Analytic finiteness for preparation

Human treatment: Guillaume Valette, On subanalytic geometry, arXiv:2507.23622v1, 31 July 2025, §1.3.2, §§1.4–1.5, the preliminaries in §1.6.1, Proposition 1.6.12, Lemma 1.6.13, Proposition 1.6.14, the dimension induction and cell/preparation theorems in §§1.7–1.8, and the one-variable and parameterized Puiseux results. Adapted by GPT-6.1 Sol (OpenAI), Ultra, October 2026. This adapted component is CC BY 4.0; Guillaume Valette remains the author of the underlying exposition. Changes: notation and Markdown/MathML formatting; the algebra prerequisites and degree-zero case are proved; the module and prescribed-jet versions of the formal linear argument are expanded; real Noetherianity is obtained from the course’s existing division proof; the normalizing parameters are allowed either sign, and the uniform derivative bound is proved. Further additions prove independent-variable convergence of the analytic split, localize on the actual compact image, and give a signed balanced-coordinate substitution under the stated cell-induction inputs. Further proof expansions give bounded-monomial unit representations, compact-domain substitution, common translations, dominance, positive and negative power-coordinate pullbacks, and the complete ordinary polynomial reduction. The simultaneous dimension induction is completed below. Further expansions specify the projective product convention, prove elementary witness operations and the bounded local comparison, and supply parameter convergence and the two-sided even-power substitution. These changes imply no endorsement.

The preparation argument for subanalytic functions needs more than the preparation of a single regular germ. A function $\psi(u,z)$ can vanish identically in $z$ on some parameter fibres. We prove that finitely many coefficient germs nevertheless control it:

$$\psi(u,z)=\sum_{i=0}^{d}c_i(u)(z-z_0)^i A_i(u,z), \qquad A_i(u_0,z_0)=1.$$

The $c_i$ are its actual Taylor coefficients in $z-z_0$, and the $A_i$ are convergent analytic units. We then normalize this finite family to obtain an analytic family whose order in $z$ is uniformly bounded. The later sections prove the unit and coordinate preliminaries and ordinary nondegenerate analytic preparation, with the lower-dimensional induction explicit. The later sections complete the two-coordinate assembly, joint dimension induction, global cell and preparation theorems, complement theorem and convergent Puiseux results.

Real analytic division and finite generation

Write

$$R_n=\mathbb R\{x_1,\ldots,x_n\},\qquad \mathfrak m_n=(x_1,\ldots,x_n).$$

These are germs at zero; translation gives the same statements at any real point. A germ is a unit precisely when its constant term is nonzero. One direction follows by evaluation. In the other direction the reciprocal of a nonvanishing analytic representative is analytic. Thus $R_n$ is local, with maximal ideal $\mathfrak m_n$. The displayed generators of this ideal follow by grouping a convergent series with zero constant term according to the first variable appearing in each monomial; the resulting $n$ series converge on a smaller polydisc.

The course already contains the full complex preparation and Cauchy division argument, adapted from Jean-Pierre Demailly, in Preparation by the zeros in one fibre and Division, finite generation and elementary algebra. Demailly’s treatment remains under his custom OpenContent grant, as identified in that lesson. We reuse that existing provider by reference.

Here is the additional real argument. Complexify the convergent real series. If a real germ is regular of order $s$ in the last variable, complex preparation gives its unique monic Weierstrass polynomial $P$ and unit. Conjugating every coefficient gives another preparation of the same germ. Uniqueness makes both factors real. Similarly, the unique division

$$f=qP+\sum_{j=0}^{s-1}r_j(x')x_n^j$$

is fixed by coefficient conjugation when $f$ and $P$ are real. It therefore supplies division in $R_n$, with $r_j\in R_{n-1}$. A nonzero real homogeneous polynomial has a real vector at which it is nonzero: successive consideration as a polynomial in one variable proves that a polynomial vanishing everywhere on $\mathbb R^n$ has every coefficient zero. Choose such a vector for the first nonzero homogeneous term. A real linear change of coordinates then makes the germ regular in the last variable.

We record the module argument used here and below. Over a Noetherian ring $R$, every submodule of $R^s$ is finitely generated. Induct on $s$. Project a submodule onto its last coordinate. The image is a finitely generated ideal; lift a finite set of its generators. The kernel is a submodule of $R^{s-1}$ and has finite generators by induction. Those lifts together with the kernel generators generate the original submodule. Any finite $R$-module is a quotient of $R^s$, so the inverse image under this quotient proves the same assertion for its submodules.

Noetherianity of real convergent germs. Induct on $n$, starting with the field $R_0=\mathbb R$. If $J\subset R_n$ is a nonzero ideal, choose a nonzero element, make it regular by a real coordinate change, and prepare it. Its unit is invertible, so its monic polynomial $P$ belongs to $J$. Division by this fixed polynomial sends $J$ to an $R_{n-1}$-submodule of $R_{n-1}^s$. Choose finitely many remainders $r_1,\ldots,r_k$ generating that submodule. Each remainder is in $J$, since it is $f-qP$ for an $f\in J$. For any $f\in J$, express its remainder as an $R_{n-1}$-linear combination of these $r_i$. The division identity then writes $f$ in the ideal generated by $P,r_1,\ldots,r_k$. The zero ideal needs no generators. Hence every ideal of $R_n$ is finitely generated. $\square$

Using a single regular polynomial and division is essential to this proof. Preparation of individual germs in a preassigned direction alone does not justify generation of an arbitrary ideal by polynomials in that direction.

Polynomial rings and stable filtrations

First we supply the algebra used in Valette’s finiteness proof.

Hilbert basis theorem. If $R$ is Noetherian, so is $R[T]$.

