What is… a blowup? · Part II of IV
A cone, its directions, and the limits of Nash blowups
A smooth cone chart, a Nash chart that repeats, and an explicit resolution that escapes the loop.
What are we trying to change?
A polynomial zero set is a geometric space: all the points satisfying its equations. Many such spaces are smooth, but a crossing has no single local sheet and a cusp has no regular local parameter in its given algebraic coordinates. These are failures of smoothness at exact solutions of exact equations.
We want a smooth space $Y$ with a map $\pi:Y\to X$ back to the original zero set. At ordinary points, the map should be invertible. Over a singular point, several new points may map to the same old one. Those extra distinctions let calculus work upstairs while keeping an exact relationship to the original problem.
Rounding the picture or moving vertices would change the solutions. A resolution instead changes the space representing them. It is a geometric method for studying equations, but it is a different goal from geometrization theorems about classifying manifolds.
Two constructions, and one theorem
An ordinary blowup needs a chosen center. At a point center, attach the direction of the line from that point to each nearby point, then include the limiting records above the center. For a smooth higher-dimensional center, record the normal direction of approach. This is the construction from Part I. The exceptional set consists of the records that project into the center.
The Nash blowup needs the tangent planes of the variety. At each smooth point $p$ of a $d$-dimensional space, attach its tangent $d$-plane $T_pX$. Include all limits of these pairs as $p$ approaches the singular set. A tangent plane here means its linear direction, translated to the origin. A new point is “old position, together with a limiting tangent plane”. Nash also has exceptional fibers: several tangent-plane records can sit above one old point.
Hironaka’s theorem supplies a terminating strategy. There is no single competing operation called “the Hironaka blowup”. The classical approach chooses smooth centers and performs a sequence of ordinary blowups, checking the remaining singularities and exceptional boundary. Constructive proofs explain how to make those choices and guarantee that the process finishes in characteristic zero. Repeating Nash’s particular rule has no such guarantee in general.
Ordinary point blowup
Record the line from the chosen center to p. On a plane curve, its slope is $y/x$.
Nash blowup
Record the curve’s own tangent at p. Where it exists, its slope is $dy/dx$.
Both constructions retain extra geometric information; neither averages neighboring points as Laplacian smoothing does. On $y=x^2$, for example, the two recorded slopes are $x$ and $2x$. They are different rules even when their limiting slopes agree.
First see a case where both succeed
For the cone $X^2+Y^2=Z^2$, put $(X,Y,Z)=(u\cos\theta,u\sin\theta,u)$. The ordinary blowup records its line direction $[\cos\theta:\sin\theta:1]$. The tangent plane away from the apex has normal $[\cos\theta:\sin\theta:-1]$, so Nash retains the same angular information in a different form. Here both give a smooth cylinder. Watch how the new circle sits over the one old apex.
Keep three objects separate: the old apex is one point; the ambient exceptional divisor $E_{\rm amb}=\{u=0\}$ is a plane in this chart; and the cylinder meets that plane in $E_C=\{u=0,\ v^2+w^2=1\}$, a circle. The circle is the exceptional divisor on the resolved cone: a one-dimensional curve inside a two-dimensional surface.
Figure 10. One old point, many new records. Press Inspect the exceptional divisor: the old apex stays a point, while a violet plane appears in the new chart. Its bold circle consists of directions used by the cone. A and B lie on that circle and both map to the apex. In the diagram below, move Q across the plane: it always maps to the same old point, but joins the resolved cone only when it reaches the circle. Then restore the surfaces and approach the apex to see how the records arise as limits.
Look directly at the exceptional plane
This diagram always shows u = 0. The pale square is a bounded patch of the ambient exceptional plane; the bold circle is its intersection with the cylinder. On this plane, (v, w) labels the old line direction [v : w : 1]. The entire plane, including Q, maps to the one old apex.
Why only the circle? For cone points with u ≠ 0, substituting into X² + Y² = Z² and dividing by u² forces v² + w² = 1. That equation persists in their limits at u = 0. A point off the circle is a valid ambient direction record, but cannot be approached by points of the lifted cone.
