} + \frac{1}{\theta}( \frac{1}{3!}\theta^3- \frac{1}{5!}\theta^5+···)(\bm{a}^{\land})^2+\bm{I} \\ & = \frac{1}{\theta}( 1-cos\theta)\bm{a}^{\land} + \frac{1}{\theta}(\theta-sin\theta)(\bm{a}\bm{a}^T-\bm{I})+\bm{I} \\ & = \frac{sin\theta}{\theta}\bm{I}+(1-\frac{sin\theta}{\theta})\bm{a}\bm{a}^T + \frac{1-cos\theta}{\theta}\bm{a}^{\land}\\ & \overset{\mathrm{def}}{=} \bm{J} \end{align*} n=0∑∞(n+1)!1(ϕ∧)n=n=0∑∞(n+1)!1(θa∧)n=I+2!1θa∧+3!1θ2(a∧)2+4!1θ3(a∧)3+5!1θ4(a∧)4+⋅⋅⋅=θ1(2!1θ2−4!1θ4+⋅⋅⋅)a∧+θ1(3!1θ3−5!1θ5+⋅⋅⋅)(a∧)2+I=θ1(1−cosθ)a∧+θ1(θ−sinθ)(aaT−I)+I=θsinθI+(1−θsinθ)aaT+θ1−cosθa∧=defJ

J = s i n θ θ I + ( 1 − s i n θ θ ) a a T + 1 − c o s θ θ a ∧ \bm{J}=\frac{sin\theta}{\theta}\bm{I}+(1-\frac{sin\theta}{\theta})\bm{a}\bm{a}^T + \frac{1-cos\theta}{\theta}\bm{a}^{\land} J=θsinθI+(1−θsinθ)aaT+θ1−cosθa∧

R = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ \bm{R}=cos\theta\bm{I}+(1-cos\theta)\bm{a}\bm{a}^T + sin\theta\bm{a}^{\land} R=cosθI+(1−cosθ)aaT+sinθa∧


t = J ρ \bm{t=Jρ} t=Jρ

平移部分发生了一次以 J \bm{J} J 为系数矩阵的 线性变换。

SO(3),SE(3),so(3),se(3)的对应关系

4.3 李代数求导与扰动模型

Baker-Campbell-Hausdorff公式(BCH公式)




4.3.2 SO(3)上的李代数求导

位姿由SO(3)上的旋转矩阵SE(3)上的变换矩阵 描述

设某时刻机器人的位姿为 T \bm{T} T, 观察到了一个世界坐标位于 p \bm{p} p 的点,产生了一个观测数据 z \bm{z} z

计算理想的观测与实际数据之间的误差: e = z − T p \bm{e = z-Tp} e=z−Tp

假设一共有 N N N 个这样的路标点和观测,对机器人进行位姿估计,相当于寻找一个最优的 T \bm{T} T ,使得整体误差最小化:
min ⁡ T J ( T ) = ∑ i = 1 N ∣ ∣ z i − T p i ∣ ∣ 2 2 \min\limits_{\bm{T}}J(\bm{T}) = \sum\limits_{i=1}^{N}||\bm{z_i-Tp_i}||_2^2 TminJ(T)=i=1∑N∣∣zi−Tpi∣∣22

求解上述问题,需要计算目标函数 J J J 关于变换矩阵 T \bm{T} T 的导数。

使用 李代数 解决 求导问题 的2种思路:

1、用李代数表示姿态,然后根据李代数加法对李代数求导。

2、对李群左乘或右乘微小扰动,然后对该扰动求导,称为左扰动右扰动模型。

4.3.3 李代数求导


推导:
∂ ( R p ) ∂ R = R 对应的李代数为 ϕ ∂ ( exp ⁡ ( ϕ ∧ ) p ) ∂ ϕ = lim ⁡ Δ ϕ → 0 exp ⁡ ( ( ϕ + Δ ϕ ) ∧ ) p − exp ⁡ ( ϕ ∧ ) p Δ ϕ 由式 ( 4.35 ) = lim ⁡ Δ ϕ → 0 exp ⁡ ( ( J l Δ ϕ ) ∧ ) exp ⁡ ( ϕ ∧ ) p − exp ⁡ ( ϕ ∧ ) p Δ ϕ = lim ⁡ Δ ϕ → 0 ( I + ( J l Δ ϕ ) ∧ ) exp ⁡ ( ϕ ∧ ) p − exp ⁡ ( ϕ ∧ ) p Δ ϕ = lim ⁡ Δ ϕ → 0 ( J l Δ ϕ ) ∧ exp ⁡ ( ϕ ∧ ) p Δ ϕ a ∧ 等效于 a × , 因此根据叉乘的性质 = lim ⁡ Δ ϕ → 0 − ( exp ⁡ ( ϕ ∧ ) p ) ∧ J l Δ ϕ Δ ϕ = − ( R p ) ∧ J l \begin{align*}\frac{\partial(\bm{Rp})}{\partial\bm{R}} &\xlongequal{R对应的李代数为\bm{\phi}}\frac{\partial(\exp(\bm{\phi}^{\land})\bm{p})}{\partial\bm{\phi}} \\ & = \lim\limits_{Δ\bm{\phi}\to0}\frac{\exp((\bm{\phi}+Δ\bm{\phi})^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{Δ\bm{\phi}} \\ & 由 式(4.35)\\ & = \lim\limits_{Δ\phi\to0}\frac{\exp((\bm{J}lΔ\bm{\phi})^{\land})\exp(\bm{\phi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{Δ\bm{\phi}} \\ & = \lim\limits{Δ\bm{\phi}\to0}\frac{(\bm{I}+(\bm{J}lΔ\bm{\phi})^{\land})\exp(\bm{\phi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{Δ\bm{\phi}} \\ & = \lim\limits{Δ\bm{\phi}\to0}\frac{(\bm{J}lΔ\bm{\phi})^{\land}\exp(\bm{\phi}^{\land})\bm{p}}{Δ\bm{\phi}} \\ & a^{\land} 等效于 a \times ,因此根据叉乘的性质 \\ & = \lim\limits{Δ\bm{\phi}\to0}\frac{-(\exp(\bm{\phi}^{\land})\bm{p})^{\land}\bm{J}_lΔ\bm{\phi}}{Δ\bm{\phi}} \\ & = -(\bm{Rp})^{\land}\bm{J}_l \end{align*} ∂R∂(Rp)R对应的李代数为ϕ ∂ϕ∂(exp(ϕ∧)p)=Δϕ→0limΔϕexp((ϕ+Δϕ)∧)p−exp(ϕ∧)p由式(4.35)=Δϕ→0limΔϕexp((JlΔϕ)∧)exp(ϕ∧)p−exp(ϕ∧)p=Δϕ→0limΔϕ(I+(JlΔϕ)∧)exp(ϕ∧)p−exp(ϕ∧)p=Δϕ→0limΔϕ(JlΔϕ)∧exp(ϕ∧)pa∧等效于a×,因此根据叉乘的性质=Δϕ→0limΔϕ−(exp(ϕ∧)p)∧JlΔϕ=−(Rp)∧Jl

