%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% %% %% Preamble of document %% %% %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Sets up the standard "article" document class \documentclass[12pt]{article} % Loads packages % Note that I typically load all packages at once (as below). % Sometimes you must load them in a certain order... \usepackage{amsmath, amsfonts, amssymb, amsthm, enumerate, enumitem, geometry, mleftright, nicefrac, mathtools, xparse, ifthen, tikz, pgfplots, floatrow, multicol, caption, xurl} % Set up PGF plots (used for graphing) \pgfplotsset{compat=newest} % Fixes spacing caused by using "\left" or "\right" \mleftright % Sets the margins of the document. % One can do this using plain TeX, but why? \geometry{margin=1in} % Lines 43-46 set up my use of the cleveref package. % This provides a "smarter" cross referencing style \usepackage[colorlinks=true]{hyperref} \usepackage[sort,capitalise]{cleveref} \newcommand{\creflastconjunction}{, and\nobreakspace} \crefname{enumi}{item}{items} % My personal command macros that I use. % Note that "\displaySpace" is use to create the correct spacing for displayed math puctuation (and can be easily updated throughout the entire document by changing line 50). \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\dpp}{\text{.}} \newcommand{\dc}{\text{,}} \newcommand{\M}{\mathcal{M}} \newcommand{\dd}{{\rm d}} \newcommand{\fX}{\mathfrak{X}} \newcommand{\fg}{\mathfrak{g}} \newcommand{\G}{\mathcal{G}} \newcommand{\dexp}{\operatorname{dexp}} \newcommand{\ad}{\operatorname{ad}} \newcommand{\lr}{\ensuremath{\mkern-1.5mu}} % Creates my paired math delimiters for easy use. \DeclarePairedDelimiter{\pr}{(}{)} \DeclarePairedDelimiter{\br}{[}{]} \DeclarePairedDelimiter{\cu}{\{}{\}} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} \DeclarePairedDelimiter{\vt}{\langle}{\rangle} % Creates desired header and footer for document. % Note that it uses the adaptively created date macro. \setlength{\headheight}{15pt} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \fancyhead[R]{} \cfoot{} \lfoot{\small{\textsc{Dissemination prohibited. \today}}} \rfoot{\thepage} % Updates equation and figure numbering. \counterwithin{equation}{section} \counterwithin{figure}{section} % Defines theorem-like environments \theoremstyle{definition} \newtheorem{definition}{Definition}[section] \newtheorem{theorem}[definition]{Theorem} \newtheorem{lemma}[definition]{Lemma} \newtheorem{problem}[definition]{Problem} \newtheorem{remark}[definition]{Remark} \newtheorem{assumption}[definition]{Assumption} \newtheorem{example}[definition]{Example} \renewcommand{\thetheorem}{\theenumi\arabic{theorem}} % Helps create better arrows in figures \usetikzlibrary{arrows.meta} %%%%%%%%%%%%%%%%%%%%%%%% %%% Automatic proof headers and endings, local labels %%%%%%%%%%%%%%%%%%%%%%%% \ExplSyntaxOn \NewDocumentEnvironment {athm} {m m} {% %\phantomsection %\addcontentsline{toc}{subsection}{\texorpdfstring{\cref{#2}}{\ref*{#2}}} \begin{#1}\label{#2}\global\def\loc{#2}% }{% \end{#1}% } \NewDocumentEnvironment{aproof} {} {% \begin{proof}[Proof~of~\cref{\loc}]% }{% \finishproofthus \end{proof}% } \ExplSyntaxOff \newcommand{\eqqref}[1]{\cref{eq:\loc.#1}} \newcommand{\eqlabel}[1]{\label{eq:\loc.#1}} \newcommand{\finishproofthus}{The proof of \cref{\loc} is thus complete.} \NewDocumentEnvironment{asol} {} {% \begin{proof}[Solution~to~\cref{\loc}]% }{% \noindent\finishsolthus \end{proof}% } \newcommand{\finishsolthus}{The solution to \cref{\loc} is thus complete.