Tameness and scope: when finite descriptions exist

The indicator-presentations chapter described the square module using two regions and a coefficient matrix. The finite-encodings chapter recovered the same module from a finite poset, its spaces and maps, and a classifier on the parameter plane. When can a module have descriptions like these?

The answer depends on controlling how its maps vary, as well as the size of its spaces. We will make that requirement precise, revisit the square, and examine one module whose stalks all have dimension one but which has no finite encoding.

For the finite constructions in these lessons, the condition comes from the construction itself. The follow-on chapter, why finite computations stay tame, explains that practical argument and the role of abelian categories. Here we establish exactly what the condition means and which descriptions it guarantees.

You need the preceding chapters' definitions of modules, finite encodings, and fringe presentations. We use a fixed coefficient field $\mathbb{k}$; take $\mathbb{k}=\mathbb{Q}$ in the examples. The general statements apply to an arbitrary parameter poset $Q$.

What should it mean for a module to be constant on regions?

Recall the square $S=[0,2]^2$ in $\mathbb{R}^2$. Its module $M$ has $M(q)=\mathbb{k}$ inside $S$ and $M(q)=0$ outside. A structure map between comparable points inside is the identity; every other structure map is the uniquely determined zero map of the appropriate shape.

There are two kinds of spaces here. Partition the plane into

\[R_1=S,\qquad R_0=\mathbb{R}^2\setminus S.\]

Use $\mathbb{k}$ as the common space for $R_1$ and $0$ for $R_0$. Once these spaces are identified with the stalks, the maps depend only on which two regions contain their endpoints:

This is more than a partition by dimensions: it describes the maps consistently too.

For a general module, choose a partition $Q=\bigsqcup_i R_i$, a vector space $V_i$ for each region, and isomorphisms $\alpha_q:V_i\to M(q)$ for $q\in R_i$. It is a constant subdivision if, for every ordered pair $(R_i,R_j)$, the map

\[\alpha_r^{-1}\,M(q\leq r)\,\alpha_q:V_i\longrightarrow V_j\]

is independent of the chosen comparable pair $q\in R_i$, $r\in R_j$. This includes $i=j$ and self-comparisons. This is the compatibility condition in Ezra Miller's Definition 2.6.

The equation starts with a vector in $V_i$, sends it to $M(q)$, applies the module map, then expresses the answer in the target region's common space $V_j$. All comparisons use the same chosen identifications. Choosing unrelated bases separately for each arrow would not establish this property. The condition is called no monodromy in the cited definition.

For points in the same region, comparison with a self-map forces the displayed map to be the identity on $V_i$. Thus any structure map between comparable points of that region must be an isomorphism. For incomparable points there is no structure map to require. Regions need not be connected, convex, or bounded; the square's entire exterior is already one region in this subdivision.

The notion of tameness used here

A module is pointwise finite-dimensional if each stalk $M(q)$ has finite dimension. Ezra Miller also calls this $Q$-finite; it does not mean that the parameter poset $Q$ is finite.

In this chapter, tame means pointwise finite-dimensional and admitting a constant subdivision with finitely many regions. Both requirements belong to Definition 2.11.

The square is tame: the two regions above work, and their common spaces have dimensions one and zero. Tameness concerns the whole stated parameter domain, here all of $\mathbb{R}^2$. Finite does not mean a bounded domain or bounded regions.

The two requirements control different things. Finite-dimensional stalks limit the size of each individual space. A finite constant subdivision limits how many descriptions of spaces and maps are needed across the domain. Together they give a uniform bound on stalk dimensions: take the largest of the finitely many dimensions $\dim V_i$. The counterexample below will show that even a uniform bound of one does not suffice by itself.

A constant subdivision is not yet an encoding

The square's two-region subdivision is valid, but those two regions cannot be the fibers of an order-preserving map to a two-element poset. Indeed,

\[(-1,1)\leq(1,1)\leq(3,1).\]

The first and last points are exterior, while the middle point is inside. If $z$ labels the exterior and $a$ labels the square, monotonicity would force $z\leq a\leq z$. Antisymmetry then gives $z=a$, incompatible with the different stalk dimensions.

The constant-subdivision condition did not require its two region names to form a poset. An encoding does. This is the extra organization that lets a finite module supply the required maps.

For this square we already have a suitable refinement. Record

\[u(x,y)=\mathbf{1}_{\{x\geq0,\ y\geq0\}},\qquad c(x,y)=\mathbf{1}_{\{x>2\ \text{or}\ y>2\}}.\]

Both bits increase with the parameters, so $\pi(x,y)=(u(x,y),c(x,y))$ is order-preserving into $\{0<1\}^2$. Its four fibers split the exterior into three pieces and retain the square as the fiber labeled $(1,0)$. Give that label the space $\mathbb{k}$ and the other three labels zero. This is the four-label encoding developed earlier.

