\input o.tex
\longmodetrue
%\proofmodetrue
\wideaccents
\refsin
\keyin
\keyout
%\noindentsecpar
%\vglue .2in
\Title
Moving Frames and
Singularities \break of Prolonged Group Actions.
\author
Peter J. Olver\\
School of Mathematics\\
University of Minnesota\\
Minneapolis, MN\quad 55455\\
U.S.A.\\
\email olver\@math.umn.edu\\
\www math.umn.edu/$\sim$olver\\
\support NSF Grant DMS 98--03154.
\printauthor
\dated
\vglue -.5in
\Abstract
The prolongation of a transformation group to jet bundles forms the geometric foundation
underlying Lie's theory of symmetry groups of differential equations, the theory of
differential invariants, and the Cartan theory of moving frames. Recent developments in
the moving frame theory have necessitated a detailed understanding of the geometry of
prolonged transformation groups. This paper begins with a basic review of moving
frames, and then focuses on the study of both regular and singular prolonged group
orbits. Highlights include a corrected version of the basic stabilization theorem, a
discussion of ``totally singular points,'' and geometric and algebraic characterizations
of totally singular submanifolds, which are those that admit no moving frame.
In addition to applications to the method of moving frames, the paper includes a
generalized Wronskian lemma for vector-valued functions, and methods for the solution to
Lie determinant equations.
\vglue .2in
AMS subject Classification Numbers: 57SXX, 58A05, 58A20, 53A55.
\bigskip
Key words: moving frame, Lie group, jet, Lie matrix, homogeneous space
\page
%\vglue .2in
\def\iso#1{G_{#1}} \def\GS{\iso S}
\def\rank{\ro{rank}\>}
\def\isa#1{\g_{#1}}
\def\giso#1{G\Upstar _{#1}}
\def\gisa#1{\g\Upstar _{#1}}
%\def\giso#1{G^{\>\displaystyle \bullet}_{#1}}
%\def\gisa#1{\g^{\>\displaystyle \bullet}_{#1}}
\def\k{\frak k}
\def\prk{\pro k}
\def\prm{\pro m}
\def\qn{q\ps n}
\def\xuno{(x,u\ps{n+1})}
\def\xui{(x,u\ps\infty )}
\def\Gn{G\ps n}
\def\gh{\widehat \g}
\def\gn{\g\ps n}
\def\z#1{z\ps {#1}}
\def\zn{\z n}
\def\zi{\z \infty}
\def\zno{\z{n+1}}
\def\atzn{\at{\zn}}
\def\j#1{{\rm j}_{#1}}
\def\jn{\j n}
\def\jk{\j k}
\def\jko{\j{k+1}}
\def\ji{\j \infty}
\def\Ji{\J \infty }
\def\V#1{\CV^{#1}}
\def\Vn{\V n}
\def\S#1{\CS^{#1}}
\def\Sn{\S n}
\def\L#1{\bo L_{#1}}
\def\Ln{\L n}
\def\M#1{\bo M_{#1}}
\def\Mn{\M n}
\def\W#1{\bo W_{#1}}
\def\Wn{\W n}
\def\hh#1{\bo h_{#1}}
\def\hht#1{\widetilde{\bo h}_{#1}}
\def\v{\bo v}
\def\w{\bo w}
\def\vh{\widehat {\bo v}}
\def\wh{\widehat {\bo w}}
\def\vn{\prn \v}
\def\vhn{\prn \vh}
\def\vk{\prk \v}
\def\vm{\prm \v}
\def\wn{\prn \w}
\def\vi{\ro{pr}\;\v}
\def\vhi{\ro{pr}\;\vh}
\def\wi{\ro{pr}\;\w}
\def\mf{\rho }
\def\KC{K}
\def\tot#1{#1_{tot}}
\def\ev#1{#1_{ev}}
\def\vtot{\tot\v}
\def\vev{\ev\v}
\def\pitot{\tot\pi}
\def\piev{\ev\pi}
\def\vsc{\bo s}
\def\No#1#2{\CN_{#1}(#2)}
\def\NH{\No H} \def\NHK{\NH K} \def\NHKH{\NH{K\cap H}}
\def\KH{(K\cdot H)\,\cap\, (H\cdot K)}
\Section i Introduction.
Given a finite-dimensional Lie transformation group acting on a manifold, there
is an induced action on the associated jet bundles known as
the prolonged group action. This action forms the foundation of the
Lie theory of symmetry groups of differential equations \rf{O,Ov}.
Differential invariants appear as invariants of the prolonged group action \rf{Liedi,Gug,E}.
Furthermore, the range of applicability of the Cartan method of moving frames \rf{Cartanrm,Cartang,Griffithsmf,Jensen}, as reformulated and extended in \rf{FOmcI,FOmcII}, is
intimately tied to the geometry of the prolonged group action on
jet spaces. A wide variety of additional applications, including
symmetries of variational problems, equivariant bifurcation problems \rf{GSS},
conservation laws and invariant differential forms, group-invariant solutions,
etc., all rely on this basic construction; see \rf{O,E}
for details. The present paper is devoted to a detailed
investigation into the geometry of the prolonged group action, with
particular emphasis on the orbit structure, and how it relates to
the group-theoretic geometry of submanifolds.
Since the dimensions of the prolonged group orbits increases with order, the
maximal prolonged orbit dimension eventually stabilizes. The singular orbits are those of
less than this maximal dimension. In the case of planar curves,
the singular variety determined by all the singular orbits can be characterized by the
vanishing of the classical Lie determinant \rf{Lieode,E} and its generalizations. A point
in the original manifold is called ``totally singular'' if every jet fiber sitting over
it consists entirely of singular orbits. The existence of totally singular points has,
apparently, not been noted in the literature before. In fact, we demonstrate that an
analytic transformation group cannot have totally singular points, although there are
elementary examples of smooth actions which allow such singularities. The basic
stabilization theorem \rf{Ov,E} states that a group that acts ``locally effectively on
subsets,'' as defined below, will act locally freely on an open subset of jet spaces of
sufficiently high order. Previous versions of the theorem omitted the phrase ``on
subsets,'' and hence do not apply at totally singular points. As an application, we
establish a generalization of the classical Wronskian lemma that characterizes linearly
dependent functions.
These results have immediate applications to Cartan's moving frame approach to the
geometry, equivalence, and symmetry of submanifolds. A submanifold is called regular of
order $n$ if its $n$-jet lies in the regular subset of jet space where the prolonged
group orbits are of maximal dimension. The newly developed method of ``moving coframes''
\rf{FOmcI,FOmcII} shows that an \nth order moving frame can be constructed if and only if the
submanifold is regular at order $n$. Most submanifolds are regular at some sufficiently high
order, and hence admit a moving frame. The exceptions are the ``totally singular'' submanifolds,
whose
$n$-jet (at a point) lies in the singular subset for all $n$.
Totally singular submanifolds are the solutions
to the classical Lie determinant equation, which is the simplest
invariant differential equation associated with the given transformation group. A simple
example is provided by the action of the special affine group that governs
the equi-affine geometry of curves in the plane \rf{Gug}. In this case, any straight line is a
totally singular submanifold at all its points, while smooth curves which have infinite order
contact with a straight line are totally singular at the point of contact. Any other curve will
admit (at least locally) an equi-affine moving frame at some order \rf{FOmcII}, even though the
classical equi-affine Frenet frame has order $3$ and only applies to curves without inflection
points \rf{Gug, FOmcI}.
The final part
of the paper is devoted to the explicit characterization, both geometrical and
algebraic, of totally singular submanifolds. In the analytic category, we prove that
a submanifold is totally singular if and only if its isotropy subgroup does not act
locally freely on it. A similar result holds for smooth submanifolds provided every point
thereon is totally singular. Finally, we analyze the most important case, which is a transitive
group action on a homogeneous space $M=G/H$. Here, the totally singular submanifolds are
characterized algebraically, being related to the existence of ``self
$H$-normalizing subgroups.'' This leads to a direct algebraic construction of the
solutions to the Lie determinant equation.
Our results are all illustrated by a number of explicit examples of independent interest.
\Section {mf} Introduction to Moving Frames.
Throughout this paper, $G$ will denote an $r$-dimensional
Lie group\fnote{The moving frame methods can be extended to infinite-dimensional
pseudo-group actions \rf{FOmcI} but for simplicity, we will only consider the
finite-dimensional case here.} acting smoothly on an
$m$-dimensional manifold $M$. In the classical Cartan theory, $G$ is a group of
geometrical significance,
\eg the Euclidean group, equi-affine group, projective group, etc. However, the methods
developed in \rf{FOmcII} apply to completely general Lie group actions.
\Df{iso} The
\is{isotropy subgroup} of a subset $S\subset M$ is defined as
$$\GS = \set{g\in G}{g\cdot S = S}.$$
The \is{global isotropy subgroup} of $S$ is
$$
\giso S = \Intersection_{z\in S} G_z = \set{g\in G}{g\cdot s = s \roh{ for all } s \in S},
$$
consisting of group elements that fix \is{all} points in $S$.
The basic definition of a moving frame is implicit in the works of Cartan \rf{Cartanrm}
and first appears explicitly in Griffiths \rf{Griffithsmf}.
\Df{mfdef} A \is{moving frame} is defined as a smooth $G$-equivariant map
$\mf \,\colon M \to G$.
In general, given actions of $G$ on $M$ and $N$, a map $\varphi \colon M\to N$ is
called \is{equivariant} if $\varphi (g\cdot z) = g\cdot \varphi (z)$ for all $g\in G$,
$z\in M$. In the case of a moving frame, there are two natural actions of $G$ on itself
--- by left multiplication $h\mapsto g\cdot h$ or by right multiplication $h\mapsto
h\cdot g^{-1} $, leading to the concepts of left and right moving frames. All
classical constructions lead to left moving frames, and so from now on we shall only deal
with them. There are, however, occasions when a right moving frame is easier to
compute. Fortunately, there is a simple connection between the two: if $\rho (z)$ is any
left moving frame, then $\widetilde\rho (z) = \rho (z)^{-1} $ is a right moving frame, and
conversely.
In practice, the construction of a globally-defined moving frame on all of $M$ is
difficult, and one is content to determine a locally-defined version. The necessary and
sufficient conditions for the local existence of a moving frame are readily established.
\Th{mf} A moving frame exists
in a neighborhood of a point $z\in M$ if and only if $G$ acts freely
and regularly near $z$.
Recall that a group is said to act \is{freely} if the isotropy subgroup of any
individual point $z_0\in M$ is trivial: $G_{z_0} = \sete$.
The group acts \is{regularly} if all its
orbits have the same dimension and, moreover, each point $z_0$ has a system of
neighborhoods such that each orbit intersects in a pathwise connected subset, \crf O. If
$G$ acts freely, then its orbits automatically all have the same dimension, namely the
dimension $r$ of $G$ itself. The regularity condition is only to avoid such pathologies
as the irrational flow on the torus, where the orbits return arbitrarily close to
themselves. In general, the group orbits all have dimension equal to $r$ if and only if
the action is \is{locally free}, which means that $G_{z_0}$ is a discrete
subgroup of $G$ for each $z_0\in M$. Locally free group actions admit locally
equivariant moving frames.
Of course, most interesting group actions, including all the geometrical examples mentioned
above, are \iz{not} (locally) free. For example, the Euclidean group $\EU2$ does not act
freely on
$\Rx2$ since any rotation fixes its center. The
problem is, of course, that the space does not have a large enough dimension to admit
$r$-dimensional orbits, which is required for freeness. Thus, a group
can only act freely on a manifold whose dimension is at least that of $G$.
Although such actions are not free, they are usually effective, which means that the
only group element which fixes \is{every} point of $M$ is the identity, and so the global
isotropy subgroup is trivial:
$\giso M =\sete$. Non-effective actions can always be made effective by the following
simple device: The global isotropy subgroup
$\giso M$ forms a normal subgroup of $G$. Moreover, the quotient group $G/\giso M$
acts effectively on $M$ in essentially the same manner as $G$ does, \crf E.
Specifically, if $g\in G$ has image $\gtilde \in G/\giso M$ then one
unambiguously defines
$\gtilde\cdot z = g\cdot z$ for $z\in M$.
For example, the projective action $x\mapsto (\alpha x+\beta )/(\gamma x+\delta )$ of
$\GLR2$ on $\RPx1$ is not effective since the multiples of the identity $\lambda \Id$ act
trivially. The effectively acting quotient group $\ro{PSL}(2,\R) = \GLR2/\{\lambda
\Id\}$ is known as the \is{projective linear group}.
