# On uniformity for the polar to partially hypoelliptic operators

Tove Dahn

##  0.1 Terminology

###  0.1.1 Introduction

The main problem in this article, is to discuss the polar to hypoellipticy and
to partial hypoellipticity relative $G_{8}$. Assume $Uf=\int fdU$, where
$dU\in G$ and f is a symbol to a (pseudo-) differential operator. A necessary
condition for hypoellipticity is absence of spirals in the polar set, for
instance $U^{\bot}/U\rightarrow 0$ in $\infty$. Consider, for $dV\in G_{8}$
([20]), $K(G_{4})=(dV(x.y),dV(y,x))$, we denote $dV(y,x)=dV^{L}(x,y)$, .
Assume $dW=xf_{x}+yf_{y}$, (that is $dW=dW^{L}$) then $G_{4}$ is given by
$f_{x},xf_{x},yf_{x},xdW$ . Consider $G_{4}\rightarrow
K(G_{4})=(G_{4},G^{L}_{4})$. With respect to conjugation $K(G)\rightarrow
K^{\bot}(G)$, a necessary condition for hypoellipticity, is
$K^{\bot}/K\rightarrow 0$ or $K/K^{\bot}\rightarrow 0$.

Assume $S_{j}(v)=(v,v^{L})$, $S_{j}^{\bot}(v)=(v,v^{\bot})$, relative
different scalar products. Thus, when $S_{j}^{\bot}:A\rightarrow B$ and
$S_{j}:C\rightarrow A$, with $\pi:C\rightarrow B$, it is necessary that we
have existence of $(S_{j}^{\bot})^{-1}$ relative $S_{j}$. Assume $\int
dS_{j}(v)=\int\rho dU_{1}$, where $dU_{1}$ denotes translation, we note that
presence of lineality (sets of translation invariance) contradicts
hypoellipticity. $S_{j}^{\bot}$ is considered as a closed extension of
$S_{j}$. We have that $S_{j}^{\bot}=S_{j}$ for large parameters, contradicts
hypoellipticity. Given a regular analytic continuation, we have that
$dS_{j}^{\bot}\rightarrow dS_{j}\rightarrow dV$ is continuous.

Consider $\Psi(t):U(x,y)\rightarrow{}^{t}U(x,y)$, so that $d{}^{t}U=\rho dU$.
Assume $L_{1}$ a scalar product to $\Psi$ and $L_{2}$ a scalar product to the
orthogonal, assume $L_{1}\rightarrow L_{2}$ algebraic. A closed set in the
plane, according to $(U,U^{\bot})\rightarrow(I,O)$ (cf. P-convexity), is
through Tarski-Seidenberg mapped onto a semi-algebraic set in
$(V,V^{L})f(x,y)=f(v,v^{L})$. Given $dI\notin G_{2}$, we do not have a closed
(relatively closed) set in $(v,v^{L})$, in particular we do not have density.
Note that given a strict condition, for instance $U^{\bot}$ algebraic, we have
that $H(G_{2})$ generates $L^{1}(G_{2})$ (cf. [13]). Thus, the problem is
completely dependent of the topology, that is if $\mathcal{H}(G_{2})$ is
generated by $H(K(G_{4}))$, then where we have symplecticity,
$\mathcal{H}(G_{2})\rightarrow H(K(G_{4}))$ is projective. We will us the
notation $H(G)$ or $G_{H}$ dependent on the context.

Assume $L^{1}\subset\mathcal{H}$. Given a set F is given by
$\parallel\cdot\parallel_{1}=0$, it can be considered as planar, that is
generated by $G_{2}$. Example: assume $p$ a semi-norm, with $p(f)\leq
C\parallel f\parallel_{1}$, that is we assume $f\in L^{1}$ but
$Vf\in\mathcal{H}\backslash L^{1}$, thus a neighborhood of $Vf-f$ is in
$\mathcal{H}\backslash L^{1}$. Example: $f(v,v^{\bot})\in H$ does not imply
$f(v)\in H$. The polar set is given by $V^{\bot}f=0$, possibly non-removable.

###  0.1.2 The extended plane

Assume $d\varphi=dU$ non-decreasing and $f\in C^{0}$, we then have existence
of $\int f(x)d\varphi(x)=\int g(u)du$, where g is undetermined over $u=const$,
but does not affect action of U ([19]).

###### The moment problem 0.1

Assume $M=H(K(G_{4}))$ and $g\in C(G_{2})$. Assume, for a positive linear
functional $B_{g}$, $B_{g}(g)=1$, for g fixed outside of $M$, we then have
$B_{g}(f)=0$ for $f\in M$ implies $B_{g}(f)=0$ for $f\in C(G_{2})$. Thus M
approximates $C(G_{2})$ uniformly.

([12]) Assume $dV\in K(G_{4})$ and $f\in H(K(G_{4}))$, we then have
$f(v)\rightarrow v$ continuous. Thus, given $g\in C$, we have that
$f(v)\rightarrow v\rightarrow g(v)$ is continuous. When $K(G_{4})$ is dense in
X, we have that $g\circ v\in C$ iff $g\in C$. In particular
$(K(G_{4}))_{L^{1}}\simeq(G_{2})_{L^{1}}$ according to Wiener. Consider
$(K(G_{4}))_{H}\in dV\rightarrow\rho dV=dU$ a continuous continuation, with
$\rho\in H$. Assume we have $\sigma dV=dI$, with $\sigma\in H$. Consider the
absolute continuous-convex closure $\mid dU(f)\mid\leq M\parallel
dV(f)\parallel$ where $dV\in(K(G_{4}))_{H}$ is allowed to vary. Assume
$g.l.b.\parallel dV(f)-g\parallel=d_{1}$, where $g\in C(G_{2})$, Then
$H(K(G_{4}))$ approximates $C(G_{2})$ uniformly iff $M\geq 1/d_{1}$. More
precisely, assume $dU/\rho=\sigma dI$, where $\mid\sigma\mid\leq d$, we then
have given $dU\geq dI$, $1/M\leq 1/\mid\rho\mid\leq\mid\sigma\mid\leq d$, that
the inequality follows. The case $0\leq dU\leq dI$ is obvious. Note that,
given $\rho$ conformal, we have existence of $1/\rho\in C$. Given $\rho\sigma$
reduced, we can represent $1/\sigma\rho$ in H, why $H(G_{2})$ approximates
$C(G_{2})$.

Consider $(K(G_{4}))_{H}\subseteq(G_{2})_{C}\in dI$, that is assume
$f\in\mathcal{H}$ with continuation $\tilde{f}\in C$, with the corresponding
$g(v)\in H$ and $dV\in K(G_{4})$, that is $g(v)\rightarrow v$ continuous.
Assume $H(G_{2})^{\bot}\simeq C(G_{2})$, in particular we assume the polar set
has correspondingly $dV=0$ with $dV\in K(G_{4})$. Absence of non-trivial polar
set for f, is motivated by density for $G_{2}$ regular. Example: $f\in
H(K(G_{4}))$, we then have $\int_{K(G_{4})}fdU=\int_{\Omega_{8}}f(u)du=0$,
when f is locally algebraic, $m\Omega_{8}=0$, (f is assumed complete).

Reflection through $L_{j}$ defines uniformities according to K(d V) and
$K^{\bot}(dV)$. Thus we have two diagonals $dK(V)=dK^{L}(V)$ and
$dK(V)=dK^{\bot}(V)$. Given convexity for translates, we have that $G_{2}$ is
decomposable to $K(G_{4})$. Given $dW=\rho(x,y)dI(x,y)\in
K(G_{4})_{\mathcal{H}}$, where $\rho$ is topologically decomposable to
$\rho=\vartheta(x)\otimes\vartheta(y)$, we have decomposability to
$(G_{8})_{\mathcal{H}}$. Assume $\Delta=\\{dK(V)=dK^{\bot}(V)\\}$ and
$\Sigma=\\{dK(V)=dK^{L}(V)\\}$, that is $\Delta\rightarrow\Sigma$ continuous,
maps spirals on to symmetric sets. Obviously $K^{L}(dV)=K(dV^{L})$. Let
$\Psi(dU)=(dU,d{}^{t}U)$, that can be assumed harmonic. Note $\Delta U=0$
implies $dU\rightarrow d{}^{t}U$ preserves type. Given $\Delta F=0$ on a
domain, there is a corresponding G symmetric on the same domain. Example:
$\phi(u,v)=c_{1}u+c_{2}v$ has $\Delta\phi=0$. Starting from two scalar
products, $L_{j}(f)=<g_{j},f>$, we have $L_{2}(L_{1}(f))=0$, for all
$f\in(I)$, means either $L_{1}(f)=0$ or $L_{1}(f)\bot g_{2}$. $g_{j}$ defines
measures regular over f. Further, $L_{1}(L_{2}(f))=0$, for all $f\in(I)$,
gives that $g_{1}\bot g_{2}$ in measure space. That is a symmetric
neighborhood of $\Sigma$ $=\\{L_{2}=0\\}$ can define a polar set for
transmission. In the same manner (geometric convexity), given
$L_{1}(L_{2})=0$, a symmetric neighborhood of $\Delta$, can define a polar set
for transmission.

###  0.1.3 A transmissions property

The transmission property (TP) that we discuss, is
$(dV,dV^{L})\rightarrow(dN,dN^{\bot})\leftrightarrow(dU,dU^{\bot})$, where the
middle element gives a rectifiable regular boundary, according to
$(dN,dN^{\bot})\rightarrow(dI,0)$. A special case of TP is
$T(\gamma_{1}\rightarrow\gamma_{2})(\zeta)\rightarrow\zeta$, that is the last
element corresponds to a transformation in the $\zeta-$ domain. P-convexity
for the group preserves hypoellipticity ([9]). Given the reversed last
implication preserves topology, this transmissions property gives in
particular an approximation property. Since our representations are based on
the Stieltjes transform, instead of $C^{\infty}_{0}(\overline{X})\rightarrow
C^{\infty}(\overline{X})$ continuous ([10]), we usually consider
$H(\overline{X})\rightarrow C^{0}(\overline{X})$ continuous. Consider
$T_{proj}(dU)=\\{\rho dU+\vartheta dU^{\bot}\\}$. It is not obvious that
$\rho\rightarrow 1/\rho$ preserves H, however if $\\{\rho=0\\}$ is of measure
zero, it is mapped on a set of measure zero, given the conditions above. Note
in the case $H\in\rho\rightarrow 1/\rho\in C$, we have in $L^{1}$, that the
order for the group is two, for the reversed last implication in TP. In order
to avoid a spiral axis in the middle element, it is sufficient to consider
$dN/dU^{\bot}=0$ at the boundary.

Rectifiable boundary: ([12]) Given $\Gamma$ with coefficients u, v, w, we have
$\Gamma$ rectifiable iff u,v,w BV (of bounded variation). For discontinuous
curves with BV, we may have absence of two-sided limits ([14]), for instance
$\lim dv\neq\lim{}^{t}dv$. Example: assume
$I^{L}:(\xi,\eta)\rightarrow(\eta,\xi)$,
$\overline{I}:(\xi,\eta)\rightarrow(\xi,-\eta)$, the harmonic conjugate
satisfies $I^{\diamondsuit}\simeq\overline{I}^{L}$. Thus given
$dI^{L}U,d\overline{I}U$ correspond to closed forms and $dI^{L}U=\rho dI$, we
have $\rho^{\prime}$ can be chosen as analytic. In particular, given $dI\notin
G$, we may have $dU\in G$, but $d\overline{I}U\notin G$.

Consider $T(\phi_{1}\rightarrow\phi_{2})$, where $\phi_{2}$ is analytic over a
domain $\Omega$, so that $U\phi_{1}(x)=\phi_{1}(u)=\phi_{2}(x)$. Thus we
assume existence of $dU\in G_{2}$, so that $U\phi_{1}\in H$, that is
$\phi_{2}$ can be reached from every element $\phi_{1}\in L^{1}$ through
planar movements (outside the polar set). Consider
$dU=dV+dV^{L}\rightarrow(x,y,z)=\tilde{x}$. Assume
$(\tilde{x})\rightarrow(u_{1},u_{2})$ surjective. Thus $f(u_{1},u_{2})=0$
implies existence of $\tilde{x}$, such that $f(\tilde{x})=0$.

###### Density in the plane 0.2

When $dU=(dV,dV^{L})$ is dense in $d\tilde{x}$-space, that is $dU^{\bot}$ have
trivial zero sets, we can reduce the two-mirror model to one scalar product.

Assume $L_{j}(f)=<f,dU_{j}>$, with $dU_{j}\in G_{\mathcal{H}}$. In particular
$L_{1}(f)\leq\Phi(f)\leq L_{2}(f)$ and given $L_{1}\rightarrow L_{2}$
projective, the same holds for $L_{1}\rightarrow\Phi\rightarrow L_{2}$. Assume
$dU^{\bot}=\rho dU$ and $\mathcal{H}\in f\rightarrow{}^{t}\rho
f\in\mathcal{H}$ continuous, then we have where $\rho\neq konst$, that $\Phi$
can be reduced to $L_{1}$, that is $\Phi(f)$ can be approximated by $\int
fd\Phi$, when $\rho\rightarrow 1$ regularly, even when $d\Phi\notin G$.

Note that UI=IU does not imply $dI\in G$, even when $dU\in G$. Assume $GW$ has
TP through completion. Example: $dI\notin G$, $GW\in dI$, $UW=I$, that is
$U^{-1}=W$, with $dW\notin G$ and given $W\rightarrow I$ regular, we have
completion to an approximation property, that is $UWI=IUW$.

Density is motivated by $f\in B_{pp}(K(v))$ ([21]). Thus, over the extended
domain, $\\{f(K(v),K^{\bot}(v))<\lambda\\}\subset\subset X$ implies
$\\{f(u,u^{\bot})<\lambda\\}\subset\subset X$. Given existence of an entire
line L in $bd\Omega$ according to $f(K(v))=f(\tilde{x})$, we do not have
hypoellipticity. P-convexity implies $K(dV)\neq dI$ on any line L.

###  0.1.4 Reduction to the boundary

Obviously, $\mid f\mid=const$ does not imply $f=const$. Assume an
approximation property, $\mid\phi\mid=const$ has a sub-sequence, such that
$\phi\rightarrow const$. This assumes $U\phi\rightarrow\delta_{0}(\phi)$
implies existence of V, such that $\phi(V\tilde{x})\rightarrow\phi(0)$, that
is $U\rightarrow I$ implies $V\rightarrow 0$.