} %%%%%%%%%%%%%%%%%%%%%%%% %%% Creating blank space with line breaks %%%%%%%%%%%%%%%%%%%%%%%% % Note that the code in lines 137-147 is simply a method to % create blank space that allows for proof environments to % break across pages. This is something that TeX does not % like by default, so this is my effort to bypass this. \newcommand{\linebreaks}[1]{{\mbox{}\par \newcount\foo \foo=#1 \loop \phantom{a} \advance \foo -1 \ifnum \foo>0 \repeat \mbox{}\par}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% %% %% Beginning of actual text. %% %% %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \title{A survey of numerical integration techniques on manifolds} \author{Joshua Lee Padgett} \date{\today} \begin{document} \maketitle \begin{abstract} The following set of notes provide a very rough overview and introduction to the ideas and concepts needed to approximate differential equations evolving on homogeneous manifolds. This document mostly serves as a means to provide some basic ideas and then allow everyone else to inform me of which concepts need additional details in order to improve understanding. \end{abstract} \tableofcontents \section{Introduction} \section{Some basic background material} The following are some basic definitions needed to discuss numerical integration on homogeneous manifolds (manifolds acted upon transitively by a Lie group). Note that the following should not be viewed as a complete list of all needed (or even relevant) topics and concepts. Any set of notes created in this way are usually incomplete, so we should either view them as a starting point from which we commence our journey to total understanding or we should discuss what further details are needed. \begin{athm}{definition}{def:manifold} Let $m,d \in \N = \{1,2,3,\ldots\}$ with $m \le d$ and let $\M \subseteq \R^d$ with $\M \neq \emptyset$. Then $\M$ is an $m$-dimensional smooth manifold if and only if for every $p \in \M$ there exist $\Omega \subseteq \R^d$, $U \subseteq \M$ with $p \in U$, and a smooth function $\varphi \colon \Omega \to \R^d$ such that $\varphi$ is a homeomorphism and $\varphi' \circ \varphi^{-1}$ is injective. See, e.g., \url{https://en.wikipedia.org/wiki/Manifold} for more details. \end{athm} \begin{athm}{definition}{def:tangent} Let $\M$ be a manifold, let $p \in \M$, and let $\rho \colon [0,1] \to \M$ be a differentiable function with $\rho(0) = p$ (cf.\ \cref{def:manifold}). Then the vector \begin{equation} \pr[\big]{ \tfrac{\dd }{\dd t} \rho } \lr (t) \Bigr\rvert_{t=0} \end{equation} is a tangent vector at the point $p$. The set of all tangent vectors at $p$ is the tangent space at $p$ and is denoted by $T\M \rvert_p$. The collection of all tangent spaces at all points $q \in \M$ is called the tangent bundle of $\M$ and is denoted by $T\M = \cup_{q\in\M} T\M\rvert_{q}$. See, e.g., \url{https://en.wikipedia.org/wiki/Tangent_space} for more details. \end{athm} \begin{athm}{definition}{def:vector_field} Let $\M$ be a manifold and let $F \colon \M \to T\M$ be a differentiable function which satisfies for all $p \in \M$ that $F(p) \in T\M\rvert_p$ (cf.\ \cref{def:manifold,def:tangent}). Then $F$ is a vector field on $\M$. The collection of all vector fields on $\M$ is denoted by $\fX(\M)$. See, e.g., \url{https://en.wikipedia.org/wiki/Vector_field} for more details. \end{athm} \begin{athm}{definition}{def:lie_algebra} Let $\fg$ be a vector space and let $[\cdot,\cdot] \colon \fg \times \fg \to \fg$ be a bilinear function which satisfies for all $a,b,c\in\fg$ that $[a,b] = -[b,a]$ and \begin{equation} [a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0 \dpp \end{equation} Then we say that $\fg$ is a Lie algebra and we call the function $[\cdot,\cdot] \colon \fg \times \fg \to \fg$ the Lie bracket on $\fg$. See, e.g., \url{https://en.wikipedia.org/wiki/Lie_algebra} for more details. \end{athm} \begin{athm}{problem}{prob:lie_alg} Let $\fg = \mathfrak{so}(3) = \{ a \in \R^{3\times 3} \colon a^* = -a \}$ and let $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ satisfy for all $a,b \in \fg$ that $[a,b] = ab - ba$. Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}). \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{problem}{prob:lie_alg2} Let $\M$ be a manifold, let $C^\infty(\M)$ be the set of smooth function on $\M$, let $\fg = \fX(\M)$, and let $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ satisfy for all $X,Y \in \fX(\M)$, $f \in C^\infty(\M)$ that $[X,Y](f) = (X\circ Y)(f) - (Y\circ X)(f)$ (cf.\ \cref{def:manifold,def:vector_field}). Show that $\fg$ is a Lie algebra (cf.