Compare gradients, distances, area, and tangent planes
| Geometric Invariant | Original Cone $X^2 + Y^2 - Z^2 = 0$ | Resolved Surface $v^2 + w^2 - 1 = 0$ |
|---|---|---|
| Space & Coordinates | Ambient $\mathbb{R}^3$, coordinates $(X, Y, Z)$ | Blowup chart $\mathbb{R}^3$, $(v, w, u)$ with $\pi(v,w,u)=(uv, uw, u)$ |
| Defining Equation & Degree | $F(X,Y,Z) = X^2 + Y^2 - Z^2 = 0$ (degree 2) | Strict transform $g(v,w,u) = v^2 + w^2 - 1 = 0$ (degree 2 in $v,w$) |
| Gradient Vector $\nabla$ & Norm | $\nabla F = (2X, 2Y, -2Z)$, $\|\nabla F\| = 2\sqrt{2}|u|$ | $\nabla g = (2v, 2w, 0)$, $\|\nabla g\| = 2$ (constant $\ne 0$ on the cylinder) |
| Regularity / Smoothness | Singular at apex $(0,0,0)$ where $\nabla F = \mathbf{0}$ | Smooth 2-manifold everywhere (one nonzero equation gradient on $g=0$) |
| Fiber Over Apex ($u = 0$) | All rays collapse to a single point: apex $(0,0,0)$ | Exceptional circle $E$: $v^2+w^2=1, u=0$ (circumference $2\pi$) |
| Point Separation ($\Delta\theta = 90^\circ$) | $\|A_{\text{orig}} - B_{\text{orig}}\| = \sqrt{2}|u|$ | $\|A_{\text{res}} - B_{\text{res}}\| = \sqrt{2} \approx 1.414$ (separated along $E$) |
| Riemannian Area Element $dA$ | $dA = \sqrt{2}|u|\,du\,d\theta \to 0$ (metric degenerates at apex) | $dA = 1\,du\,d\theta$ (standard non-degenerate cylinder metric) |
| Nash Tangent Plane Normal | $[\cos\theta : \sin\theta : -1]$ off apex; limits form circle of planes | Limiting normal family $\cos\theta\,X + \sin\theta\,Y - Z = 0$ resolves apex in 1 step |
The plane is a slice of the new coordinate space, not the old horizontal plane Z = 0. Every point of it maps to a single old point. The controls select points and sections in these two fixed spaces. The cylinder is the new space throughout. On it, u and θ remain independent coordinates at u = 0; the projection identifies that whole circle with one old point. Distances above are distances in the displayed coordinate systems.
Read the substitution and the Nash tangent-plane records
Off u = 0, dividing by u² gives g = 0. Take the closure of that equation to get the strict transform, which also includes its circle at u = 0. In this ambient chart u = 0 is a plane; the resolved cone meets it only along v² + w² = 1.
Nash records the cone’s own tangent plane. Off the apex its normal direction is [v : w : −1]. The limiting planes remain different for different θ, even though their old base point is the same. On this cone, that information also gives a smooth resolution.
Rotate or reset the view
Drag either viewport to rotate both synchronously, or use its arrow keys. The camera changes the view; it does not perform a blowup.
6. Hypersurfaces vs. varieties: keeping dimensions straight
Before exploring higher-dimensional singularities, it is vital to keep two concepts distinct: the ambient dimension (the number of coordinates in the ambient Euclidean space) and the intrinsic dimension of the geometric variety.
One nonzero equation $F(x,y,z,w)=0$ defines a hypersurface of dimension 3 in ambient dimension 4 (over $\mathbb C$, these are complex dimensions). At points where $\nabla F\ne0$, the implicit function theorem gives a tangent 3-plane. The cone shown here has intrinsic real dimension 2 in $\mathbb R^3$. Its apex is singular. The counterexamples below have intrinsic dimensions 4 and 3, with embeddings in seven and six coordinates respectively.
