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

A

A，运算记作

⋅

\cdot

⋅ ，那么群可以记作

G

=

(

A

,

⋅

)

G=(A,\cdot)

G=(A,⋅) 。群要求这个运算满足以下几个条件：

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

 

  李代数的定义

  李代数由一个集合

V

\mathbb{V}

V 、一个数域

F

\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

(

V

,

F

,

[

,

]

)

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

(V,F,[,]) 为一个李代数，记作

g

g

g。

1. 封闭性  $\forall X,Y\in \mathbb{V},[X,Y]\in \mathbb{V}.$
2. 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].

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

3. 自反性  $\forall X\in \mathbb{V},[X,X]=0.$
4. 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$SO(3)、SE(3)和Sim(3)$关于乘法成群。

  验证$SO(3)$

S

O

(

3

)

=

{

R

∈

R

3

×

3

∣

R

R

T

=

I

,

d

e

t

(

R

)

=

1

}

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

SO(3)={R∈R3×3∣RRT=I,det(R)=1}

   封闭性

  从旋转矩阵的定义正交性入手，设

R

=

R

1

R

2

∈

R

3

×

3

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

R=R1​R2​∈R3×3,

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

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

RRT=R1​R2​(R1​R2​)T=R1​R2​R2T​R1T​=R1​IR1T​=I
  所以

R

=

R

1

R

2

∈

S

O

(

3

)

R=R_1R_2\in SO(3)

R=R1​R2​∈SO(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

⋅

R

2

)

⋅

R

3

=

R

1

⋅

(

R

2

⋅

R

3

)

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

(R1​⋅R2​)⋅R3​=R1​⋅(R2​⋅R3​) 。

 满足结合律

---

   幺元

  存在

R

=

I

,

R=I,

R=I, 使得

I

R

1

=

R

1

I

=

R

1

IR_1=R_1I=R_1

IR1​=R1​I=R1​.

 满足幺元

---

   逆

  设矩阵

R

∈

S

O

(

3

)

R\in SO(3)

R∈SO(3) ，由正交性可知，

∃

R

−

1

=

R

T

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

∃R−1=RT ，使得

R

R

−

1

=

I

.

RR^{-1}=I.

RR−1=I.

 满足逆

---

  验证$SE(3)$

S

E

(

3

)

=

{

T

=

[

R

t

0

T

1

]

∈

R

4

×

4

∣

R

∈

S

O

(

3

)

,

t

∈

R

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\}

SE(3)={T=[R0T​t1​]∈R4×4∣R∈SO(3),t∈R3}

   封闭性

  设

T

1

,

T

2

∈

S

E

(

3

)

T_1,T_2\in SE(3)

T1​,T2​∈SE(3),

T

1

T

2

=

[

R

1

t

1

0

T

1

]

[

R

2

t

2

0

T

1

]

=

[

R

1

R

2

R

1

t

2

+

t

1

0

T

1

]

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]

T1​T2​=[R1​0T​t1​1​][R2​0T​t2​1​]=[R1​R2​0T​R1​t2​+t1​1​]
  由

S

O

(

3

)

SO(3)

SO(3) 的封闭性可知，

R

1

R

2

∈

S

O

(

3

)

R_1R_2\in SO(3)

R1​R2​∈SO(3) ,由矩阵的运算规律可知

R

1

t

2

+

t

1

∈

R

3

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

R1​t2​+t1​∈R3
  所以

T

1

T

2

∈

S

E

(

3

)

T_1T_2\in SE(3)

T1​T2​∈SE(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

⋅

T

2

)

⋅

T

3

=

T

1

⋅

(

T

2

⋅

T

3

)

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

(T1​⋅T2​)⋅T3​=T1​⋅(T2​⋅T3​) 。

 满足结合律

---

   幺元

  存在

T

=

I

,

T=I,

T=I, 使得

I

T

1

=

T

1

I

=

T

1

IT_1=T_1I=T_1

IT1​=T1​I=T1​.

 满足幺元

---

   逆

  设矩阵

T

∈

S

E

(

3

)

T\in SE(3)

T∈SE(3) ，易得

R

∈

S

O

(

3

)

R\in SO(3)

R∈SO(3) ,是一个满秩矩阵，所以

T

T

T 是一个满秩矩阵，所以

T

T

T 可逆，

∃
  

T

−

1

\exist\;T^{-1}

∃T−1 ，使得

T

T

−

1

=

I

.

TT^{-1}=I.

TT−1=I.

 满足逆

---

  验证$Sim(3)$

略

 
 

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

 封闭性

 设

a

⃗

,

b

⃗

∈

R

3

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

a
,b
∈R3,
 根据叉积的定义：

a

⃗

×

b

⃗

=

∣

i

⃗

j

⃗

k

⃗

a

1

a

2

a

3

b

1

b

2

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

]

∈

R

3

(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
×b
=∣∣∣∣∣∣​i
a1​b1​​j
​a2​b2​​k
a3​b3​​∣∣∣∣∣∣​=(a2​b3​−a3​b2​)i
+(a3​b1​−a1​b3​)j
​+(a1​b2​−a2​b1​)k
=⎣⎡​a2​b3​−a3​b2​a3​b1​−a1​b3​a1​b2​−a2​b1​​⎦⎤​∈R3(2.1)
 满足封闭性

---

 双线性

 设

a

⃗

,

b

⃗

,

c

⃗

∈

R

3

,

α

,

β

∈

R

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

a
,b
,c
∈R3,α,β∈R,

[

a

X

+

b

Y

,

Z

]

=

(

α

a

⃗

+

β

b

⃗

)

×

c

⃗

=

α

a

⃗

×

c

⃗

+

β

b

⃗

×

c

⃗

⋯

分

配

律

=

α

(

a

⃗

×

c

⃗

)

+

β

(

b

⃗

×

c

⃗

)

⋯

常

数

系

数

完

全

可

以

提

出

来

=

a

[

X

,

Z

]

+

b

[

Y

,

Z

]

(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}

​[aX+bY,Z]=(αa
+βb
)×c
=αa
×c
+βb
×c
=α(a
×c
)+β(b
×c
)=a[X,Z]+b[Y,Z]​⋯⋯​分配律常数系数完全可以提出来​(2.2)
 同理

[

Z

,

a

X

+

b

Y

]

=

α

(

c

⃗

×

a

⃗

)

+

β

(

c

⃗

×

b

⃗

)

=

a

[

Z

,

X

]

+

b

[

Z

,

Y

]

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

[Z,aX+bY]=α(c
×a
)+β(c
×b
)=a[Z,X]+b[Z,Y]
 满足双线性

---

 自反性

 设

a

⃗

∈

R

3

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

a
∈R3

a

⃗

×

a

⃗

=

∣

a

⃗

∣

2

sin

⁡

<

a

⃗

,

a

⃗

>

=

0

(2.3)

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

a
×a
=∣a
∣2sin<a
,a
>=0(2.3)
 满足自反性

---

 雅可比等价

 设

a

⃗

,

b

⃗

,

c

⃗

∈

R

3

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

a
,b
,c
∈R3,

[

X

,

[

Y

,

Z

]

]

+

[

Z

,

[

X

,

Y

]

]

+

[

Y

,

[

Z

,

X

]

]

=

a

⃗

×

(

b

⃗

×

c

⃗

)

+

c

⃗

×

(

a

⃗

×

b

⃗

)

+

b

⃗

×

(

c

⃗

×

a

⃗

)

(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
×(b
×c
)+c
×(a
×b
)+b
×(c
×a
)(2.4)
 将

a

⃗

,

b

⃗

,

c

⃗

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

a
,b
,c
依次轮换得

b

⃗

×

(

c

⃗

×

a

⃗

)

+

a

⃗

×

(

b

⃗

×

c

⃗

)

+

c

⃗

×

(

a

⃗

×

b

⃗

)

(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
×(c
×a
)+a
×(b
×c
)+c
×(a
×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

a

⃗

,

b

⃗

,

c

⃗

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

a
,b
,c
轮换等价。那么可令

b

⃗

=

a

⃗

,

c

⃗

=

a

⃗

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

b
=a
,c
=a
,代入

(

2.4

)