A strict condition corresponds to
$S:L^{1}(bd\Omega)\rightarrow\mathcal{O}(bd\Omega)$, (cf. [11]) through an
infinitesimal movement, for instance S can be considered as a monotropic
operator. Assume $S(P)=S(Q)$ with P,Q polynomial and Q reduced at the
boundary, then we have obviously (approximative) solvability, that is
$S(P/Q)=I$. Thus $f(u,v)$ corresponds to $P/QdI(f)$ in $C^{0}$, that is by
extending the domain, we can have solvability with respect to $P/Q$, without
having it for P. Assume $bd\Omega=\\{U=I\\}$. Given $bd\Omega$ is a subset of
the domain for G, we have $dI\in G$.

###### Non-nuclear completion 0.3

Assume $Uf-S_{1}(f)=f$, where $S_{1}$ is not nuclear, for instance a
transformation that does not have a corresponding measure $dV\in G_{H}$, that
is $S_{1}$ defines a discontinuous representation. Example: $S_{1}(f)-f\in
C^{\infty}$ defines a non-regular boundary over $S_{1}(f)=0$.

Example: the polar set is given by constant surfaces to f. Example: the polar
set is given by $(U-I)f^{N}=0$ implies $U^{\bot}f=0$. Assume
$rad(I)\simeq\overline{(I)}$, that is $(U-I)f^{N}\simeq(U-I)(f+c)=(U-I)f+c=0$
iff $(U-I)f=const$.

Assume $dU=\rho dV$, with $dV=0$ on L and $\rho\neq 0$ on L. Note that given
$\rho$ linear in V, $dU=dV^{2}$ and given d V a projective operator, $dU\simeq
dV$. Given all L are mapped on boundary cycles, we have that projective
measures preserve the boundary.

Assume $\Omega$ semi-algebraic and $\Omega\rightarrow bd\Omega$ continuous.
Example: f regular and $f\neq 0$, $0=\int f(u)du=\int f(x,y)dU(x,y)$ with
$(x,y)\in L$, for instance $u\sim 0$ (homotopic with 0), that is we consider
$If=f(0)$. Given L is defined by $U=V$, we have $f(u,v)-I$ is regular. When
$U$ is a spiral, the mapping $\Omega\rightarrow bd\Omega$ is not continuous.

Nuclear topology for $\mathcal{H}$ and $dU\in G_{\mathcal{H}}$, implies
$UI=IU$, in particular $U-I$ is algebraic, that is $(U-I)I=I(U-I)$ and given
$d(U-I)=PdI$, for P polynomial, through Hurwitz, L must be trivial.

Assume $dV(f)=f_{x}$ and $dV^{L}(f)=f_{y}$, given $f$ symmetric in x,y, we
have $f_{x}=0$ iff $f_{y}=0$, that is $f(v,v^{L})$ symmetric. Note
$d(V,V^{L})\in K(G_{4})$ iff $d(V,{V^{L}})^{\diamondsuit}\in K(G_{4})$
(harmonic conjugation). Example: $(f_{x})^{\bot}=-yf_{x}$ and
$(f_{y})^{\bot}=xf_{y}$ that is $(f_{x}+f_{y})^{\bot}=-yf_{x}+xf_{y}$.

Assume the boundary is defined by a normal operator $(dN,dN^{\bot})$, where
$dN\rightarrow dI$ iff $d{}^{t}N\rightarrow dI$, that is we consider a
symmetric neighborhood of $dI=d{}^{t}I$. Further, $dN^{\bot}\rightarrow 0$
regularly at the boundary. Consider the continuation to $(dU,dU^{\bot})$, why
$d(UN)^{\bot}\rightarrow 0$ regular. A two-sided regular limit, assumes $\lim
UN=\lim U{}^{t}N$. Consider now $dV\times dV^{L}\rightarrow dN$, so that
$dV\rightarrow dI(x)$ iff $dV^{L}\rightarrow dI(y)$, which defines a symmetric
neighborhood in (x,y) and given that translates are dense in the domain
$dV+dV^{L}\rightarrow dI(x,y)$ ([14]).

###  0.1.5 TP relative 1-parameter group

Consider $\gamma(\zeta_{T})=\gamma_{T}(\zeta)$, as a topological completion.
Assume $dU\rightarrow dV\rightarrow dV_{1}=\rho_{1}dI=0$ regular, that is dU
has an analytic base through topological completion, where through translation
$\rho_{1}\neq 0$ locally. Concerning projective mappings in this case,
$dV_{1}(\gamma_{1})=0$ in $\mathcal{D}_{L^{1}}$, implies existence of
$dV_{2}$, so that $dV_{2}(\gamma_{2})=0$ in H, when $\gamma_{j}$ are entire
lines.

Assume $Wf_{0}-f_{0}+f_{0}-I\sim 0$, then $W$ is very regular, that is
$Wf_{0}-f_{0}\sim 0$, given $f_{0}$ very regular. Note that given strict
pseudo convexity, that is given v locally algebraic and $f(u,v)\rightarrow
f(0)$, we have a transversal approximation.

Assume $dU^{\bot}=\rho dU$, where $\rho\equiv const$ gives the diagonal. In
particular, $d^{2}U^{\bot}=\rho^{\prime}dU+\rho d^{2}U$, that is given
$\rho\neq 0$, we have that $dU^{\bot}\rightarrow dU$ preserves corresponding
closed forms (projective mapping) iff $\rho^{\prime}dU=0$, thus projective
mappings include spirals.

In analogy with the fundamental theorem in algebra: $f(u,v)=0$ modulo
$C^{\infty}$, means that given f algebraic with respect to (u,v), we have
convergence in $C^{\infty}$. Assume UV=VU defines $Z=\\{(u,v)\quad
f(u,v)=0\quad u\in G\quad\exists v\in G\\}$ a domain for convergence in
$C^{\infty}$. Polynomials approximate $C^{\infty}_{c}$ ([22], Cor. 2 Lemma
15.1), that is given $U,V$ pseudo local and given $f\in C^{\infty}_{c}$, we
have existence of $P(u,v)=0$ modulo $C^{\infty}$ over Z and $\mid
f-P\mid<\epsilon$.

###  0.1.6 A weaker strict condition

Consider $X^{\prime}\subset X$, for instance $X^{\prime}$ polynomial, that is
$v\in X^{\prime}$ does not imply $1/v\in X^{\prime}$. Given $v=y/x\rightarrow
0$ or $x/y\rightarrow 0$, we have $1/(v+1/v)\rightarrow 0$ in $\infty$. Thus
we can consider $X^{\prime}\times X\in(v,1/v)$ as one sided, that is given
density for translates, we can consider $v+1/v$ as approximately in
$X^{\prime}$.

Example: Assume $\Sigma=(\frac{U}{V}+\frac{V}{U})$, with $\Sigma d\Sigma=dI$
and $d\Sigma(x,y)=d\sigma$. We then have, given $u/v\rightarrow 0$ or
$v/u\rightarrow 0$, that $1/\sigma\rightarrow 0$ and $d\Sigma^{2}\sim_{0}dI$
(geometric equivalence). Given U,V projective operators, we thus have that
$(V^{-1}U+U^{-1}V)F(x,y,z)=F(x,y,z)$ implies $U+V=UV$. Given $dU\rightarrow
dV$ projective, we have weakly $U\rightarrow V$ projective. Given d U sub-
nuclear, we have for instance $dU\rightarrow dU^{2}$ projective. Thus,
$dU\rightarrow d(I-U)^{-1}$ projective.

Given $dV^{-1}U\simeq(p/q)dI$, we have $pq/(p^{2}+q^{2})\rightarrow 0$. The
condition p/q $\rightarrow 0$ or q/p $\rightarrow 0$, is sufficient for
$L_{p^{2}+q^{2}}\subset L_{pq}$.

###### Preservation of convexity for weak Lie type 0.4

Problem: consider d U, such that the type is preserved for the convex closure
of the domain, in particular in the case where $dI\notin G$.

Consider $T_{ac}(G_{H}))=\cup_{\mid\rho\mid\leq 1}\rho dU$, where $dU\in G$
and $f\rightarrow\rho f$ continuous. $C(G_{2})$ allows cluster-sets. Example:
$\mid f\mid=1$ allows infinitely many values for $U_{2}$. A complete limit is
dependent of $U_{2}$.

Define $\int d<dE,f>=\int dE(f)=E(f)$, that is we assume $<dE,f>$ in $L_{ac}$.
Given $d<dE,f>=\rho<dI,f>$, we have that where $\rho\neq konst$, we can not
conclude $dE=dI$. Further, constant surfaces give the type, but
$<dE,f>=<dI,f>$ does not imply $dE=dI$ in regular topology, for instance E can
have a non-trivial kernel.

Given the convex closure in $C(G_{2})$, it is not obvious that the type is
preserved. Example: $Uf\rightarrow f$ iff $f(u)\rightarrow 0$ or weaker
$f(u+v)\rightarrow 0$, with $u/v\rightarrow 0$ in $\infty$. Assume $dU=\rho
dV=dWV$, then in order for $U=I$ to imply $V=I$, we have to determine W, so
that $WV=VW$. Example: $\rho$ bounded and absolute continuous, we then have
$d\rho=0$ iff $d(1/\rho)=0$, that is $\rho=const$ iff $1/\rho=const$. In
particular $\rho dV\rightarrow dV$ is absolute continuous, further projective.

Assume $(U,V)$ a continuation of projective movements through conjugation.
Given U,V of the same type, for instance $d(U+V)=\rho dU$ and U+V=I, the
conjugation gives a maximal domain for invariance. Assume U,V of different
types with U+V=I. Assume $Uf=Vf$ implies f=0. Assume $\tilde{U}f$ continuation
with 0 and in the same manner for $\tilde{V}$, we then have
$\tilde{U}+\tilde{V}=I$, with $\tilde{U}f=\tilde{V}f$ implies $f=0$, that is
given U,V with one-sided support, U,V can be continued to a projective
conjugation.

Consider $dV(\rho f)\simeq{}^{t}\rho dV(f)$, given $\rho\rightarrow{}^{t}\rho$
preserves $T_{ac}$ \- convexity we have that dV is defined on the
$T_{ac}$-closure, for instance this is the case when $\rho$ polynomial.
However, $\rho\rightarrow 1/\rho$ does not necessarily preserve convexity, a
sufficient condition is $\rho$ absolute continuous. Note that $(I)\subset(J)$
does not necessarily imply $T_{ac}(I)\subset T_{ac}(J)$. A necessary condition
for inclusion, is that the corresponding coefficients have
$\rho_{J}/\rho_{I}\rightarrow 0$ in $\infty$

In the plane, we have that $W+W_{1}+R$ according to TP, preserves convexity
(modulo removable sets). Assume W algebraic, $W_{1}$ preserves one-sidedness,
for instance reflection and R negligible. Consider the $T_{ac}$-closure,
$T_{ac}(I)=$ $\\{\rho f\\}$ of $f\in(I)$. In the case $dI\notin G_{2}$, we
consider $dV\rightarrow dV\times dV^{L}\rightarrow G_{2}$ where $dV\in G_{8}$
is the result of the inverse scheme, that is $dU\rightarrow dV$ is two-valued.
Assume the condition one-sidedness in $(dU)$ is mapped onto one-sidedness in
(dV). Then TP can be continued to $G_{8}$.

Assume d (U+V) $\rightarrow dI$, with dV compact ($C^{\infty}$), meaning the
integral operator defined by dV is compact. Then, in the same manner, d U - d
I completely continuous corresponds to dU Fredholm.
$d(U-V)^{2}=dU^{2}+dV^{2}-2dUV$, with $dUV\rightarrow dI$, thus
$2UdU+2VdV-2dI\sim d(U-V)^{2}\rightarrow 0$, that is $UdU+VdV\rightarrow dI$.
In particular $(U-V)d(U-V)\rightarrow 0$, where the diagonal gives non-regular
approximations.

Example: assume $dU\rightarrow dW$ are conjugated, according to
$dU(f)=-\\{g,f\\}$ and $dW(f)=-\\{h,f\\}$. Assume $\Psi(g)=h$, given $\Psi$
preserves linearity and order, we have that the type is preserved. In
particular starting from $dU\rightarrow d{}^{t}U$, we have
$UT(\phi)=T(U\phi)+(-\Delta g)T(\phi)$, that is the type is preserved under
transponation, given $\Psi$ preserves harmonicity.

###  0.1.7 Inverse lifting principle

###### Vector valued distributions 0.5

Consider $dU\rightarrow dV$ conjugation and $\Phi(dU)\rightarrow(dU,dV)$ and
for a second conjugation $dU_{1}\rightarrow dU_{2}$,
$\Phi^{2}(dU)=(dU_{1},dU_{2})$. With this construction, $\Phi^{2}$ can be
algebraic, without $\Phi$ being algebraic. Assume in particular,
$\Phi_{1}:dW\rightarrow d{}^{t}W$, $\Phi_{3}:dW\rightarrow dW^{\bot}$ and
$\Phi_{2}:dW\rightarrow dW^{*}$ (algebraic dual). Using ([21])
$<<T,\phi>_{1},dU>_{2}=<<T,dU>_{2},\phi>_{1}$ we have
$<T,dW^{*}>_{2}(\phi)=<T(\phi),dW>_{2}$. Consider completion to
$d\tilde{W}^{*}$ closed, so that $d\tilde{W}^{*}\rightarrow dW^{\bot}$ linear
(single valued), that is $dW^{\bot}=\rho d\tilde{W}^{*}=\rho\sigma dW^{*}$, or
$dW^{\bot}=d\tilde{W}^{*k}$.

Assume $\sigma$ regular, that is $dW^{\bot}=0$ iff $dW^{*k}=0$, we then have
$dW^{\bot}=\sigma^{k}dW^{*}$, given $\sigma^{k}$ algebraic, the continuation
has removable singularities in a regular neighborhood.

Consider $\mid\Phi(f)\mid^{2}=(\mid dV(f)\mid,\mid dV^{L}(f)\mid)$, where
$\int\mid\Phi(f)\mid^{2}$ finite, defines a normal surface, ([17]), that is an
approximations principle for $\mid\Phi(f)\mid^{2}$, does not assume $\Phi$
algebraic. Consider $d\sigma\sim\mid X\mid^{2}+\mid Y\mid^{2}$. Consider
$d\sigma=(1/v)dxdy$, where $X,Y$ are algebraic, that is $1/v=0$ removable.
Starting from $-\xi Y+\eta X$, we have that v determines the character of
movement.

Assume $dU=dV+dV^{L}$ and $dU=dI(x,y)=dI(x)\otimes dI(y)$. We then have
$dV=dI(x)$ and $dV^{L}=dI(y)$ given decomposability. That is, $dV=0$ implies
$dV^{L}=0$, that is we assume $dU$ preserves zero-lines in the sense of
$x\rightarrow y$ projective. Note, that given convexity and a strict condition
analogous with $B_{pp}$, we can assume $(v+v^{L})$ generates $(v,v^{L})$.

