0x00 前言

《视觉slam十四讲从理论到实践》第四讲习题自测解析。
借助自身知识储备和搜索引擎后完成习题，仅供参考。
部分答案会觉得没有说明的必要就会略

0x01 前置知识

李群的定义

群(Group)是一种集合加上一种运算的代数结构。我们把集合记作

$A$ ，运算记作

⋅

\cdot

$\cdot$ ，那么群可以记作

(

⋅

)

G=(A,\cdot)

$G = (A, \cdot)$ 。群要求这个运算满足以下几个条件：

封闭性: $\forall a_1,a_2\in A,\quad a_1\cdot a_2\in A.$
结合律: $\forall a_1,a_2,a_3\in A,\quad (a_1\cdot a_2)\cdot a_3=a_1\cdot(a_2\cdot a_3).$
幺元: $\exist a_0\in A,\quad s.t.\quad\forall a\in A,\quad a_0\cdot a=a\cdot a_0=a.$
逆: $\forall a\in A,\quad\exist a^{-1}\in A,\quad s.t.\quad a\cdot a^{-1}=a_0.$

李代数的定义

李代数由一个集合

\mathbb{V}

$V$ 、一个数域

\mathbb{F}

$F$ 和一个二元运算

[

]

[,]

$[,]$ 组成。如果它们满足以下几个性质，则称

(

[

]

)

(\mathbb{V},\mathbb{F},[,])

$(V, F, [,])$ 为一个李代数，记作

$g$ 。

封闭性 $\forall X,Y\in\mathbb{V},[X,Y]\in\mathbb{V}.$
双线性 $\forall X,Y,Z\in\mathbb{V},a,b\in\mathbb{F},有$

[

a

X

+

b

Y

,

Z

]

=

a

[

X

,

Z

]

+

b

[

Y

,

Z

]

,

[

Z

,

a

X

+

b

Y

]

=

a

[

Z

,

X

]

+

b

[

Z

,

Y

]

.

[aX+bY,Z]=a[X,Z]+b[Y,Z],\qquad[Z,aX+bY]=a[Z,X]+b[Z,Y].

$[a X + b Y, Z] = a [X, Z] + b [Y, Z], [Z, a X + b Y] = a [Z, X] + b [Z, Y] .$
自反性 $\forall X\in\mathbb{V},[X,X]=0.$
雅可比等价 $\forall X,Y,Z\in\mathbb{V},[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=0.$

0x02 习题部分

1、验证 $S O (3) 、 S E (3) 和 S i m (3)$ 关于乘法成群。

验证 $S O (3)$

(

)

{

∈

∣

(

)

}

SO(3)=\left\{R\in\mathbb{R}^{3\times3}|RR^T=I,det(R)=1\right\}

$S O (3) = {R \in R^{3 \times 3} ∣ R R^{T} = I, d e t (R) = 1}$

封闭性

从旋转矩阵的定义正交性入手，设

∈

R=R_1R_2\in\mathbb{R}^{3\times3}

$R = R_{1} R_{2} \in R^{3 \times 3}$ ,

(

)

RR^T=R_1R_2(R_1R_2)^T=R_1R_2R_2^TR_1^T=R_1IR_1^T=I

$R R^{T} = R_{1} R_{2} (R_{1} R_{2})^{T} = R_{1} R_{2} R_{2}^{T} R_{1}^{T} = R_{1} I R_{1}^{T} = I$
所以

∈

(

)

R=R_1R_2\in SO(3)

$R = R_{1} R_{2} \in S O (3)$

满足封闭性

结合律

由矩阵乘法规律可知，矩阵乘法满足结合律，即满足

(

⋅

)

⋅

(

⋅

)

(R_1\cdot R_2)\cdot R_3=R_1\cdot (R_2\cdot R_3)

$(R_{1} \cdot R_{2}) \cdot R_{3} = R_{1} \cdot (R_{2} \cdot R_{3})$ 。

满足结合律

幺元

存在

R=I,

$R = I,$ 使得

IR_1=R_1I=R_1

$I R_{1} = R_{1} I = R_{1}$ .

满足幺元

逆

设矩阵

∈

(

)

R\in SO(3)

$R \in S O (3)$ ，由正交性可知，

∃

−

\exist R^{-1}=R^T

$\exists R^{- 1} = R^{T}$ ，使得

−

RR^{-1}=I.