There are three important methods for converting a non-free, but effective action into a
free action. The first is to consider the Cartesian product action of $G$ on several
copies of the manifold,
$M^{\times n} = M\times\cdots \times M$, whereby
$$g\cdot \psubs zn = \psubs{g\cdot z}n. \Eq{joint}$$
For $n$ sufficiently large, the resulting action will typically become
free on a suitable open subset of $M^{\times n}$. For example, the action of the planar
Euclidean group $\SE2$ on $M = \Rx2$ becomes free on the off-diagonal part $\{z_1 \ne
z_2\}\subset M\times M$ of the second Cartesian product,
because the only planar Euclidean motion that fixes
two distinct points is the identity. Invariants of such a Cartesian product action are
known as joint invariants; \rf{FOmcI,Oji} discuss applications of moving frames to the
classification of joint invariants, which has important consequences in computer vision \rf{COSTH}.
A second method is to prolong the group action to the jet spaces $\Jn = \Jn(M,p)$ defined
by $p$-dimensional submanifolds $N\subset M$ under the equivalence relation of \nth order
contact; see the following section and \rf{E} for details. Jet space is the natural
setting for Lie's theory of symmetry groups of differential equations, \crf{O,E}, and for
Cartan's theory of moving frames in geometry \rf{Cartanrm,Jensen, FOmcII}. Since group transformations preserve contact, there is a
naturally induced prolonged group action on
$\Jn$. The basic stabilization theorem of Ovsiannikov states that, for $n$ sufficiently
large, the prolonged action is locally free on an open subset of $\Jn$, and hence a
locally equivariant moving frame can be constructed thereon. However, traditional
statements of this result \rf{Ov,E} are not quite correct, and we shall return to this
question in \section{seff}.
Finally, a new, hybrid construction that combines the joint invariants and differential
invariants in a common framework, called ``multi-space,'' has been recently proposed \rf{Onums}.
This framework is based on the classical finite difference calculus from numerical analysis
\rf{Milne}, and treats the jets of submanifolds as limits of discrete collections of points, thereby
``blowing-up'' the diagonal of the Cartesian product space to produce a jet space of the appropriate
order. Indeed, the simplest example of the multi-space construction is the algebro-geometric
blow-up, replacing a pair of coincident points by a tangent direction \rf{GH}. Application of the
moving frame construction in multi-space produces invariant
numerical approximations to differential invariants \rf{COSTH} of importance in the
theory of geometric integration of differential equations \rf{BuddIserles,Dorodfd}.
Also the theory of joint differential invariants, also known as semi-differential
invariants in computer vision \rf{MPVO}, can be included in this general construction. The details are still under
development.
In this paper, we shall only examine the second scenario, where the group acts on the jet
bundle by prolongation. The construction of a moving frame relies on the fact that, for
$n$ sufficiently large, the group acts freely and regularly on the jet space $\Jn$. In
general this is not entirely correct, and so a detailed analysis of the geometry of the
prolonged transformation group is required.
The explicit construction of moving frames relies on the method of ``normalization,''
which lies at the heart of the Cartan approach to \ic{moving frame}s and \ic{equivalence
problem}s \rf{Cartanrm,Cartanequiv,Gug,E}; an early version can be found in Killing \rf{Killingi}.
The key observation \rf{FOmcI,FOmcII} is that normalization amounts to the choice of a
cross-section to the group orbits.
\Df{crosss} Let $G$ act regularly on the
$m$-dim\-en\-s\-ional manifold $M$ with $s$-dimensional \ic{orbit}s.
A (local) \is{cross-section} is an $(m-s)\,$-dimensional submanifold
$\KC\subset M$ such that $\KC$ intersects each orbit transversally,
at most once.
The most important are the \is{coordinate cross-sections}
$$\KC = \{z_1 = c_1,\ldots ,z_s = c_s\},\Eq{ccsss}$$
obtained by equating $s$ of the coordinates (which, by relabeling if necessary, can be
taken to be the first $s$ coordinates) to constants. Of course, any cross-section can be
(locally) converted into a coordinate cross-section by a suitable choice of local
coordinates.
The elements $k\in \KC$ of the cross-section can be viewed
as ``canonical forms'' for general points $z\in M$, and their
coordinates provide the invariant \ic{moduli} for the group action. Since
$\KC$ has dimension $m-s$, precisely
$m-s$ of these invariant functions will be functionally independent
and therefore provide us with a generating system of invariants. In the case of a
coordinate cross-section
\eq{ccsss}, the first $s$ coordinates of $k=(\subs cs,z_{s+1},\ldots ,z_m)$ will be
constant, and are hence trivial invariants, while the latter $m-s$ coordinates
will provide the fundamental system of invariants, as we now describe.
\Fig{css}{Moving Frame Associated with Cross-Section}
\Th{nmf} Let $G$ act freely, regularly on $M$, and let $\KC$ be a cross-section to the
group orbits. Given $z\in M$, let $k \in \CO_z\cap \KC$ be the unique cross-section
element lying in the orbit through $z$, and let $g =
\mf(z)$ be the unique group element such that $g\cdot k = z$. Then $\mf\colon M\to G$ is
a left moving frame for the group action.
Indeed, every moving frame has this form --- the cross-section is just the inverse image
of the identity: $\KC = \mf^{-1} \{e\}$. Therefore, to determine the moving frame
associated with a chosen cross-section, one merely solves the \is{normalization equations}
$$g^{-1} \cdot z = k \in \KC \roq{for} g = \mf(z). \Eq{neq}$$
For example, if we choose the coordinate cross-section \eq{ccsss}, and write out the
group transformations in coordinates as
$w(g,z) = g^{-1} \cdot z$, then the moving frame is obtained by solving the
implicit equations
$$w_1(g,x) = c_1,\qquad \ldots \qquad w_r(g,x) = c_r, \Eq{normal}$$
for the group parameters $g$ in terms of the coordinates $x$.
Moreover, the fundamental invariants are found by substituting the
resulting moving frame formula into the remaining, unnormalized components of $w(g,z)$.
\Th{normalization} Given a free, regular Lie group action and coordinate cross-section as
described above, let $g = \mf (x)$ denote the solution to the
normalization equations \eq{normal}.
Then the functions
$$I_1(x) = w_{r+1}(\mf (x),x),\qquad \ldots \qquad
I_{m-r}(x) = w_{m}(\mf (x),x), \Eq{invs}$$
obtained by substituting the moving frame formulae into
the remaining $m-r$ components of $w$ form a complete system
of \ic{functionally independent invariants} for the action of $G$.
%\Remark If we replace $w(g,z) = g^{-1}\cdot z$ by $\wtilde(g,z) = g\cdot z$ in the
%normalization equations \eq{normal}, the result is the right moving frame. The
%fundamental invariants are constructed as in the left version.
\Section j Jet Bundles and Prolongation.
Before presenting examples of moving frames, we need to review the basics of
jet bundles and prolonged group actions. This will help prepare us for the more
detailed investigations into their geometry and singularities.
Given a manifold $M$, we let $\Jn = \Jn(M,p)$ denote the
\nth order (extended) jet bundle
consisting of equivalence classes of regular (embedded) $p$-dimensional
submanifolds $S\subset M$ under the equivalence
relation of \nth order contact, \cf \rf{O; Chapter 3}. In particular, $\J0 = M$.
The \nth jet space $\Jn$ is a fiber bundle $\pi^n_0\colon \Jn\to M$
of total dimension
$$\qn = p + q {p+n\choose n} = \dim \>\Jn(M,p) , \Eq{qn}$$
whose fibers are generalized Grassmann manifolds, \crf{Os}.
We let $\jn S\subset \Jn$ denote the $n$-jet of the submanifold $S$, which forms a
$p$-dimensional submanifold of $\Jn$. Sometimes, it is convenient to work with the
infinite jet bundle
$\Ji =\Ji(M,p)$, which is defined as the inverse limit of the finite order jet bundles
under the standard projections $\pi ^k_n\colon \Jk\to \Jn$, $k>n$, with
$\ji S$ denoting the corresponding infinite jet of $S$. An analytic submanifold is
uniquely determined by its infinite jet at a point in the same manner that an analytic
function is uniquely determined by its Taylor series at a point.
We introduce
local coordinates $z = (x,u)$ on $M$, considering the
first $p$ components $x = \psups xp$ as independent variables,
and the latter $q = m-p$ components $u = \psups uq$
as dependent variables. Splitting
the coordinates into independent and dependent variables
has the effect of locally identifying $M$ with
an open subset of a bundle $E = X\times U \simeq \R^p\times \R^q$.
Sections $u = f(x)$ of $E$ correspond to $p$-dimensional submanifolds
$S$ that are transverse to the vertical fibers $U_c = \{c\}\times U\subset E$.
The induced local coordinates on the jet bundle $\Jn$
are denoted by $\zn = \xun$, with components $u^\alpha _J$
representing the partial derivatives of
the dependent variables with respect to the independent variables up
to order $n$. Here $J = \psubs jk$, where $1\leq j_\nu \leq p$,
will denote a symmetric multi-index of order $k = \# J$.
The $\xun$ define local coordinates on
the open, dense subbundle $\Jn E\subset \Jn(M,p)$ determined by the
jets of transverse submanifolds, or, equivalently, local sections
$S = \{u = f(x)\}$ of $E$. Let $\jn f(x)$ denote the corresponding
$n$-jet of $f$ at $x$, which can be identified with the \nth
order Taylor polynomial of $f$ at $x$.
%The $n$-jet $\jn f$
%forms a section of $\Jn E$, whose graph coincides with
%the $n$-jet of the corresponding submanifold $S$, \ie
%$\jn S = \set{\xun}{u\ps n = \jn f(x)}$.
In the limit, we let $\zi = \xui $
denote the corresponding coordinates on $\Ji E \subset \Ji(M,p)$, consisting
of independent variables $x^i$, dependent variables $u^\alpha $,
and their derivatives $u^\alpha _J$, $\rg \alpha q$, of arbitrary order $\#J \geq 0$.
Any transformation group $G$ acting on $M$
preserves the order of contact between submanifolds.
Therefore, there is an induced action of $G$ on the \nth order jet
bundle $\Jn = \Jn(M,p)$ known as the \nth \is{prolongation} of $G$,
and denoted by $\Gn$.
Note that if $G$ acts globally
on $M$, then its prolonged action $\Gn$ is also a
global transformation group on $\Jn$, but, generally
only a local transformation group on the coordinate subbundles $\Jn E$
since $G$ may not preserve transversality.
\Remark We shall exclusively deal with point transformation groups,
although all our results and methods naturally extend to
contact transformation groups, \crf E.
The explicit local coordinate formulae for the prolonged group actions are found by
implicit differentiation. Let us see how this works in a simple example.
\Ex{SE2j} The
planar Euclidean group $\SE2 = \SO2 \semidirect \Rx2$ acts transitively on $M = \Rx 2$,
mapping a point
$(x,u)$ to
$$\qeq{y = x \cos \theta - u \sin \theta + a ,\\
v = x \sin \theta +u \cos \theta + b ,} \Eq{SE2}
$$
where $\theta ,a,b$ serve to parametrize the group. The second
order prolongation maps a point $(x,u,u_x,u_{xx}) \mapsto (y,v,v_y,v_{yy})$, where
$y,v$ are given by \eq{SE2} and
$$
\qeq{v_y = \od vy = \frac{\sin \theta + u_x \cos \theta }{ \cos \theta - u_x \sin
\theta },\\ v_{yy} = \odw vy = \frac{u_{xx}}{(\cos\theta - u_x \sin \theta )^3 } .}
\Eq{SE2j2}$$
\smallskip
The basic moving frame construction for a free prolonged group action proceeds as in
\th{normalization}. Solving the normalization equations for
$r = \dim G$ components of the prolonged transformation formulae
$$w\ps n(g\ps n,\zn) = (g\ps n)^{-1} \cdot \zn$$
results in a moving frame on the \nth order jet space. In classical situations, the
resulting equivariant map
$\mf\,\colon \Jn\to G$ can be identified with the standard geometric moving frame. We
will illustrate the basic procedure with two well-studied examples.
\Ex{SE2mf} The prolonged
planar \ic{Euclidean group} $\SE2$ action is (globally) free on
$\J1$, and hence also on any $\Jn$ for $n\geq 1$.