The counts measure different ingredients. They are not competing counts of persistent classes, and none of these constructions asserts minimality.

Three equivalent ways to have finite data

For a fixed field and a pointwise finite-dimensional module over any poset, the following are equivalent:

1. It admits a finite constant subdivision.
2. It admits a finite encoding: $M\cong\pi^*N$ with $\pi:Q\to P$ order-preserving, $P$ finite, and every $N(p)$ finite-dimensional.
3. It admits a finite fringe presentation: an image of a map from a finite sum of upset indicators to a finite sum of downset indicators, with one scalar on each component's intersection.

The first two conditions are related by Theorem 4.22; the third is included in Ezra Miller's syzygy theorem, Theorem 6.12. The pointwise finite-dimensional hypothesis is explicit here: a finite number of constant regions alone places no dimension bound on their common spaces.

These are existence statements about the same module. They do not identify the regions in one description with the labels or summands in another. Here is how the constructions fit together.

From an encoding to a constant subdivision

Take the nonempty fibers $\pi^{-1}(p)$. Identify each stalk in that fiber with $N(p)$ using the given module isomorphism. A map between fibers is then $N(p\leq p')$, independent of which comparable original parameters were chosen. There are only finitely many fibers, so they give a finite constant subdivision.

From a fringe presentation to an encoding

For each upset, record membership; for each downset, record membership in its complement. This produces finitely many increasing bits. Their patterns form a finite ordered set. At each pattern, select the active rows and columns of the coefficient matrix and take its image. Between comparable patterns, restrict the appropriate target-coordinate projection to the image.

The indicator-presentations chapter explained why these spaces and maps recover the presented module. The construction uses the fixed scalar coefficients as well as the regions; membership patterns alone would lose the linear algebra.

From a constant subdivision to an encoding

This is the substantive step supplied by the finite-encoding theorem: the compatible regional data admit a suitable finite ordered model. Its construction uses upsets derived from the constant regions to obtain order-preserving labels.

The new fibers need not be the original pieces. Nor should the theorem be read as saying that every arbitrarily chosen subdivision can be retained by merely splitting its pieces. It produces an encoding of the module. For the particular square subdivision above, the explicit four-fiber refinement does work.

Why a finite encoding also yields a fringe presentation

There is a finite linear-algebra construction that explains this direction. Start with the finite module $N$ on $P$. At a label $x$, form

\[F(x)=\bigoplus_{p\leq x}N(p),\qquad E(x)=\bigoplus_{r\geq x}N(r).\]

After choosing a basis for every $N(p)$, $F$ is a finite sum of principal upset indicators and $E$ a finite sum of principal downset indicators. “Principal” means the upset $\{x:p\leq x\}$ or the downset $\{x:x\leq r\}$ generated by one label.

Define maps

\[F(x)\xrightarrow{a_x}N(x)\xrightarrow{b_x}E(x),\]

where $a_x((v_p)_p)=\sum_{p\leq x}N(p\leq x)v_p$, and $b_x(v)=(N(x\leq r)v)_{r\geq x}$. The summand indexed by $x$ makes $a_x$ surjective and $b_x$ injective: in each case that coordinate uses the identity of $N(x)$. The maps commute with structure maps, so the image of $b\,a$ is isomorphic to $N$ as a module.

For a source summand indexed by $p$ and a target summand indexed by $r$, the composite at any $p\leq x\leq r$ is

\[N(x\leq r)\,N(p\leq x)=N(p\leq r).\]

Its matrix is independent of $x$. Thus every scalar component is constant on its upset–downset intersection, as a fringe presentation requires. If $p\not\leq r$, that intersection is empty and the component is zero. Pulling the construction back along $\pi$ replaces these regions by their preimages and gives a finite fringe presentation of $M$. This establishes existence; it need not give the smallest presentation. If a pulled-back intersection is empty, set its unused coefficient entries to zero.

A counterexample with dimension one everywhere

For this example only, use the one-parameter domain $\mathbb{R}$. Keep $\mathbb{k}=\mathbb{Q}$. Put

\[V(t)=\mathbb{k}\quad\text{for every }t\in\mathbb{R},\]

and, for $s\leq t$, define

\[V(s\leq t)= \begin{cases} \operatorname{id}_{\mathbb{k}},&\lfloor s\rfloor=\lfloor t\rfloor,\\ 0,&\lfloor s\rfloor\ne\lfloor t\rfloor. \end{cases}\]

Here $\lfloor t\rfloor$ is the integer $n$ satisfying $n\leq t<n+1$. The map is the identity within one unit interval and zero whenever it crosses an integer boundary. For example,

\[V(\tfrac14\leq\tfrac34)=[1],\qquad V(\tfrac34\leq1)=[0],\qquad V(1\leq\tfrac32)=[1].\]

This is a module. For $s\leq t\leq u$, the floors are nondecreasing. If the endpoint floors agree, all three agree and the maps are identities. Otherwise at least one of the two successive maps is zero, and both the composite and the direct map are zero. Self-maps are identities.