A real section or projection is a way to draw part of such a space. It can reveal a pinch, but it cannot prove that a complex variety is smooth or that an iteration terminates. For those claims we need the coordinate algebra.
7. Replacing the apex by directions
To blow up the origin of real three-space $\mathbb{R}^3$, we introduce projective direction coordinates. In the affine chart where the $Z$-direction is nonzero, we make the substitution:
Away from $u = 0$, the new coordinates simply record the slopes: $v = X/Z$ and $w = Y/Z$. The coordinate $u$ acts as a signed scaling factor along each line. Substituting into the cone equation gives the total transform:
The total transform factors into two geometric pieces: the ambient exceptional divisor $u = 0$ (the fiber sitting over the origin, with multiplicity 2 in the total transform) and the strict transform:
In $(v,w,u)$ space, this is a standard smooth circular cylinder! Its gradient $\nabla(v^2+w^2-1) = (2v, 2w, 0)$ never vanishes on the surface, meaning standard calculus is completely restored.
A natural parametrization of this cylinder is given by:
At $u = 0$, every direction $\theta \in [0, 2\pi)$ maps down to $(0,0,0)$. These are distinct points of the new cylinder, each labelled by its own angle. Their common image is the old apex. Away from $u=0$, the blowdown map is a real-analytic diffeomorphism onto the cone.
Which exceptional divisor? In the ambient $\mathbb{R}^3$ blowup, the fiber over the origin is $\mathbb{RP}^2$ (the space of all lines through the origin). Restricting to the cone, the exceptional locus is the projective conic $[X:Y:Z]$ satisfying $X^2+Y^2=Z^2$, whose real points form a circle. Because every real line on this cone has $Z \neq 0$, this single chart covers the entire real exceptional circle.
8. Why Nash iteration can fail—and how resolution still succeeds
Nash’s construction starts on the smooth locus of an embedded $d$-dimensional variety. Attach its tangent $d$-plane, a point of the Grassmannian $\operatorname{Gr}(d,N)$, and take the closure:
The construction is the closure of the tangent-space graph.[2]
The normalized Nash blowup takes one additional operation at each step: normalization, which adjoins integral algebraic functions missing from the coordinate ring. In characteristic zero, ordinary Nash iteration resolves curves; Spivakovsky proved termination for normalized surface iteration. Neither statement settles ordinary Nash iteration on all surfaces.
For the counterexample, follow one singular neighborhood through successive Nash steps. If the same singular local type returns, the rule can repeat forever. We will compare that failure with an explicit smooth model of the very same variety.
Related example: the fourfold and its real zero-level sections Figures 11–12
Castillo, Duarte, Leyton-Álvarez, and Liendo[12] exhibit a singular normal fourfold with an open chart of its Nash blowup isomorphic to the starting variety:
This is a precise obstruction to termination. Nash blowup commutes with restricting to open subsets and with isomorphisms. The same singular local type therefore survives inside every iterate. Their threefold example[11] returns after two ordinary Nash steps and starts from a non-normal variety. Normalized Nash iteration in dimension three remains open.
Optional calculation: an affine chart of the fourfold’s Nash blowup
Figure 11. The 4D counterexample. Left: equations of $X$. Right: a chart of its Nash blowup. Click the button: the blown-up coordinates satisfy the exact same equations as $X$, illustrating the recurring affine chart. The button chooses positive torus parameters $t_1,\ldots,t_4$, forms $x_i=t^{h_i}$ from the exponent vectors in the table, then computes the displayed ratios $y_i$. The resulting exponent vectors are $Uh_i$. This checks the equations numerically; the integer change of basis and the cited paper establish the chart isomorphism.