\ \cref{def:lie_algebra}). \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{definition}{def:adjoint} Let $\fg$ be a Lie algebra (cf.\ \cref{def:lie_algebra}). Then we define $\ad^n \colon \fg \times \fg \to \fg$, $n\in\N_0 = \{0\} \cup \N$, to be the functions which satisfy for all $n \in \N$, $u,v \in \fg$ that $\ad^0(u,v) = v$ and \begin{equation} \ad^n(u,v) = [u,\ad^{n-1}(u,v)] \dpp \end{equation} See, e.g., \url{https://en.wikipedia.org/wiki/Adjoint_representation} for more details. \end{athm} %\begin{athm}{definition}{def:lie_alg_homo} %Let $\fg,\mathfrak{h}$ be Lie algebras and let $\varphi \colon \fg \to \mathfrak{h}$ be a linear function which satisfies for all $u,v\in\fg$ that %\begin{equation} %\varphi\pr[\big]{ [u,v]_\fg } = \br[\big]{ \varphi(u) , \varphi(v) }_{\mathfrak{h}} %\end{equation} %(cf.\ \cref{def:lie_algebra}). %Then we say that $\varphi$ is a Lie algebra homomorphism. %\end{athm} \begin{athm}{definition}{def:lie_group} Let $\G$ be a manifold (cf.\ \cref{def:manifold}). We say that $\G$ is a Lie group if and only if there exists a function $\cdot \colon \G \times \G \to \G$ and $e \in \G$ such that \begin{enumerate}[label=(\roman*)] \item it holds for all $p,q,r \in \G$ that $p \cdot (q \cdot r) = (p \cdot r) \cdot q$, \item it holds for all $p \in \G$ that $p \cdot e = e \cdot p = p$, \item it holds for all $p \in \G$ that there exists $p^{-1} \in \G$ such that $p^{-1} \cdot p = e$, and \item it holds that $\cdot \colon \G \times \G \to \G$ and $\G \ni p \mapsto p^{-1} \in \G$ are smooth. \end{enumerate} See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group} for more details. \end{athm} \begin{athm}{problem}{prob:lie_group} Let $\G = \{ a \in \R^{3\times 3} \colon a^{-1} = a^* \}$ and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$. Show that $\G$ is a Lie group (cf.\ \cref{def:lie_group}). \emph{Note:} We have not defined matrix manifolds, but we can easily update this definition to allow for such things (e.g., via representations). \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{definition}{def:alg_of_group} Let $\G$ be a Lie group and let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$ (cf.\ \cref{def:lie_group}). Then we define the Lie algebra of a Lie group to be $\fg = T\G\rvert_{e}$ (cf.\ \cref{def:lie_algebra,def:tangent}). The Lie bracket on $\fg$ is the function $[ \cdot , \cdot ] \colon \fg \times \fg \to \fg$ which satisfies for all $u,v \in \fg$, $g,h \colon [0,1] \to \G$ with $g$ and $h$ being differentiable, $g(0) = h(0) = e$, $g'(0) = u$, and $h'(0) = v$ that \begin{equation} [u,v] = \frac{\partial^2}{\partial t \partial s} g(t) \cdot h(s) \cdot g^{-1}(t) \bigg\rvert_{t=s=0} \dpp \end{equation} See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group%E2%80%93Lie_algebra_correspondence} for more details. \end{athm} \begin{athm}{problem}{prob:lie_group2} Let $\G = \operatorname{SO}(3) = \{ a \in \R^{3\times 3} \colon a^{-1} = a^*,\ \det(a)=1 \}$ and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$. Show that $\fg = \mathfrak{so}(3)$ is the Lie algebra of $\G$ (cf.\ \cref{def:alg_of_group}). \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{definition}{def:action} Let $\M$ be a manifold, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, and let $\Lambda \colon \G \times \M \to \M$ be a smooth function which satisfies for all $p,q\in\G$, $y\in\M$ that \begin{equation} \Lambda(e,y) = y \qquad\text{and}\qquad \Lambda\pr[\big]{p,\Lambda(q,y)} = \Lambda(p\cdot q,y) \end{equation} (cf.\ \cref{def:manifold,def:lie_group}). Then we say that $\Lambda$ is an (left) action of $\G$ on $\M$. See, e.g., \url{https://en.wikipedia.org/wiki/Lie_group_action} for more details. \end{athm} \begin{athm}{definition}{def:right} Let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:alg_of_group}). Then we define the function $R_v \colon \G \to \G$, $v \in \G$, to be the function which satisfies for all $u,v\in\G$ that $R_v(u) = u\cdot v$. Moreover, we have that for all $v \in \G$ it holds that $R_v' = T\rvert_e R_v \colon \fg \to T\G\rvert_v$ (cf.