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Address review comments: negative sign and AdjointTranspose section

release/4.3a0
Gerry Chen 2021-10-31 20:53:15 -04:00
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@ -5086,6 +5086,13 @@ reference "ex:projection"
\begin_layout Subsection
Derivative of Adjoint
\begin_inset CommandInset label
LatexCommand label
name "subsec:pose3_adjoint_deriv"
\end_inset
\end_layout
\begin_layout Standard
@ -5098,7 +5105,7 @@ Consider
\end_inset
.
The derivative is notated (see
The derivative is notated (see Section
\begin_inset CommandInset ref
LatexCommand ref
reference "sec:Derivatives-of-Actions"
@ -5114,7 +5121,7 @@ noprefix "false"
\begin_layout Standard
\begin_inset Formula
\[
Df_{(T,y)}(\xi,\delta y)=D_{1}f_{(T,y)}(\xi)+D_{2}f_{(T,y)}(\delta y)
Df_{(T,\xi_{b})}(\xi,\delta\xi_{b})=D_{1}f_{(T,\xi_{b})}(\xi)+D_{2}f_{(T,\xi_{b})}(\delta\xi_{b})
\]
\end_inset
@ -5149,11 +5156,12 @@ D_{2}f_{(T,\xi_{b})}(\xi_{b})=Ad_{T}
\end_inset
To compute
\begin_inset Formula $D_{1}f_{(T,\xi_{b})}(\xi_{b})$
We will derive
\begin_inset Formula $D_{1}f_{(T,\xi_{b})}(\xi)$
\end_inset
, we'll first define
using two approaches.
In the first, we'll define
\begin_inset Formula $g(T,\xi)\triangleq T\exp\hat{\xi}$
\end_inset
@ -5194,18 +5202,30 @@ Now we can use the definition of the Adjoint representation
\begin_layout Standard
\begin_inset Formula
\begin{align*}
D_{1}f_{(T,\xi_{b})}(\xi)=D_{1}\left(Ad_{T\exp(\hat{\xi})}\hat{\xi}_{b}\right)(\xi) & =D_{1}\left(g\hat{\xi}g^{-1}\right)(\xi)\\
& =\left(D_{2}g_{(T,\xi)}(\xi)\right)\hat{\xi}g^{-1}(T,0)+g(T,0)\hat{\xi}\left(D_{2}g_{(T,\xi)}^{-1}(\xi)\right)\\
D_{1}f_{(T,\xi_{b})}(\xi)=D_{1}\left(Ad_{T\exp(\hat{\xi})}\hat{\xi}_{b}\right)(\xi) & =D_{1}\left(g\hat{\xi}_{b}g^{-1}\right)(\xi)\\
& =\left(D_{2}g_{(T,\xi)}(\xi)\right)\hat{\xi}_{b}g^{-1}(T,0)+g(T,0)\hat{\xi}_{b}\left(D_{2}g_{(T,\xi)}^{-1}(\xi)\right)\\
& =T\hat{\xi}\hat{\xi}_{b}T^{-1}-T\hat{\xi}_{b}\hat{\xi}T^{-1}\\
& =T\left(\hat{\xi}\hat{\xi}_{b}-\hat{\xi}_{b}\hat{\xi}\right)T^{-1}\\
& =-Ad_{T}(ad_{\xi_{b}}\hat{\xi})\\
& =Ad_{T}(ad_{\hat{\xi}}\hat{\xi}_{b})\\
& =-Ad_{T}(ad_{\hat{\xi}_{b}}\hat{\xi})\\
D_{1}F_{(T,\xi_{b})} & =-(Ad_{T})(ad_{\hat{\xi}_{b}})
\end{align*}
\end_inset
An alternative, perhaps more intuitive way of deriving this would be to
use the fact that the derivative at the origin
Where
\begin_inset Formula $ad_{\hat{\xi}}:\mathfrak{g}\rightarrow\mathfrak{g}$
\end_inset
is the adjoint map of the lie algebra.
\end_layout
\begin_layout Standard
The second, perhaps more intuitive way of deriving
\begin_inset Formula $D_{1}f_{(T,\xi_{b})}(\xi_{b})$
\end_inset
, would be to use the fact that the derivative at the origin
\begin_inset Formula $D_{1}Ad_{I}\hat{\xi}_{b}=ad_{\hat{\xi}_{b}}$
\end_inset
@ -5224,28 +5244,212 @@ An alternative, perhaps more intuitive way of deriving this would be to
\begin_layout Standard
\begin_inset Formula
\[
D_{1}Ad_{T}\hat{\xi}_{b}(\xi)=D_{1}Ad_{T*I}\hat{\xi}_{b}(\xi)=Ad_{T}\left(D_{1}Ad_{I}\hat{\xi}_{b}(\xi)\right)=Ad_{T}\left(ad_{\hat{\xi}_{b}}(\xi)\right)
D_{1}Ad_{T}\hat{\xi}_{b}(\xi)=D_{1}Ad_{T*I}\hat{\xi}_{b}(\xi)=Ad_{T}\left(D_{1}Ad_{I}\hat{\xi}_{b}(\xi)\right)=Ad_{T}\left(ad_{\hat{\xi}}(\hat{\xi}_{b})\right)=-Ad_{T}\left(ad_{\hat{\xi}_{b}}(\hat{\xi})\right)
\]
\end_inset
It's difficult to apply a similar procedure to compute the derivative of
\begin_inset Formula $Ad_{T}^{T}\hat{\xi}_{b}^{*}$
\end_layout
\begin_layout Subsection
Derivative of AdjointTranspose
\end_layout
\begin_layout Standard
The transpose of the Adjoint,
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\color none
\begin_inset Formula $Ad_{T}^{T}:\mathfrak{g^{*}\rightarrow g^{*}}$
\end_inset
, is useful as a way to change the reference frame of vectors in the dual
space
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(note the
\begin_inset Formula $^{*}$
\end_inset
denoting that we are now in the dual space) because
denoting that we are now in the dual space)
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.
To be more concrete, where
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as
\begin_inset Formula $Ad_{T}\hat{\xi}_{b}$
\end_inset
converts the
\emph on
twist
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\begin_inset Formula $\xi_{b}$
\end_inset
from the
\begin_inset Formula $T$
\end_inset
frame,
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\begin_inset Formula $Ad_{T}^{T}\hat{\xi}_{b}^{*}$
\end_inset
converts the
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wrench
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\begin_inset Formula $\xi_{b}^{*}$
\end_inset
from the
\begin_inset Formula $T$
\end_inset
frame
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.
It's difficult to apply a similar derivation as in Section
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:pose3_adjoint_deriv"
plural "false"
caps "false"
noprefix "false"
\end_inset
for the derivative of
\begin_inset Formula $Ad_{T}^{T}\hat{\xi}_{b}^{*}$
\end_inset
because
\begin_inset Formula $Ad_{T}^{T}$
\end_inset
cannot be naturally defined as a conjugation so we resort to crunching
cannot be naturally defined as a conjugation, so we resort to crunching
through the algebra.
The details are omitted but the result is a form vaguely resembling (but
not quite) the
The details are omitted but the result is a form that vaguely resembles
(but does not exactly match)
\begin_inset Formula $ad(Ad_{T}^{T}\hat{\xi}_{b}^{*})$
\end_inset

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