According to \th{normalization}, we can construct a left moving frame by
normalizing the inverse group transformations\fnote{Normalization of the group
transformations \eqs{SE2}{SE2j2} leads to a right moving frame. The invariants are
constructed as in the left version.}
$$\eeq{y = \cos \theta (x-a) + \sin \theta (u - b) ,&
v = -\sin \theta (x-a) + \cos \theta (u - b),\\
v_y = \frac{u_x \cos \theta - \sin \theta }{ \cos \theta + u_x \sin \theta },&
v_{yy} = \frac{u_{xx}}{(\cos\theta + u_x \sin \theta )^3 } .} \Eq{SE22}$$
Consider the simple coordinate
cross-section
$\KC = \{x=u = u_x =0\}$. The solution to the associated normalization equations
$$\qeq{y = 0,\\ v = 0,\\ v_y = 0,}$$
produces the classical moving frame $\mf \,\colon \J1 \to \SE2$, with
$$\qeq{\theta = \tan^{-1} u_x,\\ a = x,\\ b = u.}\Eq{SErmf}$$
Indeed, in its standard matrix form
$$\qeq{\rmatrix \theta = \fra{\sqrt{1+u_x^2}}\matrixb 1{-u_x}{u_x}1,\\ \vectorb ab =
\vectorb xu,}$$
the columns of the rotation component represent the unit
tangent and unit normal to the curve, while the translation component provides the point
on the curve. Substituting the normalizations
\eq{SErmf} into the final formula in \eq{SE22} for the second order derivative
$v_{yy}$, we recover the fundamental Euclidean curvature invariant
$$\kappa = {u_{xx}\over (1 + u_x ^2)^{3/2}}\>.\Eq{kappa}$$
If we prolong further, say to $\J4$, and substitute the moving frame formulae \eq{SErmf}
into the corresponding formulae for the third and fourth derivatives, we obtain the
higher order differential invariants
$$\qeq{v_{yyy} \longmapsto \od \kappa s,\\
v_{yyyy} \longmapsto \odw \kappa s - 3\kappa ^3,}\Eq{kappass}$$
where $\odo s$ denotes differentiation with respect to the Euclidean arc length element
$$ds = \sqrt {1 + u_x ^2} \> dx. \Eq{ds}$$
The Euclidean arc length \eq{ds} is characterized as the unique, up to constant
multiple, Euclidean-invariant one-form of lowest order. More correctly, it is a
``contact-invariant'' horizontal one-form on the jet bundle $\J1$, \crf{FOmcII}. It is
obtained by the general procedure of substituting the moving frame normalizations
\eq{SErmf} into the horizontal (or total) differential $dy = (\cos \theta + u_x \sin
\theta) \dx$ of the chosen independent variable.
\Remark A general result \rf{FOmcII} states that all the higher order differential
invariants are obtained by differentiating the fundamental differential invariants --- in
this case the curvature $\kappa $ --- with respect to to the invariant one-forms --- in
this case arc length $ds$. The fact that the normalized higher order invariants
\eq{kappass} do not in general agree with the differentiated invariants is of
critical important in understanding the syzygies (functional relationships) among the
differential invariants. A general algorithm for determining the ``correction terms''
--- for example, the $-3\kappa ^3$ in the second formula in
\eq{kappass} --- was found in \rf{FOmcII,FOmfmc}, and applied to provide a complete
solution to the syzygy classification problem for differential invariants.
The Euclidean group is atypical in that its first prolongation acts freely everywhere on
the jet spaces $\Jn$, $n\geq 1$. More typically, the prolonged actions are free (if at
all) only on a dense open subset of the jet space. A more representative example
follows.
\Ex{SA2} The equi-affine geometry of curves in the plane is governed by the standard
action
$$\qeq{(x,u) \longmapsto
(\alpha x+\beta u + a, \gamma x+\delta u + b),
\\ \alpha \delta - \beta \gamma =1,}\Eq{SA2}$$
of the special affine group $\SA2 = \SL2 \semidirect \R^2$ acting on $M = \Rx2$,
\crf{COT,Gug}. The
components of $w = g^{-1} \cdot (x,u)$ are
$$\qeq{y = \delta (x -a) - \beta (u-b), \\v = -\gamma (x -a) + \alpha (u-b).}\Eq{affyv}$$
Since the group has dimension $5$, we need to prolong to $\J3$ at least to have any
chance of the prolonged action being free, and the first differential invariant should
appear on $\J4$.
The explicit formulae for the fourth prolongation are
$$
\ceq{
v_y = -\frac{\gamma - \alpha u_x }{\delta - \beta u_x } , \quad
v_{yy} = - \frac{u_{xx}}{(\delta - \beta u_x )^3 },\quad
v_{yyy} = - \frac{(\delta - \beta u_x )u_{xxx} + 3\beta u_{xx} ^2}{(\delta - \beta u_x
)^5 },\\ v_{yyyy} = - \frac{ (\delta-\beta u_x)^2 u_{xxxx} + 10 \beta (\delta-\beta u_x)
u_{xx}u_{xxx} + 15 \beta^2 u_{xx}^3}{(\alpha+\beta u_x)^{7}}.
}
$$
The action is free\ignore{This fact can be checked directly,
but follows easily using the infinitesimal approach discussed below.} on
the open subset $\V4 = \{ u_{xx} \ne 0\} \subset \J4$, and hence a local moving frame can
be constructed. Thus we recover the classical result,
that a curve
$u = f(x)$ admits an equi-affine moving frame of order $3$ if and only if it does
not have an inflection point: $f''(x) \ne 0$. Indeed, choosing the normalizations
$$\qeq{y=0,\\v=0,\\v_y=0,\\v_{yy} =1,\\v_{yyy} =0,} \Eq{SA2n}$$
we obtain the classical third order equi-affine moving frame $\mf\,\colon \J3\to \SA2$,
given by
$$\qeq{\matrixb \alpha \beta \gamma \delta = \matrixb{\cuuxx}{-\f3 \uxxf{-5}3 u_{xxx}}
{ u_x \cuuxx} {\uxxf{-1}3 -\f3 \uxxf{-5}3 u_{xxx}}\>,
\\\;
\vectorb ab = \vectorb xu\>.}\Eq{SAmf}$$
The columns of the unimodular matrix can be identified as the equi-affine tangent and
normal to the curve \rf{COT}, while the translational component is the point on the curve. The first
differential invariant is found by inserting the moving frame normalizations into the
next component
$v_{yyyy}$, leading to the equi-affine curvature
$$
\kappa = \frac{ 3u_{xx}u_{xxxx} -5 u_{xxx}^2 }{3 u_{xxx}^{8/3}} \,.\Eq{affcurv}
$$
The equi-affine arc length
$$ds = \cuuxx \dx \Eq{SAds}$$
is obtained, as above, by normalizing $dy = (\alpha +\beta u_x ) \dx$. As in the
Euclidean case, all higher order differential invariants are obtained by differentiation
of the curvature with respect to the arc length.
\pari
The classical restriction of the moving frame method to equi-affine curves without
inflection points can, in fact, be circumvented by using to a higher order prolongation.
Indeed, the fifth order prolongation of \eq{SA2} acts freely on $\V5 = \{ u_{xx}
u_{xxx} \ne 0\} \subset \J5$, and hence a fifth order
equi-affine moving frame can, in principle, be constructed for any curve $u=f(x)$ without
a second order inflection point,
\ie provided either $f''(x)$ or $f'''(x) \ne 0$. (Unfortunately,
the actual normalizations require the solution of a fifth degree polynomial equation, and
hence the explicit formulae for a fifth order moving frame are not available.) More
generally,
$\SA2$ acts freely on the regular subset $\Vn= \Jn\setminus \Sn$, $n\geq 3$, where
$$\Sn = \{ u_{xx} = u_{xxx} = \cdots = u_k = 0\}, \qquad k = 1 + \left [\fr n2\right ]\>,
\Eq{SA2Sn}$$
is the singular subset, and we abbreviate the \kth order derivative of
$u$ with respect to $x$ by $u_k=(D_x)^ku$. Therefore, every curve admits a (local)
equi-affine moving frame of some sufficiently high order \is{unless} it has an inflection
point of infinite order: $f\ps n(x) = 0$, $n = 2,3,4,\ldots $. Curves that do not
admit moving frames will be called totally singular, and their
geometric characterization forms the focus of the final part of this paper. In
particular, the only analytic totally singular curves under the equi-affine group are the
straight lines
$u = mx + l$.
Once the moving frame has been constructed and the corresponding differential invariants
classified, a complete solution to the equivalence problem --- when can two submanifolds
be mapped to each other by a group transformation --- is effected by considering the
\is{signature manifold} which is parametrized by the fundamental differential
invariants. In the case of Euclidean or equi-affine curves, the signature curve is
parametrized by the first two differential invariants
$(\kappa ,\kappa _s)$. Two curves are equivalent if
and only if their signature curves are identical. We refer the reader to
\rf{FOmcII} for details, including a variety of higher dimensional cases. Applications
to classical invariant theory and the equivalence and symmetry properties of binary forms
are discussed in
\rf{I,BOpol}. Applications to computer vision and the recognition of objects in images
can be found in \rf{COSTH}. Applications to the design of invariant numerical algorithms
can be found in \rf{Onums}.
\Section o Stabilization and Orbit Dimensions.
The theoretical foundations of the moving frame method outlined in the preceding section
motivate a more detailed discussion of the geometry of the higher order prolongations
of a transformation group. The key questions that need to be addressed are:
\ritems{Does the prolonged group action become free and regular, at least on an open
subset of
$\Jn$ at sufficiently high order $n\gg 0$? As the equi-affine example makes clear, one
cannot expect the action to ever become free everywhere on $\Jn$.\\
How does one characterize the singular subset $\Sn\subset \Jn$ where the action is not
free? \\
How does one characterize the totally singular submanifolds, meaning those whose $n$-jets
have nonempty intersection with the singular subset $\Sn$ for \iz{all} $n$?}
Since these questions rely on the structure of the prolonged group orbits, the first
order of business is to understand their dimensions. In particular, the ``generic''
orbits --- those of maximal dimension --- have a distinguished status.
\Df{stable} Given $G$ acting on $M$, let $s_n$ denote the
maximal orbit dimension of the prolonged action $\Gn$ on $\Jn$.
The \is{stable orbit dimension} of $G$
is defined to be $s = \max s_n$.
The \is{stabilization order} of $G$ is the minimal
$n$ such that $s_n = s$.
\Remark Since the prolonged orbits map onto each other by
the standard jet projections $\pi _k^n\colon \Jn\to \Jk$, $k \varphi (t),\qquad \rg \kappa r,\Eq{hkappa}$$
where $\varphi \colon \R\to \R^q$ is a fixed nonzero vector function, and $a(t)$
is any scalar function such that it and its first
$r-1$ derivatives are linearly independent, \ie $a$ is \iz{not} the solution to a linear,
constant coefficient ordinary differential equation of order $r-1$. Consider
the $r$-parameter abelian group
$$(x,u) \longmapsto \left(x,u + \sum _{\kappa = 1}^r\>t_\kappa\,
h_\kappa(x^1)\,\right), \qquad \eeq{x = \psups xp \in \R^p, \\ u = \psups uq
\in \R^q,\\ t = \psubs tr \in G = \R^r.} \Eq{highstab}$$
The orbits of $G$ are the vertical lines
$\set{(x,u + s\,\varphi (x))}{s \in \R\>}$. Moreover, the prolonged orbit
dimensions of the group \eq{highstab} only increase by one at each order:
$s_k = k + 1$ for $k=0,\ldots ,r-1$. Consequently, the stabilization
order for \eq{highstab} is $n=r-1$.
\Df{singsub} The \is{regular subset} $\Vn \subset \Jn$ is
the open subset consisting of all prolonged group orbits of dimension
equal to the stable orbit dimension.
A point $\zn\in \Jn$
will be called a \is{regular jet}
provided $\zn\in \Vn$.
The remainder constitutes the \is{singular subset}
$\Sn = \Jn \setminus \Vn$, which is the union of orbits of less than maximal
dimension.
Note that, by this definition, $\Vn = \emptyset$ and $\Sn = \Jn$ if $n$ is less
than the stabilization order of $G$, so that only jets of
sufficiently high order can be regular.
If $G$ acts analytically, then
$\Vn$ is a dense open subset of $\Jn$ for $n$ greater than or equal
to the stabilization order.
\Remark Suppose $W \subset M$ is an open submanifold. For smooth actions, it
may well happen that $s(W) < s(M)$, where $s(W)$ denotes the
stable orbit dimension for the restricted action of $G$ on $W$. For
example, let
$$\hh0(x) = \mcases{e^{-1/x}, & x >0,\\ 0,& x \leq 0.} \Eq{hh0}$$
The one-parameter group
$$(x,u) \mapsto (x, u + t \,\hh0(x)), \qquad t \in G = \R, \Eq{hh0act}$$
acting on $M = \R^2$ has $s(M) =1$, while $s(W) = 0$ when $W \subset \{ x <
0\}$ because $G$ acts completely trivially on the left half plane. As we
shall see, such pathology cannot appear in the analytic category, where $s(W)
= s(M)$ for all open $W \subset M$.