(2.4)

(2.4)得

3

a

⃗

×

(

a

⃗

×

a

⃗

)

=

0

⋯

自

反

性

(2.6)

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

3a
×(a
×a
)=0⋯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$so(3)$和$se(3)$满足李代数要求的性质。

  验证$so(3)$

Φ

=

ϕ

⃗

∧

=

[

0

−

ϕ

⃗

3

ϕ

⃗

2

ϕ

⃗

3

0

−

ϕ

⃗

1

−

ϕ

⃗

2

ϕ

⃗

1

0

]

\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
​∧=⎣⎢⎡​0ϕ
​3​−ϕ
​2​​−ϕ
​3​0ϕ
​1​​ϕ
​2​−ϕ
​1​0​⎦⎥⎤​

s

o

(

3

)

=

{

ϕ

∈

R

3

,

Φ

=

ϕ

∧

∈

R

3

×

3

}

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

so(3)={ϕ∈R3,Φ=ϕ∧∈R3×3}

[

ϕ

⃗

1

,

ϕ

⃗

2

]

=

(

Φ

1

Φ

2

−

Φ

2

Φ

1

)

∨

(3.1)

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

[ϕ
​1​,ϕ
​2​]=(Φ1​Φ2​−Φ2​Φ1​)∨(3.1)

   封闭性

  设

ϕ

⃗

=

[

x

,

y

,

z

]

T

∈

R

3

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

ϕ
​=[x,y,z]T∈R3,

[

ϕ

⃗

1

,

ϕ

⃗

2

]

=

(

Φ

1

Φ

2

−

Φ

2

Φ

1

)

∨

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

[ϕ
​1​,ϕ
​2​]=(Φ1​Φ2​−Φ2​Φ1​)∨

=

(

[

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

]

)

∨

=\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

=⎝⎛​⎣⎡​0z1​−y1​​−z1​0x1​​y1​−x1​0​⎦⎤​⎣⎡​0z2​−y2​​−z2​0x2​​y2​−x2​0​⎦⎤​−⎣⎡​0z2​−y2​​−z2​0x2​​y2​−x2​0​⎦⎤​⎣⎡​0z1​−y1​​−z1​0x1​​y1​−x1​0​⎦⎤​⎠⎞​∨

=

(

[

−

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

]

−

[

−

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

]

)

∨

=\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}

=⎝⎛​⎣⎡​−z1​z2​−y1​y2​x1​y2​x1​z2​​x2​y1​−z1​z2​−x1​x2​y1​z2​​x2​z1​y2​z1​−y1​y2​−x1​x2​​⎦⎤​−⎣⎡​−z1​z2​−y1​y2​x2​y1​x2​z1​​x1​y2​−z1​z2​−x1​x2​y2​z1​​x1​z2​y1​z2​−y1​y2​−x1​x2​​⎦⎤​⎠⎞​∨

=

(

[

0

x

2

y

1

−

x

1

y

2

x

2

z

1

−

x

1

z

2

x

1

y

2

−

x

2

y

1

0

y

2

z

1

−

y

1

z

2

x

1

z

2

−

x

2

z

1

y

1

z

2

−

y

2

z

1

0

]

)

∨

=

(

ϕ

⃗

3

∧

)

∨

(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}

=⎝⎛​⎣⎡​0x1​y2​−x2​y1​x1​z2​−x2​z1​​x2​y1​−x1​y2​0y1​z2​−y2​z1​​x2​z1​−x1​z2​y2​z1​−y1​z2​0​⎦⎤​⎠⎞​∨=(ϕ
​3​∧)∨(3.2)
  由

(

3.2

)

(3.2)

(3.2)易得

ϕ

⃗

3

∧

{\vec\phi_3}^\wedge

ϕ
​3​∧ 是一个反对称矩阵，所以可以经由运算

[

∨

]

[\vee]

[∨]从反对称矩阵到向量的变换。所以

(

ϕ

⃗

3

∧

)

∨

=

ϕ

⃗

3

∈

R

3

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

(ϕ
​3​∧)∨=ϕ
​3​∈R3
 满足封闭性

---

   双线性

  设

ϕ

⃗

1

,

ϕ

⃗

2

,

ϕ

⃗

3

∈

R

3

,
  

a

,

b

∈

R

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

ϕ
​1​,ϕ
​2​,ϕ
​3​∈R3,a,b∈R ，

[

a

ϕ

⃗

1

+

b

ϕ

⃗

2

,

ϕ

⃗

3

]

=

(

(

a

ϕ

⃗

1

+

b

ϕ

⃗

2

)

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

(

a

ϕ

⃗

1

+

b

ϕ

⃗

2

)

∧

)

∨

=

(

a

ϕ

⃗

1

∧

ϕ

⃗

3

∧

+

b

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

(

a

ϕ

⃗

3

∧

ϕ

⃗

1

∧

+

b

ϕ

⃗

3

∧

ϕ

⃗

2

∧

)

)

∨

=

(

a

(

ϕ

⃗

1

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

1

∧

)

+

b

(

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

2

∧

)

)

∨

(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}

​[aϕ
​1​+bϕ
​2​,ϕ
​3​]=((aϕ
​1​+bϕ
​2​)∧ϕ
​3​∧−ϕ
​3​∧(aϕ
​1​+bϕ
​2​)∧)∨=(aϕ
​1​∧ϕ
​3​∧+bϕ
​2​∧ϕ
​3​∧−(aϕ
​3​∧ϕ
​1​∧+bϕ
​3​∧ϕ
​2​∧))∨=(a(ϕ
​1​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​1​∧)+b(ϕ
​2​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​2​∧))∨​(3.3)
  由封闭性可知，

(

ϕ

⃗

1

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

1

∧

)

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

(ϕ
​1​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​1​∧) 是反对称矩阵，所以可以得到

=

a

(

ϕ

⃗

1

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

1

∧

)

∨

+

b

(

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

2

∧

)

∨

=

a

[

ϕ

⃗

1

,

ϕ

⃗

3

]

+

b

[

ϕ

⃗

2

,

ϕ

⃗

3

]

(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​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​1​∧)∨+b(ϕ
​2​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​2​∧)∨=a[ϕ
​1​,ϕ
​3​]+b[ϕ
​2​,ϕ
​3​]​(3.4)
  同理可得

[

ϕ

⃗

3

,

a

ϕ

⃗

1

+

b

ϕ

⃗

2

]

=

a

[

ϕ

⃗

3

,

ϕ

⃗

1

]

+

b

[

ϕ

⃗

3

,

ϕ

⃗

2

]

(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​,aϕ
​1​+bϕ
​2​]=a[ϕ
​3​,ϕ
​1​]+b[ϕ
​3​,ϕ
​2​](3.5)
 满足双线性

---

   自反性

  设

ϕ

⃗

∈

R

3

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

ϕ
​∈R3 ，

[

ϕ

⃗

,

ϕ

⃗

]

=

(

ϕ

⃗

∧

ϕ

⃗

∧

−

ϕ

⃗

∧

ϕ

⃗

∧

)

∨

=

0

T

⋯

很

显

然

(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
​∧ϕ
​∧−ϕ
​∧ϕ
​∧)∨=0T⋯很显然(3.6)
 满足自反性

---

   雅可比等价

  设

ϕ

⃗

1

,

ϕ

⃗

2

,

ϕ

⃗

3

∈

R

3

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

ϕ
​1​,ϕ
​2​,ϕ
​3​∈R3 ，

[

ϕ

⃗

1

,

[

ϕ

⃗

2

,

ϕ

⃗

3

]

]

+

[

ϕ

⃗

3

,

[

ϕ

⃗

1

,

ϕ

⃗

2

]

]

+

[

ϕ

⃗

2

,

[

ϕ

⃗

3

,

ϕ

⃗

1

]

]

=

ϕ

⃗

1

∧

(

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

2

∧

)

−

(

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

2

∧

)