###### MP conjugation 0.6

Assume for $\rho$ not necessarily polynomial, $q_{1}\leq(\rho+1/\rho)\leq
q_{2}$, where $q_{1},q_{2}$ polynomial. Then $dU_{\rho}+dU_{1/\rho}$ has an
approximation property.

Obviously $q_{1}dI\leq(\rho+1/\rho)dI\leq q_{2}dI$, that is
$\int_{\Omega}q_{2}dI=0$ implies $\Omega$ a zero-set. Assume $dU\rightarrow
dU^{\bot}$ according to $dU+dU^{\bot}=(\rho+1/\rho)dI$, that is as long as
$dU+dU^{\bot}$ BV, we have convergence for d I in $\infty$.

Note that, given $\rho=e^{\phi}$, we have $\rho+1/\rho\sim 2\cos\phi$, that is
a harmonic representation.

###  0.1.8 Very regular approximations

We assume the condition $dI\notin G$ corresponds to $0\notin$ the domain for
$T\in\mathcal{H}$, that is an annular region. Consider $f(u,v)\simeq
f_{1}(u-v)$, when (u,v) is not tangent to the diagonal, we do not have a
vector field ([16]). We assume $dI\notin G$ defines discontinuous
approximations.

###### Regularity in an extended plane 0.7

Example: $dI\notin(G_{2})_{C^{\infty}}(u_{1},u_{2})$, but
$dI\in(G_{3})_{C^{\infty}}(u_{1},u_{2},du_{3})$. $f(u,v)\notin C^{\infty}$,
but we have existence of W, such that $f(u,v,w)\in C^{\infty}$. Example: f
very regular, that is $f(u,v)-I(u,v)\in C^{\infty}$ iff $f(u,v,w)\in
C^{\infty}$, that is $Wf=0$ (modulo $C^{\infty}$) over u=v.

In the last example, W corresponds to, $dW\in G_{reg}$, that is regularization
in a complete limit. Note that $f(u,v)\rightarrow f(0)$, given $dI\in G(u,v)$.
Given f algebraic, when $U\rightarrow I$, f must be algebraic in x,y,z, in
this case.

###### Uniformity 0.8

Assume U,V sub-nuclear (modulo $C^{\infty}$) and projectively conjugated, so
that $U\rightarrow I$ is regular iff $V\rightarrow 0$ regular. We then have
$E-I\in C^{\infty}(x,y,z)$ implies $E-I\in C^{\infty}(u,v)$. Conversely $f\in
C^{\infty}(u,v)$, implies $f\in C^{\infty}(x,y,z)$, where $dU(x,y,z)$ is
analytic.

The proof assumes a separation property for d I. It is sufficient for this to
use sub-nuclear (pseudo local) movements. d I is sub-nuclear since
$C^{\infty}$ is nuclear. Consider $\int fdU=\int f(u)du=Uf$, it is sufficient
that $dU\in C^{\infty}$ in x,y,z (analytic). Consider $\pi(x,y,z)=(u,v)$ and
$d\pi=(du,dv)$. Example: $dU(x,y,z)=\mu dxdydz$, with $\mu\in C^{\infty}$.
Example: $(x,y,z)\rightarrow u\rightarrow(u,v)$, that is dv regular measure in
nbhd R(d U) (the range of dU), corresponds to a separation axiom and d u=d v
corresponds to a boundary for regular continuation.

###### TP and preservation of convexity 0.9

We can determine d W, so that $F(w)=F(u,v)-I(u,v)=0$ (modulo $C^{\infty}$).
Continue W to $\tilde{W}F=P(U,V)F-Q(U,V)I$ compact and regular (P,Q linear in
u,v), that is $\tilde{W}F=0$ iff $\frac{P}{Q}F(u,v)=constI(u,v)$, that is P/Q
preserves weak constant surfaces iff $\tilde{W}F=0$.

Example: assume $dW(F)(u,v)=dI(u,v)$ (modulo $C^{\infty}$), that is given F
solution to a hypoelliptic operator d W, we have existence of W with
$F(w)=F(u,v)-I(u,v)$. Consider $\Phi:\mathcal{H}\rightarrow L^{1}$, defines a
topology that can be reduced to $L^{1}$ through two reflections. Consider
$\rho dU+\vartheta dV\in\tilde{G}_{2}$, where we choose $\rho,\vartheta$, so
that $\tilde{G}$ approximates G uniformly. Assume $P(U,V)f=g$, when $U=I$,
$P(V)f=f(P(v))$, given $f(0)=0$ we have $P(v)=0$ implies $g=0$, given $f\neq
0$ close to 0, g has algebraic zeros.

Note that $dI(u,v)$ with support on the diagonal, gives an approximation
property with cluster sets on the diagonal (a “spiral”). With this
representation the fundamental solution may not be representable in the plane.
For f phe and the corresponding parametrix E, we have that f is hypoelliptic
outside the diagonal, also on the kernel to E ([6],[8]). This implies
uniformity for $rad(I_{HE})$. For more general f, the diagonal is not
necessarily the only counter example to hypoellipticity.

##  0.2 Fundamental concepts

###  0.2.1 Convexity

For holomorphic convexity, we consider $<g,f>\rightarrow G(f)$, where G is
holomorphic, but not necessarily linear or nuclear, that is we do not assume
an approximation property. Example: $G(f)=e^{P/Q}(f)$, with for instance Q
reduced, which implies TP ([7]). Given $dI\in(G_{2})_{C}$, simultaneously
$dI\in(G_{2})_{L^{1}\cap C}$, it is sufficient to consider $\parallel
f\parallel=1$. Finite dimensional normable spaces are nuclear ([22] Prop. 50.2
Cor. 2).

Assume T the symbol for a parametrix, and $T-I=e^{E}$. Sufficient for T-I to
have negative type, is $e^{\mid x\mid}\mid T-I\mid\leq C$, when $\mid
x\mid\rightarrow\infty$. Assume $S(dIf)=xf_{x}+yf_{y}$. Note that, $\mid
S(dI)\mid\leq\mid x\mid\mid dI\mid+\mid y\mid\mid dI\mid$.
$E(dV(\phi))=<E,d{}^{t}V>(\phi)={}^{t}VE(\phi)$ and
$\mid<E,S(dI)>(\phi)\mid\leq\mid E(\phi)\mid(\mid x\mid\mid dI\mid+\mid
y\mid\mid dI\mid)$. Denote $\tilde{V}(T-I)=e^{VE}$, given V algebraic, we have
$\tilde{V}\sim V$. Given $\mid x\mid\mid E(dI)\mid\leq C$ and $\mid y\mid\mid
E(dI)\mid\leq C´$, we have $E(dI)\rightarrow 0$, that is corresponding to type
0. Thus, the kernel to $T-I$ is trivial in $\infty$. Conclusion: given T-I has
linear representation in the phase, $T-I=e^{E}$, where (T-I) has type 0 and
given $E(S(dV)(\phi))$ is locally bounded, when $dV\rightarrow dI$ regularly,
we have that $\tilde{V}(T-I)$ has negative type. Negative type gives a nuclear
topology.

###### Geometric convex closure 0.10

Consider $<f,dU>\leq C_{G}<f,dI>$, $\forall dU\in G$ and a constant $C_{G}$,
dependent on $f\in(I)\subset C^{\infty}$, that is we assume d U BV, but not
necessarily $dI\in G$.

Example: $dU=\rho dI$, where $\mid\rho(x,y)\mid\leq C_{G}$ on a compact set.
The sets $X=\\{f\\}\subset\subset C$ in the domain for G, are geometrically
convex iff $\tilde{X}\subset\subset C$. Note that $f\equiv 0$ over $G_{8}$,
implies $f\equiv 0$ over $G_{2}$. We start with two scalar products
$S(v),S^{\bot}(u)$, and $G_{8}\rightarrow G_{2}$ :
$S^{\bot}S(v)=((v,v^{L}),(v,v^{L})^{\bot})$.

###### Relative convexity 0.11

Assume $\rho$ analytic, with absence of essential singularities in $\infty$,
then there are $dU,dV$ analytic, so that $dU=\rho dV$. In particular, given
$\frac{\delta}{\delta v^{\bot}}\rho(v,v^{\bot})=0$ and
$\frac{\delta\rho}{\delta v}\geq 0$, we have that $U$ is relatively convex.
Given a strict condition, we can continue to $\rho\in L^{1}$.

Let for instance $dV^{\bot}=\sigma d(V-I)$. Assume $dU=\rho dV$, where
$\rho^{\prime}(v)\geq 0$ locally, that is $\rho$ monotonous and single valued,
thus in the plane linear. Assume $dU_{2}=\mu dU_{1}$, with $\mu$ analytic and
$\rho dU_{1}+\vartheta dU_{2}=dI$, that is $(\rho+\vartheta\mu)dU_{1}=dI$. In
particular for $1/(\rho+\mu\vartheta)\in(I)$, a multiplier ideal, we can
discuss $\rho+\vartheta\mu=const$. Example: determine $\rho,\vartheta$, so
that
$\overline{\delta}\rho+\vartheta\overline{\delta}\mu+\mu\overline{\delta}\vartheta=0$,
for instance $\overline{\delta}\rho/\overline{\delta}\vartheta=-\mu\neq 0$.
Given $\rho dU_{1}\rightarrow dI$ iff $\vartheta dU_{2}\rightarrow 0$, we get
a generalization of projectivity. Example:
$\vartheta\rho(\frac{1}{\rho}dU_{2}+\frac{1}{\vartheta}dU_{1})=\rho
dU_{1}+\vartheta dU_{2}$ for instance $\vartheta=1/\rho$. Note that
$\frac{1}{\rho}+\frac{1}{\vartheta}=1$ iff $\rho\vartheta=\rho+\vartheta$.

Concerning $G_{8}$, given for instance $dV(f)=f_{x}$, holomorphic convexity is
not preserved, consequently not geometric convexity.

###### Linearization 0.12

Assume d U (projective), completed through dW to UW algebraic (analytic), then
type is not necessarily preserved. Thus, we can have an approximation property
for $T_{ac}(dU)$, that is not present for dU.

Example: $dUW=\rho dU$, with $\rho\neq 1$, where $dU=dI$. . Example:
irreducibels: $dU=\rho dW=dVW$, we then have $dU=dI$ implies $dV=I$ or
$dW=dI$, that is $IW=VI$, corresponding to continuation under preservation of
type.

Starting with two scalar products, assume $UW$ closed, where $W^{\bot}$ is
defined relative $S_{2}$, so that $U^{\bot}W^{\bot}=W^{\bot}U^{\bot}$. Thus,
we can define $(UW)^{\bot}$ through annihilators, that is not relative type of
movements. Alternatively, consider completion to a closed movement
$(dV,dV^{L})$, further $((),()^{\bot})$ relative normed spaces and
annihilators.

Assume convergence according to $\mid Uf\mid\rightarrow\mid f\mid$, this does
not imply $\mid Uf-f\mid\rightarrow 0$, but given existence of d V, so that
$\mid VUf-f\mid\rightarrow 0$, we have mean convergence, however dependent of
$(U,V)$. Example: assume $S=UT-TU$, then a conjugated movement relative mean
convergence, is given by $VS=0$ on a given set $\Sigma$. A polar set can be
given by $VS=0$ implies $dV\notin G$.

Example: Assume $dU=\rho dV$, with $\rho\in C^{\infty}$, then we have given
monotropy (isolated singularities), existence of $v$ analytic close to
$\rho=1$. Assume $UV\rightarrow I$ regular with $dV=(1/\mu)dI$ and
$\rho/\mu=v$ analytic, that is it is not necessary to assume $1/\mu$ analytic.
The type for dV is given by $1/\mu=const$.

###  0.2.2 Symmetry

###### Symmetry mapping 0.13

Assume $\sigma$ is defined relative a scalar product, for instance
$U\rightarrow{}^{t}U$ and $\sigma:(V,V^{L})-{}^{t}(V,V^{L})$. Symmetry implies
$\sigma\sim 0$, for instance homotopic with zero. Further we assume $\sigma$
is a projective mapping, that is maps zero lines onto zero lines.

Note that $U\rightarrow{}^{t}U$ does not necessarily preserve type, that is
$U\rightarrow I$ does not imply ${}^{t}U\rightarrow I$, that is
$\Sigma(U)\neq\Sigma({}^{t}U)$ regular. Consider $\sigma_{1}$ with respect to
$dV\rightarrow dV^{L}$ and $\sigma_{2}$ with respect to
$dV+dV^{L}=dU\rightarrow d{}^{t}U$. For symmetric functions, $\sigma_{1}$ does
not define a regular symmetry. Obviously, when $\sigma$ symmetric,
$\sigma_{1}(dV)F(\phi)\simeq F(\sigma_{1}(dV)\phi)$. Given F linear and
$U,{}^{t}U$ sub-nuclear, we have
$F(\sigma_{2}(dU)\phi)\simeq\sigma_{2}(dU)F(\phi)$.

Projectivity can be compared with preservation of closed forms, these are
locally exact. Given $dU,dU^{\diamondsuit}$ correspond to closed forms, using
the involution condition, we have analyticity for d U ([1]). Assume $dU=\rho
dI$ and $dU^{\diamondsuit}=\vartheta dI$. On $\rho\neq 0$ and simultaneously
$\vartheta\neq 0$, projectivity is preserved.

###### The two-mirror model 0.14

TP : $A\rightarrow_{v}B\rightarrow_{\pi}C$, where $vA\simeq\pi^{-1}C$, that is
given $\pi$ invertible, we have a lifting principle. A strict condition is
sufficient for $(u^{\bot})\rightarrow(u)\rightarrow(x)$ continuous. Given an
approximation property, we have $(u)\rightarrow(x)$ regular.

Assume f bounded and analytic, symmetric on a planar domain with a simple
boundary L, then f is symmetric on a disk ([5]). Example: $E(u,u^{\bot})-I\sim
0$ on $L=\\{u=u^{\bot}\\}$, then E is not analytic. Further, given
$(U,U^{\bot})$ discontinuous, then $E(u,u^{\bot})$ is not analytic.

###  0.2.3 Discontinuous approximations

Assume $fdI$ BV and analytic, where $f=0$ on $L\subset bd\Omega$ implies
$f\equiv 0$ on one side of L. Consider $\pi:(x,y,z)\rightarrow(u,v)$ and
$L=\\{u=v\\}$, given $f=0$ (modulo $C^{\infty}$) on L, implies that $f\in
C^{\infty}$, for instance when $v/u\rightarrow 0$. The condition $fdI-dI=0$ on
L leaves space for a spiral.

Note that $f$ one-sided in $G_{2}$, does not imply that $f$ is one-sided in
$G_{8}$, but one-sidedness in $K(G_{4})$ implies one-sidedness in $G_{2}$.