\ \cref{def:tangent}). See, e.g., \url{https://math.stackexchange.com/questions/1740179/differential-of-the-multiplication-and-inverse-maps-on-a-lie-group} for more details. \end{athm} \begin{athm}{definition}{def:exp} Let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, and let $\sigma_a \colon [0,1] \to \G$, $a \in \fg$, be a differentiable function satisfying for all $a\in\fg$, $t \in [0,1]$ that $\sigma_a(0) = e$ and \begin{equation} \sigma_a'(t) = R_{\sigma_a(t)}'(a) \end{equation} (cf.\ \cref{def:lie_algebra,def:lie_group,def:right,def:alg_of_group}). Then we define the exponential map $\exp \colon \fg \to \G$ to be the function which satisfies for all $a \in \fg$ that $\exp(a) = \sigma_a(1)$. See, e.g., \url{https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#:~:text=In%20the%20theory%20of%20Lie,tool%20for%20studying%20Lie%20groups.}~for more details. \end{athm} \begin{athm}{problem}{prob:exp} Let $d \in \N$, let $\G = \operatorname{GL}(d;\R) = \{ a \in \R^{d\times d} \colon \det(a) \neq 0 \}$, and let $\cdot \colon \G \times \G \to \G$ satisfy for all $a,b\in\G$ that $a\cdot b = ab$. \begin{enumerate}[label=\alph*.] \item Show that $\mathfrak{gl}_d = \R^{d\times d}$ is the Lie algebra of $\G$ (cf.\ \cref{def:alg_of_group}). \item Show that for all $a \in \G$ it holds that %\begin{equation} $\exp(a) = \sum_{k=0}^\infty \tfrac{1}{k!} a^n$ %\end{equation} (cf.\ \cref{def:exp}). \item Determine an action of $\G$ on the manifold $\R^d$ (cf.\ \cref{def:action}). Is this action unique? \end{enumerate} \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{lemma}{lem:1} Let $\M$ be a manifold, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $T \in (0,\infty)$, $y_0 \in \M$, $a \in \fg$, let $\rho \colon [0,1] \to \G$ be a differentiable function satisfying that $\rho(0) = e$ and $\rho'(0) = a$, and let $y \colon [0,T] \to \M$ be a differentiable function satisfying for all $t\in [0,T]$ that $y(0) = y_0$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho(s) , y(t) } \Bigr\rvert_{s=0} \end{equation} (cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,% def:alg_of_group}). Then it holds for all $t\in[0,T]$ that \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \Lambda\pr[\big]{ \exp(ta) , y_0 } \end{equation} (cf.\ \cref{def:exp}). \end{athm} \begin{aproof} \end{aproof} \section{Approximating differential equations on manifolds} We now begin our exploration of how we may approximate solutions to differential equations evolving on (homogeneous) manifolds. To that end, let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $F \colon [0,T] \times \M \to T\M$, and let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t \in [0,T]$ that $y(0) = y_0$ and \begin{equation}\label{eq:3_1} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = F(t,y(t)) \end{equation} (cf.\ \cref{def:manifold,def:tangent}). Our goal is to approximate the solution to \cref{eq:3_1} in a manner which preserves the fact that for all $t \in [0,T]$ it holds that $y(t) \in \M$ (you should convince yourself that this is indeed true). While this may seem trivial, it is easy to see that many \emph{classical} numerical methods fail to preserve this property. For an example of such a failure, please see \cref{prob:num1} below. \begin{athm}{problem}{prob:num1} Let $T \in (0,\infty)$, $a \in \R^3$, let $y_0 = (y_0^1,y_0^2,y_0^3) \in \R^3$ satisfy that $(y_0^1)^2 + (y_0^2)^2 + (y_0^3)^2 = 1$, let $(\cdot,\cdot) \colon \R^3 \times \R^3 \to \R^3$ satisfy for all $u = (u^1,u^2,u^3) \in \R^3$, $v = (v^1,v^2,v^3)\in\R^3$ that $(u,v) = ( u^2v^3-u^3v^2 , u^3v^1 - u^1v^3 , u^1v^2 - u^2v^1 )$, and let $y \colon [0,T] \to \R^3$ be a differentiable function which satisfies for all $t \in [0,T]$ that $y(0) = y_0$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \pr[\big]{ a, y(t) } \dpp \end{equation} \begin{enumerate}[label=\alph*.] \item Show that for all $t \in [0,T]$ it holds that $y(t) \in \operatorname{SO}(3)$ (cf.