$R R^{- 1} = I .$

满足逆

验证 $S E (3)$

(

)

{

[

]

∈

∣

∈

(

)

∈

}

SE(3)=\left\{T=\left[\begin{matrix}R&t\\0^T&1\end{matrix}\right]\in\mathbb{R}^{4\times4}|R\in SO(3),t\in\mathbb{R}^3\right\}

$S E (3) = {T = [R 0^{T} t 1] \in R^{4 \times 4} ∣ R \in S O (3), t \in R^{3}}$

封闭性

设

∈

(

)

T_1,T_2\in SE(3)

$T_{1}, T_{2} \in S E (3)$ ,

[

]

[

]

[

]

T_1T_2=\left[\begin{matrix}R_1&t_1\\0^T&1\end{matrix}\right] \left[\begin{matrix}R_2&t_2\\0^T&1\end{matrix}\right]= \left[\begin{matrix}R_1R_2&R_1t_2+t_1\\0^T&1\end{matrix}\right]

$T_{1} T_{2} = [R_{1} 0^{T} t_{1} 1] [R_{2} 0^{T} t_{2} 1] = [R_{1} R_{2} 0^{T} R_{1} t_{2} + t_{1} 1]$
由

(

)

SO(3)

$S O (3)$ 的封闭性可知，

∈

(

)

R_1R_2\in SO(3)

$R_{1} R_{2} \in S O (3)$ ,由矩阵的运算规律可知

∈

R_1t_2+t_1\in\mathbb{R}^3

$R_{1} t_{2} + t_{1} \in R^{3}$
所以

∈

(

)

T_1T_2\in SE(3)

$T_{1} T_{2} \in S E (3)$

满足封闭性

结合律

由矩阵乘法规律可知，矩阵乘法满足结合律，即满足

(

⋅

)

⋅

(

⋅

)

(T_1\cdot T_2)\cdot T_3=T_1\cdot (T_2\cdot T_3)

$(T_{1} \cdot T_{2}) \cdot T_{3} = T_{1} \cdot (T_{2} \cdot T_{3})$ 。

满足结合律

幺元

存在

T=I,

$T = I,$ 使得

IT_1=T_1I=T_1

$I T_{1} = T_{1} I = T_{1}$ .

满足幺元

逆

设矩阵

∈

(

)

T\in SE(3)

$T \in S E (3)$ ，易得

∈

(

)

R\in SO(3)

$R \in S O (3)$ ,是一个满秩矩阵，所以

$T$ 是一个满秩矩阵，所以

$T$ 可逆，

∃

−

\exist\;T^{-1}

$\exists T^{- 1}$ ，使得

−

TT^{-1}=I.

$T T^{- 1} = I .$

满足逆

验证 $S i m (3)$

略

2、验证 $(\mathbb{R}^3,\mathbb{R},\times)$ 构成李代数。

封闭性

设

⃗

∈

\vec{a},\vec{b}\in\mathbb{R}^3

$a, b \in R^{3}$ ,
根据叉积的定义：

⃗

∣

⃗

∣

(

−

)

⃗

(

−

)

⃗

(

−

)

⃗

[

−

]

∈

(2.1)

\vec{a}\times\vec{b}=\left|\begin{matrix} \vec{i}&\vec{j}&\vec{k}\\ a_1& a_2&a_3\\ b_1&b_2&b_3 \end{matrix}\right|=(a_2b_3-a_3b_2)\vec{i}+(a_3b_1-a_1b_3)\vec{j}+(a_1b_2-a_2b_1)\vec{k}=\left[\begin{matrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1 \end{matrix}\right]\in\mathbb{R}^3\tag{2.1}

$a \times b = ∣ ∣ ∣ ∣ ∣ ∣ i a_{1} b_{1} j a_{2} b_{2} k a_{3} b_{3} ∣ ∣ ∣ ∣ ∣ ∣ = (a_{2} b_{3} - a_{3} b_{2}) i + (a_{3} b_{1} - a_{1} b_{3}) j + (a_{1} b_{2} - a_{2} b_{1}) k = ⎣ ⎡ a_{2} b_{3} - a_{3} b_{2} a_{3} b_{1} - a_{1} b_{3} a_{1} b_{2} - a_{2} b_{1} ⎦ ⎤ \in R^{3} (2.1)$
满足封闭性

双线性

设

⃗

∈

\vec{a},\vec{b},\vec{c}\in\mathbb{R}^3,\alpha,\beta\in\mathbb{R}

$a, b, c \in R^{3}, α, β \in R$ ,

[

]

(

⃗

)

⃗

⋯

分

配

律

(

⃗

)

(

⃗

)

⋯

常

数

系

数

完

全

可

以

提

出

来

[

]