Assume f bounded and =0 on $L\subset bd\Omega$ implies $f\equiv 0$ on a half
space (convex set), that is given I gives reflection of supp f through
$bd\Omega$, $If\cap fI=\\{0\\}$, that is when If defines linear continuation,
we have orthogonal decomposition of the corresponding measure. With these
conditions, projective movements have a decomposition $U_{+}-U_{-}=U$. Assume
f separately linear in u,v, we then have $<dV,<dU,f>>=<dU\times dV,f>$, that
is f linear in $dU\times dV$. Note that $(U+V)f=f(u+v)=f(u)+f(v)$, only when f
linear.

###### Linear phase 0.15

On a geometrically convex set, we have existence of $C_{K}$, so that
$L_{1}(L_{2}(f))\leq C_{K}$ iff $L_{2}(L_{1}(f))\leq C_{K}$.

Assume $e^{L_{1}}$ defines an algebraic symbol, why $\\{e^{L_{1}}\leq C\\}$
corresponds to a semi-algebraic set. Note that $L_{1}\equiv 0$ (or constant)
on an entire line, gives cluster sets, that is we do not have relatively
compact sets.

Assume f analytic, with $dU(f)+dU^{\bot}(f)=0$, further $dU\bot dU^{\bot}$.
Consider $F(u,v)\rightarrow u$. Assume $\Sigma(U)=\\{U-I=0\\}$. To determine
type of movement, it is necessary that $\Sigma\cap$ the domain for F. Assume
$U\rightarrow I$ implies $V\rightarrow 0$, then we have $(U+V)F\rightarrow F$
on $\Sigma$. Thus, when the conjugation is given by $\Sigma(U)=N(V)$, the
movements can be identified regularly.

###  0.2.4 The stability group

A stability group is formed by $g_{0}=Id$ and $g_{i}g_{0}=g_{0}g_{i}$, where
$g_{i}$ is a connected family, not necessarily a subgroup. ([4])

When the stability group gives two-sided regular approximations, we have that
$g_{i}I=Ig_{i}$ implies $g_{i}$ (topologically) algebraic, Example:
$F(\gamma)\in L^{2}$,
$F(\gamma\rightarrow\gamma^{\prime})\in\mathcal{D}_{L^{2}}$ and
$F(\gamma\rightarrow\gamma^{\prime}\rightarrow\gamma^{\prime\prime})\in H$,
gives a strict condition, a completion to stability.

A subgroup approximates 0, that is stability defines a (topologically)
algebraic subclass, that is not necessarily an analytic subgroup. Example:
UI=IU implies (U-I)I=I(U-I), where $dU-dI$ can define a subgroup, even when
$dU$ does not.

Two-sided limits can be motivated by regular complements, for instance $dU$
analytic with existence of d V analytic and $dU\bot dV$. Assume $dU=\rho dV$,
with $\rho\neq 0$ conformal in the plane, that is starting from $dV\rightarrow
dV\times dV\rightarrow dU\times dU^{\bot}$ regular and conformal, that $dU\bot
dU^{\bot}$, for some $dU^{\bot}$. Note that conformal mappings are locally
invertible, that is the last mapping in the two mirror model (section 0.2.2)
is reversible.

###  0.2.5 Uniformity for the polar set

###### Uniformity 0.1

Assume $E(f)=I(f)$ (modulo regularizing action) localization of the symbol,
then E(f)=0 represents the polar set. $E(f)=\int fdE$, with $dE-dI=0$ (modulo
regularizing action) over $E(f)=0$, implies that the diagonal uniquely gives
the polar for regularizing action. Thus the representation is Hausdorff
uniform ([3]).

Note that E corresponding to the parametrix to a partially hypoelliptic
differential operator, satisfies the conditions for Hausdorff uniformity
relative $K(G_{4})$ above ([8]). Concerning $G_{8}$: the six first are not
complete and thus do not preserve for instance uniformity, compactness. For
$dV_{1}=x^{2}f_{x}+xyf_{y}$, $dV_{2}=xyf_{x}+y^{2}f_{y}$, we have
$dV_{1}/dV_{2}=\frac{\frac{x}{y}f_{x}+f_{y}}{f_{x}+\frac{x}{y}f_{y}}$.
Example: $(x^{2},xy)\rightarrow(x^{2},y/x)$ and
$(xy,y^{2})\rightarrow(y^{2},x/y)$. $F(x,y)\rightarrow
F(x^{2}+y^{2})M(y/x,x/y)$, with M algebraic. This representation defines F
uniformly, where M=const.

Note $dU^{\diamondsuit}\bot dW^{\diamondsuit}$, implies $dU\bot dW$. For an
orthogonal relation, we assume the coefficients to the movements are related
through a scalar product, for instance Legendre. We do not necessarily assume
that $dU^{\bot}$ can be defined uniquely. Sufficient for $M_{1}\rightarrow
M_{2}$ to preserve algebraicity, is that
$\phi(\xi_{1}/\eta_{1})=(\xi_{2}/\eta_{2})$ a mapping between coefficients, is
algebraic. It is necessary, that the scalar product to $dU^{\bot}$ is mapped
projectively on to the scalar product to $dU^{\diamondsuit}$.

Given $dI\in(G_{2})_{C}$, we refer to $dU_{1}=dU_{2}$ as a spiral.
Correspondingly, a diagonal in $K(G_{4})$ is represented using
$F(x_{j},y_{j})M(x_{j}/y_{j},y_{j}/x_{j})$, j=1,2,3,4, where M corresponds to
a symmetric potential and $(x_{j},y_{j})$ in a convex set. In $K(G_{4})$, we
get $M=M_{1}\ldots M_{6}$, that is presence of diagonal, implies existence of
$M_{j}$ non-algebraic. The diagonal is characterized by that all of the
$M_{j}$ are symmetric, we assume the value of M constant close to
$x_{j}=y_{j}$. For the representation in $H(K(G_{4}))$, we have that $M_{1}$
algebraic implies $M_{2},M_{3}$ algebraic and conversely. Remains the case
with $xdW+ydW$. We consider $(x^{2}+xy)/(xy+y^{2})=(1+x/y)/(1+y/x)$, that is
$M_{4}$ algebraic, if $M_{1}$ algebraic and conversely. The conclusion is that
given measures with analytic coefficients and BV, we have a uniformity space
and the polar is given by the diagonals. Note that to conclude hypoellipticity
relative $K(G_{4})$, it is sufficient to exclude spirals in the polar.

Note that for non complete movements, for instance $V\otimes
If(x,y)\rightarrow f(v,0)$, we have that the condition $f(x,y)\leq f(v,0)$
does not preserve relatively compact sets and not complete movements do not
give a uniform approximation property, since the condition on a regular
neighborhood is not uniform.

Example: $U-I=I$ implies $Uf=2If$, that is invariants to U-I are given by
$Uf=2If$. Given f=log g, and U algebraic, $Ug=g^{2}$. Note that $dI\in
G(\\{g^{N}\\})$ does not imply $dI\in G(\\{g\\})$ Example: $dI\notin
G(\\{\phi\\})$, but $dI\in G(\\{\rho\phi\\})$ where $\\{\rho\phi\\}$ is a
nuclear space.

###### Interpolation 0.16

Assume $dU=\rho dV$ ($=\rho(v)dv$), for V sub-nuclear, where $\rho$ monotonous
on a planar domain $(v,v^{\bot})$. Where $\rho$ is independent of $v^{\bot}$,
we have $VdV\sim dV^{2}$, that is $(dV)\subset(dV^{2})\subset(dU)$. Thus,
given $\rho$ linear and independent of $v^{\bot}$, there are intermediate
measures.

Starting from a very regular boundary and $(dV)\subset(dV^{j})\subset(dU)$,
with $dU=\rho dV$ and $dV^{j}=\vartheta^{j-1}dV$, we can determine dV, so that
$\vartheta^{j}$ polynomial.

Example: Consider two diagonals with respect to $U,V$,
$\sigma\sim(V-V^{\bot})U^{\bot}+V^{\bot}(U^{\bot}-U)$, that is on den double
diagonal $\sigma\sim 0$. Consider $A\rightarrow B\rightarrow C$. Assume
$U^{\bot}\sim U_{+}$ and $U\sim U_{-}$ and in the same manner for V, that is
$U_{+}(A)=B_{+}$ and $V_{+}B_{-}=0$,$V_{+}U_{+}=C_{+}$, we then have
$\sigma\sim 0$. Thus, given action for U,V one-sided, there is an
approximation property.

###  0.2.6 Regular approximations

For $X_{reg}$ dimension is well-defined. Given $X_{sng}$ removable, we have a
continuable movements. Assume $UW\rightarrow I$ is regular, but
$U-I\nrightarrow 0$ regularly, that is $W\rightarrow U^{-1}$. Spectrum for the
approximation, is given by W=I, the character of W is dependent of type of
singularities. Given singularities in a bounded set, then finite iteration of
monotonous movements give a regularization.

The order of G, that regularizes (preserves) singularities, is dependent of
the order of singularities. Assume F a figure for singularities. Example:
$V,W$ pseudo local, we then have if $Vf\in C^{\infty}$, that $WVf\in
C^{\infty}$. The condition $WV=VW$ gives, when $V\bot W$, that the order is 2.

Singularities in H are given by $dUf=Uf=0$, for instance $Uf\in C\backslash
H$. Assume U not surjective, but we have existence of W, so that $UW$
surjective. Given $R(W^{\bot}U^{\bot})=R(UW)^{\bot}=\\{0\\}$ (algebraic), we
see that $U^{\bot}\bot W^{\bot}$. Further where $W^{\bot}=(I-W)$ and
$U^{\bot}=(I-U)$, we have $U^{\bot}+W^{\bot}=I$ implies $2I-U-W=I$ implies
$I=U+W$. Assume $UW$ a projection operator. Given G has order 2, we must have
$W^{\bot}=U$ or $U^{\bot}$, where $U\neq U^{\bot}$, otherwise we have in the
plane 8 possibilities ([20] ch. 4, ch. 17).

###  0.2.7 Completion and TP

Starting from ([20] chapter 5) that two equations A=c and B=$c^{\prime}$
represent the same set of curves iff B is only a function of A,
$B(x,y)=\Omega(A(x,y))$, we discuss reduction of movements to base movements.

###### Independence of involution 0.17

Consider a regular continuation $dI\rightarrow dV\times dV^{L}$, that is
completion to closedness and $dV\times dV^{L}\rightarrow dU$ linear, this can
be given as independent of involution and is not dependent of the type for d
V. Assume $dV\times dV^{L}=\mu dI$ and $(v,v^{\bot})$ dense in the domain for
$1/\mu$ regular, then the completion can be given as regular.

Consider $1/\mu(v,v^{\bot})dv\simeq dI$. Example: $(G_{2})_{C}\ni
dI\rightarrow dI\in(K(G_{4}))_{H}$ completed to $dV\times
dV^{L}\in(K(G_{4}))_{H}$. Given $dV\times dV^{L}\rightarrow dU$ single valued
in the plane, then d U can be given as linear in $(dV,dV^{L})$. Given an
elastic model, $dN\in G$, so that we have existence of $\rho,\vartheta$, with
$\rho dN+\vartheta dN^{\bot}=dI$, then d U can be determined as independent of
d V, that is we have $dI\rightarrow dU$ regular. Assume $dU_{L}=dI_{L}$, where
L denotes an entire line, and that $(dN,dN^{\bot})\rightarrow(dI,0)$ is
rectifiable. More precisely, given $d\rho/dn^{\bot}=0$ and $d\rho/dn\geq 0$,
close to the boundary. Assume $\Sigma=\\{dU=dI\\}$, we then have that as long
as $dV\times dV^{L}$ preserve $\Sigma$, the type for U is not dependent of
$dV\times dV^{L}$.

###### Normal model 0.18

Assume $(N,N^{\bot})$ surjective on X, we then have
$N\varphi_{1}+N^{\bot}\varphi_{2}=f$, for $f\in X$. Assume
$\phi:\varphi_{1}\rightarrow\varphi_{2}$, that is we have $W\varphi_{1}=f$,
where $dW$ is not necessarily in the group. Given an approximation property,
$N\rightarrow I$ regularly, implies $N^{\bot}\rightarrow 0$ and d W with
single valued (monotonous) representation in $(N,N^{\bot})$, we have that
$\phi$ does not affect the type for d W. Given $N^{\bot}\phi\rightarrow 0$
regularly, we have the approximation $W\varphi_{1}\rightarrow f$ uniformly,
without invariant sets.

Given $\psi:dN\rightarrow dN^{\bot}$, we have $\psi^{*}=\phi\rightarrow 0$
close to the boundary.

Assume $dN^{\bot}=\rho dI$, where $\rho$ can be assumed multivalent. Assume
$\phi:\varphi_{1}\rightarrow\varphi_{2}$ continuous. Assume $d\phi=\mu dI$,
then $\phi$ can be related to a normal system, $dN^{\bot}=(\rho/\mu)d\phi$.

Consider the extended system and the boundary to the domain $\Omega$, given by
for instance $(V,V^{L})$. Assume $dU_{j}=\rho_{j}dU_{1}$, that is
$\rho(u_{1},\ldots,u_{k})\simeq\rho_{1}(u_{1})\times\ldots\times\rho_{k}(u_{1})$.
Note that for $\rho\in C^{\infty}_{c}$, it is sufficient to consider
$\rho_{j}$ polynomial. Given $\rho\in L^{1}$ and $\rho\rightarrow 0$ in all
directions, we have $\rho_{j}\rightarrow 0$ $\forall j$. When the domain for
$L^{1}$ is bounded, every monotonous movement leaves the domain under
iteration.

###  0.2.8 Decompsable sets

Consider $dv\rightarrow(dv,dv^{L})\rightarrow(du_{1},du_{2})$. Assume p a
semi-norm to $(v,v^{L})$ and q a semi-norm to $(v^{\bot},v^{L,\bot})$. An
extended plane that is contractible to a domain with an approximation
property, has an approximation property. Presence of a polar set implies that
$R(G)$ is not closed. The condition $r\leq pq$ implies that it is sufficient
that a semi-norm=0, Define
$R(\Phi(G))^{\bot}=\\{(w,w^{\bot})\bot(u,u^{\bot})\\}_{2}$. $r\leq pq$ defines
the polar set, for instance q=0.

Consider an elastic barrel : $T_{ac}(dU)=\cup_{\mid\rho\mid\leq 1}\rho dU$,
with $\rho\in L^{1}$, that is continuation of the group is defined through
$\rho dU$. A necessary criterion for inclusion between weighted $L^{p}-$
spaces is for instance $\rho\rightarrow 0$, note that we do not necessarily
have existence of a subgroup $(dW)=(T_{ac}(dW))\subset(dU)$. On the other
hand, if $dU=\sigma dI$, with $\sigma\rightarrow 0$, we can determine d W
maximal, that is a regular subgroup d W, that can be continued as above, to
$T_{ac}(dW)=G$.