Proof. Let $J$ be a nonzero polynomial ideal. Its leading coefficients, together with zero, form an ideal of $R$: to add two of them, multiply the corresponding polynomials by powers of $T$ to align their degrees. Choose polynomials $f_1,\ldots,f_k\in J$ whose leading coefficients generate this ideal, and put $d=\max\deg f_i$. For $f\in J$ of degree $N\ge d$, subtract an $R$-linear combination of $T^{N-\deg f_i}f_i$ with the same leading coefficient. This lowers the degree. Repetition terminates with a polynomial of degree less than $d$. The polynomials in $J$ of degree less than $d$ form a submodule of $R^d$, and hence have finitely many module generators $h_j$. The $f_i,h_j$ generate $J$ as an ideal. If $d=0$, the reduction has zero remainder. The zero ideal is already finite. Induction proves the assertion for finitely many polynomial variables. Any quotient of such a ring is also Noetherian, by taking inverse images of ideals. $\square$

Let now $R$ be Noetherian, $I$ an ideal, and $M$ a finite $R$-module. An $I$-filtration is a decreasing sequence of submodules $M_i\subset M$, indexed by $i\ge0$, with $IM_i\subset M_{i+1}$. It is stable if $IM_i=M_{i+1}$ for all sufficiently large $i$. Introduce the Rees ring and module

$$\mathcal R(I)=\bigoplus_{i\ge0} I^iT^i,\qquad \mathcal R(M_*)=\bigoplus_{i\ge0} M_iT^i .$$

Here $I^0=R$, and each element has finitely many homogeneous terms. If $a_1,\ldots,a_\ell$ generate $I$, the map

$$R[Z_1,\ldots,Z_\ell]\longrightarrow \mathcal R(I), \qquad Z_j\longmapsto a_jT$$

is surjective. The Hilbert basis theorem makes $\mathcal R(I)$ Noetherian.

Stable-filtration criterion. $\mathcal R(M_*)$ is a finite $\mathcal R(I)$-module if and only if $M_*$ is stable.

Proof. A finite set of polynomial generators can be replaced by all its homogeneous components. Those components are still in the Rees module, include a generating set, and are finite in number. Let their degrees be at most $d$. Compare homogeneous terms of degree $i+1$, where $i\ge d$. A generator $aT^e$, $e\le d$, has coefficient in $I^{i+1-e}$ in such an expression. Because

$$I^{i+1-e}=I\,I^{i-e},\qquad I^{i-e}M_e\subset M_i,$$

its contribution is in $IM_i$. Hence $M_{i+1}\subset IM_i$, and the reverse inclusion is the filtration condition.

Conversely, if $IM_i=M_{i+1}$ for $i\ge d$, then $M_j=I^{j-d}M_d$ for $j>d$. Each $M_i$ is finite, by the module argument above. Choose its generators in every degree $0,\ldots,d$. They generate the entire Rees module. Degree zero is included also when stability starts immediately. $\square$

If $N\subset M$ and $M_*$ is stable, then

$$\mathcal R((N\cap M_i)_i)\subset\mathcal R(M_*)$$

is a submodule over the Noetherian ring $\mathcal R(I)$. It is finite, so the criterion proves that $(N\cap M_i)_i$ is stable. Taking $M_i=I^iM$ gives the following form of Artin–Rees:

$$N\cap I^{i+1}M=I(N\cap I^iM)\qquad(i\ge d)$$

for some $d$. In particular,

$$N\cap I^iM=I^{i-d}(N\cap I^dM)\qquad(i\ge d).$$

This argument works for every ideal $I$ of a Noetherian ring. Valette uses the case of the maximal ideal of a Noetherian local ring.

Krull intersection for modules

Assume now that $R$ is Noetherian and local, with maximal ideal $\mathfrak m$. If a finite module $L$ satisfies $L=\mathfrak m L$, then $L=0$. Indeed, if $L\ne0$, choose a generating set $e_1,\ldots,e_p$ of least possible size. Express

$$e_1=\sum_{j=1}^p b_je_j,\qquad b_j\in\mathfrak m.$$

The coefficient $1-b_1$ is a unit: otherwise both it and $b_1$ would belong to the unique maximal ideal, forcing $1\in\mathfrak m$. The displayed relation then eliminates $e_1$, contradicting minimality. This proves the needed form of Nakayama’s lemma.

Krull intersection. For a finite $R$-module $M$,

$$\bigcap_{i\ge0}\mathfrak m^iM=0.$$

Proof. Put $L=\bigcap_i\mathfrak m^iM$. It is a finite submodule of $M$. Artin–Rees applied to $L\subset M$ says, for sufficiently large $i$,

$$L=L\cap\mathfrak m^{i+1}M =\mathfrak m(L\cap\mathfrak m^iM)=\mathfrak m L.$$

Nakayama gives $L=0$. $\square$

Apply this to $M/N$. Its powers $\mathfrak m^i(M/N)$ are the images of $\mathfrak m^iM$, so

$$\bigcap_{i\ge0}(N+\mathfrak m^iM)=N.$$

Every submodule of a finite module is therefore closed for the $\mathfrak m$-adic topology. In particular, for an ideal $J\subset R$,

$$\bigcap_{i\ge0}(J+\mathfrak m^i)=J.$$

Locality matters in this conclusion. In $\mathbb R[x]$, the ideals $J=(x-1)$ and $I=(x)$ satisfy $J+I^i=\mathbb R[x]$ for every positive $i$, although $J$ is proper. This is why the local hypothesis cannot be dropped when interpreting the last intersection formula.

Formal linear solutions have convergent replacements

Let $R=R_n$, $\widehat R=\mathbb R[[x_1,\ldots,x_n]]$, and let

$$H:R^a\longrightarrow R^b$$

be an $R$-linear map, represented by a matrix of convergent series. Suppose $v\in R^b$, and the corresponding formal equation has a solution $\widehat f\in\widehat R^a$:

$$H\widehat f=v.$$

Linear convergence theorem. There is an $f\in R^a$ with $Hf=v$. Moreover, for each prescribed integer $q\ge0$, $f$ can be chosen with the same Taylor coefficients as $\widehat f$ in every degree less than $q$.