Equivalently, writing $I_n=[n,n+1)$,

\[V\cong\bigoplus_{n\in\mathbb{Z}}\mathbb{k}[I_n].\]

Here $\mathbb{k}[I_n]$ is the indicator module of the half-open interval $[n,n+1)$. Exactly one summand is present at each parameter. At an integer, one summand ends and the next begins, so the stalk is still $\mathbb{k}$.

Only a finite window of the construction is drawn. The unit intervals continue in both directions. Open right endpoints belong to the ending summand's convention; they do not mark zero stalks of the whole module.

Why no finite constant subdivision can work

Consider the infinitely many parameters $t_n=n+\tfrac12$ for integers $n\geq0$. In any partition into finitely many regions, two of them, say $t_n<t_m$, must belong to the same region.

Their structure map is zero, because they lie in different unit intervals. Changing bases cannot turn a zero map into an isomorphism. But two comparable points in one constant region must have an isomorphism between their stalks: their transported map must agree with the identity self-map. This is a contradiction. Hence $V$ is not tame.

There is also a direct obstruction to a finite encoding. Two of the $t_n$ would have to receive the same finite label $p$. Their map would then be recovered from $N(p\leq p)=\operatorname{id}_{N(p)}$ and would be an isomorphism, again contradicting the zero map. The equivalence theorem also rules out a finite fringe presentation.

Thus even constant dimension one everywhere does not guarantee any of these finite descriptions. A dimension plot loses precisely the distinction between the identity and zero maps in this example.

A short rule and a finite window are different claims

The floor formula describes $V$ in a few symbols. It even gives an encoding through the poset $\mathbb{Z}$, with $\mathbb{k}$ at every integer and zero maps between distinct integers. That encoding is infinite. A short program or formula is a different notion of finite description from the finite encodings considered here.

If we restrict the domain to a fixed bounded interval, only finitely many unit intervals meet it. Restricting the floor classifier to that finite set of integer labels gives a finite encoding of the restricted module. The restriction is tame, while the module on all of $\mathbb{R}$ is not. A finite drawing or a finite collection of successful queries cannot settle the global claim; the proof above does.

Figure placeholder: compare maps with the dimension profile. A pair of selected parameters will show the two stalks, their interval labels, and the identity or zero structure matrix. Widening the displayed window will reveal more intervals while the dimension stays one. The view will distinguish the finite window from the full real-line module.

Distinguish this from q-tameness

In one-parameter persistence, q-tame usually means that every map $V(s\leq t)$ with $s<t$ has finite rank; see Frédéric Chazal, William Crawley-Boevey, and Vin de Silva, Section 1.4. Our counterexample has ranks zero or one, so it is q-tame. It is not tame in the finite-constant-subdivision sense used in this chapter. The two terms impose different requirements.

From an existence theorem to an implemented construction

The theorem concerns modules over arbitrary posets and allows abstract upsets, downsets, and classifiers. To compute, we must also provide a representation of those ingredients and methods for evaluating membership and order. Finitely many sets need not come with finite geometric descriptions or an algorithm for testing membership.

The square supplies both kinds of information: its finite presentation has the coefficient $[1]$, and four explicit inequalities describe its regions. The selected encoder can construct the finite model and classify queries. The theorem alone is not an interface accepting an arbitrary mathematically specified module and deciding whether it is tame.

Some concrete TamerOp entry routes are:

The first row is not a promise of an encode(module) method. Mathematically the identity map of a finite poset is an encoding of its module, while the software can operate on that finite module directly.

For geometric presentations, consult the geometry contracts. For data, the ingestion guide explains filtration, degree, grid, truncation, and intermediate-stage choices. These determine which module is represented. A finite encoding recovers that chosen module; it does not undo a change made when constructing the input.

The coefficient field and geometric arithmetic are separate choices. Rational coefficients make the stated linear algebra exact without automatically making every geometric input exact. The exact-grade guide describes guarantees for its specific construction. Likewise, neither the existence theorem nor a successful example asserts that the returned encoding is minimal or inexpensive for every supported input.

The full syzygy theorem also relates tameness to finite indicator resolutions. A resolution organizes successive relations among simpler modules. Existence of such resolutions does not identify every computation over different base posets. In particular, finite-base Ext and Tor require the category and comparison hypotheses described in their guide.

Why do finite constructions meet this condition?

We can recognize a tame module through compatible constant regions, finite encoding data, or a finite fringe presentation. The square exhibits all three descriptions. The real-line counterexample explains why finite stalk dimensions and an informative-looking dimension plot do not replace the map compatibility requirement.

The remaining question has a practical payoff: why do our finite inputs meet this condition, and why do subsequent calculations preserve it? Continue with why finite computations stay tame for the finite-complex argument, a square-module calculation, and the closure theorems of Ezra Miller and Lukas Waas.