Given $G_{8}$ generates $R(G)^{\bot}$, we consider $G_{4}\times
G_{4}\ni(dV,dV^{L})$. Regularity for $(dV,dV^{L})$ in $G_{8}$, corresponds to
a two-sided limit. When we consider $dV-V^{L}=0$, zero-lines are mapped onto
symmetry-sets, $dU-dU^{\bot}=0$ maps zero-lines onto spiral sets.

The condition $(UF-F)+F-I\in C^{\infty}$, implies $(UF-F)\sim(F-I)$ (modulo
$C^{\infty}$), in particular over invariant sets, $F-I\rightarrow 0$ (modulo
$C^{\infty}$). Consider $(dU)=(dV,dV^{L})$, we then have that invariant sets
for d U, corresponds to a domain for UF very regular, given F very regular.
Assume $(v-v^{L})$ is dense in the domain for F and consider
$F(v^{L}-v)-I(v^{L}-v)\sim 0$, that is $(dV,dV^{L})$ is symmetric around I iff
$(v-v^{L})$ is symmetric around 0.

Assume $S=\\{f=df=0\\}$, that is given $dx\bot dy$, we have $f_{x}=f_{y}=0$.
Note that $f_{u}u_{x}=0$ does not imply $f_{u}=0$, for instance u constant in
x. Assume $\Sigma_{1}=\\{f_{u}\neq 0\quad f_{x}=0\\}$, in the same manner for
$\Sigma_{2}$, $\Sigma_{3}=\\{uf_{u}\neq 0\quad xf_{x}=0\\}$ for instance
$x/u\rightarrow 0$, that is u is reduced with respect to x. In the same manner
for $\Sigma_{4},\Sigma_{5}=\\{f_{v}\neq 0\quad yf_{x}=0\\},\Sigma_{6}$.
Starting from $dW=xf_{x}+yf_{y}=0$, with $vf_{v}\neq 0$.
$\Sigma_{7},\Sigma_{8}$ have the corresponding conditions $x^{2}/v\rightarrow
0$, $y^{2}/v\rightarrow 0,xy/v\rightarrow 0$.

Assume $\Phi(f)=\int fdU$, for instance $\Phi:L^{2}\rightarrow L^{2}$ linear
continuous, more generally $\Phi:\mathcal{H}\rightarrow\mathcal{H}$, where
$\mathcal{H}$ is not nuclear. In particular $\Phi(f)\rightarrow f$, with
$dU\notin G_{\mathcal{H}^{\prime}}$. Example: $dI\in G(du,du^{\bot})$ but
$dI\notin G(dx,dy,dz)$. Presence of non-regular approximations, does not
necessarily influence hypoellipticity, that is hypoellipticity does not imply
that all approximations are regular. Example: scaling does not necessarily
affect P-convexity, however absolute continuity is not preserved ([5], Theorem
3.4), for $dU,dU^{\bot}$ separately. Example: if we consider $dU=dI$ iff
$dV_{1}(x)\otimes dI(y)=0$, we do not have uniformity and the corresponding
group is not P-convex, for an iterate of the symbol.

Starting from $\\{f<\lambda\\}\subset\subset\Omega$, we consider regular
continuations of these sets, that is we do not have existence of L, an entire
line $\subset\\{f=const\\}$. Existence of L implies that convexity (pseudo
convexity) is not preserved.

Continuation through continuity allows an entire line in $\\{f=const\\}$, that
gives unbounded sub-level sets. Assume $dUW=dI$, where $dU\in G_{2}$. Given
$dI\in G_{2}$, we have $dW\in G_{2}$. Assume $dI\notin(G_{2})_{H}$, we then
have $dU\in(G_{2})_{H}$ implies $dU^{-1}\in(G_{8})_{H}$. Assume
$\Sigma(dU)=\\{(\xi,\eta)\sim(1,1)\\}$, gives the coefficients to dU. Given
$\Sigma(dU)=\Sigma_{1}\times\Sigma_{2}$ where $\Sigma_{1}=\\{\xi\sim 1\\}$,
that is decomposable constant surfaces, we have that invariants to $G_{8}$,
can be derived from invariants to $K(G_{4})$. Note that $(G\times
G)_{\mathcal{O}_{M}}\simeq G_{\mathcal{O}_{M}}\times G_{\mathcal{O}_{M}}$,
that is we require slow growth measures.

###  0.2.9 Preservation of convexity

###### Two-mirror-model 0.19

A movement, that can be reduced to a normal model, rectifiable at the
boundary, can be reduced to a single scalar product.

Consider $A\rightarrow_{\pi_{1}}B\rightarrow_{\pi_{2}}C$, where TP assumes
$\pi_{1}(A)\simeq\pi_{2}^{-1}(C)$. Example: $dU=\rho dV\rightarrow\rho\mu dI$,
where $dI\rightarrow dV$ is completion to closedness. For TP, it is sufficient
that $1/\rho$ is regular.

Assume $P_{1}\leq F\leq P_{2}$, where $P_{j}$ are polynomial, then the sub-
level surfaces for F can be given as semi-algebraic. Consider
$\Omega=\\{F<\lambda\\}$, with $\Omega\rightarrow bd\Omega$ continuous, where
the boundary is assumed algebraic.

Assume $S_{j}(v)=(v,{}^{t}v)=(v,L_{j}(v))$, where $L_{j}(v)$ defines ${}^{t}v$
with respect to the corresponding scalar product. Assume
$A(v,{}^{t}v)=v+{}^{t}v$. that is $S_{2}AS_{1}=(u_{1},u_{2})$. Thus, given the
translates are dense in the domain, $A(v)$ is semi-algebraic, given $S_{j}(v)$
are semi-algebraic. Assume $p_{1}(v,v^{\bot})\leq S^{2}(0)\leq
p_{2}(v,v^{\bot})$, where $S^{2}(v,v^{\bot})\sim S^{2}(0)$ (homotopical
conjugation) and $p_{j}$ polynomial. Given isolated singularities and a
separation axiom, we have that constant surfaces to $p_{j}$ are removable.
(Vp=Ip, where I is evaluation in 0).

Consider the two-mirror-model, when $d{}^{t}V\simeq q_{1}dI$ and $dV\simeq
q_{2}dI$, $dV^{\bot}/d{}^{t}V=p$ and $dV\rightarrow dV^{\bot}$ according to
$q_{2}\rightarrow 1/q_{2}$, we then have $dV^{\bot}/dV\sim 1/q_{2}^{2}$, given
$pq_{1}q_{2}\equiv 1$. We assume $d{}^{t}V\rightarrow dV^{\bot}$ algebraic
with an approximation property.

Assume $T(\phi)=f$, where $dU=\rho dI$, where $\rho=1$ describes a line.
Assume $dU^{\bot}\leq dW\leq dU$, where $dU=\rho dI$ and
$dU^{\bot}=(1/\rho)dI$ and $1/\rho\leq\sigma\leq\rho$. Cf. Nullstellensats:
$\mid(1/\rho)\rho\mid\leq C$ implies $\mid\rho\mid\leq C^{\prime}$ for large
arguments, given $1/\rho$ reduced, that is locally 1-1, or equivalently $\mid
dUdU^{\bot}\mid\leq C$ implies $\mid dU\mid\leq C^{\prime}$. Further,
$\\{\sigma\rho<\lambda\\}\subset\\{(1/\rho)\rho<\lambda\\}$ that is
$\sigma\rho\leq\lambda$ implies $\rho\leq\lambda$. Conversely, $1/\rho\equiv
1$ (positive measure) implies $dU=dU^{\bot}$ has positive measure. Assume
$U(N,N^{\bot})$ with $dN^{\bot}/dN\rightarrow 0$, where we assume
$(N,N^{\bot})$ has no invariant sets (non-trivial). Thus,
$\frac{d}{dN}U(N,N^{\bot})\simeq dU/dN$ Assume for instance $dU=\rho dN$, with
$\rho\geq 1$, that is an one-sided approximation.

###### Subnuclear movements 0.20

Consider $dV^{2}\sim VdV$, given V sub-nuclear, we have $dV=0$ implies
$dV^{2}=0$. Assume the singularities are given by $\\{Vf=dVf=0\\}$ and that
$dU=\rho dV$, we then have that $\rho=0$ does not necessarily preserve
singularities. Conversely, given $Uf=WVf$, with W sub-nuclear, we have that
regular points are preserved.

Consider V sub-nuclear modulo $C^{\infty}$, that is V pseudo local. Given
$V^{j}=I$ implies $t=0$ (parameter), there are $V^{-1}$ pseudo local, that is
over f the singularities are of finite order.

Given $(U,V)\rightarrow(I,0)$ a nuclear mapping, we do not have a nuclear
topology. Sufficient is that I sub-nuclear ([21]). We assume it is necessary
for a nuclear topology, that we have an approximation property in a convex
neighborhood of a regular approximation.

###  0.2.10 A strict condition for TP

Consider $f(\mid u\mid)-\mid f(u)\mid\sim 0$. $\mid u_{j}\mid\leq 1$,j=1,2
corresponding to a cylindrical domain, $\mid f(u)\mid\leq 1$ corresponding to
an analytic polyhedron. Let $\Omega=\\{\mid f(u)\mid\leq\lambda\\}$ and
$\tilde{\Omega}=T_{c}(\Omega)$ (convex closure), corresponds to
$(u_{1},u_{2})$ being dense in $\tilde{\Omega}$.

Consider the condition ([2]) $e^{B\mid x\mid^{k}}\mid f(1/x)\mid\leq C$
locally. Given $P(1/\mid x\mid^{k})\sim 1/Q(\mid x\mid^{k})$, we have that the
condition is $\mid f(x)\mid\leq CQ(\mid x\mid^{k})$, note that P polynomial,
defines a domain of holomorphy, where s=1/k denotes the size. Given $P(1/\mid
x\mid^{k})\mid f\mid\leq C$, we then have $\mid y\mid/\mid x\mid^{k}\leq C$
corresponds to $\mid y\mid^{s}\leq C\mid x\mid$ ($C\rightarrow 0$,
$x\rightarrow\infty$).

###### TP with strict condition 0.21

Assume $R(1/\mid x\mid^{k})\mid f(x)\mid\leq c$. Assume $\frac{Q}{P}f(x)$
analytic, with $R(1/\mid x\mid)P(\mid x\mid)\leq c^{\prime}$, then $\mid
f(x)\mid\leq P(\mid x\mid^{k})$ gives a majorisation condition corresponding
to slow growth in $\infty$. The condition Q reduced implies $1/Q\rightarrow
0$, that is a strict condition and through Nullstellensats
$\mid\frac{Q}{P}f\mid\leq c$ implies $R(1/\mid x\mid^{k})\mid f\mid\leq
c^{\prime}$.

Concerning
$\mathcal{O}_{M}(G_{8})\simeq\mathcal{O}_{M}(G_{4})\times\mathcal{O}_{M}(G_{4})$,
we consider $f\in\mathcal{O}_{M}$ according to $\mid f(x)\mid\leq CQ(\mid
x\mid^{k})$, for a polynomial Q. The condition defines a class of measures
$\mid QdU\mid$ BV and immediately regular neighborhoods of invariant sets. A
sufficient condition is a very regular boundary in $(u,v)$. Assume f
corresponding to real type, that is $\mid f(v)\mid\leq Ce^{B\mid u\mid}$, with
B close to 0, then the invariant sets can be determined in $K(G_{4})$.

##  0.3 Projective transformations

###  0.3.1 Topological degree

Assume a normal model $(N,{}^{t}N)\rightarrow(N,N^{\bot})$, that is
$N\rightarrow I$ iff $N^{\bot}\rightarrow 0$. When N is algebraic, ${}^{t}N$
is algebraic. Assume ${}^{t}N\rightarrow N^{\bot}$ completion to an
approximation property. Sufficient for an algebraic continuation is strict
pseudo convexity. Algebraicity for d N is dependent of topology. Example:
$dN(\tilde{\phi})=d\tilde{N}(\phi)$, where
$C^{\infty}\ni\phi\rightarrow\tilde{\phi}\in L^{1}$ with $d\tilde{N}$
algebraic. Starting from $N(U,U^{\bot})$, where U one-parameter, a sufficient
condition for a single valued orientation close to the boundary is
$dU^{\bot}/dN\rightarrow 0$. For hypoellipticity, it is necessary that
$N(U,U^{\bot})$ is corresponding to real type, that is determined by d N / d
U.

Projective transformations are the analogue to transformations that map closed
forms onto closed forms. For regular transformations, we assume that a closed
contour is mapped on a single Jordan curve in the plane, or a cycle in
$S^{n}$. Given a strict condition, there is to a projective mapping, a
corresponding bijective mapping. Example: $dU=pdV$, where p is reduced, we
have $\int_{L}dU=0$ implies $\int_{L}dV=0$ or $L=\\{0\\}$. A strict
homomorphism, corresponds to a local bijection $dV\rightarrow dU=\rho dV$.
Sufficient for isolated singularities, is $dU=0$ implies U=I. Note that d U
reduced does not imply U reduced. Since every cycle in $S^{n}$ can be mapped
on to a full line in $E^{n+1}$, it is to determine topological degree ([15]),
sufficient ot consider zero- lines for measures. We consider homotopy,
represented as $dU=\rho_{j}dN$ with $\rho_{1}\sim\rho_{2}$, that is given an
approximation property, the degree is preserved.

###### Order for projective transformations 0.22

Assume cycles $\rightarrow$ zero-lines, we then have that the topological
degree gives the order for projective transformations.

Consider a zero-line L to $dU$ BV, that can be mapped on $\Gamma$, a simple
Jordan curve. Assume further $L^{\prime}$ a zero line to $dU^{\diamondsuit}$
and to $dU^{\bot}=\sigma dU$, then d U=0 on $(L,L^{\prime})$, that is all
zero-lines between $L,L^{\prime}$, that can further be mapped onto cycles ([5]
Lindelöf’s theorem). Assume $dU=\rho dI$ and $dU^{\bot}=\rho^{\bot}dI$, with
$\rho-\rho^{\bot}\in H(L,L^{\prime})$, we then have given $L\rightarrow
L^{\prime}$ bijective, that $\rho-\rho^{\bot}=0$ on $L,L^{\prime}$, implies
$dU=dU^{\bot}$ on the sector between $L,L^{\prime}$. Given instead
$dU=dU^{\bot}$, that is $H\ni\rho\rightarrow\rho^{\bot}\in\mathcal{H}$ and
$\rho=\rho^{\bot}$ on $L,L^{\prime}$ and that we have existence of a radius
$\gamma$ on a convex set between $L,L^{\prime}$ (planar), so that $dU\neq
dU^{\bot}$ on $\gamma$, then, given H dense in $\mathcal{H}$, there are no
spirals (diagonal elements) in the inner of the domain between $L,L^{\prime}$.
Consider $dI\rightarrow dN\rightarrow dU$, where d N is a boundary operator.
We assume $dN^{\bot}\rightarrow 0$ regular, thus we must have for the polar
set $dU^{\bot}=\sigma^{\bot}dN^{\bot}=0$, that $\sigma^{\bot}\equiv 0$.
Consider $Y(f)=\\{u^{\bot}\quad f(u,u^{\bot})=0\\}$ a manifold for the polar.
Given that the measure for $Y(f)$ is non-zero, then when $\rho\rightarrow 1$,
we have multivalentness for the limit ([5]). Let
$\phi_{j}:dU_{j}\rightarrow\rho_{j}dU_{1}$, where $\rho_{j}\neq 0$ define
continuous mappings. The order for Y, can be given as the number of j, such
that $\delta\sigma^{\bot}/\delta u_{j}=0$.