Proof of existence. Truncate each $\widehat f_j$ to degree at most $k$, obtaining a polynomial $p_{j,k}$. Then

$$v-H(p_{1,k},\ldots,p_{a,k})\in\mathfrak m^{k+1}R^b.$$

The reason is coefficientwise: multiplying by a convergent series cannot lower the degree of the omitted formal terms. The difference on the left is convergent, and a convergent series whose terms of degree less than $k+1$ vanish belongs to $\mathfrak m^{k+1}$. To see the latter assertion, assign each monomial to a degree-$(k+1)$ monomial dividing it, and factor that monomial out. There are finitely many divisors of this degree, and the remaining series converge on a smaller polydisc.

Thus $v$ belongs to $H(R^a)+\mathfrak m^{k+1}R^b$ for every $k$. Submodule closedness gives $v\in H(R^a)$, proving existence.

Proof with a prescribed jet. For $q>0$, let $p_j$ be the Taylor polynomial of $\widehat f_j$ in degrees less than $q$. A formal series of order at least $q$ can be grouped by the finitely many degree-$q$ monomials:

$$\widehat f_j-p_j =\sum_{|\alpha|=q}x^\alpha\widehat g_{j,\alpha}.$$

Consequently the finite linear equation

$$\sum_{j=1}^a\sum_{|\alpha|=q} x^\alpha H(e_j)\,g_{j,\alpha}=v-H(p)$$

has a formal solution. The existence argument already proved gives convergent solutions $g_{j,\alpha}$. Set $f_j=p_j+\sum_{|\alpha|=q}x^\alpha g_{j,\alpha}$. This has the required jet and solves $Hf=v$. For $q=0$, use existence alone. $\square$

The theorem gives a replacement solution. It does not say that the originally chosen formal series converge. It is a statement about finite linear systems; a nonlinear approximation theorem has not been used.

A finite Taylor family with analytic units

Let $\psi$ be a real analytic germ at $y_0=(u_0,z_0)\in\mathbb R^{n-1}\times\mathbb R$. Translating, suppose $y_0=0$, and write its convergent expansion

$$\psi(u,z)=\sum_{j\ge0}c_j(u)z^j.$$

Each $c_j$ is a convergent analytic germ in $u$. Noetherianity says that the ideal generated by all the $c_j$ is generated by a finite subfamily. Enlarging that finite subfamily, choose an initial segment $c_0,\ldots,c_d$ generating the ideal. For $j>d$, choose convergent germs $b_{i,j}(u)$ with

$$c_j(u)=\sum_{i=0}^d b_{i,j}(u)c_i(u).$$

There is no assertion that the infinitely many $b_{i,j}$ share a convergence estimate. Instead, work formally. Put

$$\widehat F_i(U,Z)=\sum_{j>d}b_{i,j}(U)Z^{j-i-1}.$$

Every exponent of $Z$ is nonnegative. For any fixed total degree only finitely many $j$ can contribute, so this is a well-defined formal series. Coefficient comparison gives

$$\psi(U,Z)=\sum_{i=0}^d c_i(U)Z^i(1+Z\widehat F_i(U,Z)).$$

Equivalently, the finite linear equation

$$\sum_{i=0}^d c_i(U)Z^{i+1}F_i =\psi(U,Z)-\sum_{i=0}^d c_i(U)Z^i$$

has a formal solution. The linear convergence theorem, applied in $n$ variables, supplies convergent solutions $G_i$. Define

$$A_i(u,z)=1+zG_i(u,z).$$

Then $A_i(0,0)=1$, and

$$\psi(u,z)=\sum_{i=0}^d c_i(u)z^i A_i(u,z).$$

On a sufficiently small common neighbourhood all $A_i$ are nonzero. After undoing translation, this proves the announced finite Taylor-unit decomposition. If $\psi=0$, take $d=0,c_0=0,A_0=1$. A coefficient that happens to vanish identically may likewise be retained with unit $1$. $\square$

This is Valette’s Proposition 1.3.11, using Corollary 1.3.10. The linear-convergence step is what makes the decomposition analytic; purely formal manipulation of the infinite coefficients would not suffice.

Signed normalization and a uniform order bound

Here is the precise normalization used in the next preparation step. Suppose on a neighbourhood of $(u_0,z_0)$ that

$$\psi(u,z)=\sum_{i=0}^d c_i(u)(z-z_0)^i A_i(u,z), \qquad A_i(u_0,z_0)\ne0.$$

For $m\in\{0,\ldots,d\}$, let

$$K_m=\{t\in[-1,1]^{d+1}:t_m=1\},\qquad Q(t,u,z)=\sum_{i=0}^d t_i(z-z_0)^iA_i(u,z).$$

Uniform regularity. There is a neighbourhood $V$ of $(u_0,z_0)$ and a constant $\delta>0$ such that

$$\max_{0\le k\le m}|\partial_z^k Q(t,u,z)|\ge\delta \qquad(t\in K_m,\ (u,z)\in V).$$

In particular the order of $Q$ in $z$, at each such point, is at most $m$.

Proof. At $(u_0,z_0)$, the derivatives of $(z-z_0)^iA_i(u,z)$ of orders $k<i$ vanish, and its derivative of order $i$ is $i!A_i(u_0,z_0)\ne0$. Therefore the linear map from $(t_0,\ldots,t_m)$ to the derivatives of $Q$ of orders $0,\ldots,m$ at this point is triangular with nonzero diagonal. The higher $t_i$, $i>m$, do not enter these derivatives. Because $t_m=1$, these derivatives cannot all vanish. Their maximum is continuous and positive on the compact set $K_m$, so it has a positive minimum, say $2\delta$. Uniform continuity of the finitely many derivatives on a compact product $K_m$ times a small closed box supplies $V$ on which that maximum remains at least $\delta$. $\square$

If $\phi:B\to\mathbb R^{n-1}$ is a parameter map, set $g_i=c_i\circ\phi$. On any piece where $g_m\ne0$ and $|g_i|\le |g_m|$ for all $i$,

$$\psi(\phi(x),z) =g_m(x)\,Q\!\left( \frac{g_0(x)}{g_m(x)},\ldots, \frac{g_d(x)}{g_m(x)},\phi(x),z\right).$$

The normalizing vector lies in $K_m$, including when some ratios are negative. On the piece where every $g_i=0$, the function is identically zero and needs no normalization. Ordering these coefficient functions into suitable analytic cells belongs to the dimension induction in the full preparation proof; it has not been assumed in proving the uniform bound.