[

]

(2.2)

\begin{aligned} &\quad[aX+bY,Z]=(\alpha\vec{a}+\beta\vec{b})\times\vec{c}\\ &=\alpha\vec{a}\times\vec{c}+\beta\vec{b}\times\vec{c}&\cdots&分配律\\ &=\alpha(\vec{a}\times\vec{c})+\beta(\vec{b}\times\vec{c})&\cdots&常数系数完全可以提出来\tag{2.2}\\ &=a[X,Z]+b[Y,Z] \end{aligned}

$[a X + b Y, Z] = (α a + β b) \times c = α a \times c + β b \times c = α (a \times c) + β (b \times c) = a [X, Z] + b [Y, Z] \dots \dots 分配律常数系数完全可以提出来 (2.2)$
同理

[

]

(

⃗

)

(

⃗

)

[

]

[

]

[Z,aX+bY]=\alpha(\vec{c}\times\vec{a})+\beta(\vec{c}\times\vec{b})=a[Z,X]+b[Z,Y]

$[Z, a X + b Y] = α (c \times a) + β (c \times b) = a [Z, X] + b [Z, Y]$
满足双线性

自反性

设

⃗

∈

\vec{a}\in\mathbb{R}^3

$a \in R^{3}$

⃗

∣

⃗

∣

sin

⁡

⃗

(2.3)

\vec{a}\times\vec{a}={\left|\vec{a}\right|}^2\sin{<\vec{a},\vec{a}>}=0\tag{2.3}

$a \times a = ∣ a ∣^{2} sin < a, a > = 0 (2.3)$
满足自反性

雅可比等价

设

⃗

∈

\vec{a},\vec{b},\vec{c}\in\mathbb{R}^3

$a, b, c \in R^{3}$ ,

[

]

[

]

[

]

⃗

(

⃗

)

⃗

(

⃗

)

⃗

(

⃗

)

(2.4)

[X,[Y,Z]]+[Z,[X,Y]]+[Y,[Z,X]]=\vec{a}\times(\vec{b}\times\vec{c})+\vec{c}\times(\vec{a}\times\vec{b})+\vec{b}\times(\vec{c}\times\vec{a})\tag{2.4}

$[X, [Y, Z]] + [Z, [X, Y]] + [Y, [Z, X]] = a \times (b \times c) + c \times (a \times b) + b \times (c \times a) (2.4)$
将

⃗

\vec{a},\vec{b},\vec{c}

$a, b, c$ 依次轮换得

⃗

(

⃗

)

⃗

(

⃗

)

⃗

(

⃗

)

(2.5)

\vec{b}\times(\vec{c}\times\vec{a})+\vec{a}\times(\vec{b}\times\vec{c})+\vec{c}\times(\vec{a}\times\vec{b})\tag{2.5}

$b \times (c \times a) + a \times (b \times c) + c \times (a \times b) (2.5)$
易得式

(

2.4

)

(2.4)

$(2.4)$ 与式

(

2.5

)

(2.5)

$(2.5)$ 相等，所以可知

⃗

\vec{a},\vec{b},\vec{c}

$a, b, c$ 轮换等价。那么可令

⃗

\vec{b}=\vec{a},\vec{c}=\vec{a}

$b = a, c = a$ ,代入

(

2.4

)

(2.4)

$(2.4)$ 得

⃗

(

⃗

)

⋯

自

反

性

(2.6)

3\vec{a}\times(\vec{a}\times\vec{a})=0\qquad\cdots自反性\tag{2.6}

$3 a \times (a \times a) = 0 \dots 自反性 (2.6)$
满足雅可比等价
注：这样子验证可能不严谨，高中遗留下来的做题技巧

3、验证 $s o (3)$ 和 $s e (3)$ 满足李代数要求的性质。

验证 $s o (3)$

⃗

∧

[

−

⃗

−

⃗

−

⃗

]

\Phi={\vec\phi}^\wedge=\left[ \begin{matrix} 0&-\vec\phi_3&\vec\phi_2\\ \vec\phi_3&0&-\vec\phi_1\\ -\vec\phi_2&\vec\phi_1&0 \end{matrix}\right]

$Φ = ϕ^{\land} = ⎣ ⎢ ⎡ 0 ϕ_{3} - ϕ_{2} - ϕ_{3} 0 ϕ_{1} ϕ_{2} - ϕ_{1} 0 ⎦ ⎥ ⎤$

(

)

{

∈

∧

∈

}

so(3)=\left\{\phi\in\mathbb{R}^3,\Phi=\phi^\wedge\in\mathbb{R}^{3\times3}\right\}

