Editing 3D rotation group (section)

==Exponential map==
The exponential map for {{math|SO(3)}}, is, since {{math|SO(3)}} is a matrix Lie group, defined using the standard [[matrix exponential]] series,

:<math>\begin{cases}
\exp : \mathfrak{so}(3) \to \operatorname{SO}(3) \\
A \mapsto e^A = \sum_{k=0}^\infty \frac{1}{k!} A^k
 = I + A + \tfrac{1}{2} A^2 + \cdots.
\end{cases}</math>

For any [[skew-symmetric matrix]] {{math|''A'' ∈ 𝖘𝖔(3)}}, {{math|''e<sup>A</sup>''}} is always in {{math|SO(3)}}. The proof uses the elementary properties of the matrix exponential

:<math>\left(e^A\right)^\textsf{T} e^A = e^{A^\textsf{T}} e^A = e^{A^\textsf{T} + A} = e^{-A + A} = e^{A - A} = e^A \left(e^A\right)^\textsf{T} = e^0 = I.</math>

since the matrices {{math|''A''}} and {{math|''A''{{sup|T}}}} commute, this can be easily proven with the skew-symmetric matrix condition. This is not enough to show that {{math|𝖘𝖔(3)}} is the corresponding Lie algebra for {{math|SO(3)}}, and shall be proven separately.

The level of difficulty of proof depends on how a matrix group Lie algebra is defined. {{harvtxt|Hall|2003}} defines the Lie algebra as the set of matrices

:<math>\left\{A \in \operatorname{M}(n, \R) \left| e^{tA} \in \operatorname{SO}(3) \forall t\right.\right\},</math>

in which case it is trivial. {{harvtxt|Rossmann|2002}} uses for a definition derivatives of smooth curve segments in {{math|SO(3)}} through the identity taken at the identity, in which case it is harder.<ref>See {{harvnb|Rossmann|2002}}, theorem 3, section 2.2.</ref>

For a fixed {{math|''A'' ≠ 0}}, {{math|''e<sup>tA</sup>'', −∞ < ''t'' < ∞}} is a [[one-parameter subgroup]] along a [[geodesic]] in {{math|SO(3)}}. That this gives a one-parameter subgroup follows directly from properties of the exponential map.<ref>{{harvnb|Rossmann|2002}} Section 1.1.</ref>

The exponential map provides a [[diffeomorphism]] between a neighborhood of the origin in the {{math|𝖘𝖔(3)}} and a neighborhood of the identity in the {{math|SO(3)}}.<ref>{{harvnb|Hall|2003}} Theorem 2.27.</ref> For a proof, see [[Closed subgroup theorem]].

The exponential map is [[surjective]]. This follows from the fact that every {{math|''R'' ∈ SO(3)}}, since every rotation leaves an axis fixed ([[Euler's rotation theorem]]), and is conjugate to a [[block diagonal matrix]] of the form

:<math>D = \begin{pmatrix}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\end{pmatrix} = e^{\theta L_z},</math>

such that {{math|1=''A'' = ''BDB''<sup>−1</sup>}}, and that

:<math>Be^{\theta L_z}B^{-1} = e^{B\theta L_zB^{-1}},</math>

together with the fact that {{math|𝖘𝖔(3)}} is closed under the [[adjoint representation|adjoint action]] of {{math|SO(3)}}, meaning that {{math|''BθL<sub>z</sub>B''<sup>−1</sup> ∈ 𝖘𝖔(3)}}.

Thus, e.g., it is easy to check the popular identity

:<math>e^{-\pi L_x/2} e^{\theta L_z} e^{\pi L_x/2} = e^{\theta L_y}.</math>

As shown above, every element {{math|''A'' ∈ 𝖘𝖔(3)}} is associated with a vector {{math|1='''''ω''''' = ''θ'' '''''u'''''}}, where {{math|1='''''u''''' = (''x'',''y'',''z'')}} is a unit magnitude vector. Since {{math|'''''u'''''}} is in the null space of {{mvar|A}}, if one now rotates to a new basis, through some other orthogonal matrix {{math|''O''}}, with {{math|'''u'''}} as the {{mvar|z}} axis, the final column and row of the rotation matrix in the new basis will be zero.

Thus, we know in advance from the formula for the exponential that {{math|exp(''OAO''<sup>T</sup>)}} must leave {{math|'''''u'''''}} fixed. It is mathematically impossible to supply a straightforward formula for such a basis as a function of {{math|'''''u'''''}}, because its existence would violate the [[hairy ball theorem]]; but direct exponentiation is possible, and [[Axis–angle representation#Exponential map from 𝖘𝖔(3) to SO(3)|yields]]

:<math>\begin{align}
 \exp(\tilde{\boldsymbol{\omega}})
   &= \exp(\theta(\boldsymbol{u\cdot L}))
    = \exp\left(\theta \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}\right)\\[4pt]
   &= \boldsymbol{I} + 2cs(\boldsymbol{u\cdot L}) + 2s^2 (\boldsymbol{u \cdot L})^2 \\[4pt]
   &= \begin{bmatrix}
        2 \left(x^2 - 1\right) s^2 + 1 & 2 x y s^2 - 2 z c s & 2 x z s^2 + 2 y c s \\
        2 x y s^2 + 2 z c s & 2 \left(y^2 - 1\right) s^2 + 1 & 2 y z s^2 - 2 x c s \\
        2 x z s^2 - 2 y c s & 2 y z s^2 + 2 x c s & 2 \left(z^2 - 1\right) s^2 + 1
      \end{bmatrix},
\end{align}</math>

where <math display="inline">c = \cos\frac{\theta}{2}</math> and <math display="inline">s = \sin\frac{\theta}{2}</math>. This is recognized as a matrix for a rotation around axis {{math|'''''u'''''}} by the angle {{mvar|θ}}: cf. [[Rodrigues' rotation formula]].