For a function analytic near a compact parameter-image closure, first choose finitely many smaller boxes where the Taylor-unit decomposition and this bound hold. Such boxes exist by the local statements and compactness. Their preimages provide a finite cover. The compatible refinement for that cover is supplied by the cell induction below.

How this supplies the geometric argument

The written chain is now

$$\begin{gathered} \text{real analytic division}\ \Longrightarrow\ \text{Noetherian convergent germs},\\ \text{Rees finiteness}\ \Longrightarrow\ \text{Artin--Rees}\ \Longrightarrow\ \text{Krull submodule closedness},\\ \text{formal linear solvability}\ \Longrightarrow\ \text{convergent replacements}\ \Longrightarrow\ \text{finite Taylor units}\ \Longrightarrow\ \text{uniform signed regularity}. \end{gathered}$$

The already written curve-selection and Łojasiewicz treatment additionally needs finite analytic cell decomposition and convergent Puiseux preparation. The later sections complete its reduction, complement and Puiseux providers under the explicitly specified projective product definition.

A convergent split before restricting to a hyperbola

Let $\psi(u,z,w)$ be real analytic near $(a,0,0)$, where $u\in\mathbb R^k$. Its real series complexifies. Choose $R>0$ such that this complexification is holomorphic on a neighborhood of the closed polydisc of common radius $R$, centered at $(a,0,0)$, and bounded there by $M$. Write

$$\psi(u,z,w)= \sum_{\alpha\in\mathbb N^k,\ i,j\ge0} b_{\alpha i j}(u-a)^\alpha z^i w^j .$$

Repeated one-variable Cauchy coefficient estimates give

$$|b_{\alpha i j}|\le M R^{-|\alpha|-i-j}.$$

Define, using an independent variable $c$,

$$\begin{aligned} \psi_+(u,c,z)&= \sum_{\alpha,\ i>j\ge0} b_{\alpha i j}(u-a)^\alpha c^j z^{i-j-1},\\ \psi_-(u,c,w)&= \sum_{\alpha,\ j\ge i\ge0} b_{\alpha i j}(u-a)^\alpha c^i w^{j-i}. \end{aligned}$$

These series converge as functions of independent variables. In the first one set $i=j+\ell+1$. Its absolute terms are bounded by

$$\frac{M}{R} \left(\prod_{\nu=1}^k \left(\frac{|u_\nu-a_\nu|}{R}\right)^{\alpha_\nu}\right) \left(\frac{|c|}{R^2}\right)^j \left(\frac{|z|}{R}\right)^\ell .$$

Summing them gives the finite bound

$$\frac{M}{R} \prod_{\nu=1}^k \frac{1}{1-|u_\nu-a_\nu|/R} \frac{1}{1-|c|/R^2} \frac{1}{1-|z|/R}$$

whenever every displayed ratio is less than one. In the second series put $j=i+\ell$; its analogous bound is

$$M \prod_{\nu=1}^k \frac{1}{1-|u_\nu-a_\nu|/R} \frac{1}{1-|c|/R^2} \frac{1}{1-|w|/R}.$$

On each strictly smaller closed polydisc these bounds are uniform geometric majorants. Thus the series define holomorphic functions of $(u,c,z)$ and $(u,c,w)$, respectively, on the indicated open polydiscs. Their real restrictions are analytic. In particular, both functions are analytic on neighborhoods of compact boxes with

$$|u_\nu-a_\nu|\le R/2,\qquad |c|\le R^2/4,\qquad |z|,\ |w|\le R/2.$$

Now impose $c=zw$. An $i>j$ term becomes

$$z\,c^jz^{i-j-1}=z^iw^j,$$

and a $j\ge i$ term becomes $c^iw^{j-i}=z^iw^j$. Absolute convergence permits the two subseries to be recombined. Therefore

$$\boxed{\ \psi(u,z,w) =z\,\psi_+(u,zw,z)+\psi_-(u,zw,w)\ } .$$

For $z\ne0$, putting $w=c/z$ gives Valette’s Lemma 1.6.13 splitting identity with fully convergent functions. For $z=0$, the identity with independent $w$ still holds, so no division is needed for the limiting statement. No sign restriction on $z,w,c$ has been imposed.

For example, for $\psi(z,w)=1+z+w+zw+z^2w$, the construction gives

$$\psi_+(c,z)=1+c,\qquad \psi_-(c,w)=1+w+c.$$

Hence $z(1+c)+(1+w+c)=1+z+w+zw+z^2w$ when $c=zw$. The equality includes the mixed diagonal $zw$ in the minus series and the off-diagonal $z^2w$ in the plus series, as required by the index convention.

Localize on the actual image, rather than a larger product

Suppose $\Phi(x)=(\phi(\widetilde x),z,c(\widetilde x)/z)$ is well defined and bounded on a set $C$, and $\psi$ is analytic on an open neighborhood $U$ of

$$K=\overline{\Phi(C)}.$$

Then $K$ is compact. Write $w=c/z$, so $c=zw$ on $C$. There are three local descriptions, covering every $p=(a,z_0,w_0)\in K$.