ϕ

⃗

1

∧

+

ϕ

⃗

3

∧

(

ϕ

⃗

1

∧

ϕ

⃗

2

∧

−

ϕ

⃗

2

∧

ϕ

⃗

1

∧

)

−

(

ϕ

⃗

1

∧

ϕ

⃗

2

∧

−

ϕ

⃗

2

∧

ϕ

⃗

1

∧

)

ϕ

⃗

3

∧

+

ϕ

⃗

2

∧

(

ϕ

⃗

3

∧

ϕ

⃗

1

∧

−

ϕ

⃗

1

∧

ϕ

⃗

3

∧

)

−

(

ϕ

⃗

3

∧

ϕ

⃗

1

∧

−

ϕ

⃗

1

∧

ϕ

⃗

3

∧

)

ϕ

⃗

2

∧

=

ϕ

⃗

1

∧

ϕ

⃗

2

∧

ϕ

⃗

3

∧

−

ϕ

⃗

1

∧

ϕ

⃗

3

∧

ϕ

⃗

2

∧

−

ϕ

⃗

2

∧

ϕ

⃗

3

∧

ϕ

⃗

1

∧

+

ϕ

⃗

3

∧

ϕ

⃗

2

∧

ϕ

⃗

1

∧

+

ϕ

⃗

3

∧

ϕ

⃗

1

∧

ϕ

⃗

2

∧

−

ϕ

⃗

3

∧

ϕ

⃗

2

∧

ϕ

⃗

1

∧

−

ϕ

⃗

1

∧

ϕ

⃗

2

∧

ϕ

⃗

3

∧

+

ϕ

⃗

2

∧

ϕ

⃗

1

∧

ϕ

⃗

3

∧

+

ϕ

⃗

2

∧

ϕ

⃗

3

∧

ϕ

⃗

1

∧

−

ϕ

⃗

2

∧

ϕ

⃗

1

∧

ϕ

⃗

3

∧

−

ϕ

⃗

3

∧

ϕ

⃗

1

∧

ϕ

⃗

2

∧

+

ϕ

⃗

1

∧

ϕ

⃗

3

∧

ϕ

⃗

2

∧

=

0

\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​∧(ϕ
​2​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​2​∧)−(ϕ
​2​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​2​∧)ϕ
​1​∧+ϕ
​3​∧(ϕ
​1​∧ϕ
​2​∧−ϕ
​2​∧ϕ
​1​∧)−(ϕ
​1​∧ϕ
​2​∧−ϕ
​2​∧ϕ
​1​∧)ϕ
​3​∧+ϕ
​2​∧(ϕ
​3​∧ϕ
​1​∧−ϕ
​1​∧ϕ
​3​∧)−(ϕ
​3​∧ϕ
​1​∧−ϕ
​1​∧ϕ
​3​∧)ϕ
​2​∧=ϕ
​1​∧ϕ
​2​∧ϕ
​3​∧−ϕ
​1​∧ϕ
​3​∧ϕ
​2​∧−ϕ
​2​∧ϕ
​3​∧ϕ
​1​∧+ϕ
​3​∧ϕ
​2​∧ϕ
​1​∧+ϕ
​3​∧ϕ
​1​∧ϕ
​2​∧−ϕ
​3​∧ϕ
​2​∧ϕ
​1​∧−ϕ
​1​∧ϕ
​2​∧ϕ
​3​∧+ϕ
​2​∧ϕ
​1​∧ϕ
​3​∧+ϕ
​2​∧ϕ
​3​∧ϕ
​1​∧−ϕ
​2​∧ϕ
​1​∧ϕ
​3​∧−ϕ
​3​∧ϕ
​1​∧ϕ
​2​∧+ϕ
​1​∧ϕ
​3​∧ϕ
​2​∧=0​
 满足雅可比等价

---

 

  验证$se(3)$

s

e

(

3

)

=

{

ξ

=

[

ρ

ϕ

]

∈

R

6

,

ρ

∈

R

3

,

ϕ

∈

s

o

(

3

)

,

ξ

∧

=

[

ϕ

∧

ρ

0

T

0

]

∈

R

4

×

4

}

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\}

se(3)={ξ=[ρϕ​]∈R6,ρ∈R3,ϕ∈so(3),ξ∧=[ϕ∧0T​ρ0​]∈R4×4}

[

ξ

⃗

1

,

ξ

⃗

2

]

=

(

ξ

⃗

1

∧

ξ

⃗

2

∧

−

ξ

⃗

2

∧

ξ

⃗

1

∧

)

∨

[\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​∧ξ
​2​∧−ξ
​2​∧ξ
​1​∧)∨

   封闭性

  设

ξ

⃗

1

,

ξ

⃗

2

∈

s

e

(

3

)

\vec\xi_1,\vec\xi_2\in se(3)

ξ
​1​,ξ
​2​∈se(3) ,

[

ξ

⃗

1

,

ξ

⃗

2

]

=

(

ξ

⃗

1

∧

ξ

⃗

2

∧

−

ξ

⃗

2

∧

ξ

⃗

1

∧

)

∨

[\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​∧ξ
​2​∧−ξ
​2​∧ξ
​1​∧)∨

=

(

[

ϕ

⃗

1

∧

ρ

⃗

1

0

T

0

]

[

ϕ

⃗

2

∧

ρ

⃗

2

0

T

0

]

−

[

ϕ

⃗

2

∧

ρ

⃗

2

0

T

0

]

[

ϕ

⃗

1

∧

ρ

⃗

1

0

T

0

]

)

∨

(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​∧0T​ρ
​1​0​][ϕ
​2​∧0T​ρ
​2​0​]−[ϕ
​2​∧0T​ρ
​2​0​][ϕ
​1​∧0T​ρ
​1​0​])∨(3.7)

=

(

[

ϕ

⃗

1

∧

ϕ

⃗

2

∧

ϕ

⃗

1

∧

ρ

⃗

2

0

T

0

]

−

[

ϕ

⃗

2

∧

ϕ

⃗

1

∧

ϕ

⃗

2

∧

ρ

⃗

1

0

T

0

]

)

∨

(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​∧ϕ
​2​∧0T​ϕ
​1​∧ρ
​2​0​]−[ϕ
​2​∧ϕ
​1​∧0T​ϕ
​2​∧ρ
​1​0​])∨(3.8)

=

(

[

ϕ

⃗

1

∧

ϕ

⃗

2

∧

−

ϕ

⃗

2

∧

ϕ

⃗

1

∧

ϕ

⃗

1

∧

ρ

⃗

2

−

ϕ

⃗

2

∧

ρ

⃗

1

0

T

0

]

)

∨

(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​∧ϕ
​2​∧−ϕ
​2​∧ϕ
​1​∧0T​ϕ
​1​∧ρ
​2​−ϕ
​2​∧ρ
​1​0​])∨(3.9)
  由

s

o

(

3

)

so(3)

so(3) 的封闭性可知，

ϕ

⃗

1

∧

ϕ

⃗

2

∧

−

ϕ

⃗

2

∧

ϕ

⃗

1

∧

=

ϕ

⃗

3

∧

,
  

ϕ

⃗

3

∈

s

o

(

3

)

{\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​∧ϕ
​2​∧−ϕ
​2​∧ϕ
​1​∧=ϕ
​3​∧,ϕ
​3​∈so(3) ；由矩阵的运算可知

ϕ

⃗

1

∧

ρ

⃗

2

−

ϕ

⃗

2

∧

ρ

⃗

1

∈

R

3

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

ϕ
​1​∧ρ
​2​−ϕ
​2​∧ρ
​1​∈R3 ，所以可令

式

(

3.9

)

=

(

ξ

⃗

3

∧

)

∨

=

ξ

⃗

3

∈

s

e

(

3

)

(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​∧)∨=ξ
​3​∈se(3)(3.10)
 满足封闭性

---

   双线性

  设

ξ

⃗

1

,

ξ

⃗

2

,

ξ

⃗

3

∈

s

e

(

3

)

\vec\xi_1,\vec\xi_2,\vec\xi_3\in se(3)