∂ ( exp ⁡ ( ϕ ∧ ) p ) ∂ ϕ = − ( R p ) ∧ J l \frac{\partial(\exp(\bm{\phi}^{\land})\bm{p})}{\partial\bm{\phi}} = -(\bm{Rp})^{\land}\bm{J}_l ∂ϕ∂(exp(ϕ∧)p)=−(Rp)∧Jl


4.3.4 扰动模型(左乘)【更简单 的导数计算模型】

对 R \bm{R} R 进行一次扰动 Δ R Δ\bm{R} ΔR ,看结果对于 扰动的变化率。

设左扰动 Δ R Δ\bm{R} ΔR 对应的李代数 为 φ \bm{\varphi} φ

∂ ( R p ) ∂ φ = lim ⁡ φ → 0 exp ⁡ ( φ ∧ ) exp ⁡ ( ϕ ∧ ) p − exp ⁡ ( ϕ ∧ ) p φ = lim ⁡ φ → 0 ( I + φ ∧ ) exp ⁡ ( φ ∧ ) p − exp ⁡ ( ϕ ∧ ) p φ = lim ⁡ φ → 0 φ ∧ exp ⁡ ( ϕ ∧ ) p φ = lim ⁡ φ → 0 φ ∧ R p φ = lim ⁡ φ → 0 − ( R p ) ∧ φ φ = − ( R p ) ∧ \begin{align*}\frac{\partial(\bm{Rp})}{\partial\bm{\varphi}} &= \lim\limits_{\bm{\varphi}\to0}\frac{\exp(\bm{\varphi}^{\land})\exp(\bm{\phi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{(\bm{I} + \bm{\varphi}^{\land})\exp(\bm{\varphi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{\bm{\varphi}^{\land}\exp(\bm{\phi}^{\land})\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{\bm{\varphi}^{\land}\bm{Rp}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{-(\bm{Rp})^{\land}\bm{\varphi}}{\bm{\varphi}}\\ &= -(\bm{Rp})^{\land} \end{align*} ∂φ∂(Rp)=φ→0limφexp(φ∧)exp(ϕ∧)p−exp(ϕ∧)p=φ→0limφ(I+φ∧)exp(φ∧)p−exp(ϕ∧)p=φ→0limφφ∧exp(ϕ∧)p=φ→0limφφ∧Rp=φ→0limφ−(Rp)∧φ=−(Rp)∧

4.3.5 SE(3)上的李代数求导

假设某空间点 p \bm{p} p 经过一次变换 T \bm{T} T (对应李代数为 ξ \bm{\xi} ξ), 得到 T p \bm{Tp} Tp

现在给 T \bm{T} T 左乘一个扰动 Δ T = exp ⁡ ( Δ ξ ∧ ) Δ\bm{T} = \exp(Δ\bm{\xi}^{\land}) ΔT=exp(Δξ∧)

设扰动项的李代数为 Δ ξ = [ Δ ρ , Δ ϕ ] T Δ\bm{\xi}=[Δ\bm{\rho},Δ\bm{\phi}]^T Δξ=[Δρ,Δϕ]T,则

∂ ( T p ) ∂ Δ ξ = lim ⁡ Δ ξ → 0 exp ⁡ ( Δ ξ ∧ ) exp ⁡ ( ξ ∧ ) p − exp ⁡ ( ξ ∧ ) p Δ ξ = lim ⁡ Δ ξ → 0 ( I + Δ ξ ∧ ) exp ⁡ ( ξ ∧ ) p − exp ⁡ ( ξ ∧ ) p Δ ξ = lim ⁡ Δ ξ → 0 Δ ξ ∧ exp ⁡ ( ξ ∧ ) p Δ ξ = lim ⁡ Δ ξ → 0 [ Δ ϕ ∧ Δ ρ 0 T 0 ] [ R p + t 1 ] Δ ξ = lim ⁡ Δ ξ → 0 [ Δ ϕ ∧ ( R p + t ) + Δ ρ 0 T ] [ Δ ρ , Δ ϕ ] T = [ I − ( R p + t ) ∧ 0 T 0 T ] = d e f ( T p ) ⨀ \begin{align*}\frac{\partial(\bm{Tp})}{\partial{Δ\bm{\xi}}} &= \lim\limits_{Δ\bm{\xi}\to0}\frac{\exp(Δ\bm{\xi}^{\land})\exp(\bm{\xi}^{\land})\bm{p}-\exp(\bm{\xi}^{\land})\bm{p}}{Δ\bm{\xi}} \\ &= \lim\limits_{Δ\bm{\xi}\to0}\frac{(\bm{I} +Δ\bm{\xi}^{\land})\exp(\bm{\xi}^{\land})\bm{p}-\exp(\bm{\xi}^{\land})\bm{p}}{Δ\bm{\xi}} \\ &= \lim\limits_{Δ\bm{\xi}\to0}\frac{Δ\bm{\xi}^{\land}\exp(\bm{\xi}^{\land})\bm{p}}{Δ\bm{\xi}} \\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} Δ\bm{\phi}^{\land} & \Delta\bm{\rho}\\ \bm{0}^T & 0 \end{bmatrix}\begin{bmatrix} \bm{Rp+t}\\ 1 \end{bmatrix}}{Δ\bm{\xi}}\\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} \Delta\bm{\phi}^{\land}(\bm{Rp+t})+\Delta\bm{\rho}\\ \bm{0}^T \end{bmatrix}}{[Δ\bm{\rho},Δ\bm{\phi}]^T} \\ & = \begin{bmatrix} \bm{I} & -(\bm{Rp+t})^{\land} \\ \bm{0}^T & \bm{0}^T \end{bmatrix} \\ &\overset{\mathrm{def}}{=}(\bm{Tp})^{\bigodot} \end{align*} ∂Δξ∂(Tp)=Δξ→0limΔξexp(Δξ∧)exp(ξ∧)p−exp(ξ∧)p=Δξ→0limΔξ(I+Δξ∧)exp(ξ∧)p−exp(ξ∧)p=Δξ→0limΔξΔξ∧exp(ξ∧)p=Δξ→0limΔξ[Δϕ∧0TΔρ0][Rp+t1]=Δξ→0lim[Δρ,Δϕ]T[Δϕ∧(Rp+t)+Δρ0T]=[I0T−(Rp+t)∧0T]=def(Tp)⨀