\Section{ig} Infinitesimal Generators.
Lie's great discovery was that local geometry of a continuous, connected
transformation group is entirely determined by its
infinitesimal action.
%Each Lie algebra element gives rise to a vector field $\v$ on
%$M$, known as an infinitesimal generator of the group action. The identification of
%the Lie algebra $\g$ of the Lie group $G$ with the space of infinitesimal generators is
%unambiguous if the action is locally effective since a nonzero Lie algebra element
%gives rise to a nonzero vector field.
The space of infinitesimal generators of a transformation group
$G$ forms a Lie algebra of vector fields on the manifold $M$, which, assuming $G$ acts
locally effectively, can be identified with its Lie algebra
$\g$. In terms of local coordinates
$(x,u)$ on $M$, the infinitesimal generators have the form
$$\v = \sum_{i=1}^p\> \xi ^i (x,u) \pdo{x^i} +
\sum_{\alpha = 1}^q \>\varphi ^\alpha (x,u) \pdo{u^\alpha }.\Eq{v} $$
Let $\gn$ denote the corresponding Lie algebra of
infinitesimal generators of the \nth prolonged action $\Gn$ on $\Jn$.
Explicit local coordinate formulae for the prolonged
infinitesimal generators are given by the well-known
prolongation formula \rf {O,E}.
\Df{characteristic} The \is{characteristic} of the vector field $\v$
given in \eq v is the $q$-tuple $Q = \psups Qq$ of functions
$$\qeq{Q^\alpha(x,u\ps1) = \varphi ^\alpha(x,u)
- \sum_{i = 1}^p \xi ^i(x,u) \, u^\alpha_i,\\
\rg \alpha q.} \Eq{characteristic}$$
A vector field $\v$ is tangent to a $p$-dimensional submanifold $S$
if and only if its characteristic vanishes on $S$.
Therefore, the characteristic uniquely determines the submanifolds
that are invariant under the induced one-parameter group, \cf \rf{E; Theorem 4.9}.
%\Pr{isoc} A closed submanifold $S = \{u = f(x)\}$ is invariant under a
%one-parameter transformation group $\exp(t\v)$ if and only if $f$ is a solution to the
%\ro{invariant surface conditions} $Q(x,u\ps1) = 0$ determined by the characteristic $Q$
%of $\v$.
\Pr{isoc} The Lie algebra $\g_S$ of the isotropy subgroup $\GS $ of a closed $p$-dimensional
submanifold $S\subset M$ consists of all infinitesimal generators $\v \in \g$
whose characteristics vanish identically on $S$.
Let $\v\in\g$ generate the one-parameter
subgroup $\exp(t\v)$. The \nth prolongation $\vn$ is the %corresponding
infinitesimal generator of the prolonged one-parameter group
$\exp(t\v)\ps n$ on $\Jn$.
In terms of local coordinates $\xui$ on $\Ji$, we obtain $\vn$
by truncating the infinitely prolonged vector field
$$\vi = \sum_{i = 1}^p \> \xi ^i (x,u) \pdo{x^i} +
\sum_{\alpha = 1}^q \; \sum_{k = \# J \geq 0}
\varphi ^\alpha_{J} (x,u\ps k) \pdo{u^\alpha _J }
\Eq{prv} $$
at order $n$. The coefficient functions $\varphi ^\alpha_{J}$ of $\vi$ are explicitly constructed
as follows. Let
$$D_i = \pdo{x^i} + \sum _{\alpha =1}^q \sum _{\#J\geq 0} \,u^\alpha _{J,i} \pdo{u^\alpha _J}
\Eq{total}$$
denote the usual total derivative with respect to $x^i$. Note that we can
regard $D_i$ as a vector field on $\Ji E$, but not on any finite order jet bundle.
\Df{evtot} A \is{total vector field} on $\Ji$ is a linear combination
$\sum_i P_i\xun D_i$ of total derivatives.
\Remark In fact, the characterization of total vector fields does not depend on the choice of
independent and dependent variables. Indeed, as discussed in \rf E, they are the
sections of the annihilator of the ideal generated by the contact forms.
\Th{prof} The prolonged vector field \eq{prv} can be uniquely written as a sum
$$\vi = \vtot + \vev,\Eq{vtotev}$$
where
$$\vtot = \pitot(\vi) =
\sum_{i = 1}^p \xi ^i (x,u) D_i \Eq{V} $$
is the total vector field induced by the horizontal component of $\v$, and
$$\vev =
\piev(\vi) =
\sum_{\alpha = 1}^q \; \sum_{\# J \geq 0} D\!_JQ^\alpha \pdo{u^\alpha _J } \Eq{prw} $$
denotes the \is{evolutionary vector field} induced $\v$. Here
$D\!_J = D_{j_1} \cdots D_{j_k}$ denotes the total derivative of order $k = \#J$
corresponding to the multi-index $J = \psubs jk$.
As a consequence of this result, the coefficients of $\varphi ^\alpha_{J}$ of \eq{prv} can be
computed via the standard prolongation formula
$$\varphi ^\alpha_{J} (x,u\ps k) = D_J [Q^\alpha(x,u\ps1)]
+ \sum_{i=1}^p \xi^i(x,u) u^\alpha_{J,i},
\Eq{phiaJ}$$
which was first explicitly given in \rf{Os}.
The next result appears in \rf{O; Lemma 5.12}.
\Pr{vevD} An evolutionary vector field $\vev$ commutes with
the total derivative operators\ro:
$$\qeq{[\vev, D_i] = 0, \\ \rg ip.}\Eq{vevD}$$
\smallskip
\Remark The only other vector fields on $\Ji$ which satisfy \eq{vevD}
are the constant coefficient independent variable vector fields
$\sum c_j \partial _{x^j}$. Thus, the commutation property
\eq{vevD} almost completely characterizes evolutionary vector fields.
% Unlike total vector fields, the notion of an evolutionary vector field does depend on the
% choice of independent and dependent variables.
Let us now introduce a basis $\subs \v r$ for
the infinitesimal generators $\g$ of our transformation group.
\Df{Lm} The \is{Lie matrix} of order $n$ is defined as the $r \times \qn$ matrix
$$\Ln\xun = \pmatrix{
\xi^1_1&\ldots&\xi^p_1&\varphi^1_1&\ldots&\varphi^q_1&\ldots&\varphi^\alpha _{J,1}&\ldots&\ \cr
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots\cr
\xi^1_r&\ldots&\xi^p_r&\varphi^1_r&\ldots&\varphi^q_r&\ldots&\varphi^\alpha _{J,r}&\ldots\cr
},\Eq{Lmatrix}$$
where $0 \leq \#J \leq n$. The entries
$\xi^i_\kappa$, and $\varphi^\alpha _{J,\kappa}$
of \eq{Lmatrix} are the coefficients of the \nth
order prolongations $\vn_\kappa$, $\rg \kappa r$, of the
basis infinitesimal generators of $G$.
Similarly, the \is{Lie characteristic matrix}
of order $n$ is defined as the
$r \times \qn$ matrix
$$\Mn\xuno = \pmatrix{
\xi^1_1&\ldots&\xi^p_1&Q^1_1&\ldots&Q^q_1&\ldots&D_J Q^\alpha _1&\ldots&\ \cr
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots\cr
\xi^1_r&\ldots&\xi^p_r&Q^1_r&\ldots&Q^q_r&\ldots&D_J Q^\alpha _r&\ldots\cr
},\Eq{Cmatrix}$$
where $0 \leq \#J \leq n$. The entries
of \eq{Cmatrix} consist of the independent variable coefficients, the characteristics, and
their total derivatives up to order $n$.
\Pr{on} The dimension of the orbit of $\Gn$ through the point $\zn \in \Jn$
is the common Lie matrix rank
$$\rank \Ln(\zn) = \rank \Mn(\zno). \Eq{rankLM}$$
Here $\z{n+1}$ is any point with $\pi^{n+1}_n(\z{n+1}) = \zn$,
since the rank of $\Mn$ is independent of the
derivative coordinates $u^\alpha_K$ of order $\#K = n+1$.
\Proof
The dimension of the orbit through $\zn \in \Jn$ is equal to the dimension of
the subspace spanned by the prolonged infinitesimal generators,
\ie $\gn \atzn \subset \Jn \atzn$, and hence equals
the rank of the Lie matrix $\Ln(\zn)$. Furthermore, the
prolongation \eqf{phiaJ} implies that $\Mn(\zno)$ can be
obtained from $\Ln(\zn)$ by adding suitable multiples of the
first $p$ columns to the other columns. Since column operations do
not alter the rank of a matrix, the proof is complete. \qed
We can characterize the singular subset by the infinitesimal condition
$$\Sn = \set{\zn\in \Jn}{\dim \gn\atzn < s}.\Eq{sing}$$
Thus, the regular and singular subsets of $\Jn$
are determined by the common rank of Lie matrices \eqs{Lmatrix}{Cmatrix}.
\Co{sn} The stable orbit dimension of $G$ is given by
$$s = \max \set{\rank \Ln(\zn)}{\zn\in\Jn, \ n \geq 0}.$$
Consequently, a point $\zn \in \Sn$ is singular if and only if
$$\rank \Ln(\zn) = \rank \Mn(\z{n+1}) < s, \where \pi^{n+1}_n(\z{n+1}) = \zn. \Eq{srankLM}$$
\smallskip
In the particular case of planar curves, $p=q=1$, the Lie matrix $\L{r-2}(x,u\ps{r-2})$
of order $n = r-2 = (\dim G) - 2$ is square, and its
determinant
$$L(x,u\ps{r-2}) = \det \L{r-2}(x,u\ps{r-2}) \Eq{Liedet}$$
is known as the \is{Lie determinant}\fnote{This is not quite correct if
$G$ does not act transitively on an open subset of $\J{r-2}$, which occurs when
either $G$ acts intransitively on $\R^2$ or its prolonged orbit dimensions pseudostabilize. See \rf
E for the appropriate modifications in such cases.}
of the transformation group $G$.
See \rf{E; Chapter 6} for a discussion, as well as a complete
list of Lie determinants for the different finite-dimensional
planar transformation groups appearing in Lie's classification.
The Lie determinant equation is the differential equation of lowest order that admits the
given transformation group as a symmetry group. In \sections{ss}h we will establish an
algebraic method for constructing its general solution.
\Ex{SA2Ln} The planar equi-affine group \eq{SA2} has infinitesimal
generators
$$\qeq{\v_1 = \partial _x,\\\v_2 = \partial _u,\\\v_3 = x\partial _u,\\
\v_4 = -x\partial _x + u\partial _u,\\\v_5 = -u\partial _x .} \Eq{sa2}$$
Applying the prolongation formula \eqs{prv}{phiaJ}, we compute the \nth order Lie
matrix to be
$$\Ln\xun = \pmatrix{
1&0&0&0&0&\ldots & 0\cr
0&1&0&0&0&\ldots & 0\cr
0&x&1&0&0&\ldots & 0\cr
-x&u&2u_x &3u_{xx} &4 u_{xxx} & \ldots & (n+1) u_n\cr
-u&0&u_x^2 &3u_x u_{xx} &4 u_x u_{xxx} + 3 u_{xx} ^2& \ldots & P_n\cr},\Eq{SA2Ln}$$
where
$$P_n\xun = \fra 2 \sum _{i=1}^{n}\ {n \choose i} \> u_i u_{n-i+1}, \qquad \qquad u_n =
D_x^n u.$$ The Lie determinant is $L(x,u\ps3) =9 u_{xx} ^3$, whose vanishing describes
the singular subset $\S3$. The verification that $\Sn$ for $n\geq 3$ is given by
\eq{SA2Sn} is immediate from the form of the Lie matrix. The general connection between
the Lie determinant and the higher order singular subvarieties will be discussed below.
\Section{seff} Singularities and Effectiveness.
We are now ready to discuss the detailed geometry of the
singular subsets of the jet spaces corresponding to a given transformation
group $G$. The regular subset
$\Vn \subset\Jn$ is nonempty and open provided $n$ is
greater than or equal to the stabilization order of $G$. In the analytic
category, $\Vn$ is dense in $\Jn$. However, at any given point
$z\in M$, this does not necessarily imply that the jet fiber $\Jn\at z$
contains any regular jets.
\Df{singpt} A point $z\in M$ is \is{totally singular} if
$\Jn \at z \subset \Sn$ for all $n$. A point which is not totally singular
will be called \is{totally regular}.
Thus, a point $z\in M$ is totally regular provided $\Jn \at z$ contains a
regular jet for $n$ sufficiently large.