| Exponent of X | Vector | U sends it to | Which is the chart exponent |
|---|---|---|---|
| h₁ | (1, 0, 0, 0) | (−1, 0, 1, 0) | h₆ − h₅ |
| h₂ | (0, 1, 0, 0) | (0, −1, 0, 1) | h₄ − h₂ |
| h₃ | (0, 0, 1, 0) | (−2, −3, 2, 2) | h₄ − h₅ |
| h₄ | (0, 0, 0, 1) | (1, 0, 0, 0) | h₁ |
| h₅ | (2, 3, −2, −1) | (1, 3, −2, −1) | h₆ − h₃ |
| h₆ | (1, 3, −1, −1) | (0, 0, −1, 1) | h₄ − h₃ |
| h₇ | (1, 2, −1, 0) | (1, 1, −1, 0) | h₇ − h₂ |
A zero-set view of the actual four-dimensional example
The four-dimensional Castillo et al. counterexample lives in seven ambient coordinates $(x_1, \dots, x_7)$, so its Gauss map takes values in $\operatorname{Gr}(4, 7)$. To visualize its geometry, take a real spatial section through its defining toric relations: fix $x_5 = 1$ and $x_2 = q$, using $(p, t, r) = (x_1, x_6, x_4)$ as coordinates. The resulting zero set is:
For $q \neq 0$, the section is a smooth surface. But at $q = 0$, the equation collapses to $r t^2 = 0$, which is the union of two perpendicular planes: $r = 0$ and $t = 0$. Their intersection line lies directly inside the fourfold's singular locus.
Figure 12. Real sections of the actual 4-fold counterexample. Left: the central singular section $q = 0$ (two intersecting planes). Right: adjust $q$ to sweep through nearby smooth sections $r t^2 = q^3$. Scrubbing the slider reveals how smooth sections degenerate into intersecting planes. The singularity of a section alone does not identify the full variety’s singular locus.
r · t² = 0. Two planes intersecting along the singular line.
For q ≠ 0, the section forms a smooth section, with t ≠ 0.
The three-dimensional counterexample: see the singularity first
Three-dimensional does not mean a surface in ordinary 3D space. Here $X$ has three independent parameters but lives in six ambient coordinates. We can draw a two-dimensional section of its real points. The later lattice diagrams encode its coordinate algebra; they are a different kind of picture.
Fix $x_4=1$ and project to $(A,B,C)=(x_1,x_2,x_6)$. Direct substitution into the defining equations gives this exact real zero set:
Most of it is parametrized by $(A,B,C)=(su,s^2,u^3)$. Include the negative $B$-axis too: it consists of real solutions that this parametrization misses. Red marks the two lines belonging to the algebraic singular locus. At $B>0$, the cut contains two cubic branches with a common tangent at the origin of that cut.
This section has x₄ = 1, so it does not contain the six-coordinate origin used in the recurrence proof. It shows part of the singular locus elsewhere in the same variety. The chart calculation below tracks the full threefold, including its torus-fixed origin.
Figure 13. Rotate the real surface section, then vary $B$. The purple curves on the surface are the cut drawn at right. Try $B=0$ and a negative value. Changing $B$ changes the section; it does not perform a Nash blowup. Real projections can conceal complex tangent directions, so this picture supplies intuition; the chart calculations below establish nontermination.
Check the section against the six ambient coordinates
For a point of $A^6=B^3C^2$, put $u=\sqrt[3]{C}$. The recovered point is $(x_1,\ldots,x_6)=(A,B,Bu,1,Au,C)$. All five equations vanish:
For $B>0$, take $s=\pm\sqrt B$: the two branches are $C=\pm A^3/B^{3/2}$. For $B<0$, nonnegativity of $A^6$ forces $A=C=0$. This checks the extra negative axis. The real projection is not a replacement for the complex variety.
Compare the two routes on this same threefold
Recording tangent planes can separate some limiting positions without making the space of records smooth. That new space obeys its own algebraic relations and can have singular points of its own. Iteration asks Nash’s rule to resolve those new singularities as well.
Keep the question fixed: can every neighborhood upstairs be made smooth while retaining a map to this same $X$? These two routes have different outcomes:
Nash’s forced rule
Singular X → tangent-plane records → tangent-plane records again.
After two steps, one open chart is a copy of the original singular X. Its origin still has tangent dimension 6 instead of 3. Repeating the rule inside that chart repeats the obstruction.
A chosen resolution
The same X → normalization → a chosen subdivision → four smooth charts.