$s o (3) = {ϕ \in R^{3}, Φ = ϕ^{\land} \in R^{3 \times 3}}$

[

⃗

]

(

−

)

∨

(3.1)

[\vec\phi_1,\vec\phi_2]=(\Phi_1\Phi_2-\Phi_2\Phi_1)^\vee\tag{3.1}

$[ϕ_{1}, ϕ_{2}] = (Φ_{1} Φ_{2} - Φ_{2} Φ_{1})^{\lor} (3.1)$

封闭性

设

⃗

[

]

∈

\vec{\phi}=[x,y,z]^T\in\mathbb{R}^3

$ϕ = [x, y, z]^{T} \in R^{3}$ ,

[

⃗

]

(

−

)

∨

[\vec\phi_1,\vec\phi_2]=(\Phi_1\Phi_2-\Phi_2\Phi_1)^\vee

$[ϕ_{1}, ϕ_{2}] = (Φ_{1} Φ_{2} - Φ_{2} Φ_{1})^{\lor}$

(

[

−

]

[

−

]

−

[

−

]

[

−

]

)

∨

=\left(\left[\begin{matrix}0&-z_1&y_1\\ z_1&0&-x_1\\ -y_1&x_1&0\end{matrix}\right] \left[\begin{matrix} 0&-z_2&y_2\\ z_2&0&-x_2\\ -y_2&x_2&0\end{matrix}\right] – \left[\begin{matrix} 0&-z_2&y_2\\ z_2&0&-x_2\\ -y_2&x_2&0 \end{matrix}\right] \left[\begin{matrix} 0&-z_1&y_1\\ z_1&0&-x_1\\ -y_1&x_1&0 \end{matrix}\right] \right)^\vee

$= ⎝ ⎛ ⎣ ⎡ 0 z_{1} - y_{1} - z_{1} 0 x_{1} y_{1} - x_{1} 0 ⎦ ⎤ ⎣ ⎡ 0 z_{2} - y_{2} - z_{2} 0 x_{2} y_{2} - x_{2} 0 ⎦ ⎤ - ⎣ ⎡ 0 z_{2} - y_{2} - z_{2} 0 x_{2} y_{2} - x_{2} 0 ⎦ ⎤ ⎣ ⎡ 0 z_{1} - y_{1} - z_{1} 0 x_{1} y_{1} - x_{1} 0 ⎦ ⎤ ⎠ ⎞^{\lor}$

(

[

−

]

−

[

−

]

)

∨

=\left(\left[\begin{matrix} -z_1z_2-y_1y_2&x_2y_1&x_2z_1\\ x_1y_2&-z_1z_2-x_1x_2&y_2z_1\\ x_1z_2&y_1z_2&-y_1y_2-x_1x_2 \end{matrix}\right]- \left[\begin{matrix} -z_1z_2-y_1y_2&x_1y_2&x_1z_2\\ x_2y_1&-z_1z_2-x_1x_2&y_1z_2\\ x_2z_1&y_2z_1&-y_1y_2-x_1x_2 \end{matrix}\right]\right)^{\vee}

$= ⎝ ⎛ ⎣ ⎡ - z_{1} z_{2} - y_{1} y_{2} x_{1} y_{2} x_{1} z_{2} x_{2} y_{1} - z_{1} z_{2} - x_{1} x_{2} y_{1} z_{2} x_{2} z_{1} y_{2} z_{1} - y_{1} y_{2} - x_{1} x_{2} ⎦ ⎤ - ⎣ ⎡ - z_{1} z_{2} - y_{1} y_{2} x_{2} y_{1} x_{2} z_{1} x_{1} y_{2} - z_{1} z_{2} - x_{1} x_{2} y_{2} z_{1} x_{1} z_{2} y_{1} z_{2} - y_{1} y_{2} - x_{1} x_{2} ⎦ ⎤ ⎠ ⎞^{\lor}$

(

[

−

]

)

∨

(

⃗

∧

)

∨

(3.2)

=\left(\left[\begin{matrix} 0&x_2y_1-x_1y_2&x_2z_1-x_1z_2\\ x_1y_2-x_2y_1&0&y_2z_1-y_1z_2\\ x_1z_2-x_2z_1&y_1z_2-y_2z_1&0 \end{matrix}\right] \right)^{\vee}=\left({\vec\phi_3}^\wedge\right)^\vee \tag{3.2}