bash 复制代码
$\overset{\mathrm{def}}{=}(\bm{Tp})^{\bigodot}$

4.4 Sophus应用 【Code】

SO(3)、SE(3)

二维运动SO(2)和SE(2)

相似变换 Sim(3)

bash 复制代码
mkdir build && cd build
cmake ..
make 
./useSophus

CMakeLists.txt

bash 复制代码
cmake_minimum_required(VERSION 2.8)

project(useSophus)

find_package(Sophus REQUIRED)
include_directories( ${Sophus_INCLUDE_DIRS})

add_executable(useSophus useSophus.cpp)
target_link_libraries(useSophus ${Sophus_LIBRARIES})
cpp 复制代码
#include<iostream>
#include<cmath>
#include<Eigen/Core>
#include<Eigen/Geometry>
#include "sophus/se3.h"

using namespace std;
using namespace Eigen;

/* sophus 的基本用法 */
int main(int argc, char **argv){
    // 沿 Z轴  转90° 的旋转矩阵
    Matrix3d R = AngleAxisd(M_PI/2, Vector3d(0, 0, 1)).toRotationMatrix();

    /* 四元数 */
    Quaterniond q(R);
    Sophus::SO3 SO3_R(R);
    Sophus::SO3 SO3_q(q);

    cout << "SO(3) from matrix:\n" << SO3_R.matrix() << endl;
    cout << "SO(3) from quatenion:\n" << SO3_q.matrix() << endl;
    cout << "they are equal" << endl;

    return 0;
}
cpp 复制代码
#include<iostream>
#include<cmath>
#include<Eigen/Core>
#include<Eigen/Geometry>
#include "sophus/se3.h"

using namespace std;
using namespace Eigen;

/* sophus 的基本用法 */
int main(int argc, char **argv){
    /* 使用 对数映射 获得  李代数*/
    Matrix3d R = AngleAxisd(M_PI/2, Vector3d(0, 0, 1)).toRotationMatrix();
    Sophus::SO3 SO3_R(R);
    Vector3d so3 = SO3_R.log();
    cout << "so3 = " << so3.transpose() << endl;
    
    // hat  向量 ------> 反对称矩阵
    cout << "so3 hat = \n" << Sophus::SO3::hat(so3)<< endl;

    // vee  反对称 ------> 向量
    cout << "so3 hat vee = " << Sophus::SO3::vee(Sophus::SO3::hat(so3)).transpose() << endl;

    Vector3d update_so3(1e-4, 0, 0);// 假设更新量为这么多
    Sophus::SO3 SO3_updated =Sophus::SO3::exp(update_so3) * SO3_R;
    cout << "SO3 updated = \n" << SO3_updated.matrix() << endl;
    return 0;
}
cpp 复制代码
#include<iostream>
#include<cmath>
#include<Eigen/Core>
#include<Eigen/Geometry>
#include "sophus/se3.h"

using namespace std;
using namespace Eigen;

/* sophus   SE(3) 的基本用法 */
int main(int argc, char **argv){
    
    Matrix3d R = AngleAxisd(M_PI/2, Vector3d(0, 0, 1)).toRotationMatrix();  // 沿 Z轴 旋转 90° 的旋转矩阵
    Vector3d t(1, 0, 0);  // 沿 X 轴平移1
    Sophus::SE3 SE3_Rt(R, t);  // 从R,t 构造 SE(3)
    Quaterniond q(R);
    Sophus::SE3 SE3_qt(q, t); // 从q, t 构造 SE(3)
    cout << "SE3 from R, t = \n" << SE3_Rt.matrix() << endl;
    cout << "SE3 from q, t = \n" << SE3_qt.matrix() << endl; 
    
    
    /* 李代数se(3) 是一个 6 维 向量*/
    typedef Eigen::Matrix<double, 6, 1> Vector6d;
    Vector6d se3 = SE3_Rt.log();
    cout << "se3 = " << se3.transpose() << endl;
    
    // hat  向量 ------> 反对称矩阵
    cout << "se3 hat = \n" << Sophus::SE3::hat(se3)<< endl;
    // vee  反对称 ------> 向量
    cout << "se3 hat vee = " << Sophus::SE3::vee(Sophus::SE3::hat(se3)).transpose() << endl;
    
    // 更新
    Vector6d update_se3;// 更新量
    update_se3.setZero();
    update_se3(0, 0) = 1e-4;
    Sophus::SE3 SE3_updated =Sophus::SE3::exp(update_se3) * SE3_Rt;
    cout << "SE3 updated = \n" << SE3_updated.matrix() << endl;
    return 0;
}

4.4.2 评估轨迹的误差 【Code】


考虑一条估计轨迹 T e s t i , i \bm{T}{esti,i} Testi,i 和真实轨迹 T g t , i \bm{T}{gt,i} Tgt,i ,其中 i = 1 , ⋅ ⋅ ⋅ , N i= 1,···,N i=1,⋅⋅⋅,N

