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SLAM学习-李群和李代数相互转换公式

作者:互联网

在学习 十四讲时,李群和李代数相互转换公式为
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 ∧ exp(\phi^{\wedge})=exp(\theta a^{\wedge})=cos\theta I+(1-cos\theta)aa^{T}+sin\theta a^{\wedge} exp(ϕ∧)=exp(θa∧)=cosθI+(1−cosθ)aaT+sinθa∧
其证明过程可以参考四十讲

而在选在IMU 预积分中,李群和李代数相互转换公式为
e x p ( ϕ ∧ ) = I + s i n ( ∣ ∣ ϕ ∣ ∣ ) ∣ ∣ ϕ ∣ ∣ ϕ ∧ + 1 − c o s ( ∣ ∣ ϕ ∣ ∣ ) ∣ ∣ ϕ ∣ ∣ 2 ( ϕ ∧ ) 2 exp(\phi^{\wedge})= I+\frac{sin(||\phi ||)}{||\phi||}\phi^{\wedge}+\frac{1-cos(||\phi ||)}{||\phi||^{2}}(\phi^{\wedge})^{2} exp(ϕ∧)=I+∣∣ϕ∣∣sin(∣∣ϕ∣∣)​ϕ∧+∣∣ϕ∣∣21−cos(∣∣ϕ∣∣)​(ϕ∧)2

这是两种等价的表示方式

下边我们简单证明下:

ϕ = θ a ⃗ a ∧ a ∧ = a a T − I \phi = \theta \vec{a} \\ a^{\wedge}a^{\wedge} = aa^{T}-I\\ ϕ=θa a∧a∧=aaT−I
则有
∣ ∣ ϕ ∣ ∣ = θ ϕ ∧ = θ a ∧ = ∣ ∣ ϕ ∣ ∣ a ∧ a ∧ = ϕ ∧ ∣ ∣ ϕ ∣ ∣ ( ϕ ∧ ) 2 = θ 2 ( a a T − I ) = ∣ ∣ ϕ ∣ ∣ 2 ( a a T − I ) a a T = ( ϕ ∧ ) 2 ∣ ∣ ϕ ∣ ∣ 2 + I ||\phi||=\theta\\ {\phi}^{\wedge} =\theta a^{\wedge}=||\phi||a^{\wedge}\\ a^{\wedge}=\frac{{\phi}^{\wedge} }{||\phi||}\\ ({\phi}^{\wedge})^{2} = \theta^{2}(aa^{T}-I)=||\phi||^{2}(aa^{T}-I)\\ aa^{T}=\frac{({\phi}^{\wedge})^{2}}{||\phi||^{2}}+I ∣∣ϕ∣∣=θϕ∧=θa∧=∣∣ϕ∣∣a∧a∧=∣∣ϕ∣∣ϕ∧​(ϕ∧)2=θ2(aaT−I)=∣∣ϕ∣∣2(aaT−I)aaT=∣∣ϕ∣∣2(ϕ∧)2​+I
带入
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 ∧ exp(\phi^{\wedge})=exp(\theta a^{\wedge})=cos\theta I+(1-cos\theta)aa^{T}+sin\theta a^{\wedge} exp(ϕ∧)=exp(θa∧)=cosθI+(1−cosθ)aaT+sinθa∧
则有
e x p ( ϕ ∧ ) = c o s ∣ ∣ ϕ ∣ ∣ I + ( 1 − c o s ∣ ∣ ϕ ∣ ∣ ) ∗ ( ( ϕ ∧ ) 2 ∣ ∣ ϕ ∣ ∣ 2 + I ) + s i n ∣ ∣ ϕ ∣ ∣ ∗ ϕ ∧ ∣ ∣ ϕ ∣ ∣ = I + s i n ( ∣ ∣ ϕ ∣ ∣ ) ∣ ∣ ϕ ∣ ∣ ϕ ∧ + 1 − c o s ( ∣ ∣ ϕ ∣ ∣ ) ∣ ∣ ϕ ∣ ∣ 2 ( ϕ ∧ ) 2 exp(\phi^{\wedge})=cos||\phi|| I+(1-cos||\phi||)*(\frac{({\phi}^{\wedge})^{2}}{||\phi||^{2}}+I)+sin||\phi|| *\frac{{\phi}^{\wedge} }{||\phi||}\\ =I+\frac{sin(||\phi ||)}{||\phi||}\phi^{\wedge}+\frac{1-cos(||\phi ||)}{||\phi||^{2}}(\phi^{\wedge})^{2} exp(ϕ∧)=cos∣∣ϕ∣∣I+(1−cos∣∣ϕ∣∣)∗(∣∣ϕ∣∣2(ϕ∧)2​+I)+sin∣∣ϕ∣∣∗∣∣ϕ∣∣ϕ∧​=I+∣∣ϕ∣∣sin(∣∣ϕ∣∣)​ϕ∧+∣∣ϕ∣∣21−cos(∣∣ϕ∣∣)​(ϕ∧)2

标签:wedge,phi,李群,cos,SLAM,exp,theta,代数,sin
来源: https://blog.csdn.net/m0_54238665/article/details/122805107