\ \cref{prob:lie_group2}). \item Let $w \colon [0,T] \to \R^3$ satisfy for all $t \in [0,T]$ that $w(0) = y_0$ and \begin{equation}\label{eq:3_3} w(t) = w_0 + t \pr[\big]{ a, w_0 } \dpp \end{equation} Show that for all $t \in (0,T]$ it holds that $w(t) \not\in \operatorname{SO}(3)$. \emph{Note:} The function in \cref{eq:3_3} is actually known as Euler's method and is the basic tangent line approximation to a vector field. \end{enumerate} \end{athm} \begin{asol} \linebreaks{30} \end{asol} In order to circumvent the issues observed in \cref{prob:num1} above, we assume that there exist a Lie group $\G$ with Lie algebra $\fg$ and $e \in \G$ satisfying for all $p \in \G$ that $p \cdot e = e \cdot p = p$, an action $\Lambda \colon \G \times \M \to \fg$, a function $a \colon [0,T] \times \M \to \fg$, and a differentiable function $\rho_{t,x} \colon [0,1] \to \G$, $t\in [0,T]$, $x \in \fg$, which satisfies for all $s \in [0,1]$, $t\in[0,T]$, $x\in\fg$ that $\rho_{t,x}(0) = e$, $\rho_{t,x}'(0) = a(t,x)$, and \begin{equation}\label{eq:2_2} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t,y(t))}(s) , y(t) } \Bigr\rvert_{s=0} \end{equation} (cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,% def:alg_of_group,eq:3_1}). \begin{athm}{remark}{rem:1} While the above assumptions may seem restrictive, it is important to note that Sophus Lie's third fundamental theorem guarantees that \cref{eq:2_2} will always hold \emph{locally} (cf., e.g., \url{https://en.wikipedia.org/wiki/Lie%27s_third_theorem}). Thus, we may construct such actions locally throughout the interval of interest and then use the semigroup property of the exponential map to ``piece these solutions together.'' However, in practice, one rarely runs into such pathological issues (at least, when trying to utilize structure-preserving methods). \end{athm} \begin{athm}{definition}{def:dexp} Let $\G$ be a Lie group and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:lie_algebra}). Then we define $\dexp \colon \fg \times \fg \to \fg$ to be the function which satisfies for all $u,v \in \fg$ that \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} \exp } \lr (u)(v) = R_{\exp(u)}' \circ \dexp_u(v) \end{equation} (cf.\ \cref{def:exp,def:right}). See, e.g., \url{https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map} for more details. \end{athm} \begin{athm}{lemma}{lem:2} Let $\G$ be a Lie group and let $\fg$ be the Lie algebra of $\G$ (cf.\ \cref{def:lie_group,def:lie_algebra}). Then it holds for all $u,v \in \fg$ that \begin{equation} \dexp_u(v) = \pr[\big]{ \ad^1(u,v) }^{-1} \br[\Big]{ \exp\pr[\big]{ \ad^1(u,v) } - v } = \sum_{k=0}^\infty \frac{1}{(k+1)!} \ad^k(u,v) \end{equation} (cf.\ \cref{def:dexp,def:exp,def:adjoint}). \end{athm} \begin{aproof} \end{aproof} \begin{athm}{lemma}{lem:2a} Let $\G$ be a Lie group, let $\fg$ be the Lie algebra of $\G$, and let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that $B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$ (cf.\ \cref{def:lie_group,def:lie_algebra}). Then it holds for all $u,v \in \fg$ that \begin{equation} \begin{split} \dexp_u^{-1}(v) & = \sum_{k=0}^\infty \frac{B_k}{k!} \ad^k(u,v) \\ & = v - \tfrac{1}{2} \ad^1(u,v) + \tfrac{1}{12} \ad^2(u,v) - \tfrac{1}{720} \ad^4(u,v) + \tfrac{1}{30240} \ad^6(u,v) - \ldots \end{split} \end{equation} (cf.\ \cref{def:dexp,def:adjoint}). \end{athm} \begin{aproof} \end{aproof} \begin{athm}{remark}{rem:b} The sequence of rational numbers $B_0,B_1,B_2,\ldots \in \R$ which satisfy for all $n \in \N_0$ that \begin{equation}\label{eq:3_8} B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \frac{j^n}{(i+1)} \end{equation} are known as \emph{Bernoulli's numbers} (cf., e.g., \url{https://en.wikipedia.org/wiki/Bernoulli_number}). \cref{eq:3_8} is just one of many representations of these numbers. \end{athm} \begin{athm}{lemma}{lem:3} Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \times \M \to \fg$, let $\rho_{t,x} \colon [0,1] \to \G$, $t\in [0,T]$, $x \in \fg$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$, $x \in \fg$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t,x)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t,x)}(s) , y(t) } \Bigr\rvert_{s=0} \dc \end{equation} and