Each final chart has three independent coordinates. Their monomial maps satisfy the original equations exactly. The exceptional surfaces retain distinctions above the old singular set.
This second route uses the special monomial structure of this example. It demonstrates the endpoint promised by resolution, with every chart calculated. It is not an execution of the general Hironaka or ATW algorithm; the general center-selection procedure was explained in Part I.
The result of Castillo, Duarte, Leyton-Álvarez, and Liendo[11] concerns the ordinary, non-normalized Nash blowup. A selected open chart after two iterations is isomorphic to the starting singular threefold. It does not say that the entire blowup is unchanged, or settle normalized Nash resolution in dimension three.
At $(x_1,\ldots,x_6)=0$, all five defining equations have zero linear part. The tangent space there has dimension 6, while $X$ has dimension 3: this fixed point is singular. We will see a point of this same local type return after two selected Nash charts.
How exponent vectors encode the threefold and its tangent planes
The three-dimensional construction[11] begins with an affine semigroup $S \subset \mathbb{Z}^3$ generated by six vectors $h_i$, the columns of the matrix:
Each column vector represents the exponent of a monomial. The associated 3-fold variety $X(S)$ is the closure of the image of:
How tangent planes enter this calculation. On the dense torus, the tangent-plane coordinates are Jacobian minors. Dividing by a chosen nonzero minor gives an affine Nash chart. Ratios of monomials subtract their exponent vectors: for example $x_5/x_1=\chi^{h_5-h_1}$. That is why the new lattice generators are differences, rather than arbitrarily chosen arrows.
What Figure 14 is checking
A resolution has to make every neighborhood smooth. To disprove termination, it is enough to follow one bad neighborhood. Here a “chart” means an open piece of the new space, described by its own coordinates and equations.
The test below uses which algebraic functions those coordinates can express. A chart’s coordinate algebra consists of polynomial expressions in its coordinates, subject to its defining equations. On the cusp $(t^2,t^3)$, for comparison, the parameter $t$ is missing from its coordinate ring although $t^2$ and $t^3$ are present. Such a missing integral function witnesses non-normality, which implies singularity. The threefold has a similar obstruction. An exponent plot is bookkeeping for these functions, not a drawing of the variety.
Figure 14. Click through the two Nash steps, then identify the returning chart. Read what changed and what still fails. The first step fixes one missing function but leaves another. After the second, the whole coordinate algebra is the original one in a new integer basis. That is why this singular neighborhood can recur indefinitely. Expand the lattice inspection only to check that arithmetic.
0 blowups · singular at the torus-fixed origin
Inspect the exponent arithmetic 3D algebra diagram
A dot (i, j, k) means the monomial aⁱbʲcᵏ, not a spatial position. Multiplying monomials adds their exponents. Nonnegative integer sums of the solid generators give the allowed monomials. For example, q = c/b has exponent (0, −1, 1); its cube has exponent (0, −3, 3).
u = (0, −1, 1) is missing; 3u = h₄ + h₆ is present in S.
X(S_A) has dimension 3 and 8 monomial coordinates.
The first chart uses A = {h₁, h₄, h₆}, whose determinant is 3.
Drag either plot to rotate the 3D lattice, or use the rotation slider. The coordinate axes represent integer exponent powers.
Verify the missing function and the returning coordinate algebra
Why there is a singularity to begin with: The lattice point $u = (0, -1, 1)$ is absent from the semigroup ($u \notin S$), yet its third multiple $3u = h_4 + h_6 = (0, -3, 3)$ is present in $S$! Using the grading functional $L(x,y,z) = x + 2y + 3z$, all generators have positive integer grades, and only $h_1$ and $h_6$ have grade 1 (neither of which equals $u$). Because $S$ has missing lattice points, the semigroup is not saturated: its coordinate ring is not integrally closed, and the variety is singular.