$= ⎝ ⎛ ⎣ ⎡ 0 x_{1} y_{2} - x_{2} y_{1} x_{1} z_{2} - x_{2} z_{1} x_{2} y_{1} - x_{1} y_{2} 0 y_{1} z_{2} - y_{2} z_{1} x_{2} z_{1} - x_{1} z_{2} y_{2} z_{1} - y_{1} z_{2} 0 ⎦ ⎤ ⎠ ⎞^{\lor} = (ϕ_{3}^{\land})^{\lor} (3.2)$
由

(

3.2

)

(3.2)

$(3.2)$ 易得

⃗

∧

{\vec\phi_3}^\wedge

$ϕ_{3}^{\land}$ 是一个反对称矩阵，所以可以经由运算

[

∨

]

[\vee]

$[\lor]$ 从反对称矩阵到向量的变换。所以

(

⃗

∧

)

∨

⃗

∈

\left({\vec\phi_3}^\wedge\right)^\vee=\vec\phi_3\in\mathbb{R}^3

$(ϕ_{3}^{\land})^{\lor} = ϕ_{3} \in R^{3}$
满足封闭性

双线性

设

⃗

∈

\vec\phi_1,\vec\phi_2,\vec\phi_3\in\mathbb{R}^3,\;a,b\in\mathbb{R}

$ϕ_{1}, ϕ_{2}, ϕ_{3} \in R^{3}, a, b \in R$ ，

[

⃗

]

(

⃗

)

∧

⃗

∧

−

⃗

∧

(

⃗

)

∧

)

∨

(

⃗

∧

⃗

∧

⃗

∧

⃗

∧

−

(

⃗

∧

⃗

∧

⃗

∧

⃗

∧

)

∨

(

⃗

∧

⃗

∧

−

⃗

∧

⃗

∧

)

(

⃗

∧

⃗

∧

−

⃗

∧

⃗

∧

)

∨

(3.3)

\begin{aligned} &\qquad[a\vec\phi_1+b\vec\phi_2,\vec\phi_3]\\ &=\left((a\vec\phi_1+b\vec\phi_2)^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge(a\vec\phi_1+b\vec\phi_2)^\wedge\right)^\vee\\ &=\left(a{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge+b{\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-(a{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge+b{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge)\right)^\vee\\ &=\left(a({\vec\phi_1}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge)+b({\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge)\right)^\vee \end{aligned}\tag{3.3}

$[a ϕ_{1} + b ϕ_{2}, ϕ_{3}] = ((a ϕ_{1} + b ϕ_{2})^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} (a ϕ_{1} + b ϕ_{2})^{\land})^{\lor} = (a ϕ_{1}^{\land} ϕ_{3}^{\land} + b ϕ_{2}^{\land} ϕ_{3}^{\land} - (a ϕ_{3}^{\land} ϕ_{1}^{\land} + b ϕ_{3}^{\land} ϕ_{2}^{\land}))^{\lor} = (a (ϕ_{1}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{1}^{\land}) + b (ϕ_{2}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{2}^{\land}))^{\lor} (3.3)$
由封闭性可知，

(

⃗

∧

⃗

∧

−

⃗

∧

⃗

∧

)

({\vec\phi_1}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge)

$(ϕ_{1}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{1}^{\land})$ 是反对称矩阵，所以可以得到

(

⃗

∧

⃗

∧

−

⃗

∧

⃗

∧

)

∨

(

⃗

∧

⃗

∧

−

⃗

∧

⃗

∧

)

∨

[

⃗

]

[

⃗

]

(3.4)

\begin{aligned} &=a({\vec\phi_1}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge)^\vee+b({\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge)^\vee\\ &=a[\vec\phi_1,\vec\phi_3]+b[\vec\phi_2,\vec\phi_3] \end{aligned}\tag{3.4}

$= a (ϕ_{1}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{1}^{\land})^{\lor} + b (ϕ_{2}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{2}^{\land})^{\lor} = a [ϕ_{1}, ϕ_{3}] + b [ϕ_{2}, ϕ_{3}] (3.4)$
同理可得

[

⃗

]

[

⃗

]

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(3.5)

[\vec\phi_3,a\vec\phi_1+b\vec\phi_2]=a[\vec\phi_3,\vec\phi_1]+b[\vec\phi_3,\vec\phi_2]\tag{3.5}

$[ϕ_{3}, a ϕ_{1} + b ϕ_{2}] = a [ϕ_{3}, ϕ_{1}] + b [ϕ_{3}, ϕ_{2}] (3.5)$
满足双线性

自反性

设

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\vec\phi\in\mathbb{R}^3

$ϕ \in R^{3}$ ，

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很

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(3.6)