let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\Big]{ a\pr[\big]{ t,y(t) } } = \dexp_{\Theta(t)}^{-1} \pr[\bigg]{ a\pr[\Big]{ t, \Lambda\pr[\big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } } } \end{equation} (cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,% def:alg_of_group,def:tangent,def:dexp,def:exp}). Then there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that \begin{equation} y(t) = \Lambda\pr[\Big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } \dpp \end{equation} \end{athm} \begin{aproof} \end{aproof} Note how the assumptions employed in \cref{lem:4} below differ from, e.g., \cref{lem:3} above. Due to the particular structure we have encountered in our actual project, \cref{lem:4} below utilizes a slightly simplified setting in order to reduce the cumbersome of the ensuing notation. \begin{athm}{lemma}{lem:4} Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \to \fg$, let $\rho_{t} \colon [0,1] \to \G$, $t\in [0,T]$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t)}(s) , y(t) } \Bigr\rvert_{s=0} \dc \end{equation} let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that $B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$, let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } \dc \end{equation} and let $\Theta^{[k]} \colon [0,T] \to \fg$, $k \in \N_0$, satisfy for all $k \in \N$, $t \in [0,T]$ that $\Theta^{[0]}(t) = O$ and \begin{equation} \Theta^{[k]}(t) = \int_0^t \dexp_{\Theta^{[k-1]}(\xi)}^{-1} \pr[\big]{ a(\xi) } \, \dd \xi \end{equation} (cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,% def:alg_of_group,def:tangent,def:dexp,def:exp}). Then \begin{enumerate}[label=(\roman*)] \item it holds for all $k \in \N$, $t \in [0,T]$ that \begin{equation} \Theta^{[k]}(t) = \sum_{i=0}^\infty \frac{B_i}{i!} \int_0^t \ad^i\pr[\big]{ \Theta^{[k-1]}(\xi), a(\xi) } \, \dd \xi \end{equation} and \item there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that \begin{equation} \Theta(t) = \lim_{k\to\infty} \Theta^{[k]}(t) \end{equation} \end{enumerate} (cf.\ \cref{def:adjoint}). \end{athm} \begin{aproof} \end{aproof} \begin{athm}{remark}{rem:3} Armed with \cref{lem:4}, we are now in a position to construct structure-preserving numerical approximations to \cref{eq:3_1}. These approximations will proceed via two steps. First, we will choose some $m \in \N_0$ and utilize $\Theta^{[m]}$ in our approximation method: this requires proving a so-called \emph{convergence result} for the Picard iterates. Next, we will construct an appropriate approximation to the exponential map: this will be accomplished via certain \emph{rational function} approximation techniques. These two issues will be discussed further as these notes continue. \end{athm} We now (informally) continue the line of thinking developed in \cref{lem:4} above. First, note that direct calculations yield that for all $t \in [0,T]$ it holds that \begin{equation}\label{expand1} \Theta^{[1]}(t) = \int_0^t a(\xi_1) \, \dd \xi_1 \dc \end{equation} \begin{equation}\label{expand2} \begin{split} \Theta^{[2]}(t) & = \int_0^t a(\xi_1) \, \dd \xi_1 - \tfrac{1}{2} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\ & \qquad + \tfrac{1}{12} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \right] \, \dd \xi_1 + \ldots \dc \end{split} \end{equation} and \begin{equation}\label{expand3} \begin{split} \Theta^{[3]}(t) & = \int_0^t a(\xi_1) \, \dd \xi_1 - \tfrac{1}{2} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\ & \qquad + \tfrac{1}{12} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , a(\xi_1) \right] \right] \, \dd \xi_1 \\ & \qquad + \tfrac{1}{4} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd\xi_2, a(\xi_1) \right] \, \dd\xi_1 \\ & \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \right] \, \dd\xi_2 , a(\xi_1) \right] \, \dd \xi_1 \\ & \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd \xi_2 , \left[ \int_0^{\xi_1} a(\xi_2) \, \dd\xi_2 , a(\xi_1) \right]\right] \, \dd\xi_1 \\ & \qquad - \tfrac{1}{24} \int_0^t \left[ \int_0^{\xi_1} a(\xi_2) \, \dd \xi_2 , \left[ \int_0^{\xi_1} \left[ \int_0^{\xi_2} a(\xi_3) \, \dd\xi_3 , a(\xi_2) \right] \, \dd \xi_2 , a(\xi_1) \right] \right ] \, \dd \xi_1 + \ldots \dpp \end{split} \end{equation} Observe that we could have continued these calculations, but things become increasingly tedious (however, things can be simplified considerably via the use of graph theory). %%%Next, note that \cref{expand1,expand2,expand3} seem to indicate that one could construct $H_k \colon [0,T] \to \fg$, $k\in\N_0$, which satisfy for all $k \in \N_0$, $t\in[0,T]$ that $H_k$ is a linear combination of terms including exactly $k+1$ integrals (or, equivalently, $k$ Lie brackets) and %%%\begin{equation} %%%\Theta(t) = \sum_{k=0}^\infty H_k(t) \dpp %%%\end{equation} Next, note that even when we keep all (infinitely many) terms in \cref{expand2} or \cref{expand3}, we are only able to approximate the function $\Theta$ with limited accuracy (this is a direct consequence of the Picard-Lindel{\"o}f theorem (cf., e.g., \url{https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem}). That is, let $\norm{\cdot} \colon \fg \to [0,\infty)$ be an appropriate norm. Then, there exists $C_k \in \R$, $k\in\N_0$, and $f \colon \N_0 \to [0,\infty)$ such that for all $k\in\N_0$, $t \in [0,T_*]$ it holds that \begin{equation} \norm[\big]{ \Theta(t) - \Theta^{[k]}(t) } \le C_k t^{f(k)} \dpp \end{equation} This indicates that keeping \emph{all} terms in the low-level expansions is not ideal. This motivates the result in \cref{lem:5} below. \begin{athm}{lemma}{lem:5} Let $\M$ be a manifold, let $T \in (0,\infty)$, $y_0 \in \M$, let $\G$ be a Lie group, let $e \in \G$ satisfy for all $p \in \G$ that $p \cdot e = e \cdot p = p$, let $\fg$ be the Lie algebra of $\G$, let $\Lambda \colon \G \times \M \to \fg$ be an action, let $a \colon [0,T] \to \fg$, let $\rho_{t} \colon [0,1] \to \G$, $t\in [0,T]$, be a differentiable function which satisfies for all $s \in [0,1]$, $t\in[0,T]$ that $\rho_{t}(0) = e$ and $\rho_{t}'(0) = a(t)$, let $y \colon [0,T] \to \M$ be a differentiable function which satisfies for all $t\in[0,T]$ that $y(0) = y_0$ and \begin{equation}\label{lem:5_ode} \pr[\big]{ \tfrac{\dd}{\dd t} y } \lr (t) = \tfrac{\dd}{\dd s} \Lambda\pr[\big]{ \rho_{t,a(t)}(s) , y(t) } \Bigr\rvert_{s=0} \dc \end{equation} let $B_0,B_1,B_2,\ldots \in \R$ satisfy for all $n \in \N_0$ that $B_n = \sum_{i=0}^n \sum_{j=0}^i (-1)^j \binom{i}{j} \nicefrac{j^n}{(i+1)}$, let $\Theta \colon [0,T] \to \fg$ satisfy for all $t \in [0,T]$ that $\Theta(0) = O$ and \begin{equation} \pr[\big]{ \tfrac{\dd}{\dd t} \Theta } \lr (t) = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } = \dexp_{\Theta(t)}^{-1} \pr[\big]{ a\pr[]{ t } } \dc \end{equation} and let $\Omega^{[k]} \colon [0,T] \to \fg$, $k \in \N_0$, satisfy for all $k \in \N$, $t \in [0,T]$ that $\Omega^{[0]}(t) = O$ and \begin{equation}\label{eq:3_23} \Omega^{[k]}(t) = \sum_{i=0}^{k-1} \frac{B_i}{i!} \int_0^t \ad^i\pr[\big]{ \Omega^{[k-1]}(\xi), a(\xi) } \, \dd \xi \end{equation} (cf.\ \cref{def:manifold,def:lie_algebra,def:lie_group,def:action,% def:alg_of_group,def:tangent,def:dexp,def:exp,def:adjoint}). Then \begin{enumerate}[label=(\roman*)] \item\label{i1} there exists $T_* \in (0,T]$ such that for all $t \in [0,T_*]$ it holds that \begin{equation} \Theta(t) = \lim_{k\to\infty} \Omega^{[k]}(t) \end{equation} and \item\label{i2} it holds for all $k \in \N_0$, $t \in [0,T_*]$, and $\norm{\cdot} \colon \fg \to [0,\infty)$ which satisfy for all $u,v \in \fg$, $c \in \R$ that $\norm{u+v} \le \norm{u} + \norm{v}$, $\norm{cu} = \abs{c}\norm{u}$, and $\norm{u} = 0$ if and only if $u=O$ that there exists $C \in \R$ such that \begin{equation}\label{eq:3_25} \norm[\big]{ \Theta(t) - \Omega^{[k]}(t) } \le C t^{k+1} \dpp \end{equation} \end{enumerate} \end{athm} \begin{aproof} \end{aproof} It is now important to emphasize what is implied by \cref{lem:5} above. \begin{enumerate}[label=\arabic*.] \item \Cref{i1} in \cref{lem:5} above demonstrates that we can truncate the Picard iterates and still obtain the desired convergence to the true solution. \emph{Note:} We