Consider the $T_{ac}$-closure, $T_{ac}(dV)(\mathcal{H})=\\{\rho
dV\quad\rho\in\mathcal{H}\\}$, that is $dV\rightarrow\rho dV$ is projective.
Given $dV=\mu dI$, $\mu\rho=\mu+\rho$ iff $1/\mu+1/\rho=1$, that is the
condition implies a dual multiplier. Given $\rho(v,v^{\bot})$, where $\rho$
single valued, then $\rho=\alpha\rho_{1}(v)+\beta\rho_{2}(v^{\bot})$ and
$\rho=\alpha\rho_{1}(x)\mu(x)+\beta\rho_{2}(x)\mu^{\bot}(x)$. Example: Assume
conjugation according to $dU=dI$ iff $dU^{\bot}=0$, that is $dU^{\bot}\in G$
iff $dU-dI\in G$ but $dI\notin G$. A necessary condition for $dI\in G$, is a
regular neighborhood of $dU=dI$.

Assume $\Phi(f)=\int fdU$, assume $f=T_{x}(\phi)$, why given that the topology
for $\Phi$ is nuclear, we have $<T_{x},dU>(\phi)=\Phi(f)$, where we assume the
topology invariant for U. In a non-nuclear topology, we do not necessarily
have $\Phi(f)-f\rightarrow 0$ implies $(T_{x}-I)\rightarrow 0$. Example:
assume
$\mathcal{H}_{1}\subset\mathcal{H}\subset\mathcal{H}_{2}=\mathcal{D}^{\prime}$.
That is assume $\Phi(f)-f\rightarrow 0$ in $\mathcal{H}$ and
$(T_{x}-I)\rightarrow 0$ in $\mathcal{H}_{1}$, that is we assume
$\mathcal{H}_{1}$ nuclear. Further, $\Phi(f)-f\rightarrow 0$ implies
$T_{x}-I\rightarrow 0$ in $\mathcal{D}^{\prime}$, but not in $\mathcal{H}$.

Assume existence of $\Phi$ (ähnlich), with $dU=\Phi(dI)\in G$, but
$\Phi^{-1}(dU)=dI\notin G$. Consider $K(G)$, that is $(dV,dV^{L})\in G\times
G$. Starting from $G_{8}$, consider $G_{8}^{\prime}=\\{dV\quad dV+dV^{L}\in
G_{2}\\}$. Consider $G_{8}^{\prime\prime}=\\{dV\quad\Phi(dV,dV^{L})\in
G_{2}\times G_{2}\\}$ and consider $G_{8}^{ac}=\\{dV\quad\Phi(dV,dV^{L})\in
G_{2}^{ac}\times G_{2}^{ac}\\}$, where we assume $dU\in G_{2}^{ac}$ with
$dU=0$ implies $U=I$ (invariants). Further, $H(K(G_{4}))\subseteq C(G_{2})$,
where $dI\notin(G_{2})_{H}$, but $dI\in(G_{2})_{C}$. Given $\Phi\rightarrow
dI$ regularly, we have an approximation property. Assume $\\{\Phi<\lambda\\}$
is semi-algebraic (and BV). We then have $dI\notin G_{2}$ (in H), but
$\Phi(dV,dV^{L})\in G_{2}\times G_{2}$, that is we consider
$G_{8}^{\prime\prime}$ above, that is not necessarily semi-algebraic in
$G_{8}$.

Consider $f(u,v)$ with isolated singularities, that is $\\{f(u,v)=df=0\\}$ are
isolated. Obviously $\\{(u,v)\quad f=df=0\\}\subset\\{(x,y)\quad f=df=0\\}$.
$f(u,v)\in L^{1}$ is related to $f_{0}$ very regular in $L^{1}$ and when we
assume $f\rightarrow f_{0}$, we have $dI\in G_{2}$.

Consider $dV\rightarrow dV\times dV\rightarrow dU\times dU^{\bot}$, in
particular where d U is determined by d V only, ex $dU\rightarrow
dV\rightarrow dI$ simultaneously $dU^{\bot}/dV\rightarrow 0$, for instance d U
of type 0, real dominant. Note that $dV\times dV\rightarrow dU\times
dU^{\bot}$ can be given as projective, given $dU^{\bot}\bot dU$. Assume that
the support for f includes $\\{dU=dI\\}$ and given existence of $dV=0$ over
invariant sets to dU (projective conjugation), we have that a neighborhood of
invariant sets can be given as a regular analytic domain. Consider $dV(f)=\xi
f_{x}+\eta f_{y}$, we then have $dV(f)=dV(u)f_{u}+dV(u^{\bot})f_{u^{\bot}}$.

###  0.3.2 Partial conditions

###### Partial regularization 0.23

Assume $f(u_{1},\ldots,u_{k})$ have regular approximations in some directions.
Given a condition that invariant sets in irregular directions are subsets of a
domain of holomorphy and given $f$ partially hypoelliptic, $f^{N}$ has regular
approximations in all directions.

Assume $dU+dW=dI$ in nbhd $\Sigma(dU)=\\{dU=dI\\}$, where dU in $G$. Thus, the
polar set is subset of $dW=0$. Note that $WI=IW$ does not imply $dW=0$. Linear
phase for W acting on algebraic elements, is sufficient for an algebraic polar
set.

Consider $H(G_{2})\subset H(K(G_{4}))$, with the condition that every domain
$\Omega$, has a movement in $H(K(G_{4}))$, with $\Omega$ as a domain for
analyticity. Example: $f\in C(G_{2})$ with $dU_{3}(f)=0$, where $dU_{3}\in
K(G_{4})$. Then given $f(u_{1},u_{2},u_{3})$ partially hypoelliptic, there are
N, with $f^{N}\in H(\Omega)$.

Consider $U\rightarrow U^{\diamondsuit}\rightarrow U^{\bot}$, where the last
mapping is trivial. ([1]), given $dU=0$ iff $dU^{\diamondsuit}=0$, we have an
analytic representation. Pure mappings preserve analyticity in the plane, that
is symmetry with respect to $(x,y)\rightarrow(y,x)$, further
$(x,y)\rightarrow(x,-y)$. Given $(dU)^{\bot}\simeq dU^{\bot}$, with the
conditions above, we have an analytic orthogonal. Assume partial P-convexity,
with symmetric domain otherwise, we then have $f^{N}$ has a corresponding
P-convex G, in particular an approximation property in all variables
separately.

The condition on invariant sets in a domain of holomorphy, implies
$\\{U-I\\}\subset\\{dW=0\\}$, where $dW$ analytic, that is U has an analytic
projective conjugate, that is d W with d U + d W=d I, that can be continued to
$U^{\bot}$. Consider $\pi:(x,y,z)\rightarrow(u,v)$, so that
$\pi^{-1}(0)\sim(x,y,z)$. Example: $(dU,dU^{\bot})$ can be reduced to a normal
system $(dN,dN^{\bot})$, so that $dN(U,U^{\bot})$ is only dependent of d U
close to the boundary. Alternatively, consider ($\rho dU_{1},\vartheta
dU_{2})$ with $\rho dU_{1}\rightarrow dI$ and $\vartheta dU_{2}\rightarrow 0$,
as regular approximations. We can assume, in particular
$(dU_{1},dU_{2})\rightarrow(dI,0)$ in a normal model. We have the case
$dI\notin G$, when $1/\rho$ is not regular.

Example: assume $f(v_{1},v_{2})\rightarrow 0$, when $v_{1}\rightarrow 0$ is
regular, that is $v_{1}=0$ has a regular neighborhood. In the converse case,
assume that $v_{2}$ can be continuously deformed to a non-regular direction,
for instance $f\nrightarrow 0$, $v_{2}\rightarrow 0$. Sufficient for a regular
neighborhood, is that f is algebraic in $v_{2}$.

Example: assume u,v one-parameter, with $f(u/v)\rightarrow 0$ regularly, but
not $f(u)\rightarrow 0$ regularly. If $f(u,v)\rightarrow 0$ regularly, when
$u\in L$ and $u\rightarrow 0$, that is L gives a regular neighborhood of 0, at
the same time, $f(u,v)\rightarrow g(u)$, when $v\rightarrow 0$, that is
$\lim_{u\rightarrow 0}\neq\lim_{v\rightarrow 0}$. Then, we do not have regular
continuations according to a normal model. However, there may be $dW\in
G_{8}$, with $f(u,v,w)\rightarrow 0$ regularly. Assume the boundary is given
by $U=I$ and that $R(dU)^{\bot}=R(dV)$ and that the polar is defined by
$Vf=0$. A very regular boundary over $K(G)$, can be represented by
$f^{(k)}(u,v)=const$ implies $u=v=0$. For a normal model, we require that
$\lim_{u\rightarrow 0}\frac{\delta f}{\delta v}(u,v)=0$ regularly. Note that
$f^{N}$ is corresponding to real type, does not imply that f is corresponding
to real type.

###  0.3.3 Reduction to normal model

###### The group for the inverse mapping 0.24

Assume $F(\gamma_{T})(\zeta)=F(\gamma)(\zeta_{T})$, then $U\gamma_{T}(\zeta)$
is linear in $\gamma_{T}$, but only corresponds to a transformation in
$\zeta_{T}$.

Assume $\zeta_{T}$ through a transformation in $C(G_{2})$, simultaneously
analytic in $K(G_{4})$, that is $G_{2}\gamma(\zeta)=\gamma(K(G_{4}))$. Note
that, for $\gamma(\zeta)=(\gamma_{1}(\zeta_{1}),\gamma_{2}(\zeta_{2}))$, $\xi
f_{\zeta_{1}}+\eta f_{\zeta_{2}}$
$=\xi\gamma_{1,\zeta_{1}}f_{\gamma_{1}}+\eta\gamma_{2,\zeta_{2}}f_{\gamma_{2}}$.
Thus, given a movement in $\zeta$, we can derive a movement in
$\gamma_{1},\gamma_{2}$, according to $\zeta\rightarrow\gamma\rightarrow
F(\gamma)(\zeta)=f(\zeta)$. Assume $F(U\phi_{T})(\zeta)=F(\phi)(\zeta_{T})$,
then given d U is analytic, $\zeta_{T}$ is defined through a continuous
transformation. Example: $dU_{1}(f)(\zeta)=\xi(\zeta)f_{x}+\eta(\zeta)f_{y}$.
Given $\zeta\rightarrow(x,y)$ analytic, through Cauchy-Riemann and if we
consider $\zeta_{2}$ as parameter, $\xi=(y+x)_{\zeta_{2}}$ and
$\eta=(y-x)_{\zeta_{2}}$. Thus $\xi+\eta$ is only dependent of y and
$\xi-\eta$ is only dependent of x.

Assume $U$ sub-nuclear , that is $dU=\rho dV$, where $dV\in G_{r}$ and $dU\in
G$. Given $\rho\rightarrow 1$ regular and $\mathcal{H}(dV)$ nuclear, we have
$\mathcal{H}(dU)$ nuclear. A special case of TP is $U\rightarrow
U_{r}\rightarrow\zeta$, where $\mathcal{H}(dU)$ nuclear. Thus given $dU=0$, we
have existence of $dU_{r}=0$, that is $\rho\neq 0$ and $1/\rho\in\mathcal{H}$.

Assume $dU(x,y)$ analytic over supp f with $f\in(I)$. We then have
$f(u,u^{\bot})$ analytic implies $f(x,y,z)$ analytic. Starting with
$F(\gamma_{1}\rightarrow\gamma_{2})(\zeta)$ analytic, it is necessary that
$\gamma_{2}\rightarrow\gamma_{1}\rightarrow F(\gamma_{1})$ continuous. We
consider $\gamma_{1}\rightarrow\gamma_{2}$ as topological completion.

###### Reducibles 0.25

Assume $dN=\rho dU$ a reduction, with $(dN,dN^{\bot})\rightarrow(dI,0)$
regular and $dN=dI$ rectifiable. Then $\rho(u,u^{\bot})\rightarrow 1$ as an
absolute continuous function and the reduction preserves zero-sets ([18]).

More precisely $\rho:\\{dU=dI\\}\rightarrow\\{dN=dI\\}$ is absolute continuous
and maps sets of measure zero on sets of measure zero. We assume $\rho$
analytic, where $N^{\bot}\neq 0$. In particular the limit of $\rho$ is not
dependent of path for the approximation. Given $\rho$ monotonous, that is
single valued, we have close to the boundary, $\rho(u,u^{\bot})\sim
c_{1}u+c_{2}$, that is we do not have a spiral axes. Example:
$dU^{\bot}=\sigma dN^{\bot}$, with $\sigma^{k}$ polynomial, alternatively
$dU^{\bot}=\mu dU$, with $\mu$ polynomial. We can obviously determine a domain
for absolute continuity in $\\{(u,u^{\bot})\\}$, where $\delta\rho/\delta
u^{\bot}\rightarrow 0$ and we have two-sided limits, when $\pm
u^{\bot}\rightarrow 0$ regularly, cf sharp fronts ([14]).

Example: $dU=\rho dV=\rho\sigma dI$. Given $\rho$ bounded and absolute
continuous, with $Vf=If$ iff $f(v)=f(0)$ $(=f(x,y,z)$), we have where
$d\rho=0$ that $dU=dI$ implies $dV=dI$, that is we have reduction of dU to for
instance a normal model. Example: $dU^{\bot}=dV^{\bot}W^{\bot}$, where
$dV^{\bot}\rightarrow 0$ regularly. Assume $dW=pdI$ and $dW^{\bot}=1/pdI$.
Then we must have $pdU^{\bot}\rightarrow 0$ regularly.