The set $M_0\subset M$
consisting of all totally regular points forms
a nonempty open submanifold. Totally singular points are rather
pathological. The following examples show that they can occur for
non-analytic actions.
\Ex{hx} Consider the one-parameter group \eq{hh0act} acting on $M = \R^2$. The
stable orbit dimension is $s(M) = 1$, and so every point in the
right half plane $\{x \leq 0\}$ is totally singular.
On the other hand, the restricted action on the open submanifold
$W = \{x < 0\}$ is completely trivial, so $s(W) = 0$, and every
point in $W$ now becomes totally regular.
Thus, for smooth actions, the notion of a totally
singular point can change upon restriction of the group action to an open
submanifold.
\Ex{ax} Let
$$\hh1(x) = \mcases{e^{-1/|x|}, & x \ne 0,\\ 0,& x = 0.} \Eq{hh1}$$
Consider the two-parameter abelian transformation group
$$(x,u)\longmapsto (x + a \hh1(u) + b,u), \where (a,b)\in G = \R^2.$$
The orbits are the horizontal lines, and so $G$ acts regularly.
Since the infinitesimal generators are $\v_1 = \hh1(u)\partial _x$, $\v_2 = \partial _x$,
the \nth order Lie matrix is
$$\seq{\Ln\xun = \pmatrix{
\hh1(u)&0& -\hh1'(u) u_x^2 &-\hh1''(u) u_x^3 - 3\hh1'(u) u_x u_{xx} &
%-\hh1'''(u) u_x^3 - 3\hh1''(u) u_x u_{xx}- \hh1'(u) u_{xxx} &
\cdots&\cr
1&0&0&0&\cdots\cr},}$$
with the $(1,k+2)$ entry being $\hh1(u) D_x^{k+1}u - D_x^{k}[\hh1(u) u_x]$ for $0\leq
k\leq n$. Consequently, the singular subset has two components,
$$\Sn = \{u = 0\} \>\cup\> \{ u_x = u_{xx} = \cdots = u_n = 0\} \subset \Jn,$$
%where $u_n = (D_x)^nu$.
and every point on the $x$ axis %$\{u=0\}$
is totally singular.
Such pathologies cannot occur in the analytic category.
\Th{analpt} If $G$ acts analytically, then there are no totally singular points.
\Proof
The proof rests on a series of lemmas. Let $z_0\in M$.
We introduce
local coordinates $z = (x,u)$ near $z_0 = (0,0)$, which we take
as the origin for simplicity. A vector field
$$\v = \sum_{i=1}^m \zeta^i(z) \pdo{z^i}$$
is said to be of \is{degree $n$ at $z_0$} if all
its coefficients and their partial derivatives
up to order $n-1$ vanish at $z_0=0$, so
$\partial_J \zeta^i(0) = 0$ for $0\leq \#J < n$.
In particular, the isotropy subalgebra of $z_0$ consists of
all infinitesimal generators that have degree $1$ (or more) at $z_0$.
The condition that a vector field have a certain degree at a point is, in fact, independent of the
choice of local coordinates.
\Lm{prn1}
If a vector field $\v$ has degree $n+1$ at $z_0$, then its \nth order prolongation $\prn \v$
vanishes on $\Jn\at{z_0}$.
A vector field $\v$ is called a \is{polynomial vector field}
of \is{degree $n$} if all its
coefficients are polynomials of degree $\leq n$ in the
given local coordinates $z=(x,u)$.
A polynomial vector field $\v$ is said to be \is{homogeneous}
of degree $n$ if its coefficients $\zeta^i$ are homogeneous polynomials
of degree $n$. For example, the elementary
scaling vector field
$$\vsc = \sum_{i=1}^m \>z^i \pdo{z^i} = \sum_{j=1}^p \>x^j \pdo{x^j}
+ \sum_{\alpha =1}^q \>u^\alpha \pdo{u^\alpha } \Eq{vsc}$$
is homogeneous of degree $1$.
%A constant vector field is homogeneous of degree $0$.
Any analytic vector field
can thus be written as a convergent power series whose \nth term is a homogeneous
polynomial vector field of degree $n$ at the point $z_0 = 0$.
\Lm{pvf} Let $\v$ be a polynomial vector field of degree $n$.
Then $\vn\at{\zn} = 0$ for all $\zn \in \Jn\at{z_0}$ if and only if
$\v = p_{n-1}(z) \,\vsc$ is a multiple of the scaling vector field
\eq{vsc} where the coefficient $p_{n-1}(z)$ is a scalar-valued homogeneous polynomial
of degree $n-1$.
\Proof
Unfortunately, at present the only proof I know is a straightforward, but tedious
computation, based on the local coordinate formulae \eqs{prv}{phiaJ} for the
prolongations. For brevity, I shall omit the details. \qed
%Putting these two results together, we have:
\Lm{vn} Let $\v$ be a smooth vector field on $M$.
Suppose that $\vn\at{\zn} = 0$ for all $\zn \in \Jn\at{z_0}$.
Then $\v = p_{n-1}(z) \,\vsc + \v_{n+1}$, where
$p_{n-1}(z)$ is a scalar-valued homogeneous polynomial
of degree $n-1$ and $\v_{n+1}$ is a vector field of degree $n+1$.
\Proof
We use Taylor's theorem to write $\v = \w_n + \v_{n+1}$, where
$\w_n$ is a polynomial vector field of degree $n$,
and the remainder $\v_{n+1}$ has degree $n+1$. The lemma then follows
immediately from \lms{prn1}{pvf}. \qed
A simple induction on the order $n$ now demonstrates that
the only analytic vector field that vanishes on all jet spaces at
the point $z_0$ is the trivial one.
\Lm{vi} Let $\v$ be an analytic vector field on $M$.
Suppose that $\vn\at{\zn} = 0$ for all $\zn \in \Jn\at{z_0}$ and all\fnote{With
a little more work, this condition
can be relaxed to just require $n > n_0$ for some $n_0$.} $n > 0$. Then $\v \equiv 0$.
\th{analpt} now follows immediately from this final lemma.
\qed
Ovsiannikov's stabilization theorem \rf{Ov}, \rf{E; Theorem 5.11}, completely
characterizes the stable orbit dimension. The usual statement of this result contains
the hypothesis that the group acts \is{locally effectively}, and the conclusion is that
the stable orbit dimension equals the dimension of $G$. However,
this is not quite justified, as the following example makes clear.
\Ex{hx0} \
Consider the two-parameter
abelian transformation group $G\simeq \R^2$ acting on $M = \R^2$
via $(x,u) \mapsto (x,u+a \,\hh0(x) + b \,\hh0(-x))$,
where $(a,b) \in G$, $(x,u)\in M$.
The $C^\infty$ function $\hh0$ is defined in \eq{hh0}.
The only group transformation which fixes every point $(x,u)$ is the identity $a=b=0$,
and so $G$ acts effectively on
$M$. However, it is not hard to see that the stable orbit dimension is $s=1$, which is
strictly less than the dimension of $G$. Note also that the $u$-axis consists entirely
of totally singular points.
The problem with this example is that $G$ acts effectively on $M$, but does \is{not} act
effectively on either the right half plane $\{x>0\}$ or left half plane $\{x<0\}$. In
other words, at least in the smooth category, a group may act effectively on a manifold,
but not effectively on certain open submanifolds, and this would cause the traditional
statement of the stabilization theorem to be invalid. In order to state a correct smooth
version of the theorem, we need a slight refinement of the notion of effectiveness.
\Df{effsub} A transformation group $G$ acts \is{effectively on subsets} if $\giso W =
\sete$ for every open $W\subset M$. The group acts \is{locally effectively on subsets} if
$\giso W$ is a discrete subgroup of $G$ for every open $W\subset M$.
Clearly, if $G$ acts effectively on subsets, then $G$ acts effectively. While the converse
does not hold for general smooth actions, it \is{is} true in the
analytic category.
\Lm{analyticeff} If $G$ acts analytically on $M$ and acts
\ro(locally\ro) effectively, then $G$ acts \ro(locally\ro) effectively on subsets.
The proof is a straightforward application of analytic continuation,
based on the fact that the only analytic function $\psi\colon M \to M$ which
agrees with the identity map on an open subset of $M$ is the identity.
We can now state a correct version of the stabilization theorem.
\Th{stab} A Lie
group $G$ acts locally effectively on subsets of $M$ if and only if
for every open submanifold $W \subset M$, the stable orbit dimension of the restricted
action of
$G$ on $W$ equals the dimension of the group\ro: $s(W) = s(M) = r = \dim G$.
\Proof
The proof is adapted from that in \rf E, which works as long as we are not at a totally
singular point. First, if $G$ does not act locally effectively on
$M$, then $\Gn$ does not act locally effectively on $\Jn(M,p)$, and hence
the maximal orbit dimension $s_n$ is strictly less than the dimension of
$G$. This implies that the stable orbit dimension is strictly less than the
dimension of $G$, so $s(M)2\pi ),
\qquad \varepsilon \in \R,$$
where $\hh1$ is given in \eq{hh1} acts effectively on subsets. The orbits are the
one-dimensional vertical circles, and
$G = \R$ acts locally freely on $M$, and freely on $M\setminus U_0$, where $U_0 = \{x = 0\}$
denotes the circle through the origin. Therefore, every point in $\Jn = \Vn$ is regular.
However, the action is only locally free on
$\Jn\,|\,U_0$ for any $n\geq 0$. Indeed, since all the derivatives of $\hh1(x)$ vanish at the
origin the prolonged action on $\Jn \,|\,U_0$ reduces to
$$(0,u,u_x ,\ldots ,u_n)\longmapsto (0,u+\varepsilon \ \mod\>2\pi ,u_x,\ldots ,u_n),$$
and so the isotropy subgroup for any $\zn= (0,u,u_x ,\ldots ,u_n)\in \Jn \,|\,U_0$ is the
discrete subgroup
$2\pi \,\Z \subset \R$ consisting of integral multiples of $2\pi $.
Note particularly that although $\Gn$ acts freely on a dense open subset of
$\Jn$, the set of points where the prolonged action is not free contains the
entire jet fiber over the subvariety $U_0$.
An important application of \th{stab}
is the following ``Wronskian lemma,'' generalizing the
classical result on the linear dependence of functions with vanishing
Wronskian determinant. Applications to the problem of characterizing differential
operators, both linear and nonlinear, which leave a prescribed finite-dimensional
subspace of functions invariant, and the consequent method of constructing explicit
solutions to partial differential equations by reduction to dynamical systems are
discussed in \rf{KMO}.
\Df{li} A set of vector-valued functions
$h_\kappa \colon X \to \R^q$, $\rg \kappa r$,
is \is{linearly dependent} on a subset $W\subset X$
if there exist constants $c = \psubs cr \ne 0$, \is{not all zero},
such that
$$\sum _{\kappa =1}^r c_\kappa h_\kappa (x) \equiv 0, \forall x \in W. \Eq{li}$$
The functions are \is{linearly independent} on $W$ if the only
constant coefficient linear combination that vanishes identically
is the trivial one $c_1 = \cdots = c_r =0$. The functions
$\subs hr$ are \is{linearly independent on subsets of $X$} if they are
linearly independent on every open subset $W \subset X$.
In the analytic category, linear independence implies
linear independence on subsets.
On the other hand, the smooth functions $\hh0(x)$ and $\hh0(-x)$, \cf \eq{hh0},
are linearly independent on $\R$ but not
linearly independent on subsets. Standard existence and uniqueness theorems
imply that linearly independent solutions
of a linear system of ordinary differential equations are always
linearly independent on subsets.
Classically, one checks the linear independence of
scalar functions by analyzing their Wronskian determinant.
In the multivariate version, one needs to analyze the ranks of general
rectangular Wronskian matrices.
\Df{Wronskian} The \nth order \is{Wronskian matrix} of smooth
functions $h_\kappa \colon X \to \R^q$, $\rg \kappa r$, where $X \subset \R^p$,
is the $r \times q{p+n \choose n}$ matrix
$$\Wn(x) = \pmatrix{
h^1_1(x)&\ldots&h^q_1(x)&\ldots&\partial_J h^\alpha _1(x)&\ldots&\ \cr
\vdots&\ddots&\vdots&\ddots&\vdots&\ddots\cr
h^1_r(x)&\ldots&h^q_r(x)&\ldots&\partial_J h^\alpha _r(x)&\ldots\cr
}\,,\Eq{Wmatrix}$$
whose entries are the partial derivatives of the $h$'s with respect
to the $x$'s of all orders $0 \leq \#J \leq n$.