1、绝对误差轨迹(Absolute Trajectory Error, ATE) 旋转和平移误差
A T E a l l = 1 N ∑ i = 1 N ∣ ∣ l o g ( T g t , i − 1 T e s t i , i ) ∨ ∣ ∣ 2 2 ATE_{all}=\sqrt{\frac{1}{N}\sum\limits_{i=1}^{N}||log(\bm{T}{gt,i}^{-1}\bm{T}{esti,i})^{\vee}||_2^2} ATEall=N1i=1∑N∣∣log(Tgt,i−1Testi,i)∨∣∣22

  • 每个位姿 李代数 的均方根误差 (Root-Mean-Squared Error,RMSE)

2、绝对平移误差(Average Translational Error)

A T E t r a n s = 1 N ∑ i = 1 N ∣ ∣ t r a n s ( T g t , i − 1 T e s t i , i ) ∣ ∣ 2 2 ATE_{trans}=\sqrt{\frac{1}{N}\sum\limits_{i=1}^{N}||trans(\bm{T}{gt,i}^{-1}\bm{T}{esti,i})||_2^2} ATEtrans=N1i=1∑N∣∣trans(Tgt,i−1Testi,i)∣∣22

其中 trans 表示 取括号内部变量的平移部分。

  • 从整条轨迹上看,旋转出现偏差后,随后的轨迹在平移上也会出现误差。

3、相对误差

考虑 i i i 时刻到 i + Δ t i+\Delta t i+Δt 的运动,相对位姿误差(Relative Pose Error, RPE)

R P E a l l = 1 N − Δ t ∑ i = 1 N − Δ t ∣ ∣ l o g ( ( T g t , i − 1 T g t , i + Δ t ) − 1 ( T e s t i , i − 1 T e s t i , i + Δ t ) ) ∨ ∣ ∣ 2 2 RPE_{all}=\sqrt{\frac{1}{N-\Delta t}\sum\limits_{i=1}^{N-\Delta t}||log((\bm{T}{gt,i}^{-1}\bm{T}{gt,i+\Delta t})^{-1}(\bm{T}{esti,i}^{-1}\bm{T}{esti,i+\Delta t}))^{\vee}||_2^2} RPEall=N−Δt1i=1∑N−Δt∣∣log((Tgt,i−1Tgt,i+Δt)−1(Testi,i−1Testi,i+Δt))∨∣∣22

R P E t r a n s = 1 N − Δ t ∑ i = 1 N − Δ t ∣ ∣ t r a n s ( ( T g t , i − 1 T g t , i + Δ t ) − 1 ( T e s t i , i − 1 T e s t i , i + Δ t ) ) ∣ ∣ 2 2 RPE_{trans}=\sqrt{\frac{1}{N-\Delta t}\sum\limits_{i=1}^{N-\Delta t}||trans((\bm{T}{gt,i}^{-1}\bm{T}{gt,i+\Delta t})^{-1}(\bm{T}{esti,i}^{-1}\bm{T}{esti,i+\Delta t}))||_2^2} RPEtrans=N−Δt1i=1∑N−Δt∣∣trans((Tgt,i−1Tgt,i+Δt)−1(Testi,i−1Testi,i+Δt))∣∣22


bash 复制代码
mkdir build && cd build
cmake ..
make 
./trajectoryError

CMakeLists.txt

bash 复制代码
cmake_minimum_required(VERSION 2.8)

project(trajectoryError)


find_package(Sophus REQUIRED)
include_directories( ${Sophus_INCLUDE_DIRS})

option(USE_UBUNTU_20 "Set to ON if you are using Ubuntu 20.04" OFF)
find_package(Pangolin REQUIRED)
if(USE_UBUNTU_20)
    message("You are using Ubuntu 20.04, fmt::fmt will be linked")
    find_package(fmt REQUIRED)
    set(FMT_LIBRARIES fmt::fmt)
endif()
include_directories(${Pangolin_INCLUDE_DIRS})


add_executable(trajectoryError trajectoryError.cpp)
target_link_libraries(trajectoryError ${Sophus_LIBRARIES})
target_link_libraries(trajectoryError ${Pangolin_LIBRARIES} ${FMT_LIBRARIES})

trajectoryError.cpp

cpp 复制代码
#include <iostream>
#include <fstream>
#include <unistd.h>
#include <pangolin/pangolin.h>
#include <sophus/se3.h>

using namespace Sophus;
using namespace std;

string groundtruth_file = "../groundtruth.txt";
string estimated_file = "../estimated.txt";

typedef vector<Sophus::SE3, Eigen::aligned_allocator<Sophus::SE3>> TrajectoryType;

void DrawTrajectory(const TrajectoryType &gt, const TrajectoryType &esti);

TrajectoryType ReadTrajectory(const string &path);

int main(int argc, char **argv) {
  TrajectoryType groundtruth = ReadTrajectory(groundtruth_file);
  TrajectoryType estimated = ReadTrajectory(estimated_file);
  assert(!groundtruth.empty() && !estimated.empty());
  assert(groundtruth.size() == estimated.size());

  // compute rmse
  double rmse = 0;
  for (size_t i = 0; i < estimated.size(); i++) {
    Sophus::SE3 p1 = estimated[i], p2 = groundtruth[i];
    double error = (p2.inverse() * p1).log().norm();
    rmse += error * error;
  }
  rmse = rmse / double(estimated.size());
  rmse = sqrt(rmse);
  cout << "RMSE = " << rmse << endl;