Why two Nash blowups cannot escape: The first Nash chart introduces eight generators $g_i$. A subsequent Nash blowup produces a chart with six generators $t_i$. Their relation to the original six generators $h_i$ is an exact integer linear transformation:
Because $U_3 \in \operatorname{GL}(3, \mathbb{Z})$ is an unimodular integer matrix, it is an isomorphism of the integer lattice $\mathbb{Z}^3$. It identifies the full semigroup $T = U_3 S$ with $S$, proving that the resulting affine chart is algebraically isomorphic to the starting variety: $X(T) \cong X(S)$. Iterating Nash blowups simply reproduces this exact singular chart indefinitely.
An explicit resolution that does terminate
The Nash rule fails on this example, but resolution is still possible. Here is a computed toric resolution of this same threefold, using normalization and a regular fan refinement. It gives all four smooth charts and their maps to the original six coordinates. This is a resolution of the variety; it is not presented as the canonical sequence of ambient weighted centers produced by the general Abramovich–Temkin–Włodarczyk algorithm.
What does the smooth space look like?
Use ordinary coordinates (r, s, t) upstairs. They can vary independently, so this source chart is smooth. The map below sends each such point to six coordinates satisfying the original equations. Hold r and s fixed and reduce t: the image approaches the old origin, while the source approaches a particular point of the plane t = 0.
In chart D, the map is particularly easy to inspect:
The new ray $n_5$ gives the scaling exponents $(1,1,2,1,2,2)$ of the six coordinates as $t\to0$. At $t=0$, the entire smooth coordinate plane $(r,s,0)$ maps to the old singular origin. Here the exceptional divisor is a surface, because the resolved space is three-dimensional. In a surface blowup, the analogous divisor is a curve.
Figure 15. Where did the singular point go? Compare several level sets of $t=x_4$ at once. Each colored plane on the left maps to a matching colored patch on the right. The violet plane $t=0$ is different: every one of its points maps to the old origin. The black path holds $r,s$ fixed while $t$ varies; the white dot marks your selected point. Animate the approach, then change $r$ or $s$ at $t=0$: the resolved point moves while its image stays put.
The four charts describe a complex algebraic resolution. The earlier section's extra negative B-axis has no real lift to the normalization: V² = x₂x₄ = B < 0 there. Those points lift over ℂ, through non-real values of V. This is another reason a real surface picture cannot display the whole complex resolution.
Compare the level sets numerically along the black path
Hold the current r and s fixed. Each row is a point on the corresponding colored slice; the right-hand coordinates are its exact image, before the three-coordinate projection.
| t = x₄ | x₁ | x₂ | x₃ | x₅ | x₆ |
|---|
These are coordinate level sets of t = x₄. All colored images satisfy the same five defining equations of X; they are not different error levels. The right view projects six coordinates to three, so apparent overlaps need not mean the full points coincide. Only a bounded real patch is drawn. This is chart D of the computed toric resolution; all four charts are needed globally. Drag either view or use its arrow keys to rotate both.
Verify all four charts: normalization and the fan calculation
Now work in the dual lattice $N$, where a vector $n$ records the vanishing order $\langle h_i,n\rangle$ of the coordinate monomial $x_i$. Solving the six inequalities $\langle h_i,n\rangle\ge0$ gives the four primitive rays:
The polygon below is a flat cross-section of this cone, not the physical variety. First normalize $S$ to $\operatorname{cone}(S)\cap\mathbb Z^3$. Then triangulate along $\langle n_1,n_4\rangle$. One cone has determinant 1 in absolute value; the other has determinant 2. The latter contains the primitive lattice vector:
Subdivide that cone around $n_5$. The final cones are $124$, $135$, $145$, and $345$ (indices of the $n_i$); their determinants are $-1,+1,-1,+1$. Thus every chart is a smooth $\mathbb A^3$. This is the regular-fan resolution criterion[13], applied here with explicit integer arithmetic.
Compute the normalization, not just its name
On the original dense torus, adjoin $U=c/b$ and $V=c/a$, with exponents $u=(0,-1,1)$ and $v=(-1,0,1)$. Both are integral: $U^3=x_4x_6$ and $V^2=x_2x_4$. The saturated semigroup has generators $h_1,h_2,h_4,h_6,u,v$. In particular $x_3=x_2U$ and $x_5=x_1U$, so the original ring sits inside this six-generator normal ring.