ξ
​1​,ξ
​2​,ξ
​3​∈se(3) ,

[

a

ξ

⃗

1

+

b

ξ

⃗

2

,

ξ

⃗

3

]

=

(

(

a

ξ

⃗

1

+

b

ξ

⃗

2

)

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

(

a

ξ

⃗

1

+

b

ξ

⃗

2

)

∧

)

∨

=

(

(

a

ξ

⃗

1

∧

+

b

ξ

⃗

2

∧

)

ξ

⃗

3

∧

−

ξ

⃗

3

∧

(

a

ξ

⃗

1

∧

+

b

ξ

⃗

2

∧

)

)

∨

=

(

a

ξ

⃗

1

∧

ξ

⃗

3

∧

+

b

ξ

⃗

2

∧

ξ

⃗

3

∧

−

(

a

ξ

⃗

3

∧

ξ

⃗

1

∧

+

b

ξ

⃗

3

∧

ξ

⃗

2

∧

)

)

∨

=

(

a

(

ξ

⃗

1

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

1

∧

)

+

b

(

ξ

⃗

2

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

2

∧

)

)

∨

=

a

(

ξ

⃗

1

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

1

∧

)

∨

+

b

(

ξ

⃗

2

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

2

∧

)

∨

=

a

[

ξ

⃗

1

,

ξ

⃗

3

]

+

b

[

ξ

⃗

2

,

ξ

⃗

3

]

(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}

​[aξ
​1​+bξ
​2​,ξ
​3​]=((aξ
​1​+bξ
​2​)∧ξ
​3​∧−ξ
​3​∧(aξ
​1​+bξ
​2​)∧)∨=((aξ
​1​∧+bξ
​2​∧)ξ
​3​∧−ξ
​3​∧(aξ
​1​∧+bξ
​2​∧))∨=(aξ
​1​∧ξ
​3​∧+bξ
​2​∧ξ
​3​∧−(aξ
​3​∧ξ
​1​∧+bξ
​3​∧ξ
​2​∧))∨=(a(ξ
​1​∧ξ
​3​∧−ξ
​3​∧ξ
​1​∧)+b(ξ
​2​∧ξ
​3​∧−ξ
​3​∧ξ
​2​∧))∨=a(ξ
​1​∧ξ
​3​∧−ξ
​3​∧ξ
​1​∧)∨+b(ξ
​2​∧ξ
​3​∧−ξ
​3​∧ξ
​2​∧)∨=a[ξ
​1​,ξ
​3​]+b[ξ
​2​,ξ
​3​]​(3.11)
  同理可得

[

ξ

⃗

3

,

a

ξ

⃗

1

+

b

ξ

⃗

2

]

=

a

[

ξ

⃗

3

,

ξ

⃗

1

]

+

b

[

ξ

⃗

3

,

ξ

⃗

2

]

(3.12)

[\vec\xi_3,a\vec\xi_1+b\vec\xi_2]=a[\vec\xi_3,\vec\xi_1]+b[\vec\xi_3,\vec\xi_2]\tag{3.12}

[ξ
​3​,aξ
​1​+bξ
​2​]=a[ξ
​3​,ξ
​1​]+b[ξ
​3​,ξ
​2​](3.12)
 满足双线性

---

   自反性

  设

ξ

⃗

∈

R

6

\vec\xi\in\mathbb{R}^6

ξ
​∈R6 ，

[

ξ

⃗

,

ξ

⃗

]

=

(

ξ

⃗

∧

ξ

⃗

∧

−

ξ

⃗

∧

ξ

⃗

∧

)

∨

=

0

T

⋯

很

显

然

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

[ξ
​,ξ
​]=(ξ
​∧ξ
​∧−ξ
​∧ξ
​∧)∨=0T⋯很显然
 满足自反性

---

   雅可比等价

  设

ξ

⃗

1

,

ξ

⃗

2

,

ξ

⃗

3

∈

R

6

\vec\xi_1,\vec\xi_2,\vec\xi_3\in\mathbb{R}^6

ξ
​1​,ξ
​2​,ξ
​3​∈R6 ，

[

ξ

⃗

1

,

[

ξ

⃗

2

,

ξ

⃗

3

]

]

+

[

ξ

⃗

3

,

[

ξ

⃗

1

,

ξ

⃗

2

]

]

+

[

ξ

⃗

2

,

[

ξ

⃗

3

,

ξ

⃗

1

]

]

=

ξ

⃗

1

∧

(

ξ

⃗

2

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

2

∧

)

−

(

ξ

⃗

2

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

2

∧

)

ξ

⃗

1

∧

+

ξ

⃗

3

∧

(

ξ

⃗

1

∧

ξ

⃗

2

∧

−

ξ

⃗

2

∧

ξ

⃗

1

∧

)

−

(

ξ

⃗

1

∧

ξ

⃗

2

∧

−

ξ

⃗

2

∧

ξ

⃗

1

∧

)

ξ

⃗

3

∧

+

ξ

⃗

2

∧

(

ξ

⃗

3

∧

ξ

⃗

1

∧

−

ξ

⃗

1

∧

ξ

⃗

3

∧

)

−

(

ξ

⃗

3

∧

ξ

⃗

1

∧

−

ξ

⃗

1

∧

ξ

⃗

3

∧

)

ξ

⃗

2

∧

=

ξ

⃗

1

∧

ξ

⃗

2

∧

ξ

⃗

3

∧

−

ξ

⃗

1

∧

ξ

⃗

3

∧

ξ

⃗

2

∧

−

ξ

⃗

2

∧

ξ

⃗

3

∧

ξ

⃗

1

∧

+

ξ

⃗

3

∧

ξ

⃗

2

∧

ξ

⃗

1

∧

+

ξ

⃗

3

∧

ξ

⃗

1

∧

ξ

⃗

2

∧

−

ξ

⃗

3

∧

ξ

⃗

2

∧

ξ

⃗

1

∧

−

ξ

⃗

1

∧

ξ

⃗

2

∧

ξ

⃗

3

∧

+

ξ

⃗

2

∧

ξ

⃗

1

∧

ξ

⃗

3

∧

+

ξ

⃗

2

∧

ξ

⃗

3

∧

ξ

⃗

1

∧

−

ξ

⃗

2

∧

ξ

⃗

1

∧

ξ

⃗

3

∧

−

ξ

⃗

3

∧

ξ

⃗

1

∧

ξ

⃗

2

∧

+

ξ

⃗

1

∧

ξ

⃗

3

∧

ξ

⃗

2

∧

=

0

\begin{aligned} &\qquad[\vec\xi_1,[\vec\xi_2,\vec\xi_3]]+[\vec\xi_3,[\vec\xi_1,\vec\xi_2]]+[\vec\xi_2,[\vec\xi_3,\vec\xi_1]]\\ &=\quad{\vec\xi_1}^\wedge({\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge)-({\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge){\vec\xi_1}^\wedge\\ &\quad+{\vec\xi_3}^\wedge({\vec\xi_1}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge)-({\vec\xi_1}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge){\vec\xi_3}^\wedge\\ &\quad+{\vec\xi_2}^\wedge({\vec\xi_3}^\wedge{\vec\xi_1}^\wedge-{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge)-({\vec\xi_3}^\wedge{\vec\xi_1}^\wedge-{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge){\vec\xi_2}^\wedge\\ &=\quad{\vec\xi_1}^\wedge{\vec\xi_2}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge+{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge\\ &\quad+{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge{\vec\xi_2}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge-{\vec\xi_1}^\wedge{\vec\xi_2}^\wedge{\vec\xi_3}^\wedge+{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge\\ &\quad+{\vec\xi_2}^\wedge{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge-{\vec\xi_2}^\wedge{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge-{\vec\xi_3}^\wedge{\vec\xi_1}^\wedge{\vec\xi_2}^\wedge+{\vec\xi_1}^\wedge{\vec\xi_3}^\wedge{\vec\xi_2}^\wedge\\ &=0 \end{aligned}