  DrawTrajectory(groundtruth, estimated);
  return 0;
}

TrajectoryType ReadTrajectory(const string &path) {
  ifstream fin(path);
  TrajectoryType trajectory;
  if (!fin) {
    cerr << "trajectory " << path << " not found." << endl;
    return trajectory;
  }

  while (!fin.eof()) {
    double time, tx, ty, tz, qx, qy, qz, qw;
    fin >> time >> tx >> ty >> tz >> qx >> qy >> qz >> qw;
    Sophus::SE3 p1(Eigen::Quaterniond(qw, qx, qy, qz), Eigen::Vector3d(tx, ty, tz));
    trajectory.push_back(p1);
  }
  return trajectory;
}

void DrawTrajectory(const TrajectoryType &gt, const TrajectoryType &esti) {
  // create pangolin window and plot the trajectory
  pangolin::CreateWindowAndBind("Trajectory Viewer", 1024, 768);
  glEnable(GL_DEPTH_TEST);
  glEnable(GL_BLEND);
  glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);

  pangolin::OpenGlRenderState s_cam(
      pangolin::ProjectionMatrix(1024, 768, 500, 500, 512, 389, 0.1, 1000),
      pangolin::ModelViewLookAt(0, -0.1, -1.8, 0, 0, 0, 0.0, -1.0, 0.0)
  );

  pangolin::View &d_cam = pangolin::CreateDisplay()
      .SetBounds(0.0, 1.0, pangolin::Attach::Pix(175), 1.0, -1024.0f / 768.0f)
      .SetHandler(new pangolin::Handler3D(s_cam));


  while (pangolin::ShouldQuit() == false) {
    glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);

    d_cam.Activate(s_cam);
    glClearColor(1.0f, 1.0f, 1.0f, 1.0f);

    glLineWidth(2);
    for (size_t i = 0; i < gt.size() - 1; i++) {
      glColor3f(0.0f, 0.0f, 1.0f);  // blue for ground truth
      glBegin(GL_LINES);
      auto p1 = gt[i], p2 = gt[i + 1];
      glVertex3d(p1.translation()[0], p1.translation()[1], p1.translation()[2]);
      glVertex3d(p2.translation()[0], p2.translation()[1], p2.translation()[2]);
      glEnd();
    }

    for (size_t i = 0; i < esti.size() - 1; i++) {
      glColor3f(1.0f, 0.0f, 0.0f);  // red for estimated
      glBegin(GL_LINES);
      auto p1 = esti[i], p2 = esti[i + 1];
      glVertex3d(p1.translation()[0], p1.translation()[1], p1.translation()[2]);
      glVertex3d(p2.translation()[0], p2.translation()[1], p2.translation()[2]);
      glEnd();
    }
    pangolin::FinishFrame();
    usleep(5000);   // sleep 5 ms
  }

}

4.5 相似变换群 与 李代数

单目视觉 相似变换群Sim(3)

尺度不确定性 与 尺度漂移

对位于空间的点 p \bm{p} p ,在相机坐标系下要经过一个相似变换。
p ′ = [ s R t 0 T 1 ] p = s R p + t \bm{p}^{\prime}=\begin{bmatrix}s\bm{R} & \bm{t}\\ \bm{0}^T& 1 \end{bmatrix}\bm{p} = s\bm{Rp+t} p′=[sR0Tt1]p=sRp+t





对于尺度因子,李群中的 s s s 即为李代数中 σ \sigma σ 的指数函数

4.6 小结

李群 SO(3) 和 SE(3) 以及对应的李代数 s o ( 3 ) \mathfrak{so}(3) so(3) 和 s e ( 3 ) \mathfrak{se}(3) se(3)

BCH 线性近似, 对位姿进行扰动并求导

习题


题1

验证SO(3)、SE(3)、Sim(3)关于乘法成群

特殊正交群SO(n) 旋转矩阵群

特殊欧式群SE(n) n维欧式变换



题2

题4

√ 题5

证明:
原式等效于证明 R p ∧ R T R = ( R p ) ∧ R R p ∧ I = ( R p ) ∧ R R p ∧ = ( R p ) ∧ R 对于向量 v ∈ R 3 R p ∧ v = ( R p ) ∧ R v 上式等号左边 表示向量 p , v 叉乘后所得向量根据 R 旋转。 等号右边表示 向量 p , v 分别根据 R 旋转后叉乘,显然得到同一个向量。 证毕。 \begin{align*} 原式等效于证明\\ \bm{Rp^{\land}R^TR} & = \bm{(Rp)}^{\land}\bm{R} \\ \bm{Rp^{\land}\bm{I}} & = \bm{(Rp)}^{\land}\bm{R} \\ \bm{Rp^{\land}} & = \bm{(Rp)}^{\land}\bm{R} \\ 对于向量 \bm{v} \in \mathbb{R}^3\\ \bm{Rp^{\land}}\bm{v} & = \bm{(Rp)}^{\land}\bm{R}\bm{v} \\ 上式等号左边& 表示向量\bm{p,v}叉乘后所得向量根据 \bm{R}旋转。\\ 等号右边表示& 向量\bm{p,v}分别根据 \bm{R}旋转后叉乘,显然得到同一个向量。\\ 证毕。 \end{align*} 原式等效于证明Rp∧RTRRp∧IRp∧对于向量v∈R3Rp∧v上式等号左边等号右边表示证毕。=(Rp)∧R=(Rp)∧R=(Rp)∧R=(Rp)∧Rv表示向量p,v叉乘后所得向量根据R旋转。向量p,v分别根据R旋转后叉乘,显然得到同一个向量。