[\vec\phi,\vec\phi]=\left({\vec\phi}^\wedge{\vec\phi}^\wedge-{\vec\phi}^\wedge{\vec\phi}^\wedge\right)^\vee=0^T\qquad\cdots很显然\tag{3.6}

$[ϕ, ϕ] = (ϕ^{\land} ϕ^{\land} - ϕ^{\land} ϕ^{\land})^{\lor} = 0^{T} \dots 很显然 (3.6)$
满足自反性

雅可比等价

设

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\vec\phi_1,\vec\phi_2,\vec\phi_3\in\mathbb{R}^3

$ϕ_{1}, ϕ_{2}, ϕ_{3} \in R^{3}$ ，

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\begin{aligned} &\qquad[\vec\phi_1,[\vec\phi_2,\vec\phi_3]]+[\vec\phi_3,[\vec\phi_1,\vec\phi_2]]+[\vec\phi_2,[\vec\phi_3,\vec\phi_1]]\\ &=\quad{\vec\phi_1}^\wedge({\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge)-({\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge){\vec\phi_1}^\wedge\\ &\quad+{\vec\phi_3}^\wedge({\vec\phi_1}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge)-({\vec\phi_1}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge){\vec\phi_3}^\wedge\\ &\quad+{\vec\phi_2}^\wedge({\vec\phi_3}^\wedge{\vec\phi_1}^\wedge-{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge)-({\vec\phi_3}^\wedge{\vec\phi_1}^\wedge-{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge){\vec\phi_2}^\wedge\\ &=\quad{\vec\phi_1}^\wedge{\vec\phi_2}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge+{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge\\ &\quad+{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge-{\vec\phi_1}^\wedge{\vec\phi_2}^\wedge{\vec\phi_3}^\wedge+{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge\\ &\quad+{\vec\phi_2}^\wedge{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge-{\vec\phi_3}^\wedge{\vec\phi_1}^\wedge{\vec\phi_2}^\wedge+{\vec\phi_1}^\wedge{\vec\phi_3}^\wedge{\vec\phi_2}^\wedge\\ &=0 \end{aligned}

$[ϕ_{1}, [ϕ_{2}, ϕ_{3}]] + [ϕ_{3}, [ϕ_{1}, ϕ_{2}]] + [ϕ_{2}, [ϕ_{3}, ϕ_{1}]] = ϕ_{1}^{\land} (ϕ_{2}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{2}^{\land}) - (ϕ_{2}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{2}^{\land}) ϕ_{1}^{\land} + ϕ_{3}^{\land} (ϕ_{1}^{\land} ϕ_{2}^{\land} - ϕ_{2}^{\land} ϕ_{1}^{\land}) - (ϕ_{1}^{\land} ϕ_{2}^{\land} - ϕ_{2}^{\land} ϕ_{1}^{\land}) ϕ_{3}^{\land} + ϕ_{2}^{\land} (ϕ_{3}^{\land} ϕ_{1}^{\land} - ϕ_{1}^{\land} ϕ_{3}^{\land}) - (ϕ_{3}^{\land} ϕ_{1}^{\land} - ϕ_{1}^{\land} ϕ_{3}^{\land}) ϕ_{2}^{\land} = ϕ_{1}^{\land} ϕ_{2}^{\land} ϕ_{3}^{\land} - ϕ_{1}^{\land} ϕ_{3}^{\land} ϕ_{2}^{\land} - ϕ_{2}^{\land} ϕ_{3}^{\land} ϕ_{1}^{\land} + ϕ_{3}^{\land} ϕ_{2}^{\land} ϕ_{1}^{\land} + ϕ_{3}^{\land} ϕ_{1}^{\land} ϕ_{2}^{\land} - ϕ_{3}^{\land} ϕ_{2}^{\land} ϕ_{1}^{\land} - ϕ_{1}^{\land} ϕ_{2}^{\land} ϕ_{3}^{\land} + ϕ_{2}^{\land} ϕ_{1}^{\land} ϕ_{3}^{\land} + ϕ_{2}^{\land} ϕ_{3}^{\land} ϕ_{1}^{\land} - ϕ_{2}^{\land} ϕ_{1}^{\land} ϕ_{3}^{\land} - ϕ_{3}^{\land} ϕ_{1}^{\land} ϕ_{2}^{\land} + ϕ_{1}^{\land} ϕ_{3}^{\land} ϕ_{2}^{\land} = 0$
满足雅可比等价