​[ξ
​1​,[ξ
​2​,ξ
​3​]]+[ξ
​3​,[ξ
​1​,ξ
​2​]]+[ξ
​2​,[ξ
​3​,ξ
​1​]]=ξ
​1​∧(ξ
​2​∧ξ
​3​∧−ξ
​3​∧ξ
​2​∧)−(ξ
​2​∧ξ
​3​∧−ξ
​3​∧ξ
​2​∧)ξ
​1​∧+ξ
​3​∧(ξ
​1​∧ξ
​2​∧−ξ
​2​∧ξ
​1​∧)−(ξ
​1​∧ξ
​2​∧−ξ
​2​∧ξ
​1​∧)ξ
​3​∧+ξ
​2​∧(ξ
​3​∧ξ
​1​∧−ξ
​1​∧ξ
​3​∧)−(ξ
​3​∧ξ
​1​∧−ξ
​1​∧ξ
​3​∧)ξ
​2​∧=ξ
​1​∧ξ
​2​∧ξ
​3​∧−ξ
​1​∧ξ
​3​∧ξ
​2​∧−ξ
​2​∧ξ
​3​∧ξ
​1​∧+ξ
​3​∧ξ
​2​∧ξ
​1​∧+ξ
​3​∧ξ
​1​∧ξ
​2​∧−ξ
​3​∧ξ
​2​∧ξ
​1​∧−ξ
​1​∧ξ
​2​∧ξ
​3​∧+ξ
​2​∧ξ
​1​∧ξ
​3​∧+ξ
​2​∧ξ
​3​∧ξ
​1​∧−ξ
​2​∧ξ
​1​∧ξ
​3​∧−ξ
​3​∧ξ
​1​∧ξ
​2​∧+ξ
​1​∧ξ
​3​∧ξ
​2​∧=0​
 满足雅可比等价

---

 
 

4、证明性质$(4.20)$和$(4.21)$。

  证明性质 $a^{\wedge }{a}^{\wedge }=a{a}^T-I$

  设

a

⃗

=

[

a

1

,

a

2

,

a

3

]

T

\vec{a}=[a_1,a_2,a_3]^T

a
=[a1​,a2​,a3​]T.

  已知

a

∧

=

[

0

−

a

3

a

2

a

3

0

−

a

1

−

a

2

a

1

0

]

,

a

1

2

+

a

2

2

+

a

3

2

=

1.

a^\wedge=\left[\begin{matrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{matrix}\right],{a_1}^2+{a_2}^2+{a_3}^2=1.

a∧=⎣⎡​0a3​−a2​​−a3​0a1​​a2​−a1​0​⎦⎤​,a1​2+a2​2+a3​2=1.

a

∧

a

∧

=

[

0

−

a

3

a

2

a

3

0

−

a

1

−

a

2

a

1

0

]

[

0

−

a

3

a

2

a

3

0

−

a

1

−

a

2

a

1

0

]

(4.1)

a^\wedge a^\wedge=\left[\begin{matrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{matrix}\right]\left[\begin{matrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{matrix}\right]\tag{4.1}

a∧a∧=⎣⎡​0a3​−a2​​−a3​0a1​​a2​−a1​0​⎦⎤​⎣⎡​0a3​−a2​​−a3​0a1​​a2​−a1​0​⎦⎤​(4.1)

=

[

−

a

2

2

−

a

3

2

a

1

a

2

a

1

a

3

a

1

a

2

−

a

1

2

−

a

3

2

a

2

a

3

a

1

a

3

a

2

a

3

−

a

1

2

−

a

2

2

]

+

I

−

I

(4.2)

=\left[\begin{matrix} -{a_2}^2-{a_3}^2&a_1a_2&a_1a_3\\ a_1a_2&-{a_1}^2-{a_3}^2&a_2a_3\\ a_1a_3&a_2a_3&-{a_1}^2-{a_2}^2 \end{matrix}\right]+I-I\tag{4.2}

=⎣⎡​−a2​2−a3​2a1​a2​a1​a3​​a1​a2​−a1​2−a3​2a2​a3​​a1​a3​a2​a3​−a1​2−a2​2​⎦⎤​+I−I(4.2)

=

[

1

−

a

2

2

−

a

3

2

a

1

a

2

a

1

a

3

a

1

a

2

1

−

a

1

2

−

a

3

2

a

2

a

3

a

1

a

3

a

2

a

3

1

−

a

1

2

−

a

2

2

]

−

I

(4.3)

=\left[\begin{matrix} 1-{a_2}^2-{a_3}^2&a_1a_2&a_1a_3\\ a_1a_2&1-{a_1}^2-{a_3}^2&a_2a_3\\ a_1a_3&a_2a_3&1-{a_1}^2-{a_2}^2 \end{matrix}\right]-I\tag{4.3}

=⎣⎡​1−a2​2−a3​2a1​a2​a1​a3​​a1​a2​1−a1​2−a3​2a2​a3​​a1​a3​a2​a3​1−a1​2−a2​2​⎦⎤​−I(4.3)

=

[

a

1

2

a

1

a

2

a

1

a

3

a

1

a

2

a

2

2

a

2

a

3

a

1

a

3

a

2

a

3

a

3

2

]

−

I

=

a

a

T

−

I

,

得

证

(4.3)

=\left[\begin{matrix} {a_1}^2&a_1a_2&a_1a_3\\ a_1a_2&{a_2}^2&a_2a_3\\ a_1a_3&a_2a_3&{a_3}^2 \end{matrix}\right]-I=aa^T-I , 得证\tag{4.3}

=⎣⎡​a1​2a1​a2​a1​a3​​a1​a2​a2​2a2​a3​​a1​a3​a2​a3​a3​2​⎦⎤​−I=aaT−I,得证(4.3)
 

  证明性质 $a^{\wedge }{a}^{\wedge }{a}^{\wedge }=-a^{\wedge }$

  设

a

⃗

=

[

a

1

,

a

2

,

a

3

]

T

\vec{a}=[a_1,a_2,a_3]^T

a
=[a1​,a2​,a3​]T.
  由

(

4.2

)

(4.2)

(4.2)可得：

a

∧

a

∧

a

∧

=

a

∧

(

a

∧

a

∧

)

(4.4)

a^\wedge a^\wedge a^\wedge=a^\wedge (a^\wedge a^\wedge)\tag{4.4}

a∧a∧a∧=a∧(a∧a∧)(4.4)

=

[

0

−

a

3

a

2

a

3

0

−

a

1

−

a

2

a

1

0

]

[

−

a

2

2

−

a

3

2

a

1

a

2

a

1

a

3

a

1

a

2

−

a

1

2

−

a

3

2

a

2

a

3

a

1

a

3

a

2

a

3

−

a

1

2

−

a

2

2

]

(4.5)

=\left[\begin{matrix} 0&-a_3&a_2\\ a_3&0&-a_1\\ -a_2&a_1&0 \end{matrix}\right] \left[\begin{matrix} -{a_2}^2-{a_3}^2&a_1a_2&a_1a_3\\ a_1a_2&-{a_1}^2-{a_3}^2&a_2a_3\\ a_1a_3&a_2a_3&-{a_1}^2-{a_2}^2 \end{matrix}\right]\tag{4.5}

=⎣⎡​0a3​−a2​​−a3​0a1​​a2​−a1​0​⎦⎤​⎣⎡​−a2​2−a3​2a1​a2​a1​a3​​a1​a2​−a1​2−a3​2a2​a3​​a1​a3​a2​a3​−a1​2−a2​2​⎦⎤​(4.5)

=

[

−

a

1

a

2

a

3

+

a

1

a

2

a

3

a

1

2

a

3

+

a

3

3

+

a

2

2

a

3

−

a

2

a

3

2

−

a

1

2

a

2

−

a

2

3

−

a

2

2

a

3

−

a

3

3

−

a

1

2

a

3

a

1

a

2

a

3

−

a

1

a

2

a

3

a

1

a

3

2

+

a

1

3

+

a

1

a

2

2

a

2

3

+

a

2

a

3

2

+

a

1

2

a

2

−

a

1

a

2

2

−

a

1

3

−

a

1

a

3

2

−

a

1

a

2

a

3

+

a

1

a

2

a

3

]