If $z_0\ne0$, define

$$F_z(u,c,z)=\psi(u,z,c/z)$$

on the inverse image of $U$ under the analytic map $(u,c,z)\mapsto(u,z,c/z)$, with $z\ne0$. It is analytic on a neighborhood of $(a,z_0w_0,z_0)$. Choose a small closed box $B_z$ about that point whose neighborhood lies in this inverse image, and on which $|z|\ge |z_0|/2$. By continuity, there is a neighborhood $V_p$ of $p$ such that

$$(u,z,w)\in V_p \quad\Longrightarrow\quad (u,zw,z)\in\operatorname{int}B_z.$$

Consequently, on the portion of $C$ mapping into $V_p$, the original function is $F_z(\phi,c,z)$, analytic in its last coordinate $z$ on the required compact box.

If $w_0\ne0$, instead define

$$F_w(u,c,w)=\psi(u,c/w,w).$$

The same argument gives a compact box $B_w$ where $F_w$ is analytic on a neighborhood and $|w|\ge |w_0|/2$, and a neighborhood $V_p$ whose images $(u,zw,w)$ lie in its interior. Here the original function is analytic in the coordinate $w=c/z$. Notice that $c\ne0$ on this portion of the original domain because $z\ne0$ there and $w$ is bounded away from zero; the fibre-coordinate change can therefore be inverted.

If $z_0=w_0=0$, then $(a,0,0)$ is an actual point of $K$. The function $\psi$ is analytic near this point, so the convergent splitting just proved applies. Choose a smaller neighborhood $V_p$ such that its points satisfy the splitting bounds and both compact maps

$$(u,z,w)\longmapsto(u,zw,z),\qquad (u,z,w)\longmapsto(u,zw,w)$$

fit in the interiors of the respective boxes of analyticity.

The $V_p$ cover $K$; extract a finite subcover. Their preimages under $\Phi$ cover $C$. This proves the finite analytic localization without any cell theorem. In the full induction, the finite two-coordinate localization below turns these finitely many preimages into compatible cells before applying cell reductions.

The existence of $\psi$ near $K$ does not imply analyticity near $(a,0,0)$ for every $a\in\overline{\phi(B)}$. Those points can lie outside $K$. The proof uses the split only at the third type of point, and uses the two nonsingular-coordinate descriptions at the other points. No analytic extension to a larger parameter product is required.

The signed balanced-coordinate calculation

The following calculation is the remaining analytic mechanism in Valette’s balanced case, conditional on the stated nondegenerate reductions and translation dichotomy from the preceding cell induction.

Work on a piece where $z\ne0,c\ne0$ and both split contributions

$$X_+=z\,\psi_+(\phi,c,z),\qquad X_-=\psi_-(\phi,c,c/z)$$

are nonzero and comparable in absolute value, with constants uniform on the piece. Suppose that the ordinary analytic preparation step has supplied, with nonzero base coefficients $a_+,a_-$, nonnegative rational exponents $r,s$, and uniformly bounded nonvanishing units,

$$\begin{aligned} |X_+|&\asymp |z|\,|a_+|\, |z-\theta_+|^r,\\ |X_-|&\asymp |a_-|\, |c/z-\theta_-|^s. \end{aligned}$$

Here the coefficients and translations depend only on the base variables. The preceding translation dichotomy has three possibilities.

If $|z|\asymp|\theta_+|$, take $b=|\theta_+|>0$. If $|c/z|\asymp|\theta_-|$, take $b=|c/\theta_-|>0$. In either case $|z|\asymp b$.

If neither comparison holds and the dichotomy supplies reductions with both translations zero, comparability gives

$$|a_+|\,|z|^{r+1} \asymp |a_-|\,|c|^s|z|^{-s}.$$

Therefore take

$$b= \left(\frac{|a_-|\,|c|^s}{|a_+|}\right)^{1/(r+s+1)}>0.$$

The denominator exponent is positive because $r,s\ge0$. Again $|z|\asymp b$. No positive-sign assumption on $X_+,X_-,c,z$ was needed.

Set

$$q=z/b,\qquad d=c/b,\qquad \Psi(x)=(\phi(\widetilde x),b(\widetilde x), d(\widetilde x),q(x)).$$

The ratio $q$ is bounded and bounded away from zero. Since $z$ and $c/z$ are bounded, comparability bounds $b$, and

$$d=(c/z)q$$

is bounded as well. Thus $\Psi(C)$ is bounded. Define

$$\widetilde\psi(u,b,d,q)=\psi(u,bq,d/q).$$

On the open set $q\ne0$, this is analytic wherever the inner map lands in $U$. That condition holds on a neighborhood of $\overline{\Psi(C)}$: at each of its points, $q\ne0$, and continuity maps it into $K$, since the equality

$$(\phi,bq,d/q)=(\phi,z,c/z)$$

holds on $C$. The open inverse image of $U$ therefore contains the compact closure. Hence the original function is a restricted analytic composition in the last coordinate $q$, with base parameters $(\phi,b,d)$. Pulling back the resulting preparation through $q=z/b$ uses the power-coordinate result proved below, including its zero-scale and sign cases and its explicit lower-dimensional induction input.

If $c=0$, the original function is $\psi(\phi,z,0)$ and belongs directly to the ordinary analytic-composition case. If one split contribution vanishes, the other is the whole function. The balanced calculation is used only after these cases are separated.

This proves the analytic splitting, actual-compact-image localization and signed balanced substitution. The sections below prove the compatible endpoint refinements, ordinary nondegenerate reduction, translation dichotomy, analytic unit representations and coordinate pullbacks using the lower-dimensional induction. The later sections complete the full two-coordinate assembly and joint dimension induction.

Analytic units and the lower-dimensional induction

This section adapts Valette’s definitions in §§1.4–1.5 and the preliminaries in §1.6.1. The additional bounded-monomial construction and compact-image arguments supply the unit representations used in the source’s abbreviated calculations.

An $\mathcal L$-function is obtained by finitely many sums, products and compositions from restricted real analytic functions and rational powers, on domains where these expressions are defined. A restricted analytic function is analytic on a neighborhood of a compact cube. Its restriction and all the expressions below retain their stated domains. We consider such functions on $\mathcal L$-cells: inductively, graphs and open bands of analytic $\mathcal L$-functions over lower-dimensional cells, with infinite band endpoints allowed.