验证 $s e (3)$

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se(3)=\left\{\xi=\left[\begin{matrix}\rho\\\phi\end{matrix}\right]\in\mathbb{R}^6,\rho\in\mathbb{R}^3,\phi\in so(3),\xi^\wedge=\left[\begin{matrix}\phi^\wedge&\rho\\0^T&0\end{matrix}\right]\in\mathbb{R^{4\times4}}\right\}

$s e (3) = {ξ = [ρ ϕ] \in R^{6}, ρ \in R^{3}, ϕ \in s o (3), ξ^{\land} = [ϕ^{\land} 0^{T} ρ 0] \in R^{4 \times 4}}$

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[\vec\xi_1,\vec\xi_2]=\left({\vec\xi_1}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge\right)^\vee

$[ξ_{1}, ξ_{2}] = (ξ_{1}^{\land} ξ_{2}^{\land} - ξ_{2}^{\land} ξ_{1}^{\land})^{\lor}$

封闭性

设

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$ξ_{1}, ξ_{2} \in s e (3)$ ,

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[\vec\xi_1,\vec\xi_2]=\left({\vec\xi_1}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge\right)^\vee

$[ξ_{1}, ξ_{2}] = (ξ_{1}^{\land} ξ_{2}^{\land} - ξ_{2}^{\land} ξ_{1}^{\land})^{\lor}$

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(3.7)

=\left( \left[\begin{matrix} {\vec\phi_1}^\wedge&\vec\rho_1\\0^T&0 \end{matrix}\right] \left[\begin{matrix} {\vec\phi_2}^\wedge&\vec\rho_2\\0^T&0 \end{matrix}\right]- \left[\begin{matrix} {\vec\phi_2}^\wedge&\vec\rho_2\\0^T&0 \end{matrix}\right] \left[\begin{matrix} {\vec\phi_1}^\wedge&\vec\rho_1\\0^T&0 \end{matrix}\right] \right)^\vee\tag{3.7}

$= ([ϕ_{1}^{\land} 0^{T} ρ_{1} 0] [ϕ_{2}^{\land} 0^{T} ρ_{2} 0] - [ϕ_{2}^{\land} 0^{T} ρ_{2} 0] [ϕ_{1}^{\land} 0^{T} ρ_{1} 0])^{\lor} (3.7)$

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(3.8)

=\left(\left[\begin{matrix} {\vec\phi_1}^\wedge{\vec\phi_2}^\wedge&{\vec\phi_1}^\wedge\vec\rho_2\\0^T&0 \end{matrix}\right]- \left[\begin{matrix} {\vec\phi_2}^\wedge{\vec\phi_1}^\wedge&{\vec\phi_2}^\wedge\vec\rho_1\\0^T&0 \end{matrix}\right] \right)^\vee\tag{3.8}

$= ([ϕ_{1}^{\land} ϕ_{2}^{\land} 0^{T} ϕ_{1}^{\land} ρ_{2} 0] - [ϕ_{2}^{\land} ϕ_{1}^{\land} 0^{T} ϕ_{2}^{\land} ρ_{1} 0])^{\lor} (3.8)$

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(3.9)

=\left( \left[\begin{matrix} {\vec\phi_1}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge&{\vec\phi_1}^\wedge\vec\rho_2-{\vec\phi_2}^\wedge\vec\rho_1\\0^T&0 \end{matrix}\right] \right)^\vee\tag{3.9}

$= ([ϕ_{1}^{\land} ϕ_{2}^{\land} - ϕ_{2}^{\land} ϕ_{1}^{\land} 0^{T} ϕ_{1}^{\land} ρ_{2} - ϕ_{2}^{\land} ρ_{1} 0])^{\lor} (3.9)$
由

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{\vec\phi_1}^\wedge{\vec\phi_2}^\wedge-{\vec\phi_2}^\wedge{\vec\phi_1}^\wedge={\vec\phi_3}^\wedge,\;\vec\phi_3\in so(3)

$ϕ_{1}^{\land} ϕ_{2}^{\land} - ϕ_{2}^{\land} ϕ_{1}^{\land} = ϕ_{3}^{\land}, ϕ_{3} \in s o (3)$ ；由矩阵的运算可知

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{\vec\phi_1}^\wedge\vec\rho_2-{\vec\phi_2}^\wedge\vec\rho_1\in\mathbb{R}^3

$ϕ_{1}^{\land} ρ_{2} - ϕ_{2}^{\land} ρ_{1} \in R^{3}$ ，所以可令

式

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(3.10)

