例四和例五：The exponential map of SO(3)+

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Example 4:The exponential map of SO(3)

We have seen in Ex.3 that $\dot{R}=R[\omega {]}_{\times }\in T_{R}SO(3).$
For ω constant, this is an ordinary differential equation(ODE), whose solution is
$R(t)=R0\ast exp([\omega {]}_{\times }t)$. At the origin $R0=I$ we have the exponential map,
$$R(t)=exp([\omega {]}_{\times }t)\in SO(3)$$ .
We now define the vector $\theta \triangleq u\theta \triangleq \omega t\in R^3$ as the integrated rotation in angle-axis form, with angle θ and unit axis u. Thus $[\theta {]}_{\times }\in so(3)$ is the total rotation expressed in the Lie algebra. We substitute it above. Then write the exponential as a power series,
$$R=exp([\theta {]}_{\times })=\sum _k\frac{{\theta }^k}{k!}([u{]}_{\times })k.$$
In order to find a closed-form expression, we write down a few powers of [u]×,
$$\begin{gathered} [u{]}_{\times }^0=I,[u{]}_{\times }^1=[u{]}_{\times } \\ [u{]}_{\times }^2=u{u}^t-I,[u{]}_{\times }^3=-[u{]}_{\times }, \\ [u]4\times =-[u{]}_{\times }^2,\cdot \cdot \cdot \end{gathered}$$
and realize that all can be expressed as multiples of I,[u]_× or [u]^2_×. We thus rewrite the series as,
$$\begin{gathered} R=I+[u{]}_{\times }(\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\dots \dots ) \\ +[u{]}_{\times }^2(\frac{1}{2}{\theta }^2-\frac{1}{4!}{\theta }^4+\frac{1}{6!}{\theta }^6-\dots \dots ) \end{gathered}$$
where we identify the series of sin θ and cos θ, yielding the closed form
$$R=exp([u\theta ]\times )=I+[u{]}_{\times }sin\theta +[u{]}_{\times }^2(1-cos\theta )$$
This expression is the well known Rodrigues rotation formula. It can be used as the capitalized exponential just by doing $R=Exp(u\theta )=exp([u\theta {]}_{\times })$。

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我们已经在 Ex3 中看到了$\dot{R}=R[\omega {]}_{\times }\in T_{R}SO(3)$
对于 ω 常数，这是一个ODE (常微分方程)，其解为$R(t)=R0\ast exp([\omega {]}_{\times }t)$。 在
原点 R0 = I 我们有指数映射，
$$R(t)=exp([\omega {]}_{\times }t)\in SO(3)$$
我们现在定义向量 $\theta \triangleq u\theta \triangleq \omega t\in R^3$ 为角度轴形式的积分旋转，角度 θ 和单位轴 u。 因此 $[\theta {]}_{\times }\in so(3)$是李代数中表达的总旋转。 我们在上面替换它。 然后将指数写成幂级数，
$$R=exp([\theta {]}_{\times })=\sum _k\frac{{\theta }^k}{k!}([u{]}_{\times })k$$
为了找到一个封闭形式的表达式，我们写下 [u]× 的几个幂，
$$\begin{gathered} [u{]}_{\times }^0=I,[u{]}_{\times }^1=[u{]}_{\times } \\ [u{]}_{\times }^2=u{u}^t-I,[u{]}_{\times }^3=-[u{]}_{\times }, \\ [u]4\times =-[u{]}_{\times }^2,\cdot \cdot \cdot \end{gathered}$$
并意识到所有都可以表示为 I,[u]_× 或 [u]^2_× 的倍数。 因此，我们将其重写为，
$$\begin{gathered} R=I+[u{]}_{\times }(\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\dots \dots ) \\ +[u{]}_{\times }^2(\frac{1}{2}{\theta }^2-\frac{1}{4!}{\theta }^4+\frac{1}{6!}{\theta }^6-\dots \dots ) \end{gathered}$$
通过 sin θ 和 cos θ 的序列形成的封闭形式
$$R=exp([u\theta ]\times )=I+[u{]}_{\times }sin\theta +[u{]}_{\times }^2(1-cos\theta )$$
这个表达式就是众所周知的罗德里格斯旋转公式。 只需执行以下操作即可将其用作大写指数 $R=Exp(u\theta )=exp([u\theta {]}_{\times })$。

Example 5: The unit quaternions group S3(cont.)

单位四元数群 S3(续)

In the group S3(recall Ex. 2 and see e.g. [8]), the time derivative of the unit norm condition $q^{\ast }q=1$ yields
$$q^{\ast }\dot{q}=-(q^{\ast }\dot{q}{)}^{\ast }.$$
This reveals that $q^{\ast }\dot{q}$ is a pure quaternion (its real part is zero). Pure quaternions uv ∈ Hp have the form
$$uv=(i{u}_x+j{u}_y+k{u}_z)v=i{v}_x+j{v}_y+k{v}_z$$
where$u\triangleq i{u}_x+j{u}_y+k{u}_z$is pure and unitary, v is the norm, and i, j, k are the generators of the Lie algebra $s^3=H_p$. Re-writing the condition above we have,
$$\dot{q}=quv\in T_q{S}^3$$
which integrates to $q=q0\ast exp(uvt)$. Letting q0 = 1 and defining $$\phi \triangleq u\phi \triangleq uvt$$ we get the exponential map,
$$q=exp(u\phi )\triangleq \sum \frac{{\phi }^k}{k!}{u}^k\in S^3$$
The powers of u follow the pattern 1, u, −1, −u, 1, · · · .
Thus we group the terms in 1 and u and identify the series of cos φ and sin φ. We get the closed form
$$q=exp(u\phi )=cos(\phi )+usin(\phi ),$$
which is a beautiful extension of the Euler formula,exp(iφ) = cos φ+isin φ.
The elements of the Lie algebra φ = uφ ∈ s3 can be identified with the rotation vectorθ ∈ R3 trough the mappings hat and vee,
$$\begin{gathered} Hat:R^3\to s^3;\theta \to {\theta }^{\wedge }=2\phi \\ Vee:s^3\to R^3;\phi \to {\phi }^{\vee }=\theta /2 \end{gathered}$$ ,
where the factor 2 accounts for the double effect of the quaternion in the rotation action, x’ = q x q∗. With this choice of Hat and Vee, the quaternion exponential
$$q=Exp(u\theta )=cos(\theta /2)+usin(\theta /2)$$
is equivalent to the rotation matrix R = Exp(uθ)

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在群 S3（回忆例 2 并参见例 [8]）中，单位范数条件 $q^{\ast }q=1$ 的时间导数要求
$$q^{\ast }\dot{q}=-(q^{\ast }\dot{q}{)}^{\ast }.$$
这表明 $q^{\ast }\dot{q}$ 是一个纯四元数（它的实部为零）。 纯四元数 uv ∈ Hp 具有形式
$$uv=(i{u}_x+j{u}_y+k{u}_z)v=i{v}_x+j{v}_y+k{v}_z$$
其中$u\triangleq i{u}_x+j{u}_y+k{u}_z$是纯幺正的，v是范数，i, j, k是李代数 $s^3=H_p$的生成元。 重写我们上面的条件，

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