Example: Assume improperly $VIf=f(v-0)$ and $IVf=f(v+0)$, given $VI=IV$ over
the support to f and given one-sidedness for R(V), we do not have traces. We
conclude that one-sidedness for R(V) preserve convexity in the plane. An
irreducible continuation $dU=\rho dV$, is unique. When $\rho$ is linear in V,
the continuation can be given through iteration. Given $dU=dI$ implies d V =
dI, the continuation preserves type. Assume for real numbers $\alpha,\beta$,
$f(\alpha u+\beta u^{\bot})\leq\alpha f(u)+\beta f(u^{\bot})$, where
$\alpha+\beta=1$, that is f is convex (absolute continuous). Given
$f(u)\rightarrow f(0)$, we have $f(u^{\bot})\rightarrow 0$.

Assume $f(u)=f(0)$ implies $u\in L$, for instance f 1-homogeneous over
invariant sets. Obviously, we have existence of $dV$ conjugated, so that f
(-1) homogeneous with respect to v. We assume f independent of v on L. For
irreducibels, we have $f(u,v)=f(0)$ implies $U=I$ or $V=I$, for instance a
projective model, invariant on L, where $L\sim 0$.

###### Volume preserving conjugation 0.26

It is not clear that the reduction model above, is independent of scaling
$f\rightarrow\tilde{f}$. However, given $dU\rightarrow dU^{\bot}$ conjugated
according to $t\rightarrow 1/t$ (scaling parameter), we have that
$\tilde{f}(u,u^{\bot})$ has the same volume as $f(u,u^{\bot})$. P-convexity
for the group implies further, that $U^{\bot}f=0$ implies t finite.

Consider $f(x,y)M(y/x,x/y)\simeq f(x,y/x)f(x/y,y)$, with M=1, we then have
$f(x,y)=0$ implies $f(x,y/x)=0$ or $f(y,x/y)=0$ (or both). Irreducibility
excludes the case both. The condition M=0 for irreducibels implies $U=U_{1}$
or $U=U_{2}$, that is a decomposition into irreducibels excludes spirals.

###  0.3.4 The polar group

Example: consider $E$, the localization of f, $C=\\{(u,u^{\bot})\quad
E(f)(u,u^{\bot})=0\\}$, where $E-I=0$ modulo regularizing action, C gives the
polar set to hypoellipticity. Assume $(u,u^{\bot})$ is dense in the domain for
f, E has type 0 over $u=u^{\bot}$ and does not correspond to a regularizing
action. Assume $E_{k}$ corresponds to the localization of $f^{k}$, where f is
assumed partially hypoelliptic, that is $f^{k}$ locally hypoelliptic, for
large k. We then have ker $E_{k}\downarrow\\{0\\}$ (modulo regularizing
action), when $k\uparrow\infty$.

Consider $C(G_{2})\rightarrow H(K(G_{4}))$, where we assume $H(K(G_{4}))$ with
an approximation property, that is uniform convergence on convex, compact
equilibrated sets. Assume further a strict condition, that
$(v,v^{L})\rightarrow v+v^{L}$ continuous and surjective. Then we have that
$C(G_{2})$ has an approximation property.

$\Omega(x,y,y^{\prime},y^{\prime\prime})=y^{\prime\prime}-\omega(x,y,y^{\prime})=0$
represents at most 8 infinitesimal transformations in the plane ([20]).
Maximal for $\omega\equiv 0$. $y^{\prime\prime}=0$ has as integral curves
$\infty^{2}$ straight lines in the plane. $\Omega=0$ can be deformed to
$y^{\prime\prime}=0$. In the plane there is a domain $\Sigma$, such that
$p,q\in\Sigma$ implies existence of a unique integral curve through $p,q$.

Example: projective transformations map lines in the plane on lines in the
plane. Through Theorem 35 ([20], cf. ch. 17 prgf. 6)
$-\xi_{yy}\rho^{3}+(\eta_{yy}-2\xi_{xy})\rho^{2}+(2\eta_{xy}-\xi_{xx})\rho+\eta_{xx}=0$,
where $\xi,\eta$ is only dependent of $x,y$. Since $\xi_{yy}=0$ implies
$\xi=X(x)y+X_{0}(x)$ and in the same manner for $\eta_{xx}=0$. Simultaneously,
$\eta_{y}-2\xi_{x}=X_{1}(x)$, $\xi_{x}-2\eta_{y}=Y_{1}(y)$. Thus
$3\xi_{x}=-2X_{1}(x)-Y_{1}(y)$ and $3\eta_{y}=-2Y_{1}(y)-X_{1}(x)$. Further,
$3X^{\prime}y+3X_{0}^{\prime}=-2X_{1}(x)-Y_{1}(y)$,
$3Y^{\prime}x+3Y_{0}^{\prime}=-2Y_{1}(y)-X_{1}(x)$. The conclusion is,
$Uf=(a+cx+dy+hx^{2}+kxy)\frac{\delta f}{\delta
x}+(b+cx+gy+hxy+ky^{2})\frac{\delta f}{\delta y}$. Thus, there are 8
independent transformations that are generated by
$f_{x},f_{y},xf_{x},yf_{x},xf_{y},yf_{y}$,$x^{2}f_{x}+xyf_{y}$ and
$xyf_{x}+y^{2}f_{y}$.

Note $Wf=g$ with $Wf=\int f_{1}dW$, for some $f_{1}$, where g=$Ig=\int gdI$,
that is solvability does not imply an approximation property for the solution.
The condition that $dU=0$ over $\gamma_{1},\gamma_{2}$, implies that the
mapping $\gamma_{1}(u)\rightarrow\gamma_{2}(u)$ is projective. That is dU
defines a geometric ideal, $u\in N(I)$. Assume $L_{1},L_{2}$ are invariant
lines to $dU_{1},dU_{2}$, further $L_{1}\rightarrow L_{2}$ projective, we then
have if $dU_{1}=0$ over $L_{2}$, $\gamma_{1}(0)=0$ implies $\gamma_{2}(0)=0$.

##  0.4 TP and P-convexity

Given strict pseudo convexity, we have $H(G_{2})\subset L^{1}$, with an
approximation property, in the converse case, we have that $H(K(G_{4}))$
approximates $L^{1}$ uniformly.

We consider the transmission property over $T_{ac}(dI)=\cup_{\mid\rho\mid\leq
1}\rho dI$, where $\rho\rightarrow 1/\rho$ have the same constant surfaces,
that are assumed trivial. Example: $\rho=p/q$, where either $p/q\rightarrow 0$
in $\infty$ or $q/p\rightarrow 0$ in $\infty$.

Assume for every $dU(x.y.z)$, there are $\Omega$, so that $dU$ are analytic
over $\Omega$. Consider $dU=\rho dI$, given $\rho\in L^{1}$, given a strict
condition, there are $\rho\in H$ for which $dU$ is analytic. Assume
$\phi\rightarrow\phi_{1}$ translation of the support to $\Omega$, so that $dU$
is analytic over the support to $\phi_{1}$, then modulo translation, we can
assume dU analytic.

Consider $U\rightarrow{}^{t}U\rightarrow U^{\bot}$, under a strict condition,
with $dU^{\bot}=0$ (polar set). The TP model (special case) gives
$F(\gamma_{1}\rightarrow\gamma_{2})(\zeta)\rightarrow\zeta$, where
$\gamma_{2}$ analytic, can now be continued to $\mid\zeta\mid^{\sigma}\leq
C\mid\gamma_{3}\mid\leq\mid F(\gamma_{1})\mid$, that is $\gamma_{3}$ is
locally reduced. Sufficient is $\mid\zeta\mid^{\sigma}\leq
C\mid\gamma_{3}\mid^{N}\leq\mid F(\gamma_{1})\mid$, that is a sufficient
condition is $\gamma_{3}$ partially hypoelliptic. In this case the polar set
can contain spiral sets, that is we do not have a strict condition. Thus,
given TP, we can prove that P-convexity implies hypoellipticity.

Topological singularities: Consider $\gamma_{1}\rightarrow\gamma_{2}$
according to $\mathcal{H}\rightarrow H$ or $C^{\infty}$, that is we have
completion to $C^{\infty}$ through a subgroup $G_{reg}$. Note that $T(\phi)\in
C^{0}$ is necessary for a representation with a non-decreasing Stieltjes
measure.

Assume $h(f/g)\rightarrow 0$ defines convergence in norm, that is an
approximation property with respect to norm. We assume existence of W, so that
$W\mid f\mid\simeq f$ (rotation), relative topology. Thus, convergence in norm
with respect to U, implies mean convergence with respect to U,W. W is assumed
pseudo local with $W\rightarrow I$ in $C^{\infty}$, thus (topologically)
algebraic.

###### Condition for mean convergence 0.27

Assume $dU=\rho dV=\rho\mu dI$, where $1/\rho\notin\mathcal{H}^{\prime}$.
Assume $h(f/g)=0$ implies $\mid\mid f\mid-\mid g\mid\mid\rightarrow 0$ that is
ker h defines a symmetric ideal (complex conjugation). However, $\mid g\mid$
polynomial, does not imply that f is approaching a polynomial, but given
isolated singularities for h, we can assume $f\rightarrow g$ in the mean.

Existence of d W, so that $W(\mid f\mid-\mid g\mid)\sim\mid f-g\mid$, is
motivated by the convergence being preserved under $S:(x,y)\rightarrow(x,-y)$
and ${}^{t}Id:(x,y)\rightarrow(y,x)$.

Assume $dUV=\rho\mu dI$, where $dU\in C(G_{2})$ and $dV\in H(K(G_{4}))$.
Sufficient for $f\rightarrow\mu f$ to preserve analyticity, is that $\mu$ is
analytic. Given $1/\mu\sim\rho$, we require that $f\rightarrow\rho f$ preserve
continuity. Example: $d\mu\neq 0$ (conformal) gives $\rho\in C$. We have
isolated singularities for h, only if the limit involves both $U_{1}$ and
$U_{2}$, that is completeness for the limits.

###  0.4.1 Representations

Assume $dU\rightarrow dV$ contractible, in particular a restriction without
changing type, that is without affecting invariant sets. Note that two scalar
products are involved. Consider
$(dU,dU^{\bot})\rightarrow(dU,dU^{\bot}/dU)\rightarrow(dI,0)$, defines a weak
orthogonal in $\infty$, corresponding to a strict condition for convergence.
Alternatively, determine $\Gamma$ with $dN\neq 0$ on $\Gamma$ and $dN\bot dU$,
that is dU is a continuation of d N, analogously with P-convexity.
Simultaneously, $dN\rightarrow dI$ on a component (contractible). Note, using
Schwartz kernel theorem, $<<T(\phi),dU>,dU^{\bot}>=<T(\phi),<dU,dU^{\bot}>>$.

Assume $C(G_{2})\backslash H(G_{2})\simeq H(K(G_{4}))$, where H approximates
$L^{1}$, given strict pseudo convexity. The condition $dU\bot dU^{\bot}$, is a
weak condition with respect to $C(G_{2})$, that is it is defined by Stieltjes
measures. The completion to $L^{1}$ defines $dU^{\bot}$ ([1]). Consider
$<T_{x.y},V^{L}\phi_{1}\otimes V\phi_{2}>=<T_{x,y},dV^{L}\otimes
dV>(\phi_{1}\otimes\phi_{2})$. Assume $T_{x,y}$ very regular and consider
$T(\phi_{1}\otimes\phi_{2})=0$, that is movements $dV^{L}\otimes dV$, that
preserves a zero space. Thus, $dV^{L}\otimes dV$ describes the orthogonal to T
over $\phi_{1}\otimes\phi_{2}$.

Assume $dUdU^{\bot}=\mu dNdN^{\bot}$, where the boundary is defined according
to P-convexity. Given $dUdU^{\bot}$ gives the volume for $\Omega$,
$dNdN^{\bot}$, when $N^{\bot}\rightarrow 0$, gives the measure for $bd\Omega$.
A necessary condition for TP is that $\Omega\rightarrow bd\Omega$ continuous,
that is we assume $dUdU^{\bot}$ invertible, that is $\mu\neq 0$. Example:
assume $dU=\rho\sigma dI$, with $dU\rightarrow dN$ regular, that is consider
$\rho\sigma$ regular. Given $dU=\rho dN$ and $dU^{\bot}=dN/\rho$, we have
$dN(U,U^{\bot})=(1/\rho)dU+\rho dU^{\bot}$, where $\sigma/\rho\geq 0$ close to
the boundary.

Consider $U(N,N^{\bot})\rightarrow N(U,U^{\bot})$. When $N\rightarrow I$,
$U,U^{\bot}$ describes the polar set. According to Gauss-Greeen
$-\int_{S}-YdU-XdU^{\bot}=\int\int_{A}(X_{U}+Y_{U^{\bot}})dUdU^{\bot}$. Assume
for instance Y,X are given by $dN/dU$, $dN/dU^{\bot}$. Given N symmetric in
$U,U^{\bot}$, we have $\int_{S}dN=0$, when $N\rightarrow I$ regularly, why A
must be trivial in a normal model.

Example: $dU\rightarrow dI$ with $dU/dN^{\bot}\rightarrow 0$, simultaneously
$dU^{\bot}\rightarrow 0$. An algebraic continuation preserves reducedness, in
particular $dN\rightarrow dU$ is reversible, when $dN^{\bot}/dN\rightarrow 0$.

###  0.4.2 A transmissions group

Starting from the condition, that $U\rightarrow{}^{t}U$ preserves type, we can
assume a Dirichlet problem, that is $\eta_{x}-\xi_{y}=0$
($\xi_{x}+\eta_{y}=0$), for
$\mathcal{O}_{M}(G_{8})\simeq\mathcal{O}_{M}(G_{4})\times\mathcal{O}_{M}(G_{4})$
([21]). Consider $A\simeq dV\times dV^{L}\rightarrow dU\times dU^{\bot}=C$ BV,
as a continuous continuation. Assume $B=dN\times dN^{\bot}$ defines a
rectifiable boundary, with $N^{\bot}/N\rightarrow 0$ and $dN\rightarrow dI$.
With these conditions, we assume TP is given by $U\times U^{\bot}\rightarrow
N\times N^{\bot}$ continuous (absolute continuous). With the condition that
$dN\times dN^{\bot}$ is P-convex, we have that hypoellipticity is preserved.
Sufficient for preservation of $T_{ac}$-convexity, is that $dU\rightarrow
1/\rho dU$ is absolute continuous.

###### A TP subgroup 0.28

Assume Y : $\gamma_{1}\rightarrow\gamma_{2}$ projective on test-functions and
for instance $F(\gamma_{1})$ continuous, but $F(\gamma_{2})$ analytic.
Example: $A\rightarrow B\rightarrow_{Y}C$ and $A\rightarrow_{U}C$, that is
$Y^{-1}U:A\rightarrow B$ and $U^{-1}Y:B\rightarrow A$.

Note that given strict pseudo convexity, we can define $L^{1}(bdX)\rightarrow
H(bdX)$, that is a monotropic operator, further the scheme above can be given
as symplectic.