√ 题6


根据题 5 的结论 : ( R p ) ∧ = R p ∧ R T exp ⁡ ( ( R p ) ∧ ) = exp ⁡ ( R p ∧ R T ) 级数展开 = ∑ n = 0 ∞ ( R p ∧ R T ) n N ! = ∑ n = 0 ∞ R p ∧ R T ⋅ R p ∧ R T ⋅ , ⋅ ⋅ ⋅ , ⋅ R p ∧ R T ⋅ R p ∧ R T N ! 其中 R T R = I = ∑ n = 0 ∞ R ( p ∧ ) n R T N ! = R ∑ n = 0 ∞ ( p ∧ ) n N ! R T = R exp ⁡ ( p ∧ ) R T \begin{align*}根据题5 的结论&:\bm{(Rp)}^{\land} = \bm{Rp^{\land}R^T} \\ \exp((\bm{Rp})^{\land})& = \exp(\bm{Rp^{\land}R^T}) \\ 级数展开\\ &= \sum\limits_{n=0}^{\infty}\frac{(\bm{Rp^{\land}R^T}) ^n}{N!} \\ &= \sum\limits_{n=0}^{\infty}\frac{\bm{Rp^{\land}R^T}·\bm{Rp^{\land}R^T}·,···,·\bm{Rp^{\land}R^T}·\bm{Rp^{\land}R^T}}{N!} \\ 其中 \bm{R^TR=I}\\ & = \sum\limits_{n=0}^{\infty}\frac{\bm{R(p^{\land})^nR^T}}{{N!} }\\ & = \bm{R} \sum\limits_{n=0}^{\infty}\frac{\bm{(p^{\land})^n}}{{N!} } \bm{R}^T \\ & = \bm{R}\exp(\bm{p}^{\land})\bm{R}^T \end{align*} 根据题5的结论exp((Rp)∧)级数展开其中RTR=I:(Rp)∧=Rp∧RT=exp(Rp∧RT)=n=0∑∞N!(Rp∧RT)n=n=0∑∞N!Rp∧RT⋅Rp∧RT⋅,⋅⋅⋅,⋅Rp∧RT⋅Rp∧RT=n=0∑∞N!R(p∧)nRT=Rn=0∑∞N!(p∧)nRT=Rexp(p∧)RT

证毕。


6.2 SE(3)伴随性质

T exp ⁡ ( ξ ∧ ) T − 1 = T ∑ n = 0 ∞ ( ξ ∧ ) n n ! T − 1 由于 T − 1 T = I = ∑ n = 0 ∞ T ξ ∧ T − 1 ⋅ T ξ ∧ T − 1 ⋅ T ξ ∧ T − 1 ⋅ T ξ ∧ T − 1 ⋅ , ⋅ ⋅ ⋅ , T ξ ∧ T − 1 ⋅ T ξ ∧ T − 1 n ! = ∑ n = 0 ∞ ( T ξ ∧ T − 1 ) n n ! = exp ⁡ ( T ξ ∧ T − 1 ) \begin{align*} \bm{T}\exp(\bm{\xi}^{\land})\bm{T}^{-1} &= \bm{T}\sum\limits_{n=0}^{\infty}\frac{(\bm{\xi}^{\land})^n}{n!}\bm{T}^{-1} \\ 由于 \bm{T^{-1}T=I} \\ & = \sum\limits_{n=0}^{\infty}\frac{\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}·\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}·\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}·\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}·,···, \bm{T}\bm{\xi}^{\land}\bm{T}^{-1}·\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}}{n!} \\ & = \sum\limits_{n=0}^{\infty}\frac{(\bm{T}\bm{\xi}^{\land}\bm{T}^{-1})^n}{n!} \\ &= \exp(\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}) \\ \end{align*} Texp(ξ∧)T−1由于T−1T=I=Tn=0∑∞n!(ξ∧)nT−1=n=0∑∞n!Tξ∧T−1⋅Tξ∧T−1⋅Tξ∧T−1⋅Tξ∧T−1⋅,⋅⋅⋅,Tξ∧T−1⋅Tξ∧T−1=n=0∑∞n!(Tξ∧T−1)n=exp(Tξ∧T−1)

ξ = [ ρ ϕ ] , ξ ∧ = [ ϕ ∧ ρ 0 T 0 ] , T = [ R t 0 T 1 ] , T − 1 = [ R T − R T t 0 T 1 ] \bm{\xi=}\begin{bmatrix}\bm{\rho}\\ \bm{\phi} \end{bmatrix},\bm{\xi}^{\land}=\begin{bmatrix}\bm{\phi}^{\land} & \bm{\rho} \\ \bm{0}^T & 0 \end{bmatrix},\bm{T}=\begin{bmatrix}\bm{R} & \bm{t}\\ \bm{0}^T & 1 \end{bmatrix},\bm{T}^{-1}=\begin{bmatrix}\bm{R}^T & -\bm{R}^T\bm{t}\\ \bm{0}^T & 1 \end{bmatrix} ξ=[ρϕ],ξ∧=[ϕ∧0Tρ0],T=[R0Tt1],T−1=[RT0T−RTt1]

则:
T ξ ∧ T − 1 = [ R t 0 T 1 ] [ ϕ ∧ ρ 0 T 0 ] [ R T − R T t 0 T 1 ] = [ R ϕ ∧ R ρ 0 T 0 ] [ R T − R T t 0 T 1 ] = [ R ϕ ∧ R T − R ϕ ∧ R T t + R ρ 0 T 0 ] 由题 5 的结论 R p ∧ R T = ( R p ) ∧ = [ ( R ϕ ) ∧ − ( R ϕ ) ∧ t + R ρ 0 T 0 ] 对比 ξ , ξ ∧ 进行转换 = [ − ( R ϕ ) ∧ t + R ρ R ϕ ] ∧ 由叉乘性质, = [ t ∧ R ϕ + R ρ R ϕ ] ∧ = [ R t ∧ R 0 R ] [ ρ ϕ ] ) ∧ = ( A d ( T ) ξ ) ∧ \begin{align*}\bm{T}\bm{\xi}^{\land}\bm{T}^{-1}&= \begin{bmatrix}\bm{R} & \bm{t}\\ \bm{0}^T & 1 \end{bmatrix}\begin{bmatrix}\bm{\phi}^{\land} & \bm{\rho} \\ \bm{0}^T & 0 \end{bmatrix}\begin{bmatrix}\bm{R}^T & -\bm{R}^T\bm{t}\\ \bm{0}^T & 1 \end{bmatrix}\\ &= \begin{bmatrix}\bm{R}\bm{\phi}^{\land} & \bm{R\rho}\\ \bm{0}^T & 0 \end{bmatrix}\begin{bmatrix}\bm{R}^T & -\bm{R}^T\bm{t}\\ \bm{0}^T & 1 \end{bmatrix}\\ &= \begin{bmatrix}\bm{R}\bm{\phi}^{\land}\bm{R}^T & -\bm{R}\bm{\phi}^{\land}\bm{R}^T\bm{t + \bm{R\rho}}\\ \bm{0}^T & 0 \end{bmatrix} \\ 由题5的结论& \bm{Rp^{\land}R^T} = \bm{(Rp)}^{\land} \\ &= \begin{bmatrix}(\bm{R}\bm{\phi})^{\land} & -(\bm{R}\bm{\phi})^{\land}\bm{t + \bm{R\rho}}\\ \bm{0}^T &0 \end{bmatrix}\\ & 对比 \bm{\xi, {\xi}^{\land}}进行转换\\ &= \begin{bmatrix}-(\bm{R}\bm{\phi})^{\land}\bm{t + \bm{R\rho}}\\ \bm{R}\bm{\phi} \end{bmatrix}^{\land}\\ 由叉乘性质, \\ &= \begin{bmatrix}\bm{t}^{\land}\bm{R}\bm{\phi}+ \bm{R\rho}\\ \bm{R}\bm{\phi} \end{bmatrix}^{\land}\\ &= \begin{bmatrix}\bm{R}& \bm{t}^{\land}\bm{R} \\ \bm{0} & \bm{R} \end{bmatrix}\begin{bmatrix}\bm{\rho}\\ \bm{\phi} \end{bmatrix})^{\land}\\ & = (Ad(\bm{T})\bm{\xi})^{\land} \end{align*} Tξ∧T−1由题5的结论由叉乘性质,=[R0Tt1][ϕ∧0Tρ0][RT0T−RTt1]=[Rϕ∧0TRρ0][RT0T−RTt1]=[Rϕ∧RT0T−Rϕ∧RTt+Rρ0]Rp∧RT=(Rp)∧=[(Rϕ)∧0T−(Rϕ)∧t+Rρ0]对比ξ,ξ∧进行转换=[−(Rϕ)∧t+RρRϕ]∧=[t∧Rϕ+RρRϕ]∧=[R0t∧RR][ρϕ])∧=(Ad(T)ξ)∧

