version I gave to Jean Gallier
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@ -2042,15 +2042,11 @@ q(\omega)=Re^{\Skew{\omega}}p\]
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hence the derivative is:
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\begin_inset Formula \[
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\deriv{q(\omega)}{\omega}=R\deriv{}{\omega}\left(e^{\Skew{\omega}}p\right)=R\deriv{}{\omega}\left(\Skew{\omega}p\right)=RH_{p}\]
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\deriv{q(\omega)}{\omega}=R\deriv{}{\omega}\left(e^{\Skew{\omega}}p\right)=R\deriv{}{\omega}\left(\Skew{\omega}p\right)=R\Skew{-p}\]
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\end_inset
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To calculate
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\begin_inset Formula $H_{p}$
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\end_inset
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we make use of
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To show the last equality note that
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\begin_inset Formula \[
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\Skew{\omega}p=\omega\times p=-p\times\omega=\Skew{-p}\omega\]
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@ -2323,16 +2319,9 @@ We would now like to know what an incremental rotation parameterized by
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\end_inset
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hence the derivative (following the exposition in Section
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\begin_inset CommandInset ref
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LatexCommand ref
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reference "sec:Derivatives-of-Actions"
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\end_inset
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):
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hence the derivative is
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\begin_inset Formula \[
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\deriv{\hat{q}(\xi)}{\xi}=T\deriv{}{\xi}\left(\xihat\hat{p}\right)=TH_{p}\]
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\deriv{\hat{q}(\xi)}{\xi}=T\deriv{}{\xi}\left(\xihat\hat{p}\right)\]
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\end_inset
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@ -2387,7 +2376,7 @@ By only taking the top three rows, we can write this as a velocity in
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\omega\times p+v=-p\times\omega+v=\left[\begin{array}{cc}
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-\Skew p & I_{3}\end{array}\right]\left[\begin{array}{c}
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\omega\\
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v\end{array}\right]=H_{p}\xi\]
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v\end{array}\right]\]
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\end_inset
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