To check completeness, split the exponent cone into $\langle h_1,h_2,h_4\rangle$ and $\langle h_1,h_4,h_6\rangle$. Their determinants are 2 and 3. The only nonzero lattice points in their half-open fundamental parallelepipeds are $v$, $u$, and $2u$. Every lattice point in either cone is one of these representatives plus a nonnegative integer sum of its ray generators. Thus adjoining $U,V$ generates the whole saturation.
Figure 16. Play the computed steps or advance manually. Normalization changes the semigroup even though the polygon stays fixed. Triangulation alone leaves an index-2 singular chart. Adding $n_5$ removes that defect. At the last step, select any of the four charts to see all six monomial coordinate maps.
Input: the non-normal threefold
Each step changes the fan by the displayed integer calculation. The four determinant-one cones certify smoothness.
Why the chart cover is a resolution
For a final cone with basis vectors $(n_i,n_j,n_k)$ and coordinates $(r,s,t)$, set $x_\ell=r^{\langle h_\ell,n_i\rangle}s^{\langle h_\ell,n_j\rangle}t^{\langle h_\ell,n_k\rangle}$. All exponents are nonnegative integers. Substitution satisfies the five original equations identically. Determinant $\pm1$ makes the torus map invertible with integer monomial exponents. The cones cover the original cone, with unchanged boundary faces, so the charts glue to a proper birational morphism from a smooth variety, unchanged over the original smooth locus.
How this relates to the ATW weighted algorithm
Abramovich, Temkin, and Włodarczyk[9] construct a functorial embedded resolution using stack-theoretic weighted blowups. Their centers come from recursively computed coefficient ideals and maximal contact, not just the exponents of one polynomial in arbitrary coordinates. Smooth stack charts can have finite quotient identifications; passing to a smooth scheme requires further work such as destackification.
The calculation above instead exploits this example's toric presentation. It supplies a complete smooth model and its maps explicitly, but does not compute or assert the ATW center sequence in the six-dimensional ambient space. Nor does it assert termination of normalized Nash iteration: our subdivisions were chosen by the toric resolution procedure.
| Dimension of Variety | Ordinary Nash Blowup | Normalized Nash Blowup | Key Mechanism |
|---|---|---|---|
| 1 · Curves | Terminates (1 step for cusp) | Terminates (Normal curves smooth) | Iteration terminates; one step suffices for the ordinary cusp. |
| 2 · Surfaces | Open in full generality | Terminates (Spivakovsky, 1990) | Normalization is included at each step; this does not settle ordinary Nash iteration. |
| 3 · Threefolds | Fails (Castillo et al., 2025) | Open problem | Lattice hole $u \notin S, 3u \in S$ recurs under $U_3 \in \operatorname{GL}(3, \mathbb{Z})$. |
| 4+ · Fourfolds+ | Fails (Castillo et al., 2024/2026) | Fails (Castillo et al., 2024/2026) | Unimodular recurrence holds even for normal starting varieties. |
Why this proves failure. Nash blowup restricts to open subsets and respects isomorphisms. Once an open chart of the starting singular type returns after two steps, the same calculation applies inside that chart again. Thus every even iterate contains a singular open subset. The ordinary Nash iteration cannot become smooth. Hironaka's theorem uses a different, controlled sequence of smooth centers; it is not contradicted.
Why the other resolution succeeds. The Nash construction above is forced to reproduce the same semigroup. Our toric procedure instead normalizes it and subdivides its cone until each chart has an integer coordinate basis. Determinant $\pm1$ is a verifiable stopping test. General Hironaka-style algorithms use richer invariants to choose centers and prove termination without requiring a toric presentation.
A blowup can improve a singularity without finishing it; a canonical modification can even repeat it. Hironaka’s achievement is the existence of a controlled finite sequence that reaches the required endpoint. The next chapter follows one tensor-fitting problem into learning theory, and where the analogy stops.