\Th{Wr} A set of functions $h_\kappa \colon X \to \R^q$, $\rg \kappa r$,
is linearly independent on subsets of $X$ if and only if
their \xsts{r-1}order Wronskian matrix $\W{r-1}(x) = \W{r-1}\csubs hr(x)$ has maximal
rank $r$ for all $x$ in a dense open subset of $X$.
\Proof
Consider the action
$$\qeq{(x,u) \longmapsto \left( x, u + \sum_{\kappa=1}^r t_\kappa\, h_\kappa(x) \right),
\\ t = \psubs tr \in \R^r,} \Eq{hab}$$
of the abelian $r$-parameter Lie group $G = \R^r$. The infinitesimal generators are the
vector fields
$$\v_\kappa =\sum_{\alpha =1}^q \ h^\alpha _\kappa (x) \,\pdo{u^\alpha }.$$
The Lie matrix $\Ln(\zn)$ of this group consists of $p$ initial columns of
zeros, since the group acts trivially on the independent variables,
followed by the Wronskian
matrix $\Wn(x)$ of the functions $h_\kappa$. Therefore, the prolonged orbit dimension
$s_n$ equals the maximal rank of $\Wn(x)$.
It is not hard to see that $G$ acts locally effectively on subsets
if and only if the functions $h_\kappa$ are linearly independent on
subsets. Therefore, \th{stab} implies that, on any open subset $Y\subset
X$, we have $r = \max \rank \Wn(x)$ for $x \in Y$ and any $n$ greater than or
equal to the stabilization order of the transformation group \eq{hab}. Moreover, it is not
hard to prove that the group \eq{hab} does not pseudostabilize, and hence
\th{staborder} implies that the stabilization order is at most $r-1$. \qed
For example, in the case $p=q=1$, so that $h_1 (x), \ldots ,h_r(x)$ are scalar-valued
functions of a single variable $x$, the matrix $\W{r-1}\csubs hr$
becomes the usual $r\times r$ Wronskian
of the functions $h_\kappa $. Therefore, \pr{Wr} reduces
to the statement that a set of scalar functions is linearly independent
on subsets of $\R$ if and only if its Wronskian determinant does not vanish
identically, thus providing a converse to the standard Wronskian
lemma in the theory of linear ordinary differential equations.
\Remark Kolchin \rf{Kolchin; p. 86} proves a version of \th{Wr} assuming
that the functions belong to a differential field, \eg the field of
meromorphic functions. Further extensions appear in \rf{KMO}.
Since the group \eq{hab} does not pseudostabilize, its stabilization
order is the minimal $n$ for which $\max \rank \Wn(x) = \max \rank
\W{n+1}(x) = s$. Thus $s = r$ if and only if the functions are linearly
independent on subsets. In the smooth category, they may be linearly
independent even when $s < r$. In the analytic category, $s$ equals the
dimension of the vector space spanned by $\subs hr$.
Suppose that $h_1(x),\ldots h_r(x)$ are linearly independent on subsets. By
definition, a \is{totally singular point} is a point $x\in X$ such that
$\rank \Wn(x) < r$ for all $n \geq 0$; the remaining points are called
\is{totally regular}. \th{Wr} states that the set of totally regular points
is open, dense in $X$. Moreover, if the functions are analytic, then linear
independence implies linear independence on subsets, and every
point is totally regular. At a totally regular point $x\in X$, we have, by
definition, $\rank \Wn(x) = r$ for $n$ sufficiently large (and not necessarily
$\leq r$), which, by continuity, also holds in a neighborhood of $x$. However,
even in the analytic category, it may be impossible to find a finite value of
$n$ for which $\rank \Wn(x) = r$ for all $x \in X$. For example, let $h_1(x)$
be an analytic function which has a zero of order $k$ at $x = a_k$, for $k =
1,2,3,\ldots\ $, where $a_k \to \infty$ as $k \to \infty$; such a function can
be constructed using a Weierstrass product expansion, \cf \rf{Ahlfors; p.~194}.
Then $\rank \Wn[h_1](a_k) = 0$ for $n < k$, while $\rank \Wn[h_1](x) = 1$ for
$n \geq k$ and $x$ in any neighborhood of $a_k$ that does not contain
$a_{k+1}, a_{k+2}, \ldots\ $.
\Section{ss} Singular Submanifolds.
We now turn to the study of singular submanifolds.
From now on we make the blanket assumption that $G$ is an $r$-dimensional Lie group
that acts locally effectively on subsets of $M$.
\th{stab} implies that its stable orbit dimension equals its dimension,
$s = r = \dim G$. The regular subset $\Vn \subset \Jn$ consists of all orbits whose dimension
is the same as the dimension of $G$.
\Df{ss} A submanifold $S\subset M$ is \is{order $n$ regular}
if $\jn S\subset \Vn$. A submanifold $S$ is called \is{regular}
if it is regular at some finite order $n$.
A submanifold $S\subset M$ is \is{totally singular at a point $z\in S$}
if $\jn S\at z\subset \Sn$ for all $n=0,1,\ldots \ $ A \is{totally singular}
submanifold consists entirely of totally singular points.
In other words, if $S$ is an order $n$ regular submanifold, then
$\Gn$ acts locally freely in a neighborhood of $\jn S\subset \Vn$.
Note that if $S$ is order $n$ regular, then it is also order $m$ regular
for any $m\geq n$.
The set of regular points on a submanifold forms
an open submanifold $S_0 \subset S$, which is a dense subset if $G$ acts
analytically and $S$ is an analytic submanifold.
The regular submanifolds of order $n$ are precisely those that admit (locally
equivariant) moving frames of that order. Vice versa, a totally singular submanifold
admits \is{no} moving frame of any order.
The goal of this section is to provide a direct,
geometrical characterization of totally singular submanifolds.
The first easy result demonstrates that non-freeness of
the isotropy subgroup of a submanifold at a point implies singularity.
\Pr{regfree} Let $S$ be a submanifold, and $\iso S\subset G$ its
isotropy subgroup. If $S$ is regular at $z\in S$, then $\iso S$
acts locally freely on $S$ at $z$.
\Proof
Assume that $\iso S$ does not act locally freely at a point $z \in S$.
This implies that there exists a nontrivial
infinitesimal generator $0 \ne \v \in \isa S$ that
vanishes at $z$, so $\v\at{z} = 0$. Therefore, $\exp(t\v)z = z$ and
$\exp(t\v) S \subset S$.
Together, these imply that, for every $n$,
the $n$-jet $\zn = \jn S \at {z}$ is a fixed point for
the prolonged one-parameter group, so $\exp(t\v)\ps n \cdot \zn = \zn$.
But this implies $\vn \at{\zn} = 0$, and so $\zn \in \Sn$ for all $n$,
proving that $S$ is totally singular at $z$. \qed
\Ex{tspt} Consider the analytic two-parameter group
$(x,u) \mapsto (\lambda x, u + b)$. The infinitesimal generators
are $\v_1 = \partial_u$, $\v_2 = x \partial_x$, and
hence the Lie matrix is
$$\pmatrix{0&1&0&0&0&\cdots&\cr
x&0& - u_x & - 2 u_{xx} & -3 u_{xxx} & \cdots &\cr}.$$
This shows that the singular subset is
$$\Sn = \{x = u_x = u_{xx} = \cdots = u_n = 0\}.$$
Therefore, any horizontal line $L = \{u = c\}$ is totally singular
when $x=0$, but is regular at nonzero $x$. The isotropy subgroup $G_L$ consists
of the scaling $x \mapsto \lambda x$, which does not act locally freely at $x=0$.
On the other hand, the curve $C$ defined by the function
$$u = \hh2(x) = \mcases{e^{-1/x}, & x >0,\\ 0,& x = 0,\\
xe^{1/x}, & x <0,} \Eq{hht1}$$
has infinite order contact
with the $x$ axis at the origin, and is totally singular at
$z_0 = (0,0)$. However, $C$ has trivial isotropy subgroup $G_C = \sete$.
This shows that the converse to \pr{regfree}
is not necessarily true, at least in the smooth category.
\Ex{sct} Slightly generalizing the preceding example,
we consider the analytic four-parameter group
$(x,u) \mapsto (\lambda x + a, \mu u + b)$. The infinitesimal generators
are
$$\qeq{\v_1 = \partial_x,\\
\v_2 = x \partial_x,\\
\v_3 = \partial_u,\\
\v_4 = u \partial_u,}$$
and so the Lie matrix is
$$\pmatrix{\;1&0&0&0&0&0&\cdots&\cr
\;x&0& - u_x & - 2 u_{xx} & -3 u_{xxx} &-4 u_{xxxx} & \cdots &\cr
\;0&1&0&0&0&0&\cdots&\cr
\;0&u& u_x & u_{xx} & u_{xxx} &u_{xxxx} & \cdots &\cr}.$$
Therefore, the singular subset occurs where all but one of
the derivatives $u_x, u_{xx}, u_{xxx}, \ldots $ vanish. Therefore,
any straight line $u = a x + b$ is totally singular at every point thereon.
Furthermore, the polynomial curves $u = a (x-x_0)^n + u_0$ for $n \geq 2$,
are totally singular at the point $(x_0,u_0)$. Also, by reversing
the roles of $x$ and $u$, the curves
$x = (u-u_0)^n + x_0$ are also totally singular at $(x_0,u_0)$.
Thus, each point $z_0\in \R^2$ has
an interesting collection of totally singular analytic curves passing through it.
\Th{asing} Suppose $G$ acts analytically. An analytic submanifold $S$ is
regular at a point $z_0\in S$ if and only if its isotropy subgroup $\GS $
acts locally freely on $S$ at $z_0$.
\Proof
Assume that $S$ is
totally singular at a point $z_0 \in S$.
This assumption means that we can find a nontrivial infinitesimal generator $0\ne \v\in \g$ such
that $\vn \at{z_0} = 0$ for all $n\geq 0$. Choose local coordinates $(x,u)$ such that $S =
\{u = f(x)\}$ is transverse at $z_0 = (x_0,f(x_0))$. The prolongation formula \eq{phiaJ}
implies that the total derivatives of the characteristic $Q(x,u\ps1)$ of $\v$ all vanish
at $z_0$, so that $D_J Q(x_0,\jko f(x_0)) = 0$ for all $0\leq k = \#J$, where
$(x_0,\jko f(x_0)) = \jko S\at{z_0}$. This means that the analytic function
$F(x) = Q(x,\j1f(x))$ satisfies $\partial_J F(x) = 0$ for all $J$, and hence, by
analyticity,
$F(x) \equiv 0$ for all $x$. Consequently, $u = f(x)$
is a solution to the invariant surface conditions $Q(x,u\ps1) = 0$
for the vector field $\v$.
\th{isoc} implies that $\v$ is an
infinitesimal generator of the isotropy subgroup $\GS $. Moreover, since we have
$\v\at{z_0} = 0$, the isotropy subgroup $\GS $ cannot act
locally freely on $S$ at $z_0$.\qed
As we saw, in the smooth category,
freeness of the isotropy subgroup action on $S$ is not enough to
prevent isolated totally singular points from occurring.
The next result shows that if all the points on $S$ are totally singular, then
the isotropy subgroup must act non-freely, even in the smooth category.
The method of proof differs from that of the
analytic \th{asing}, and is based on the proof of the stabilization \th{stab}.
\Th{sing} Let $G$ act smoothly and locally effectively on subsets of $M$.
A submanifold $S\subset M$ is totally singular if and only if its
isotropy subgroup $\iso S$ does not act locally freely on $S$.
\Proof
Assume that $\GS $ does act locally freely.
Let
$$t_n = \max \set{\dim \gn \at{\zn}}{\zn\in \jn S}.$$
Clearly,
$t_0 \leq t_1 \leq \cdots \;$. Thus $t_m = t = \max t_n$ for all
$m$ sufficiently large. Since $\jn S\subset \Sn$ for all $n$,
we have $t \{\vm_1 \at{\z m},\ldots ,\vm_t \at{\z m}\},$$
for all $\z m \in \j m R$ and all $m\geq n$, and hence the remaining generators
can be written as linear combinations of the independent generators.
Therefore, letting $m\to\infty$, we obtain the linear relations
$$\vi_{t+\nu } = \sum_{\kappa =1}^t \;\lambda _\nu ^\kappa \>\vi_\kappa ,
\qquad \rg \nu {r-t}.\Eq{vv}$$
Projecting the higher order linear dependencies
\eq{vv} on $\Jn$ implies that the coefficients $\lambda _\nu ^\kappa
= \lambda _\nu ^\kappa (\zn)$ are
functions of only the \nth order jet $\zn \in \jn R$. \pari
The remainder of the proof is almost identical to that of \th{stab}.