Fix a dimension $n$, write $x$ for the first $n-1$ coordinates and $z$ for the last, and work in the dimension induction. The lower-dimensional package says that finitely many $\mathcal L$-functions on supplied lower-dimensional cells can be made analytic, of constant sign and pairwise ordered on a common finite cell refinement. Their domains and any finitely many supplied cells can be respected by that refinement. These are the induction hypotheses in dimensions less than $n$; the full preparation and arbitrary-set cell theorem in dimension $n$ are not being assumed here.

For a cell $C$ with base $B$, take an analytic base translation $\theta:B\to\mathbb R$ with its graph disjoint from $C$, and put $y=|z-\theta(x)|>0$. An analytic $\mathcal L$-unit in $y$ has the form

$$U(x,z)=\Psi\bigl( \beta(x),u(x)y^{1/s},v(x)y^{-1/s}\bigr), \qquad s\in\mathbb Z_{>0}.$$

The base functions are analytic $\mathcal L$-functions, the whole argument map is bounded, and $\Psi$ is analytic and nonzero on a neighborhood of its compact image closure. Shrinking that neighborhood if necessary is permitted. A reduced function is

$$f(x,z)=a(x)y^rU(x,z),\qquad r\in\mathbb Q,$$

with analytic $\mathcal L$-coefficient $a$. The zero coefficient is allowed. A unit is uniformly bounded and bounded away from zero on $C$, by compactness. In the arguments below we prove the requisite analytic representation; uniform bounds alone are not used as a substitute for it.

Cells are connected. For a graph this follows from connectedness of the base. A finite band has a continuous parametrization from $B\times(0,1)$ using its endpoints. A one-sided band is parametrized from $B\times(0,\infty)$, and a band with both endpoints infinite is $B\times\mathbb R$. Induction proves connectedness in all cases. Hence a continuous nonzero unit has constant sign on a cell.

Finite endpoint refinements

Given finitely many cells in $\mathbb R^n$, together with finitely many analytic $\mathcal L$-functions on their bases, there is a finite cylindrical refinement respecting the supplied cells and any graphs or bands cut out by these functions.

Proof in the dimension induction. Refine the bases with the lower-dimensional package so that every relevant function is analytic and every pairwise difference has constant sign. Include all finite endpoints defining the original cells. On a fixed base cell, two of these functions are either identical or strictly ordered everywhere. Remove repeated identical endpoints and list the others in order. Their graphs and the intervening bands, including the two outer infinite bands, partition that whole cylinder. They are analytic $\mathcal L$-cells. Each supplied cell is a union of these cells, because its base and all its endpoints were respected. There are finitely many bases and endpoints. $\square$

This also supplies a common refinement of finitely many existing cell decompositions. It does not assume that an arbitrary globally subanalytic set has a cell decomposition.

A graph cell $z=\zeta(x)$ can be handled entirely in the lower dimension. Substituting $\zeta$ into a well-defined finite $\mathcal L$-expression gives an $\mathcal L$-function of $x$. Make it analytic on a lower-dimensional refinement. On the graph choose $\theta=\zeta-1$, so $y=1$, take $r=0,U=1$, and use that base function as coefficient. In particular graph cells require no inverse last-coordinate map.

A base function on a band is also reduced after cutting at $z=0$: on the nonzero bands take $\theta=0,r=0,U=1$; handle the possible graph $z=0$ as just described. The finite endpoint refinement supplies these pieces even when the original band has both endpoints infinite.

If $f=a|z-\theta|^rU$ is already reduced, refine the base so $a$ has constant sign. The distance is positive, and the unit is continuous and nonzero. Thus $f$ has constant sign on each resulting cell. All these sign refinements use only the lower-dimensional package.

Collecting bounded monomials into two coordinates

Let $t>0$ on a cell with base $B$. Suppose that a finite family

$$M_i(x,t)=b_i(x)t^{q_i},\qquad q_i\in\mathbb Q,$$

is bounded on the cell, with $\mathcal L$-coefficients $b_i$. There may also be finitely many bounded base parameters. After a lower-dimensional refinement, all these arguments can be written as polynomials of bounded base parameters and just two bounded arguments

$$Z=u(x)t^{1/s},\qquad W=v(x)t^{-1/s},$$

where $s>0$ is an integer and $u,v$ are nonnegative analytic $\mathcal L$-functions. Either of $u,v$ may be identically zero.

Proof. Choose $s$ so every $a_i=sq_i$ is an integer. Refine the base so each $b_i$ is analytic and identically zero, strictly positive or strictly negative. For $a_i>0$, put $u_i=|b_i|^{1/a_i}$; order these finitely many functions on the base and choose their largest, $u$. If this family is empty or all its coefficients vanish, put $u=0$. Otherwise $u>0$. In that case

$$M_i=\operatorname{sign}(b_i) \left(\frac{u_i}{u}\right)^{a_i}Z^{a_i} \qquad(a_i>0),$$

where $u_i/u\in[0,1]$. Moreover

$$Z=\max_{a_i>0}|M_i|^{1/a_i},$$

so $Z$ is bounded. With $u=0$, all these $M_i$ are zero and their representation is immediate.

For $a_i=-d_i<0$, similarly put $v_i=|b_i|^{1/d_i}$, choose the largest $v$, and obtain

$$M_i=\operatorname{sign}(b_i) \left(\frac{v_i}{v}\right)^{d_i}W^{d_i}.$$

The ratios lie in $[0,1]$, and $W=\max_{a_i<0}|M_i|^{1/d_i}$ is bounded. If all negative-exponent coefficients vanish, put $v=0$. For $a_i=0$, the argument is the base function $b_i$, which is bounded on $B$: every base point has a nonempty fibre. Include these and the original bounded base parameters among the new base parameters. All roots, ratios and choices are $\mathcal L$-functions and are analytic after the lower-dimensional refinement. This proves the claim. $\square$

There is a useful compact-domain consequence. Let $\Gamma=(\eta(x),Z,W)$ be the new bounded argument map, and let $P$ be the polynomial map recovering all old arguments. Then

$$P\bigl(\overline{\Gamma(C)}\bigr) \subset\overline{P(\Gamma(C))}.$$

Indeed every point of the left-hand domain is the limit of a sequence of new argument values, and polynomial continuity recovers the limit of the old ones. Thus, if an outer analytic function was given only near the old compact image, its composite with $P$ is analytic near the new compact image: take the open inverse image of its domain. No extension to a larger parameter box has been assumed. If the old function was nonzero on that image closure, the composite is nonzero on the new one as well.