式(3.9)=\left({\vec\xi_3}^\wedge\right)^\vee=\vec\xi_3\in se(3)\tag{3.10}

$式 (3.9) = (ξ_{3}^{\land})^{\lor} = ξ_{3} \in s e (3) (3.10)$
满足封闭性

双线性

设

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$ξ_{1}, ξ_{2}, ξ_{3} \in s e (3)$ ,

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(3.11)

\begin{aligned} &\qquad[a\vec\xi_1+b\vec\xi_2,\vec\xi_3]\\ &=\left((a\vec\xi_1+b\vec\xi_2)^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge(a\vec\xi_1+b\vec\xi_2)^\wedge\right)^\vee\\ &=\left((a{\vec\xi_1}^\wedge+{b\vec\xi_2}^\wedge){\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge(a{\vec\xi_1}^\wedge+{b\vec\xi_2}^\wedge)\right)^\vee\\ &=\left(a{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge+b{\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-(a{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge+b{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge)\right)^\vee\\ &=\left(a({\vec\xi_1}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge)+b({\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge)\right)^\vee\\ &=a({\vec\xi_1}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge)^\vee+b({\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge)^\vee\\ &=a[\vec\xi_1,\vec\xi_3]+b[\vec\xi_2,\vec\xi_3] \end{aligned}\tag{3.11}

$[a ξ_{1} + b ξ_{2}, ξ_{3}] = ((a ξ_{1} + b ξ_{2})^{\land} ξ_{3}^{\land} - ξ_{3}^{\land} (a ξ_{1} + b ξ_{2})^{\land})^{\lor} = ((a ξ_{1}^{\land} + b ξ_{2}^{\land}) ξ_{3}^{\land} - ξ_{3}^{\land} (a ξ_{1}^{\land} + b ξ_{2}^{\land}))^{\lor} = (a ξ_{1}^{\land} ξ_{3}^{\land} + b ξ_{2}^{\land} ξ_{3}^{\land} - (a ξ_{3}^{\land} ξ_{1}^{\land} + b ξ_{3}^{\land} ξ_{2}^{\land}))^{\lor} = (a (ξ_{1}^{\land} ξ_{3}^{\land} - ξ_{3}^{\land} ξ_{1}^{\land}) + b (ξ_{2}^{\land} ξ_{3}^{\land} - ξ_{3}^{\land} ξ_{2}^{\land}))^{\lor} = a (ξ_{1}^{\land} ξ_{3}^{\land} - ξ_{3}^{\land} ξ_{1}^{\land})^{\lor} + b (ξ_{2}^{\land} ξ_{3}^{\land} - ξ_{3}^{\land} ξ_{2}^{\land})^{\lor} = a [ξ_{1}, ξ_{3}] + b [ξ_{2}, ξ_{3}] (3.11)$
同理可得

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(3.12)

[\vec\xi_3,a\vec\xi_1+b\vec\xi_2

《视觉slam十四讲从理论到实践》第四讲习题自测解答

0x00 前言

0x01 前置知识

李群的定义

李代数的定义

0x02 习题部分

1、验证 S O ( 3 ) 、 S E ( 3 ) 和 S i m ( 3 ) SO(3)、SE(3)和Sim(3) SO(3)、SE(3)和Sim(3)关于乘法成群。

验证 S O ( 3 ) SO(3) SO(3)

封闭性

结合律

幺元

逆

验证 S E ( 3 ) SE(3) SE(3)

封闭性

结合律

幺元

逆

验证 S i m ( 3 ) Sim(3) Sim(3)

2、验证 ( R 3 , R , × ) (\mathbb{R}^3,\mathbb{R},\times) (R3,R,×)构成李代数。

封闭性

双线性

自反性

雅可比等价

3、验证 s o ( 3 ) so(3) so(3)和 s e ( 3 ) se(3) se(3)满足李代数要求的性质。

验证 s o ( 3 ) so(3) so(3)

封闭性

双线性

自反性

雅可比等价

验证 s e ( 3 ) se(3) se(3)

封闭性

双线性

1、验证 $S O (3) 、 S E (3) 和 S i m (3)$ 关于乘法成群。

验证 $S O (3)$

验证 $S E (3)$

验证 $S i m (3)$

2、验证 $(\mathbb{R}^3,\mathbb{R},\times)$ 构成李代数。

3、验证 $s o (3)$ 和 $s e (3)$ 满足李代数要求的性质。

验证 $s o (3)$

验证 $s e (3)$