综上:
T exp ⁡ ( ξ ∧ ) T − 1 = exp ⁡ ( ( A d ( T ) ξ ) ∧ ) \bm{T}\exp(\bm{\xi}^{\land})\bm{T}^{-1}=\exp((Ad(\bm{T})\bm{\xi})^{\land}) Texp(ξ∧)T−1=exp((Ad(T)ξ)∧)

其中
A d ( T ) = [ R t ∧ R 0 R ] Ad(\bm{T})=\begin{bmatrix}\bm{R}& \bm{t}^{\land}\bm{R} \\ \bm{0} & \bm{R} \end{bmatrix} Ad(T)=[R0t∧RR]

证毕


√ 题7



SO(3):

设右扰动 Δ R Δ\bm{R} ΔR 对应的李代数 为 φ \varphi φ

∂ ( R p ) ∂ φ = lim ⁡ φ → 0 exp ⁡ ( ϕ ∧ ) exp ⁡ ( φ ∧ ) p − exp ⁡ ( ϕ ∧ ) p φ = lim ⁡ φ → 0 exp ⁡ ( ϕ ∧ ) ( I + φ ∧ ) p − exp ⁡ ( ϕ ∧ ) p φ = lim ⁡ φ → 0 exp ⁡ ( ϕ ∧ ) φ ∧ p φ = lim ⁡ φ → 0 R φ ∧ p φ 由题 5 : R p ∧ = ( R p ) ∧ R = lim ⁡ φ → 0 ( R φ ) ∧ R p φ 由叉乘性质 = lim ⁡ φ → 0 − ( R p ) ∧ R φ φ = − ( R p ) ∧ R \begin{align*}\frac{\partial(\bm{Rp})}{\partial\bm{\varphi}} &= \lim\limits_{\bm{\varphi}\to0}\frac{\exp(\bm{\phi}^{\land})\exp(\bm{\varphi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{\exp(\bm{\phi}^{\land})(\bm{I} + \bm{\varphi}^{\land})\bm{p}-\exp(\bm{\phi}^{\land})\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{\exp(\bm{\phi}^{\land})\bm{\varphi}^{\land}\bm{p}}{\bm{\varphi}}\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{\bm{R}\bm{\varphi}^{\land}\bm{p}}{\bm{\varphi}}\\ 由题5 : \bm{Rp^{\land}} & = \bm{(Rp)}^{\land}\bm{R} \\ &= \lim\limits_{\bm{\varphi}\to0}\frac{(\bm{R}\bm{\varphi})^{\land}\bm{Rp}}{\bm{\varphi}}\\ 由叉乘性质\\ &= \lim\limits_{\bm{\varphi}\to0}\frac{-(\bm{Rp})^{\land}\bm{R}\bm{\varphi}}{\bm{\varphi}}\\ & = -(\bm{Rp})^{\land}\bm{R} \end{align*} ∂φ∂(Rp)由题5:Rp∧由叉乘性质=φ→0limφexp(ϕ∧)exp(φ∧)p−exp(ϕ∧)p=φ→0limφexp(ϕ∧)(I+φ∧)p−exp(ϕ∧)p=φ→0limφexp(ϕ∧)φ∧p=φ→0limφRφ∧p=(Rp)∧R=φ→0limφ(Rφ)∧Rp=φ→0limφ−(Rp)∧Rφ=−(Rp)∧R

SE(3):

假设某空间点 p \bm{p} p 经过一次变换 T \bm{T} T (对应李代数为 ξ \bm{\xi} ξ), 得到 T p \bm{Tp} Tp

现在给 T \bm{T} T 右乘一个扰动 Δ T = exp ⁡ ( Δ ξ ∧ ) Δ\bm{T} = \exp(Δ\bm{\xi}^{\land}) ΔT=exp(Δξ∧)

设扰动项的李代数为 Δ ξ = [ Δ ρ , Δ ϕ ] T Δ\bm{\xi}=[Δ\bm{\rho},Δ\bm{\phi}]^T Δξ=[Δρ,Δϕ]T,则