We can introduce local coordinates
$(x,u)$ on $M$ such that $R$ is (locally) the
the graph of a section $u = f(x)$.
Applying the evolutionary projection $\piev$ to \eq{vv} gives
$$
\w_{t+\nu } = \sum_{\kappa =s+1}^t \lambda _\nu ^\kappa \,\w_\kappa ,
\qquad \rg \nu {r-t},\Eq{ww}
$$
on $R$, where $\w_\kappa = \piev(\vi_\kappa)$ denotes
the evolutionary part of $\vi_\kappa$.
\pr{isoc} implies that the first $s$ evolutionary vector fields $\subs \w s$
vanish on $R$ because they lie in its isotropy subalgebra.
Taking the
commutator of \eq{ww} with the total derivative $D_i$ and using the
property \eq{vevD} of evolutionary vector fields yields
$$
0 = \sum_{\kappa =s+1}^t D_i(\lambda _\nu^\kappa ) \,\w_\kappa ,
\qquad \rg \nu {r-t},
$$
which holds on $R$.
Linear independence of the vector fields $\displaystyle\vi_\kappa $
on the submanifold $R$ implies linear independence of their evolutionary
representatives $\displaystyle\w_{s+1}, \ldots ,\w_t $ on $R$. Therefore
$$
D_i(\lambda _\nu^\kappa ) = 0,\qquad \rg ip,
$$
on $R$, which implies that all the coefficients $\lambda _\nu^\kappa $ are constant on
$R$. Consequently, the nonzero vector field
$$
\widetilde \v_\nu = \v_\nu - \sum_{\kappa =s+1}^t \lambda _\nu^\kappa \, \v_\kappa
$$
vanishes on $R$. This contradicts our original
assumption that $\GS $ acts locally freely on $S$, since
$\widetilde \v_\nu $ belongs to its isotropy subalgebra.
\qed
\Ex{A2} Consider the action $z\mapsto Az + b$ of the
full affine group $\AG 2$ on $M = \Rx2$. Here the
totally singular curves are the parabolas and the straight lines.
The isotropy group of a parabola, say $u = x^2$, is the
two-dimensional non-abelian subgroup
$$(x,u)\mapsto (\lambda x + \mu ,
\lambda ^2 u + 2 \lambda \mu x + \mu ^2). \Eq{parabolasub}$$
In this case the singular subset of $\Jn$ is governed by the total
derivatives of the affine Lie determinant
$$\Vn = \set{\xun}
{D_x^{n-4}\left [ u_{xx} u_{xxxx} - \fr53 u_{xxx} ^2 \right ] = 0}.
\Eq{Asing}$$
Note that parabolas and straight lines comprise the general
solution to the affine Lie determinant equation
$$u_{xx} u_{xxxx} = \fr53 u_{xxx} ^2. \Eq{affLiedet}$$
\medskip
\Remark In both of the special and full affine groups, if an analytic curve contains one
totally singular point, then it is totally singular at all points. However,
the scaling group in \ex{sct} does not enjoy this property.
This is related to the factorizability of the Lie determinant associated
with the group, although
I do not know a general geometrical characterization of this phenomenon.
In the planar case, the Lie determinant can be used to completely
characterize the totally singular curves. This theorem also serves to
explain why higher order singular subsets are usually obtained by
total differentiation (or prolongation) of the Lie determinant equation.
\Th{Ldet} Suppose $G$ is an $r$-dimensional Lie group that acts transitively and locally
effectively on subsets on a two-dimensional manifold. Assume that the prolonged group
actions do not pseudostabilize. Then, locally, the set of totally singular curves form the general
solution to the Lie determinant equation $L(x,u\ps{r-2}) = 0$.
\Remark The result remains valid in the intransitive and/or
pseudostabilization cases if one suitably modifies the definition of the Lie
determinant, as in \rf E.
\Proof
Since the singular subset $\S{r-2} = \{L(x,u\ps{r-2}) = 0\}$ coincides
with the Lie determinant ordinary differential equation, the latter admits $G$ as a symmetry
group. Let $u = f(x)$ be a solution, and suppose the jet $\z{r-2}_0 = (x_0, f\ps{r-2}(x_0))$
is a regular (nonsingular) point for the Lie determinant differential equation, meaning that we
can uniquely and smoothly solve the equation $L(x,u\ps{r-2}) = 0$ for the highest order derivative
$u_{r-2}$ in a neighborhood of $\z{r-2}_0$. (Under the hypotheses, the regular points form a
dense open subset of
$\S{r-2}$ --- this can be seen from the Lie classification of Lie determinants given in \rf E.)
Let $0\ne \v\in \g$ be an infinitesimal generator
whose prolongation $\pro{r-2}\v$ vanishes at $\z{r-2}_0$; such a generator
exists because $\z{r-2}_0$ lies in $\S{r-2}$.
The corresponding one-parameter subgroup
$\exp t\v$ then fixes $\z{r-2}_0$. On the other hand, since
$G$ is a symmetry group of the Lie determinant equation, it
maps solutions to solutions. By the uniqueness of solutions to
ordinary differential equations, $\exp t\v$ also fixes the entire solution
$u = f(x)$, and hence $G$ does not act locally freely on the curve.
\pr{regfree} completes the proof. \qed
The same proof will extend to curves in higher dimensional spaces, or
more general submanifolds, provided we have a uniqueness theorem
for the solutions to the differential equation defined by the singular
subset $\Sn$ for some $n$. This would follow if we knew the differential system
$\Sn\subset \Jn$ was of finite type for $n$ sufficiently large; see \rf{GJ}.
This seems a reasonable conjecture, although I do not know
any results along these lines. The following example shows
that there are some subtleties involved in trying to formulate a general theorem.
\Ex{s2} Consider the four parameter group
$$(x,u,v) \longmapsto (\lambda x + a, \lambda ^3 u + \mu x + b, \lambda ^2 v),$$
acting on
$M=\R^3$ with infinitesimal generators
$$\qeq{\v_1 = \partial _x,\\\v_2 = \partial _u,\\\v_3 = x\partial _u,\\
\v_4 = x\partial _x + 3u\partial _u + 2v\partial _v.} \Eq{s2}$$
We consider curves in $M$, viewing $x$ as
the independent variable and $u,v$ as dependent variables.
The second order Lie matrix is
$$\pmatrix{
1&0&0&0&0&0&0\cr
0&1&0&0&0&0&0\cr
0&x&0&1&0&0&0\cr
x&3u&2v&2u_x &v_x& u_{xx} &0\cr}.$$
Therefore, the first two singular subsets are
$$\qeq{\S1 = \{v = v_x = 0\}\subset \J1,\\ \S2 = \{v = v_x = u_{xx} = 0\}
\subset \J2.}$$
The differential system $\S1$ is underdetermined, and so solutions to its initial value problem
are not unique.
However, $\S2$ is a normal system of ordinary differential equations,
and hence, applying the proof of \th{Ldet}, we
conclude that the totally singular curves are the solutions to
the system $u_{xx} = 0$, $v=0$, which implies that $u = ax + b$, $v=0$.
Interestingly, since $\dim \J1 > \dim G$,
the first order Lie matrix already has more
columns than rows, and one might have guessed, in analogy with
the scalar case, that it would govern the totally singular curves.
This example indicates that the finite type conjecture is a bit more
subtle than might be initially expected.
\Section h Homogeneous Spaces.
In the case $G$ acts transitively on $M$, then we can locally\fnote{Recall that $G$ might only
be a local transformation group.} identify
$M$ with the homogeneous space $M \simeq G/H$. Here $H = G_{z_0}$
is the isotropy subgroup of the base point $z_0 = \pi(e)$, which is the
image of the identity group element under the natural projection
$\pi \colon G\to G/H$. In this final section, we conduct a more detailed investigation
into the totally singular submanifolds $S\subset G/H$. We shall relate
them to the existence of certain Lie subgroups of the transformation group $G$.
Note that we can always move any submanifold $S\subset M$ by a group transformation
without affecting its intrinsic geometric properties.
By transitivity, then, we can assume that, without loss of generality, the
submanifold passes through the base point $z_0 = \pi (e)$.
We first analyze the easier case when the totally singular submanifold
has an isotropy subgroup that acts transitively on it.
\Df{transitive} A submanifold $S\subset M$ is \is{transitively totally singular}
if its isotropy subgroup $\GS $ acts transitively and non-freely on $S$.
\Remark In the formulation of this definition, we are
excluding those smooth submanifolds
which contain a totally singular point, but nevertheless have freely acting isotropy
subgroups.
\th{asing} shows that such pathology does not occur in the analytic category.
If $K=\GS $ is the isotropy subgroup of a submanifold $S\subset G/H$ passing through $z_0 =
\pi (e)$, and $K$ acts transitively on $S$, then we can identify $S = \pi (K)$.
Conversely, if $S=\pi (K)$ for some subgroup $K$, then $K$ acts transitively on $S$, and hence
is a subgroup of the full isotropy subgroup of $S$. We require a general characterization of the
full isotropy subgroup, which relies on the following purely group-theoretic constructions.
Recall that a subgroup $K\subset G$ is called
\is{self-normalizing} provided $g^{-1}\cdot K\cdot g= K$ if and only
if $g\in K$. The Cartan subgroups of a semisimple Lie group are particularly important
examples \rf{Vara}. The characterization of transitively totally singular submanifolds relies
on a generalization of the notion of a self-normalizing subgroup.
\Df{Hsn} Let $H\subset G$ be a subgroup of a group $G$.
The $H$-normalizer of a subgroup $K\subset G$, denoted $\NHK$, consists of all
group elements $g\in \KH$ such that
$$\qeq{g^{-1}\cdot K\cdot g \subset K\cdot H, \andq g\cdot K\cdot g^{-1} \subset K\cdot
H.} \Eq{Hnormalizer}$$
The subgroup $K$ is called \is{self
$H$-normalizing} if it equals its $H$-normalizer: $K = \NHK$.
Note that we are not assuming that $K\cdot H$ is a subgroup here, which would require
$K\cdot H = H \cdot K$. Unfortunately, I have been unable to find a purely Lie
algebraic characterization of the $H$-normalizer of a Lie subgroup.
\ignore{On the
infinitesimal level, the
\is{$\h$-normalizer} of a subalgebra $\k\subset \g$ consists of all $\v\in \k + \h$ such
that $[\v,\w] \in \k+\h$ for all $\w\in \k$. The subalgebra $\k$ is
\is{self $\h$-normalizing} if the only $\v$ for which this holds are those
already in $\k$.}
\Lm{hsn} Consider the homogeneous space
$\pi \colon G\to G/H$. Let $K\subset G$ be a subgroup, and
let $S = \pi (K)\subset G/H$. Then the isotropy subgroup of
$S$ equals the $H$-normalizer of $K$.
\Proof
A group element $g\in G$ is a symmetry of $S = \pi (K)$ if and only if for every $k\in K$,
we have
$$\eeq{g\cdot k = \ktilde \cdot \htilde \\g^{-1} \cdot k = \khat \cdot \hhat }
\roq{for some} \seq{\ktilde\in K, \quad \htilde \in H,\\
\khat \in K, \quad \hhat \in H.} \Eq{GSeq}$$
%$$\eeq{g\cdot k = \ktilde \cdot \htilde \roq{for some} \ktilde\in K, \quad \htilde \in H,\\
%g^{-1} \cdot k = \khat \cdot \hhat \roq{for some} \khat \in K, \quad \hhat \in H.} \Eq{GSeq}$$
(The proof does rely on both conditions.) If we set $k=e$ in \eq{GSeq}, we find
$$g = k_0\cdot h_0 = h_1\cdot k_1 \roq{for some} k_0,k_1\in K, \quad h_0,h_1\in H,$$
so that $g\in\KH$. Therefore, given $k\in K$,
$$\seq{g\cdot k\cdot g^{-1} = g\cdot (k\cdot k_1^{-1} )\cdot h_1^{-1}
= \kbar\cdot \hbar\cdot h_1^{-1} \in K\cdot H,\\
g^{-1} \cdot k\cdot g = g^{-1} \cdot (k\cdot k_0)\cdot h_0
= k\Upstar \cdot h\Upstar \cdot h_0 \in K\cdot H,}$$
for certain $\kbar,k\Upstar \in K$ and $\hbar,h\Upstar \in H$. In the second
equalities, we use \eq{GSeq} with $k$ replaced by $k\cdot k_1^{-1}, k\cdot k_0$, respectively.
This proves that the isotropy subgroup $\GS \subset \NHK$ is contained in the $H$-normalizer of
$K$.