Consequently a finite list of bounded monomial arguments can be used in a unit representation without requiring one separate forward or inverse coordinate for each exponent.

Closure and substitution for analytic units

Products, reciprocals and powers. Finitely many units with the same distance $y$ have a common two-coordinate argument map. Apply the bounded-monomial construction to their finitely many original forward/inverse arguments and concatenate the bounded base parameters. The recovered old maps are polynomial maps of the new arguments. Compose the outer analytic functions with these maps. Their product is analytic and nonzero near the new compact closure. The same holds for reciprocals.

For a rational power of the absolute value of a unit, first fix its sign on the cell. The signed analytic outer function is positive on the compact image and hence on a smaller neighborhood. Its real rational power is analytic there. Thus $|U|^\lambda$ is a unit for every $\lambda\in\mathbb Q$. Integer powers retain the corresponding constant sign if desired.

Change of distance. Suppose $h>0$ is another distance, and on a cell it has a reduction

$$h=A(x)y^eW(x,y),$$

where $A>0$, $e\in\mathbb Q$, and $W$ is a positive unit in $y$. Every unit originally expressed in $h$ becomes a unit in $y$.

To prove this rather than infer it from bounds, write each original bounded monomial argument as

$$b(x)h^q=b(x)A(x)^q y^{eq}W(x,y)^q.$$

Since $W$ and its reciprocal are uniformly bounded, the monomial $bA^qy^{eq}$ is bounded whenever the original argument is bounded. Collect these finitely many monomials, together with the bounded monomial arguments representing $W$, into two coordinates. Recover $W$ by its analytic outer function, take its positive analytic rational powers, and multiply by the recovered monomials. These operations define an analytic map near the new compact closure. On the actual cell the resulting map is exactly the old argument map, so its closure maps into the old compact closure. The old analytic outer function can therefore be composed on a neighborhood of that new compact closure. It remains nonzero there. This is the required unit representation.

In particular if $a h^r U(h)$ was reduced in $h$, it becomes

$$(aA^r)y^{er}\bigl(W^rU(h)\bigr)$$

reduced in $y$. The unit in parentheses has just been proved to have the required analytic form.

A common translation for finitely many reductions

Suppose $f_1,\ldots,f_k$ are reduced on a cell $C$, with translations $\theta_1,\ldots,\theta_k$, each disjoint from $C$. After a finite endpoint refinement, all $f_i$ are reduced with one common translation.

Proof. First fix signs and order the base functions as necessary. Comparing the distances $|z-\theta_i|$ and $|z-\theta_j|$ only requires cutting at their midpoint: the squared-distance difference is

$$(z-\theta_i)^2-(z-\theta_j)^2 =(\theta_j-\theta_i)(2z-\theta_i-\theta_j).$$

The finite endpoint refinement therefore makes these distance comparisons constant on each piece. Choose an index $j$ with

$$y=|z-\theta_j|\le |z-\theta_i|\qquad(1\le i\le k)$$

throughout the piece. This distance is strictly positive because $\theta_j$’s graph is outside the original cell.

For a fixed $i$, put $d=z-\theta_j$ and $\delta=\theta_j-\theta_i$. If $\delta=0$ identically, the two distances are equal. Otherwise refine at the endpoints $\theta_j\pm|\delta|$, and fix the sign of $d$.

On $|d|\le|\delta|$, put $\tau=d/\delta$. The nearest-distance condition says $|\tau|\le|1+\tau|$. Squaring gives $\tau\ge-1/2$. Hence

$$-\tfrac12\le\tau\le1,\qquad |z-\theta_i|=|\delta|(1+\tau).$$

The argument $\tau=(\operatorname{sign}d/\delta)y$ is a bounded forward monomial, and $1+\tau$ is positive and at least $1/2$ on its compact closure. Thus this is a reduced distance with exponent zero and an explicit unit in $y$.

On $|d|\ge|\delta|$, put $\tau=\delta/d$. Now $\tau\in[-1,1]$, and the nearest-distance condition gives $1\le|1+\tau|$. Since $1+\tau\ge0$, it follows that $0\le\tau\le1$. Therefore

$$|z-\theta_i|=y(1+\tau), \qquad \tau=\operatorname{sign}(d)\,\delta\,y^{-1}.$$

This is a bounded inverse monomial and a positive analytic unit. The case $\delta=0$ is included by taking the unit $1$.

We have expressed each old distance as $A_i y^{e_i}W_i$, with $e_i=0$ or $1$, positive $A_i$, and an explicit positive unit. The distance-substitution result converts each complete reduction, including its old unit, to the common translation $\theta_j$. There are only finitely many indices and endpoints. $\square$

Products of reducible functions are now reducible: use a common refinement and translation, multiply the base coefficients, add the rational exponents, and multiply the units. A quotient is handled similarly on its stated domain where the denominator is nonzero. Its nonzero base coefficient can be inverted after the base sign refinement. No zero denominator is silently included.

Dominant and comparable pieces

Let $f,g$ be reduced, and let $\epsilon>0$. There is a finite endpoint refinement on which one of

$$|f|\le\epsilon|g|,\qquad |g|\le\epsilon|f|,\qquad |f|\asymp|g|$$

holds on every piece, with uniform positive comparability constants in the last case.