∂ ( T p ) ∂ Δ ξ = lim ⁡ Δ ξ → 0 exp ⁡ ( ξ ∧ ) exp ⁡ ( Δ ξ ∧ ) p − exp ⁡ ( ξ ∧ ) p Δ ξ = lim ⁡ Δ ξ → 0 exp ⁡ ( ξ ∧ ) ( I + Δ ξ ∧ ) p − exp ⁡ ( ξ ∧ ) p Δ ξ = lim ⁡ Δ ξ → 0 exp ⁡ ( ξ ∧ ) Δ ξ ∧ p Δ ξ = lim ⁡ Δ ξ → 0 [ R t 0 1 ] [ Δ ϕ ∧ Δ ρ 0 T 0 ] p Δ ξ 把 p 前的 矩阵当做旋转矩阵,结合向量旋转的性质, = lim ⁡ Δ ξ → 0 [ R t 0 1 ] [ Δ ϕ ∧ p + Δ ρ 0 ] Δ ξ = lim ⁡ Δ ξ → 0 [ R Δ ϕ ∧ p + R Δ ρ 0 T ] [ Δ ρ , Δ ϕ ] T 将 Δ ρ , Δ ϕ 提出来,方便求导 根据 R p ∧ = ( R p ) ∧ R = lim ⁡ Δ ξ → 0 [ ( R Δ ϕ ) ∧ R p + R Δ ρ 0 T ] [ Δ ρ , Δ ϕ ] T = lim ⁡ Δ ξ → 0 [ − ( R p ) ∧ R Δ ϕ + R Δ ρ 0 T ] [ Δ ρ , Δ ϕ ] T = [ R − ( R p ) ∧ R 0 T 0 T ] \begin{align*}\frac{\partial(\bm{Tp})}{\partial{Δ\bm{\xi}}} &= \lim\limits_{Δ\bm{\xi}\to0}\frac{\exp(\bm{\xi}^{\land})\exp(Δ\bm{\xi}^{\land})\bm{p}-\exp(\bm{\xi}^{\land})\bm{p}}{Δ\bm{\xi}} \\ &= \lim\limits_{Δ\bm{\xi}\to0}\frac{\exp(\bm{\xi}^{\land})(\bm{I} +Δ\bm{\xi}^{\land})\bm{p}-\exp(\bm{\xi}^{\land})\bm{p}}{Δ\bm{\xi}} \\ &= \lim\limits_{Δ\bm{\xi}\to0}\frac{\exp(\bm{\xi}^{\land})Δ\bm{\xi}^{\land}\bm{p}}{Δ\bm{\xi}} \\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} \bm{R} & \bm{t}\\ 0 & 1 \end{bmatrix}\begin{bmatrix} Δ\bm{\phi}^{\land} & \Delta\bm{\rho}\\ \bm{0}^T & 0 \end{bmatrix}\bm{p}}{Δ\bm{\xi}}\\ 把\bm{p}前的&矩阵当做旋转矩阵,结合向量旋转的性质,\\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} \bm{R} & \bm{t}\\ 0 & 1 \end{bmatrix}\begin{bmatrix} Δ\bm{\phi}^{\land}\bm{p} + \Delta\bm{\rho}\\ 0 \end{bmatrix}}{Δ\bm{\xi}}\\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} \bm{R}\Delta\bm{\phi}^{\land}\bm{p}+ \bm{R}\Delta\bm{\rho}\\ \bm{0}^T \end{bmatrix}}{[Δ\bm{\rho},Δ\bm{\phi}]^T} \\ 将&Δ\bm{\rho},Δ\bm{\phi} 提出来,方便求导\\ & 根据 \bm{Rp^{\land}} = \bm{(Rp)}^{\land}\bm{R} \\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} (\bm{R}\Delta\bm{\phi})^{\land}\bm{R}\bm{p}+\bm{R}\Delta\bm{\rho}\\ \bm{0}^T \end{bmatrix}}{[Δ\bm{\rho},Δ\bm{\phi}]^T} \\ &=\lim\limits_{Δ\bm{\xi}\to0} \frac{\begin{bmatrix} -(\bm{R}\bm{p})^{\land}\bm{R}\Delta\bm{\phi}+ \bm{R}\Delta\bm{\rho}\\ \bm{0}^T \end{bmatrix}}{[Δ\bm{\rho},Δ\bm{\phi}]^T} \\ & = \begin{bmatrix} \bm{R} & -(\bm{R}\bm{p})^{\land}\bm{R} \\ \bm{0}^T & \bm{0}^T \end{bmatrix} \\ \end{align*} ∂Δξ∂(Tp)把p前的将=Δξ→0limΔξexp(ξ∧)exp(Δξ∧)p−exp(ξ∧)p=Δξ→0limΔξexp(ξ∧)(I+Δξ∧)p−exp(ξ∧)p=Δξ→0limΔξexp(ξ∧)Δξ∧p=Δξ→0limΔξ[R0t1][Δϕ∧0TΔρ0]p矩阵当做旋转矩阵,结合向量旋转的性质,=Δξ→0limΔξ[R0t1][Δϕ∧p+Δρ0]=Δξ→0lim[Δρ,Δϕ]T[RΔϕ∧p+RΔρ0T]Δρ,Δϕ提出来,方便求导根据Rp∧=(Rp)∧R=Δξ→0lim[Δρ,Δϕ]T[(RΔϕ)∧Rp+RΔρ0T]=Δξ→0lim[Δρ,Δϕ]T[−(Rp)∧RΔϕ+RΔρ0T]=[R0T−(Rp)∧R0T]


√ 题8

cmake的 find_package指令:

官方文档

bash 复制代码
find_package(包名 版本 REQUIRED)

一些包需要 sudo make install后才能被找到


LaTex

bash 复制代码
$\mathfrak{g}$

g \mathfrak{g} g

bash 复制代码
$\mathbb{R}$

R \mathbb{R} R

bash 复制代码
$\times$

× \times ×

bash 复制代码
$\varphi$

φ \varphi φ

bash 复制代码
$\Phi$

Φ \Phi Φ

bash 复制代码
$\overset{\mathrm{def}}{=}$
$\xlongequal{def}$

= d e f \overset{\mathrm{def}}{=} =def
= d e f \xlongequal{def} def

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