(4.6)

=\left[\begin{matrix} -a_1a_2a_3+a_1a_2a_3&a_1^2a_3+a_3^3+a_2^2a_3&-a_2a_3^2-a_1^2a_2-a_2^3\\ -a_2^2a_3-a_3^3-a_1^2a_3&a_1a_2a_3-a_1a_2a_3&a_1a_3^2+a_1^3+a_1a_2^2\\ a_2^3+a_2a_3^2+a_1^2a_2&-a_1a_2^2-a_1^3-a_1a_3^2&-a_1a_2a_3+a_1a_2a_3 \end{matrix}\right]\tag{4.6}

=⎣⎡​−a1​a2​a3​+a1​a2​a3​−a22​a3​−a33​−a12​a3​a23​+a2​a32​+a12​a2​​a12​a3​+a33​+a22​a3​a1​a2​a3​−a1​a2​a3​−a1​a22​−a13​−a1​a32​​−a2​a32​−a12​a2​−a23​a1​a32​+a13​+a1​a22​−a1​a2​a3​+a1​a2​a3​​⎦⎤​(4.6)

=

[

0

a

3

(

a

1

2

+

a

2

2

+

a

3

2

)

−

a

2

(

a

1

2

+

a

2

2

+

a

3

2

)

−

a

3

(

a

1

2

+

a

2

2

+

a

3

2

)

0

a

1

(

a

1

2

+

a

2

2

+

a

3

2

)

a

2

(

a

1

2

+

a

2

2

+

a

3

2

)

−

a

1

(

a

1

2

+

a

2

2

+

a

3

2

)

0

]

(4.7)

=\left[\begin{matrix} 0&a_3(a_1^2+a_2^2+a_3^2)&-a_2(a_1^2+a_2^2+a_3^2)\\ -a_3(a_1^2+a_2^2+a_3^2)&0&a_1(a_1^2+a_2^2+a_3^2)\\ a_2(a_1^2+a_2^2+a_3^2)&-a_1(a_1^2+a_2^2+a_3^2)&0 \end{matrix}\right]\tag{4.7}

=⎣⎡​0−a3​(a12​+a22​+a32​)a2​(a12​+a22​+a32​)​a3​(a12​+a22​+a32​)0−a1​(a12​+a22​+a32​)​−a2​(a12​+a22​+a32​)a1​(a12​+a22​+a32​)0​⎦⎤​(4.7)

=

[

0

a

3

−

a

2

−

a

3

0

a

1

a

2

−

a

1

0

]

=

−

a

∧

(4.8)

=\left[\begin{matrix} 0&a_3&-a_2\\ -a_3&0&a_1\\ a_2&-a_1&0 \end{matrix}\right]=-a^\wedge\tag{4.8}

=⎣⎡​0−a3​a2​​a3​0−a1​​−a2​a1​0​⎦⎤​=−a∧(4.8)
 
 

5、证明: $R{p}^{\wedge }{R}^T=(Rp{)}^{\wedge }$ . [参考]

  设

R

=

[

r

1

,

r

2

,

r

3

]

∈

S

O

(

3

)

,
  

r

x

∈

R

3

,
  

p

=

[

a

1

,

a

2

,

a

3

]

T

R=[r_1,r_2,r_3]\in SO(3),\;r_x\in\mathbb{R}^3,\;p=[a_1,a_2,a_3]^T

R=[r1​,r2​,r3​]∈SO(3),rx​∈R3,p=[a1​,a2​,a3​]T

求

证

R

p

∧

R

T

=

(

R

p

)

∧

即

证

p

∧

=

R

T

(

R

p

)

∧

R

(5.1)

\begin{aligned} &求证&\qquad Rp^\wedge R^T&=&&(Rp)^\wedge\\ &即证&p^\wedge&=&&R^T(Rp)^\wedge R\tag{5.1} \end{aligned}

​求证即证​Rp∧RTp∧​==​​(Rp)∧RT(Rp)∧R​(5.1)
  考虑等式右边

(

R

p

)

∧

=

(

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

[

a

1

a

2

a

3

]

)

∧

(Rp)^\wedge=\left([\vec{r_1},\vec{r_2},\vec{r_3}]\left[\begin{matrix}a_1\\a_2\\a_3\end{matrix}\right]\right)^\wedge

(Rp)∧=⎝⎛​[r1​
​,r2​
​,r3​
​]⎣⎡​a1​a2​a3​​⎦⎤​⎠⎞​∧

=

(

a

1

r

1

⃗

+

a

2

r

2

⃗

+

a

3

r

3

⃗

)

∧

=

a

1

r

1

⃗

∧

+

a

2

r

2

⃗

∧

+

a

3

r

3

⃗

∧

(5.2)

=\left(a_1\vec{r_1}+a_2\vec{r_2}+a_3\vec{r_3}\right)^\wedge=a_1{\vec{r_1}}^\wedge+a_2{\vec{r_2}}^\wedge+a_3{\vec{r_3}}^\wedge\tag{5.2}

=(a1​r1​
​+a2​r2​
​+a3​r3​
​)∧=a1​r1​
​∧+a2​r2​
​∧+a3​r3​
​∧(5.2)

  可得

R

T

(

R

p

)

∧

R

=

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

(

a

1

r

1

⃗

∧

+

a

2

r

2

⃗

∧

+

a

3

r

3

⃗

∧

)

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

R^T(Rp)^\wedge R=\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right](a_1{\vec{r_1}}^\wedge+a_2{\vec{r_2}}^\wedge+a_3{\vec{r_3}}^\wedge)[\vec{r_1},\vec{r_2},\vec{r_3}]

RT(Rp)∧R=⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​(a1​r1​
​∧+a2​r2​
​∧+a3​r3​
​∧)[r1​
​,r2​
​,r3​
​]

=

a

1

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

1

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

+

a

2

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

2

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

+

a

3

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

3

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

(5.3)

=a_1\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_1}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]+ a_2\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_2}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]+ a_3\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_3}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]\tag{5.3}

=a1​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r1​
​∧[r1​
​,r2​
​,r3​
​]+a2​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r2​
​∧[r1​
​,r2​
​,r3​
​]+a3​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r3​
​∧[r1​
​,r2​
​,r3​
​](5.3)
  考虑第一项

a

1

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

1

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

=

a

1

[

r

1

⃗

T

r

1

⃗

∧

r

1

⃗

r

1

⃗

T

r

1

⃗

∧

r

2

⃗

r

1

⃗

T

r

1

⃗

∧

r

3

⃗

r

2

⃗

T

r

1

⃗

∧

r

1

⃗

r

2

⃗

T

r

1

⃗

∧

r

2

⃗

r

2

⃗

T

r

1

⃗

∧

r

3

⃗

r

3

⃗

T

r

1

⃗

∧

r

1

⃗

r

3

⃗

T

r

1

⃗

∧

r

2

⃗

r

3

⃗

T

r

1

⃗

∧

r

3

⃗

]

(5.4)

a_1\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_1}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]=a_1\left[\begin{matrix} {\vec{r_1}}^T{\vec{r_1}}^\wedge\vec{r_1}&{\vec{r_1}}^T{\vec{r_1}}^\wedge\vec{r_2}&{\vec{r_1}}^T{\vec{r_1}}^\wedge\vec{r_3}\\ {\vec{r_2}}^T{\vec{r_1}}^\wedge\vec{r_1}&{\vec{r_2}}^T{\vec{r_1}}^\wedge\vec{r_2}&{\vec{r_2}}^T{\vec{r_1}}^\wedge\vec{r_3}\\ {\vec{r_3}}^T{\vec{r_1}}^\wedge\vec{r_1}&{\vec{r_3}}^T{\vec{r_1}}^\wedge\vec{r_2}&{\vec{r_3}}^T{\vec{r_1}}^\wedge\vec{r_3} \end{matrix}\right]\tag{5.4}

a1​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r1​
​∧[r1​
​,r2​
​,r3​
​]=a1​⎣⎡​r1​
​Tr1​
​∧r1​
​r2​
​Tr1​
​∧r1​
​r3​
​Tr1​
​∧r1​
​​r1​
​Tr1​
​∧r2​
​r2​
​Tr1​
​∧r2​
​r3​
​Tr1​
​∧r2​
​​r1​
​Tr1​
​∧r3​
​r2​
​Tr1​
​∧r3​
​r3​
​Tr1​
​∧r3​
​​⎦⎤​(5.4)
  因为