Proof. Give the functions a common translation. Their zero base coefficients are separated first. Where both are nonzero, write

$$f=a y^rU,\qquad g=b y^sV.$$

The compact unit bounds give constants $0<C_1\le C_2$ such that

$$C_1h\le |f/g|\le C_2h,\qquad h=|a/b|y^{r-s}.$$

Choose $\eta=\min(\epsilon,1/2)$. Cut at the two levels

$$h=\eta/C_2,\qquad h=1/(\eta C_1).$$

They are ordered because $C_1\le C_2$. Below the first, $|f/g|\le\eta\le\epsilon$. Above the second, $|g/f|\le\eta\le\epsilon$. Between them the displayed unit bounds give a fixed positive lower bound and a fixed finite upper bound on $|f/g|$.

If $r=s$, these cuts involve only base functions and use the lower-dimensional package. Otherwise each positive level $L$ corresponds to

$$y=(L|b/a|)^{1/(r-s)}.$$

This is a positive analytic $\mathcal L$-function of the base on the nonzero-coefficient pieces. The endpoints $\theta\pm(L|b/a|)^{1/(r-s)}$ give the needed finite refinement. A negative exponent merely reverses the ordering of the inequalities. The threshold graphs are included as separate cells. If one function is zero, the corresponding dominance assertion is immediate; if both vanish, both inequalities hold. $\square$

Zero translation or comparison with the coordinate

After a finite refinement, a given reduction with translation $\theta$ can be arranged so that either $\theta=0$, or $|z|\asymp|\theta|$ and $z,\theta$ have the same sign.

Proof. Graph cells are handled in the lower dimension. On nonzero-coordinate bands, cut at $z=0$, fix the sign of $\theta$, and compare $|z|$ with $|\theta|/2$ and $2|\theta|$. These are finite base endpoints. If $\theta=0$, there is nothing to change.

If $|\theta|\le |z|/2$, put $t=|z|$. Then

$$|z-\theta|=t(1-\theta/z), \qquad |\theta/z|\le1/2.$$

The ratio is a bounded inverse monomial in $t$, and its analytic unit is positive. If $|z|\le|\theta|/2$, then

$$|z-\theta|=|\theta|(1-z/\theta), \qquad |z/\theta|\le1/2.$$

This is a base factor times a unit of a bounded forward monomial in $t$. In both cases the distance-substitution theorem changes the full reduction to translation zero.

In the remaining comparable case, if signs are opposite then

$$|z-\theta|=t(1+|\theta|/t), \qquad 1/2\le|\theta|/t\le2.$$

Again the inverse argument and analytic unit are explicit, so use translation zero. With the same sign retain the old translation, which satisfies the claimed comparison. $\square$

Pullback through a power coordinate

Consider

$$H(x,z)=\bigl(x,c(x)|z|^{1/p}\bigr), \qquad p\in\mathbb Z\setminus\{0\},$$

on a cell where it is well defined. If a function is reducible on the image cells of $H$, its composite with $H$ is reducible after a finite refinement.

First handle graph cells as above. Refine the base into constant-sign pieces for $c$, and cut at $z=0$. If $c=0$, the image lies in the graph of zero; evaluating the expression on that graph is a lower-dimensional $\mathcal L$-function. Its pullback is a base function, handled by the graph/band rule. If $z=0$ is in the domain, it is already a graph cell. For $p<0$, such points are excluded by the stated domain.

It remains to work with $c\ne0$ and $z\ne0$, with both signs fixed. The fibre map is analytic and strictly monotone. With $v=c|z|^{1/p}$ and $\sigma=\operatorname{sign}z$, its inverse on the branch is

$$z=\sigma(v/c)^p,\qquad v/c>0.$$

Thus the image of a graph or band is a graph or band, and the preimage of an image cell is again such a cell. Its finite endpoints are analytic $\mathcal L$-functions after lower-dimensional refinement; zero and infinite endpoints are interpreted by one-sided limits. In particular a supplied finite image refinement pulls back to finitely many cells.

On one such image piece a reduction has an old positive distance $|v-\theta(x)|$. Its pullback is

$$h=|c|\,\bigl|t^{1/p}-\xi(x)\bigr|, \qquad t=|z|,\quad \xi=\theta/c.$$

We give explicit reductions of this distance.

For $p=m>0$, put $T=t^{1/m}$.

If $\xi=0$, then $h=|c|t^{1/m}$. If $\xi<0$, put $a=|\xi|>0$ and cut at $t=a^m$. On $T\ge a$,

$$h=|c|t^{1/m}(1+a/T);$$

on $T\le a$,

$$h=|c|a(1+T/a).$$

In each case the ratio lies in $[0,1]$, is a bounded monomial in $t$, and $1+$ that ratio is a positive analytic unit. These are reductions with translation zero.

If $\xi>0$, cut at $T=\xi/2$ and $T=2\xi$, using the base endpoints $t=(\xi/2)^m,(2\xi)^m$. The two outer regions give respectively

$$h=|c|\xi(1-T/\xi),\qquad h=|c|t^{1/m}(1-\xi/T),$$

with the ratios in $[0,1/2]$. On the middle region $1/2\le T/\xi\le2$, use

$$T-\xi= \frac{t-\xi^m} {\xi^{m-1}P_m(T/\xi)},\qquad P_m(v)=1+v+\cdots+v^{m-1}.$$

This polynomial is positive on $[1/2,2]$. The numerator vanishes precisely at $z=\sigma\xi^m$, which is outside the pulled-back image piece because the old distance was positive there. Put

$$y=|z-\sigma\xi^m|,\qquad \kappa=\operatorname{sign}(t-\xi^m),\qquad \delta=y/\xi^m.$$

Split at that numerator’s graph if necessary. On each nonzero piece,

$$T/\xi=(1+\kappa\delta)^{1/m}.$$

For $\kappa=1$, $0\le\delta\le2^m-1$. For $\kappa=-1$, $0\le\delta\le1-2^{-m}<1$. Thus $1+\kappa\delta$ stays strictly positive on its compact argument interval, and