have purposely avoided the issue of discussing\slash defining what a limit means in a Lie algebra, but hopefully this is clear. \item \Cref{i2} in \cref{lem:5} above shows that we may choose our Picard iterate based on the amount of accuracy we desire. This can be seen in the right-hand side of \cref{eq:3_25}. \item It should be clear that \cref{i2} in \cref{lem:5} holds true if we can prove the result for one such $\norm{\cdot} \colon \fg \to [0,\infty)$ as we are in a finite-dimensional setting. \item Something which is hidden in the details (and often misunderstood) is the constant ``$C$'' in \cref{i2} in \cref{lem:5}. This constant is a function of the underlying smoothness of the function $a$, as it will be the result of trying to bound the nested commutators within the Picard iterates. For our project we will not need to concern ourselves with this constant too much; we only need to understand its role and how it can be controlled. \end{enumerate} \begin{athm}{problem}{prob:1} Verify \cref{i2} in \cref{lem:5} for the case $k=1$. Feel free to impose as much regularity (i.e., increase the assumptions on the objects introduced in \cref{lem:5}) if you feel this will help. \end{athm} \begin{asol} \linebreaks{30} \end{asol} \begin{athm}{problem}{prob:2} Verify \cref{i2} in \cref{lem:5} in the case where $k=2$, $T = 1$, $\M = \G = \operatorname{SO}(3)$ and $\Lambda \colon \G \times \G \to \fg$ satisfies for all $a,b \in \G$ that $\Lambda(a,b) = ab$. If you want to simplify things further, you may let $\alpha \in \fg$ and choose $a \colon [0,T] \to \fg$ to satisfy for all $t \in [0,T]$ that $a(t) = t\alpha$. \end{athm} \begin{asol} \linebreaks{30} \end{asol} \subsection{A first attempt at a numerical approximation} We will now (somewhat informally) motivate the construction of a numerical approximation to, e.g., \cref{lem:5_ode}. To that end, we will commence by fixing $k=1$. This and \cref{eq:3_23} yield that for all $t \in [0,T_*]$ it holds that \begin{equation}\label{eq:3_26} \Omega^{[1]}(t) = B_0 \int_0^t \ad^0\pr[\big]{ \Omega^{[0]}(\xi) , a(\xi) } \, \dd \xi = \int_0^t a(\xi) \, \dd \xi \dpp \end{equation} We now apply an approximation to the integral in \cref{eq:3_26} (since, in general, we will not be able to compute the integral exactly) to obtain that for all $t\in[0,T_*]$ it holds that \begin{equation}\label{eq:3_27} \Omega^{[1]}(t) \approx t a(0) \dpp \end{equation} \begin{athm}{problem}{prob:3} Let $T,L \in (0,\infty)$, let $\fg$ be a Lie algebra, let $\norm{\cdot} \colon \fg \to [0,\infty)$ satisfy for all $u,v \in \fg$, $c \in \R$ that $\norm{u+v} \le \norm{u} + \norm{v}$, $\norm{cu} = \abs{c}\norm{u}$, and $\norm{u} = 0$ if and only if $u=O$, and let $a \colon [0,T] \to \fg$ satisfy for all $s,t \in [0,T]$ that $\norm{a(t) -a(s)} \le L\abs{t-s}$ (cf.\ \cref{def:lie_algebra}). Show that for all $t\in[0,T]$ it holds that \begin{equation} \norm*{ \int_0^t a(\xi) \, \dd \xi - t a(0) } \le \frac{Lt^2}{2} \dpp \end{equation} \end{athm} \begin{asol} \linebreaks{30} \end{asol} Combining \cref{lem:3,lem:5,eq:3_27} demonstrate that for all $t\in[0,T_*]$ it holds that \begin{equation} y(t) = \Lambda\pr[\Big]{ \exp\pr[\big]{ \Theta(t) } , y_0 } \approx \Lambda\pr[\Big]{ \exp\pr[\big]{ \Omega^{[1]}(t) } , y_0 } = \Lambda\pr[\Big]{ \exp\pr[\big]{ t a(0) } , y_0 } \dpp \end{equation} Provided that the group action, $\Lambda$, and $\exp(ta(0))$ can be computed with relative ease, we have arrived at a reasonable numerical approximation. Moreover, the approximations employed thus far have been linear approximations performed on objects in the Lie algebra---thus, our approximation will still lie in the original manifold for all $t\in[0,T_*]$. \begin{athm}{problem}{prob:4} Develop a similar approximation using $\Omega^{[2]}$ (cf.\ \cref{lem:5}). Note that in this case, you will likely need to impose additional regularity assumptions (be sure to only impose precisely enough to derive your result). \end{athm} \begin{asol} \linebreaks{30} \end{asol} \subsection{Approximating the exponential map} \end{document}