R

R

R 是正交阵，所以

R

R

R 矩阵的列向量(或行向量)两两垂直且模长为1，
  又因为

”

∧

”

“\wedge”

“∧“ 可以看成向量叉乘，即

a

⃗

∧

b

⃗

=

a

⃗

×

b

⃗

{\vec{a}}^\wedge\vec{b}=\vec{a}\times\vec{b}

a
∧b
=a
×b
，根据右手坐标系可得

r

1

⃗

∧

r

1

⃗

=

0

,
  

r

1

⃗

∧

r

2

⃗

=

r

3

⃗

,
  

r

1

⃗

∧

r

3

⃗

=

−

r

2

⃗

{\vec{r_1}}^\wedge\vec{r_1}=0,\;{\vec{r_1}}^\wedge\vec{r_2}=\vec{r_3},\;{\vec{r_1}}^\wedge\vec{r_3}=-\vec{r_2}

r1​
​∧r1​
​=0,r1​
​∧r2​
​=r3​
​,r1​
​∧r3​
​=−r2​
​.

(

5.4

)

=

a

1

[

0

r

1

⃗

T

r

3

⃗

−

r

1

⃗

T

r

2

⃗

0

r

2

⃗

T

r

3

⃗

−

r

2

⃗

T

r

2

⃗

0

r

3

⃗

T

r

3

⃗

−

r

3

⃗

T

r

2

⃗

]

(5.5)

(5.4)=a_1\left[\begin{matrix} 0&{\vec{r_1}}^T\vec{r_3}&-{\vec{r_1}}^T\vec{r_2}\\ 0&{\vec{r_2}}^T\vec{r_3}&-{\vec{r_2}}^T\vec{r_2}\\ 0&{\vec{r_3}}^T\vec{r_3}&-{\vec{r_3}}^T\vec{r_2} \end{matrix}\right]\tag{5.5}

(5.4)=a1​⎣⎡​000​r1​
​Tr3​
​r2​
​Tr3​
​r3​
​Tr3​
​​−r1​
​Tr2​
​−r2​
​Tr2​
​−r3​
​Tr2​
​​⎦⎤​(5.5)
  

a

⃗

T

b

⃗

{\vec{a}}^T\vec{b}

a
Tb
可以看成向量点积

a

⃗

⋅

b

⃗

\vec{a}\cdot\vec{b}

a
⋅b
,可得

(

5.5

)

=

a

1

[

0

0

0

0

0

−

1

0

1

0

]

(5.5)=a_1\left[\begin{matrix} 0&0&0\\ 0&0&-1\\ 0&1&0 \end{matrix}\right]

(5.5)=a1​⎣⎡​000​001​0−10​⎦⎤​
  同理可得

a

2

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

2

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

=

a

2

[

0

0

1

0

0

0

−

1

0

0

]

,

a

3

[

r

1

⃗

T

r

2

⃗

T

r

3

⃗

T

]

r

3

⃗

∧

[

r

1

⃗

,

r

2

⃗

,

r

3

⃗

]

=

a

3

[

0

−

1

0

1

0

0

0

0

0

]

a_2\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_2}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]=a_2 \left[\begin{matrix} 0&0&1\\ 0&0&0\\ -1&0&0 \end{matrix}\right],\quad a_3\left[\begin{matrix}{\vec{r_1}}^T\\{\vec{r_2}}^T\\{\vec{r_3}}^T\end{matrix}\right]{\vec{r_3}}^\wedge[\vec{r_1},\vec{r_2},\vec{r_3}]=a_3 \left[\begin{matrix} 0&-1&0\\ 1&0&0\\ 0&0&0 \end{matrix}\right]

a2​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r2​
​∧[r1​
​,r2​
​,r3​
​]=a2​⎣⎡​00−1​000​100​⎦⎤​,a3​⎣⎡​r1​
​Tr2​
​Tr3​
​T​⎦⎤​r3​
​∧[r1​
​,r2​
​,r3​
​]=a3​⎣⎡​010​−100​000​⎦⎤​
  带入

(

5.3

)

(5.3)

(5.3) 可得

(

5.3

)

=

[

0

−

a

3

a

2

a

3

0

−

a

1

−

a

2

a

1

0

]

=

p

∧

,
  

得

证

.

(5.3)=\left[\begin{matrix} 0&-a_3&a_2\\ a_3&0&-a_1\\ -a_2&a_1&0 \end{matrix}\right]=p^\wedge,\;得证.

(5.3)=⎣⎡​0a3​−a2​​−a3​0a1​​a2​−a1​0​⎦⎤​=p∧,得证.
 
 

6、证明: $Rexp(p^{\wedge })R^T=exp\left((Rp{)}^{\wedge }\right)$ . [参考]

  设

R

∈

S

O

(

3

)

,
  

p

∈

R

3

R\in SO(3),\;p\in\mathbb{R}^3

R∈SO(3),p∈R3

求

证

R

e

x

p

(

p

∧

)

R

T

=

e

x

p

(

(

R

p

)

∧

)

即

证

R

e

x

p

(

p

∧

)

R

T

=

e

x

p

(

R

p

∧

R

T

)

⋯

由

题

5

结

论

得

\begin{aligned} &求证&Rexp(p^\wedge)R^T&=&&exp\left((Rp)^\wedge\right)\\ &即证&Rexp(p^\wedge)R^T&=&&exp(Rp^\wedge R^T)\qquad\cdots 由题5结论得 \end{aligned}

​求证即证​Rexp(p∧)RTRexp(p∧)RT​==​​exp((Rp)∧)exp(Rp∧RT)⋯由题5结论得​
  可将矩阵的指数映射写成一个泰勒展开

e

x

p

(

A

)

=

∑

n

=

0

∞

(

A

)

n

n

!

exp(A)=\sum_{n=0}^\infty\frac{(A)^n}{n!}

exp(A)=∑n=0∞​n!(A)n​, 于是可得

e

x

p

(

R

p

∧

R

T

)

=

∑

n

=

0

∞

(

R

p

∧

R

T

)

n

n

!

(6.1)

exp(Rp^\wedge R^T)=\sum_{n=0}^\infty\frac{(Rp^\wedge R^T)^n}{n!}\tag{6.1}

exp(Rp∧RT)=n=0∑∞​n!(Rp∧RT)n​(6.1)
  易得

(

R

p

∧

R

T

)

n

=

R

p

∧

R

T

R

p

∧

R

T

R

p

∧

R

T

⋯

R

p

∧

R

T

=

R

(

p

∧

)

n

R

T

\begin{aligned} (Rp^\wedge R^T)^n&=&&Rp^\wedge R^TRp^\wedge R^TRp^\wedge R^T\cdots Rp^\wedge R^T\\ &=&&R(p^\wedge)^nR^T \end{aligned}

(Rp∧RT)n​==​​Rp∧RTRp∧RTRp∧RT⋯Rp∧RTR(p∧)nRT​
  代入得

(

6.1

)

=

∑

n

=

0

∞

R

(

p

∧

)

n

R

T

n

!

=

R

(

∑

n

=

0

∞

(

p

∧

)

n

n

!

)

R

∧

=

R

e

x

p

(

p

∧

)

R

T

,
  

得

证

.

(6.1)=\sum_{n=0}^\infty\frac{R(p^\wedge)^nR^T}{n!}=R\left(\sum_{n=0}^\infty\frac{(p^\wedge)^n}{n!}\right)R^\wedge=Rexp(p^\wedge)R^T,\;得证.

(6.1)=n=0∑∞​n!R(p∧)nRT​=R(n=0∑∞​n!(p∧)n​)R∧=Rexp(p∧)RT,得证.
 