Conversely, if $g= k_0\cdot h_0 = k_1\cdot h_1\in \KH$ satisfies the
$H$-normalizer equations \eq{Hnormalizer}, then for any $k\in K$ we have
$$\seq{g\cdot k = g\cdot (k\cdot k_1)\cdot k_1^{-1}
= [g\cdot (k\cdot k_1)\cdot g^{-1} ]\cdot g\cdot k_1^{-1}
= \kbar\cdot \hbar\cdot h_0^{-1},\\
g^{-1} \cdot k = g^{-1} \cdot (k\cdot k_0^{-1} )\cdot k_0
= [g^{-1} \cdot (k\cdot k_0^{-1} )\cdot g]\cdot g^{-1} \cdot k_0
= k\Upstar \cdot h\Upstar \cdot h_0^{-1} ,}$$
and hence $g$ satisfies \eq{GSeq},
which implies that $g\in \GS $. \qed
\Ex{affH} Consider the action of the planar affine group $\AG2$ discussed in \ex{A2},
which we identify with the homogeneous space $\pi \colon \AG2 \to \AG2/\GL2 \simeq \R^2$
corresponding to the linear subgroup $\GL2\subset \AG2$. Consider the two-dimensional
subgroup $K$ given by the horizontal transformations $(x,u)\mapsto (\lambda x+\mu ,u)$. Let
us compute its $\GL2$-normalizer using the symmetry criterion \eq{GSeq}. We
identify affine group elements $(A,b)\in \AG2$ with $3\times 3$ matrices $\matrixb Ab01\in
\GL3$ in the usual manner. The subgroup
$K$ consists of all matrices of the form $\matrixc \lambda 0\mu 010001$. Then $g
=\matrixc\alpha \beta a\gamma
\delta 0001 \in K\cdot \,\GL2$ belongs to $\No{\GL2}K$ provided
$$\matrixc \alpha \beta a\gamma \delta 0001 \cdot \matrixc \lambda 0\mu 010001
= \matrixc {\alpha \lambda }\beta {\alpha \mu +a}{\gamma \lambda }\delta {\gamma \mu }001
\in K\cdot \GL2,$$
which requires that its $(2,3)$ entry vanishes, and hence $\gamma =0$. Therefore,
the $\GL2$-normalizer of $K$ is the subgroup $(x,u)\mapsto (\alpha x+\beta u + a,\delta u)$,
which forms the full four-dimensional isotropy subgroup of the $x$ axis which is the image
$S = \pi (K)$.
\pari
On the other hand, the subgroup $\Ktilde$ given in \eq{parabolasub} is self $H$-normalizing.
Using the matrix version
$\ktilde = \matrixc \lambda 0\mu {2\lambda \mu}{\lambda ^2}{\mu ^2}001\in \Ktilde$ for its
elements, we see that
$g = \matrixc \alpha \beta a\gamma
\delta {a^2}001 \in \Ktilde\cdot \GL2$ belongs to $\No{\GL2}\Ktilde$ provided the
product $g\cdot \ktilde$ belongs to $ \Ktilde\cdot \GL2$, which requires that its $(2,3)$ entry
equals the square of its $(1,3)$ entry:
$$(\alpha \mu +\beta \mu ^2 + a)^2 = \gamma \mu + \delta \mu ^2 + a^2.$$
Clearly this holds if and only if $\beta =0$, $\delta =\alpha ^2$, $\gamma =2\alpha a$, and hence
$g\in \Ktilde$ already. This reconfirms that the isotropy subgroup of the parabola
$\pi (\Ktilde) = \{u = x^2\}$ equals $\Ktilde=G_P$.
\Th{tts} There is a one-to-one correspondence\fnote{We identify two
submanifolds passing through a point if they are the same in a neighborhood
thereof, \ie they have the same germ.} between transitively
totally singular $p$-dimensional
submanifolds $S\subset G/H$ passing through $z_0 = \pi (e)$
and connected $k$-dim\-en\-sional self $H$-normalizing Lie subgroups $K\subset G$ such that
the intersection $K\cap H$ has dimension $t = k - p \geq 1$.
\Remark In view of \th{Ldet}, these constructions provide a
purely algebraic approach to the solution of the Lie determinant differential equation.
Vice-versa, the Lie determinant equation provides a differential equation approach to
the algebraic problem of determining the $H$-normalizers of subgroups $K\subset G$.
\Proof
Given such a subgroup $K$, let $S = \pi (K)$.
It is not hard to prove that $S$ is, in a neighborhood of $z_0$, a smooth submanifold
of the correct dimension; see also \lm{project} below. Moreover, $K$ is clearly contained in
the isotropy subgroup of $S$, and acts non-freely since it
has a nontrivial intersection with $H$. According to \lm{hsn}, $K$ is the
entire isotropy subgroup of $S$. Conversely, given a transitively totally singular submanifold $S$,
passing through $z_0$, we let $K=\GS$ denote its
isotropy subgroup. Transitivity implies that $S = \pi (K)$.
Moreover, any $k\in K\cap H$ satisfies $k\cdot z_0 = z_0$,
and hence $K$ will act freely on $S$ at $z_0$ if and only if $K\cap H = \sete$.
\qed
The intransitive case is a little harder. We begin with a lemma
that characterizes those submanifolds of $G$ which project
to smooth submanifolds of the homogeneous space $G/H$.
Let $L_g(h) = g\cdot h$ and $R_g(h) = h\cdot g$ denote
the right and left multiplication maps on $G$.
We let $\h \subset \g$ denote the Lie subalgebra for the subgroup $H$.
%Recall that the (right) Lie algebra $\g$ consists of all right-invariant
%vector fields on $G$, and hence $\g \at g \simeq TG\at g$
%can be identified with the tangent space at each group element $g\in G$.
\Lm{project} If $N\subset G$ is a smooth $n$-dimensional submanifold, then
$S = \pi (N)$ is, locally, a smooth $p$-dimensional submanifold of
$M = G/H$ if and only if
$$\dim \Bigl [ \>\h \at e\cap (dL_g)^{-1} \bigl( TN \at g \bigl ) \>\Bigr ] = n - p,
\forall g\in N. \Eq{dimpr}$$
In other words, near $e$, a submanifold $N\subset G$ will project to a submanifold
$S\subset G/H$ provided
the subspace $\h \at e\cap (dL_g)^{-1} \bigl( TN \at g \bigl ) $ has constant dimension. Note
that, globally, the projected submanifold $S = \pi(N)$ may have self-intersections.
\Proof
By the Implicit Function Theorem, we need only to prove that $(\ker d\pi) \cap TN$
has constant dimension $n-p$ over $N$. \ignore{Given a Lie algebra element $\v\in \g$,
let $\v_R$ denote the corresponding right-invariant vector field on $G$,
and let $\widetilde \v$ denote the associated infinitesimal generator
for the action of $G$ on $G/H$.} Recall that the isotropy subgroup
of a point $z = \pi (g) \in M$ is $G_z = g\cdot H \cdot g^{-1} $, with Lie algebra
$\g_z = \Ad g\,(\h)$, where $\Ad$ denote the adjoint representation
of $G$ on its Lie algebra \rf{Vara}. Since we are identifying Lie algebra elements with
right-invariant vector fields on $G$, we have
$$\zeq{\ker d\pi \at g
%=\set{\v\at g}{\v\in \Ad g\,(\h)}\\
= \Ad g\,(\h) \at g \\= dL_g\comp (dR_g)^{-1} \,\h \at g\\
= dL_g(\h \at e).}$$
In other words, the kernel of $d\pi $ equals the subspace of
$TG\at g$ spanned by the \is{left}-invariant vector fields
corresponding to elements of the subalgebra $\h$. Therefore,
$$(\ker d\pi \at g) \cap TN \at g = dL_g(\h \at e) \cap TN \at g =
dL_g\Bigl[\>\h \at e\cap (dL_g)^{-1} \bigl( TN \at g \bigr )\,\Bigr]$$
will have the correct dimension if and only if \eq{dimpr} holds.
\qed
\Remark If $N = K$ is a Lie subgroup with Lie algebra $\k$, then $$(dL_g)^{-1} ( TK \at g
) = \k \at e,$$ and hence condition \eq{dimpr} is automatic. This justifies our earlier
claim during the proof of \th{tts}.
Now, suppose $S = \pi (N)$ is a totally singular submanifold passing through $z_0 = \pi (e)$.
Let $K = \GS $ be the intransitively acting isotropy subgroup of
$S$. In a neighborhood of $e\in G$, we can identify $N = K\cdot A\subset G$, where
$e\in A\subset G$ is a submanifold that is transverse to
$K\cdot H$, meaning that
$$TA \at e \cap \bigl(\,\k \at e+ \h\at e\,\bigr) = \szero.\Eq{Atr}$$
Now, for $g = k\cdot a\in N$ near $e$,
$$TN \at g = \k \at g + dL_k(TA\at a).$$
Since$$(dL_g)^{-1} (\k \at g)
= \Ad g^{-1} (\k\at e)
= \Ad a^{-1} (\k\at e),$$
we have
$$\eeq{(dL_g)^{-1} (TN \at g ) = (dL_g)^{-1} (\k \at g) + (dL_a)^{-1} (TA\at a)\\
= \Ad a^{-1} (\k\at e) + (dL_a)^{-1} (TA\at a).}$$
Now, by continuity, \eq{Atr} implies that $(dL_a)^{-1} (TA\at a)$ remains
transverse to the subspace $\k\at e+\h\at e\subset TG\at e$ for $a$
sufficiently close to the identity.
Therefore
$$\dim \Bigl[ \>(dL_g)^{-1} (TN \at g) \cap \h\at e\>\Bigr] =
\dim \Bigl[ \>\Ad a^{-1} (\k) \cap \h\>\Bigr], \where g = k\cdot a.$$
Consequently, \lm{project} implies that $S = \pi (N)$ will be a smooth submanifold near $z_0$ if
and only if
$$\dim \Bigl[ \>\Ad a^{-1} (\k) \cap \h\>\Bigr] = \dim \k \cap \h, \Eq{khdim}$$
for $a$ near $e$. Continuous dependence of
$\Ad a(\v)$ on $a$ for fixed $\v\in \k\cap \h$ implies that \eq{khdim} will hold if and only if
$A\subset \NHKH$, where
$$\NHKH = \set{g\in G}{g^{-1} k g \in H\roh{ for all }k\in K\cap H} \Eq{NKH}$$
is the $H$-normalizer of the subgroup $K\cap H$.
Finally, the group elements in $K = \GS $ that fix $z_0$ are those
belonging to $K\cap H$, and hence the requirement that the isotropy
subgroup does not act locally freely necessitates that the dimension \eq{khdim} be at least
$1$. We have therefore proved the following characterization of
general totally singular submanifolds.
\Th{totally} Every submanifold $S\subset G/H$ which is totally singular at
a point $z = \pi(g) \in S$
has the form $S = g\cdot S_0$, where $z = g\cdot z_0$, and
we can locally identify $S_0 = \pi (K\cdot A)$, where
$K\subset G$ is a $k$-dimensional subgroup of $G$, with
$t = \dim K\cap H \geq 1$, and $A\subset \NHKH$
is an $l$-dimensional submanifold. The dimension of
$S$ is $p = k+l-t$, and $K\subset \GS $ is a non-freely acting subgroup of its isotropy group.
\Ex{scale} Consider the group $G = \R^+ \semidirect \R^2$
with multiplication rule
$$(\lambda,a,b) \cdot (\mu,c,d) = (\lambda \mu, a + \lambda c, b + \lambda^2 c).$$
The group acts on $M = \R^2 = G/H$ according to
$(x,u) \mapsto (\lambda x + a, \lambda^2 u + b)$; the isotropy
subgroup is $H = G_0 = \{(\lambda,0,0)\}$.
The straight line $L = \{u = 0\}$ is totally singular. We can
realize $L = \pi(K)$, where the isotropy subgroup $\iso L = K = \{(\lambda, a,0)\}$
acts transitively on $L$ and has one-dimensional intersection with $H$,
confirming \th{tts} in this case. Similarly, the
parabola $P = \{u = x^2\}$ has the scaling subgroup $H$ as its
isotropy subgroup, $\iso P = H$, which is not transitive,
and only the origin is a totally singular point. Indeed,
$P = \pi(H\cdot A)$, where the curve $A = \{(1,a,a^2)\} \subset G$
is transverse to $H$, reconfirming \th{totally}.
%\vglue .5in
\Ack It is a pleasure to thank Mark Fels, Niky Kamran, Juha Pohjanpelto, and
Michael Singer for encouragement, critical comments, references, and
suggestions. I would also like to acknowledge the support of the Mathematical Sciences
Research Institute, in Berkeley, where the final parts of this research were completed.
%\page
\vglue .5in
\References
\ends