Assume F defines singularities and $WF\cap F=\\{0\\}$, that is dW is in the
regularity group, then G can be defined as irreducibels relative W, that is
$U\neq W^{-1}$ over F. Given $dW^{2}$ has isolated singularities, when $Wf\in
C^{\infty}$ for $f\in C^{\infty}$, we have that W can be represented as
algebraic, analogous with an approximation property.

Consider $dV\subset dU\subset dV$, with $dU=\rho dV$ and $\mid\rho\mid\leq 1$,
that is $dU\in T_{ac}(dV)$. We assume $\rho\rightarrow 1$ at the boundary,
further $dV=\sigma dU$, with $\sigma\rightarrow 1$ at the boundary. Example:
$dU=dV$ outside compact, and otherwise one-parameter $pdU=qdV$, for
polynomials p,q, that is we have $p/q\in H$, except for finitely many terms.
Example: assume $dU^{\bot}=\rho^{\bot}dV^{\bot}=\rho^{\bot}\mu^{\bot}dI$ and
that $\rho^{\bot}\mu^{\bot}\rightarrow 0$, as $t\rightarrow\infty$,
simultaneously $\rho\rightarrow 1$, as $t\rightarrow 0$.

###  0.4.3 Functional transformations

P-convexity implies that $U^{\bot}f\neq 0$ in the infinity. Further, assume
existence of d N with $Nf\neq 0$ and $dU\bot{}^{t}pdN$ on the boundary. $dI\in
G(R(N))$ (range) implies $dUN=\rho dN$, with $1/\rho$ regular. Given
$U(N,N^{\bot})$ single valued, we have $dU/dN=1$ at the boundary, that is
$dU\rightarrow dN\rightarrow dI$ regular. Note that, $U^{\bot}\phi=0$ implies
$u^{\bot}$ finite, that is $u^{\bot}\nrightarrow 0$, but
$u^{\bot}/u\rightarrow 0$.

Starting from $H(K(G_{4}))\subseteq C(G_{2})$, we have in H, that $f\equiv
const$ on a boundary of positive measure, implies $f\equiv const$ on a convex
neighborhood. Through the homotopy mapping we can prove, given f has a zero in
the neighborhood, that $f\equiv 0$, for instance
$f(K(G_{4}))\sim_{0}g(G_{2})$, gives that g reduced to a normal model is
$\equiv 0$, and can be continued to a convex neighborhood.

###### Normal model 0.29

Assume $\Phi(u)=u-\mathcal{F}(u)$, for instance $\mathcal{F}(u)=u^{\bot}$
(conjugation), starting from a strict condition, for instance $u\rightarrow 0$
iff $U^{\bot}\rightarrow 0$, we have that $\Phi$ bijective, corresponds to
density for $(u,\Phi)$.

Assume $\Phi$ defined by $d\sigma$, for instance Dirichlet measure, L a zero-
line to $d\sigma$ in the sense that $\Phi(L)\equiv 0$, implies that $\Phi$ is
projective on the x-domain. When $Id=\delta_{x}(y)$, we have that Id has
support on the diagonal $x=y$, that is a local operator. Given Id has point
support and x fixed, we have $dI\notin G$, when $x\neq y$. In particular, when
the domain is G, $u\neq v$ implies Id has no support, when u varies. This
means that an approximation property is only possible, by including the
diagonal. Further, $\Phi\equiv 0$ on the diagonal (spirals). Example:
$u\rightarrow v$ as continuation, the diagonal gives trivial continuations.
Orthogonal continuations can be given as regular.

Consider $dV\rightarrow dV\times dV\rightarrow dU\times dU^{\bot}$. Starting
from a strict condition $dU^{\bot}/dU\rightarrow 0$, we get
$dU^{\bot}(V,W)\mid_{W=V}\rightarrow 0$, when $dU\rightarrow dV$. Example:
consider $dU^{\bot}$ as a measure in $\\{dU\\}^{\bot}$, we have
$dU^{\bot}/dV\rightarrow 0$, that is a strictly weaker continuation, in
particular $dV$ and $dU^{\bot}$ are not of the same type. When V is Lagrange,
we assume $\Delta V=0$, that is $(dU,dU^{\bot})$ corresponding to real type
relative d V. The condition $dU^{\bot}/dV\rightarrow 0$ regularly, implies
that $(U,U^{\bot})$ has a trivial polar set relative (V,V). Example: assume
$dV/dI=\rho\rightarrow 1$ regularly. Further, $dV\rightarrow dU$ according to
$\rho\rightarrow 1/\rho$, a sufficient condition for $dU^{\bot}/dI\rightarrow
0$, is given by $dU^{\bot}/dU=\sigma\prec\prec\rho$, that is given
$\rho\rightarrow\infty$, $\sigma(1/\rho^{2})\rho\rightarrow 0$. Further,
$\rho\rightarrow 1$, as $t\rightarrow 0$, gives the condition for a diagonal,
$\sigma/\rho\rightarrow 1$ regular.

Example: assume $\Phi$ projective, according to $dU=\rho dV$, with $\rho\neq
0$ on L. Further, $\rho\rightarrow 0$ on L, with $\rho\neq 0$ outside
$\infty$. The corresponding condition $dU-dI=\rho(dV-dI)$, gives that the type
is preserved, the condition is dependent of topology.

###### Topological completion 0.30

When $\frac{d\mathcal{F}}{dx}$ is not bijective, consider $v$ so that
$\frac{d\mathcal{F}}{dv}$ is bijective. We get density through topological
completion to R(V).

Note that $d\mathcal{F}(u)/du\rightarrow 0$ regularly, when
$u\rightarrow\infty$, that is $dv/du=0$ implies $1/u=0$, sufficient is $\mid
du/dv\mid\leq\mid u\mid$.

Example: Vf=0 iff Wf=f, that is start from invariant sets. Assume
$dI\in(K(G_{4}))_{H}\subseteq(G_{2})_{C}$. Given $f=T(\phi)$ we assume
$f(u,u^{\bot})\rightarrow f(0)$ in $C^{0}$ iff
$f(u_{1},\ldots,u_{8})\rightarrow f(0)$ in H. The principle for preservation
of approximation property is $\mathcal{H}_{1}\subset\mathcal{H}_{2}$ implies
$G_{\mathcal{H}_{2}}\subset G_{\mathcal{H}_{1}}$. Note that
$(U,U^{\bot})f\rightarrow f(u,u^{\bot})$ is not necessarily continuous.
However $(U,U^{\bot})f=f(u_{1},\ldots,u_{8})$, that is the domain for regular
convergence in $G_{2}$ has the corresponding domain for the arguments in
$G_{8}$. Isolated singularities in $G_{8}$ are according to
$\\{g(v)=\delta_{j}g(v)=0\\}$ trivial, for $j=1,\ldots,8$. The corresponding
condition in $G_{2}$ is not continuous derivatives, but possible existence of
w, so that $g(u,u^{\bot},w)\in C^{1}$.

Assume $dI\notin G$, but $dU-dI\in G$, along invariant sets in the domain, we
do not assume an approximation property, thus not all approximations are
regular. d U - d I ($\simeq dU^{\bot}$) bijective corresponds to
symplecticity. Given completion through $V(f)$, we have $\int f(dU-dI)=0$
implies $Vf=0$, that is trivial when V algebraic

N considered as a functional over d U, gives that $N\in T^{*}(bdX)$, where
$X=R(dU)$. Consider $N(dU)=dN_{1}(U)$, given $bdX$ strictly pseudo convex, we
have that $\Sigma=\subset T^{*}(bdX)$ is symplectic.

###### Semi-algebraic sets 0.31

Consider $dV\in G_{8}$ with $dV+dV^{L}=dU_{1}\in G_{2}$, with $dI\in G_{2}$,
Then we have $\\{f(v,v^{L})<\lambda\\}$ semi-algebraic, implies
$\\{f(v,\cdot)<\lambda\\}$ semi-algebraic.

Assume $dI\in G_{2}$. For a geometrically convex closure of $If$ , we consider
$dV+dV^{L}\in G_{2}$. For holomorphic convex closure, we assume $dV+dV^{L}=dI$
iff $dW=0$ (analytic), with $\mid f(w)\mid\leq\sup_{K}\mid f\mid$, for a fixed
compact K, defining a regular neighborhood of $dV+dV^{L}=dI$. Example: $\mid
f(w)\mid\leq e^{\epsilon\mid v+v^{L}\mid}$ close to invariants. Given $dW$
projective, zero-lines L are preserved. Given $dW$ algebraic according to =Pd
I, for a polynomial P, we must have $L=\\{0\\}$. Given d V projective, PdI
algebraic corresponds to a strict condition.

Assume $dU^{\bot}=\rho dU$, where $\rho\rightarrow 1$ in $T_{ac}(H(G_{2}))$,
that is the disk closure when $\rho\rightarrow 1$ in $C(G_{2})\supset
H(K(G_{4}))$, that is we assume $\rho\in H$ implies $\rho\neq 1$. In
particular, $\\{G_{4}\quad\rho<\lambda\\}$ semi-algebraic, but not necessarily
relatively compact.

Weak type is defined through $<T,dU>(\phi)=<T,dI>(\phi)$, $\forall T\in(I)$,
where (I) is a class of symbols, dependent of base $\phi$. Given
$U\rightarrow{}^{t}U$ preserve invariant sets $\Sigma(U)$ over the base, that
is $U,{}^{t}U$ of the same type, we can define invariant sets over the base.

Consider, given $\\{dN=dI\\}$ rectifiable, $dV\times dV^{L}\rightarrow
dN\times dN^{\bot}\rightarrow dU\times dU^{\bot}$, so that $\\{dV\times
dV^{L}=dI\\}\rightarrow\\{dN=dI\\}$ absolute continuous and preserves sets of
measure zero. We assume $(N,N^{\bot})$ defines a normal operator, so that for
instance $N^{\bot}\phi=0$ implies $\phi$ has bounded support. Given
$dV\rightarrow d{}^{t}V$ preserves type, we get decomposable sets
$\\{dU=dI\\}$ according to $(dV\otimes dI)(dI\otimes dV^{L})=dI\otimes dI$.
Note that given $(dV,dV^{L})$ is dense in the domain for $dU$ and projective
as operators, we have $(dV\otimes dI)(dI\otimes dV^{L})\simeq dV+dV^{L}$.
Further, when $V^{L}\simeq{}^{t}V$ (harmonic), $UT(\phi)=VT(V\phi)$. Invariant
sets for strong topology, have corresponding invariant sets for weak topology.
Convergence $dV\rightarrow dI$, implies $d{}^{t}V\rightarrow dI$, that is we
have two-sided convergence. In particular, $dV\times dV^{L}$ preserve
relatively compact sub-level surfaces. Conversely, the mapping
$\\{dU^{\bot}=0\\}\rightarrow\\{dN^{\bot}=0\\}$ is absolute continuous and
preserve sets of measure zero. Thus, $N(U,U^{\bot})$ is at the boundary given
by N(U), that is not a spiral axis.

###  0.4.4 Order of reducible movements

###### Order for constant surfaces 0.32

Assume $A_{1}A_{2}A_{3}=I$ under the condition $A_{1}=A_{2}^{-1}$, then we
have order 1, given all three $=I$, we have order 3 (maximal order)

Note $f(u_{1},u_{2})=const$ in C iff $f(u_{1},\ldots,u_{8})=const$ in H, that
is the order is dependent of topology. Consider $M(x/y,y/x)$, with dense
domain and M algebraic, the condition $x/y\rightarrow 0$ or $y/x\rightarrow 0$
in $\infty$ implies not on the reflection axes, that is M is assumed algebraic
outside invariant sets. Irreducibels: $(A-I)(B-I)=I$ iff $AB=A+B$, that is
irreducibels give decomposable constant surfaces.

AB=I with $A\bot B$ non-linear, implies A=B=I that is of order 2. Assume AB=I,
given $A=1/B$ single valued in the plane, then A can be given linearly.
Example: choose B monotonous and minimal, according to $dB\neq 0$. Example:
$dU^{-1}\notin(G_{2})_{H}$ and UV=I, with $dU,dV\in(G_{2})_{H}$ implies
$U=V=I$, that is irreducibels relative $(G_{2})_{H}$.

Example: $dU=\rho dI$, with $\rho\rightarrow c$ in $\infty$, $dW=\mu dI$ and
$dU=\sigma dW$, with $\sigma\rightarrow 0$ in $\infty$, we then have
$\sigma=\rho/\mu$, sufficient is $\mu\rightarrow\infty$ in $\infty$.

###  0.4.5 Continuation according to TP

Continuation according to TP can be seen as a linear problem, given
independence of $y^{\prime}$. We have independence of $y^{\prime}$ ([20]),
given $\eta_{x}=0$, $(\eta_{y}-\xi_{x})=0$,$\xi_{y}=0$, for instance
$(\xi,\eta)$ is given by $(1,1),(x,y)$. Assume $dI\rightarrow dN\rightarrow
dU=\rho dN$, where $dN\rightarrow dU$ continuous, in particular $dN\rightarrow
dI$ implies $dN^{\bot}\rightarrow 0$, thus N has order 2, that is translation,
rotation. We further assume $dU^{\bot}$ dense in $R(dU)^{\bot}$, in the sense
that $\forall g\in\mathcal{H}$, we have existence of $f\in R(N)$, so that
$(U+U^{\bot})f=g$.

###### Reduction to a normal model 0.33

Consider $f_{L}(u,v)$, where $u,v$ are one-parameter, for instance $f_{L}$ is
considered over a normal model. Given $\\{f_{L}\\}$ has an approximation
property, with $u\rightarrow 0$,$V\rightarrow 0$ (or $\delta f/\delta
v\rightarrow 0$), we have that $\\{f\\}$, a continuation according to TP, has
an approximation property. In particular, $If_{L}=f_{L}I$ corresponds to
$f_{L}$ is topologically algebraic in $(u,v)$. Assume $f=f_{L}+R$, where R
corresponds to multivalentness for the representation of f, consider for
instance $\\{R<\lambda\\}\subset\subset X$, a semi-algebraic set for R and
$X\subset C^{\infty}$. Given a compact neighborhood of this set, we can assume
continuation to $C^{\infty}$, that is $f-f_{L}$ can be continued to
$C^{\infty}$.

When $(u,v)$ convex and dense in the domain, we assume for a normal model, an
approximation property $(u,v)\rightarrow 0$ regular, that is $f_{L}=\int
f(u,v)dI(u,v)\rightarrow f(0)$ regularly, where we assume $u=v$ implies $v=0$.

In general, we have $\int_{X}(df)\omega=\int_{bdX}f\omega-\int_{X}fd\omega$.
Topological degree gives the order for G that preserve bd X. Using the
logarithm mapping, we have that lines are mapped on to closed contours.
Absence of entire lines in R(U=I), correspond to absence of closed curves
(non-trivial), that preserve $dU=dI$.

## References

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