 

7、仿照左扰动的推导，推导$SO(3)$和$SE(3)$在右扰动下的导数

  推导$SO(3)$在右扰动下的导数

  设右扰动

Δ

R

\Delta R

ΔR 所对应的李代数为

φ

\varphi

φ .
  对

φ

\varphi

φ求导:

∂

(

R

p

)

∂

φ

=

lim

⁡

φ

→

0

e

x

p

(

ϕ

∧

)

e

x

p

(

φ

∧

)

p

−

e

x

p

(

ϕ

∧

)

p

φ

=

lim

⁡

φ

→

0

e

x

p

(

ϕ

∧

)

(

I

+

φ

∧

)

p

−

e

x

p

(

ϕ

∧

)

p

φ

⋯

泰

勒

展

开

并

舍

去

高

阶

项

=

lim

⁡

φ

→

0

e

x

p

(

ϕ

∧

)

φ

∧

p

φ

=

lim

⁡

φ

→

0

e

x

p

(

ϕ

∧

)

(

−

p

∧

φ

)

φ

⋯

叉

乘

的

性

质

=

−

e

x

p

(

ϕ

∧

)

p

∧

=

−

R

p

∧

\begin{aligned} \frac{\partial(Rp)}{\partial\varphi}&=&&\lim_{\varphi\rightarrow 0}\frac{exp(\phi^\wedge)exp(\varphi^\wedge)p-exp(\phi^\wedge)p}{\varphi}\\ &=&&\lim_{\varphi\rightarrow 0}\frac{exp(\phi^\wedge)(I+\varphi^\wedge)p-exp(\phi^\wedge)p}{\varphi}\quad&\cdots& 泰勒展开并舍去高阶项\\ &=&&\lim_{\varphi\rightarrow 0}\frac{exp(\phi^\wedge)\varphi^\wedge p}{\varphi}\\ &=&&\lim_{\varphi\rightarrow 0}\frac{exp(\phi^\wedge)(-p^\wedge \varphi)}{\varphi}&\cdots& 叉乘的性质\\ &=&&-exp(\phi^\wedge)p^\wedge\\ &=&&-Rp^\wedge \end{aligned}

∂φ∂(Rp)​​======​​φ→0lim​φexp(ϕ∧)exp(φ∧)p−exp(ϕ∧)p​φ→0lim​φexp(ϕ∧)(I+φ∧)p−exp(ϕ∧)p​φ→0lim​φexp(ϕ∧)φ∧p​φ→0lim​φexp(ϕ∧)(−p∧φ)​−exp(ϕ∧)p∧−Rp∧​⋯⋯​泰勒展开并舍去高阶项叉乘的性质​

 

  推导$SE(3)$在右扰动下的导数

  设右扰动

Δ

T

\Delta T

ΔT 所对应的李代数为

δ

ξ

\delta\xi

δξ .
  对

δ

ξ

\delta\xi

δξ 求导:

∂

(

T

p

)

∂

δ

ξ

=

lim

⁡

δ

ξ

→

0

e

x

p

(

ξ

∧

)

e

x

p

(

δ

ξ

∧

)

p

−

e

x

p

(

ξ

∧

)

p

δ

ξ

=

lim

⁡

δ

ξ

→

0

e

x

p

(

ξ

∧

)

(

I

+

δ

ξ

∧

)

p

−

e

x

p

(

ξ

∧

)

p

δ

ξ

⋯

泰

勒

展

开

并

舍

去

高

阶

项

=

lim

⁡

δ

ξ

→

0

e

x

p

(

ξ

∧

)

δ

ξ

∧

p

δ

ξ

=

lim

⁡

δ

ξ

→

0

[

R

t

0

T

1

]

[

δ

ϕ

∧

δ

ρ

0

T

0

]

p

δ

ξ

⋯

这

里

的

p

是

齐

次

的

形

式

=

lim

⁡

δ

ξ

→

0

[

R

t

0

T

1

]

[

δ

ϕ

∧

p

+

δ

ρ

0

]

δ

ξ

⋯

这

里

的

p

是

非

齐

次

的

形

式

=

lim

⁡

δ

ξ

→

0

[

R

δ

ϕ

∧

p

+

R

δ

ρ

0

T

]

[

δ

ρ

,

δ

ϕ

]

T

=

lim

⁡

δ

ξ

→

0

[

−

R

p

∧

δ

ϕ

+

R

δ

ρ

0

T

]

[

δ

ρ

,

δ

ϕ

]

T

⋯

叉

乘

的

性

质

=

[

R

−

R

p

∧

0

T

0

T

]

\begin{aligned} \frac{\partial(Tp)}{\partial\delta\xi}&=&&\lim_{\delta\xi\rightarrow 0}\frac{exp(\xi^\wedge)exp(\delta\xi^\wedge)p-exp(\xi^\wedge)p}{\delta\xi}\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{exp(\xi^\wedge)(I+\delta\xi^\wedge)p-exp(\xi^\wedge)p}{\delta\xi}\quad&\cdots&泰勒展开并舍去高阶项\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{exp(\xi^\wedge)\delta\xi^\wedge p}{\delta\xi}\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{ \left[\begin{matrix}R&t\\0^T&1\end{matrix}\right] \left[\begin{matrix}\delta\phi^\wedge&\delta\rho\\0^T&0\end{matrix}\right]p }{\delta\xi}&\cdots&这里的p是齐次的形式\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{ \left[\begin{matrix}R&t\\0^T&1\end{matrix}\right] \left[\begin{matrix}\delta\phi^\wedge p+\delta\rho\\0\end{matrix}\right] }{\delta\xi}&\cdots&这里的p是非齐次的形式\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{ \left[\begin{matrix}R\delta\phi^\wedge p+R\delta\rho\\0^T\end{matrix}\right] }{[\delta\rho,\delta\phi]^T}\\ &=&&\lim_{\delta\xi\rightarrow 0}\frac{ \left[\begin{matrix}-Rp^\wedge\delta \phi+R\delta\rho\\0^T\end{matrix}\right] }{[\delta\rho,\delta\phi]^T}&\cdots&叉乘的性质\\ &=&&\left[\begin{matrix}R&-Rp^\wedge\\0^T&0^T\end{matrix}\right] \end{aligned}

∂δξ∂(Tp)​​========​​δξ→0lim​δξexp(ξ∧)exp(δξ∧)p−exp(ξ∧)p​δξ→0lim​δξexp(ξ∧)(I+δξ∧)p−exp(ξ∧)p​δξ→0lim​δξexp(ξ∧)δξ∧p​δξ→0lim​δξ[R0T​t1​][δϕ∧0T​δρ0​]p​δξ→0lim​δξ[R0T​t1​][δϕ∧p+δρ0​]​δξ→0lim​[δρ,δϕ]T[Rδϕ∧p+Rδρ0T​]​δξ→0lim​[δρ,δϕ]T[−Rp∧δϕ+Rδρ0T​]​[R0T​−Rp∧0T​]​⋯⋯⋯⋯​泰勒展开并舍去高阶项这里的p是齐次的形式这里的p是非齐次的形式叉乘的性质​

  最后一步的矩阵求导是按照以下规则求的

d

[

a

b

]

d

[

x

y

]

=

[

d

a

d

x

d

a

d

y

d

b

d

x

d

b

d

y

]

⋯

矩

阵

中

各

元

素

可

以

看

成

偏

导

的

结

果

\frac{d\left[\begin{matrix}a\\b\end{matrix}\right]} {d\left[\begin{matrix}x\\y\end{matrix}\right]}= \left[\begin{matrix} \frac{da}{dx}&\frac{da}{dy}\\ \frac{db}{dx}&\frac{db}{dy} \end{matrix}\right]\qquad\cdots 矩阵中各元素可以看成偏导的结果

d[xy​]d[ab​]​=[dxda​dxdb​​dyda​dydb​​]⋯矩阵中各元素可以看成偏导的结果
 
 

8、搜索cmake的find_package指令是如何运作的。它有哪些可选的参数？为了让cmake找到某个库，需要哪些先决条件？

  略