Looking for clarification of C-H Sah's definition of abstract scissors congruence

Question

In C-H Sah's book Hilbert's third problem: scissors congruence, the author defines the data for abstract scissors congruence in order to prove Zylev's theorem by combinatorial means in great abstraction. I expect that most of the book is over my head as an undergrad but I'm looking for clarification as to wether the definition Sah gives is actually correct. My advisor and I could not figure out how to resolve the apparent discrepancies.

The definition is given as follows on page 5:

The abstract scissors congruence data consists of a distinguished family of nonempty subsets (to be called $n$-simplices where $n$ is to be interpreted as the dimension) of a nonempty set $X$ and a specified equivalence relation (called congruence) among the $n$-simplices. We need a few definitions before stating the addition condition to be satisfied by the $n$-simplices.

Two $n$-simplices $A$ and $B$ are said to be interior disjoint if the following conditions hold:

• (D1) $A \cap B$ contains no $n$-simplices, and
• (D2) if $C$ is an $n$-simplex contained in $A \cup B$, then $C \subset A$ if and only if $C \cap B$ contains no $n$-simplices.

A polyhedron $P$ is understood to be a finite pairwise interior disjoint union of $n$-simplices. The concept of interior disjoint union can then be extended to polyhedra with $n$-simplices replaced by nonempty polyhedra. We will use $\coprod$ to denote pairwise interior disjoint unions (as well as direct sum when there is no chance of confusion). We omit the proof of the following elementary result:

Lemma 2.1. If $P$, $Q$ and $R$ are polyhedra with $P \coprod R = Q \coprod R$, then $P = Q$.

If $A$, $B$, $C$ are $n$-simplices with $A = B \coprod C$, we say $A$ is simply subdivided into $B$ and $C$. If $P = \coprod P_i$ is a polyhedron where each $P_i$ is an $n$-simplex, then a subdivision of $P$ is understood to be a finite succession of simple subdivisions such that each simple subdivision is performed on one of the $n$-simplices exhibited in the preceeding step. For example, the first step may be a simple subdivision of $P_1$ into $Q$ and $R$; the second step may be a simple subdivision of $Q$ or $R$ or any $P_j$ with $j > 1$; and so on.

To complete the abstract scissors congruence data, we impose the following condition:

• (S) Let $A$ and $B$ be $n$-simplices. Then there is at least one subdivision of $A$, say $A = \coprod_{1 \le i \le t} A_i$, such that $A \cap B = \coprod_{j \in J} A_j$ for some $J \subset \{ 1,\dots,t \}$.

My issue is with the axiom (S). It seems that the expression $A \cap B = \coprod_{j \in J} A_j$ is either not well-defined or implicitly states the idea that $A \cap B$ must be a polyhedron. Taking the second interpretation, if we consider the usual case of Euclidean scissors congruence then this axiom does not hold, as the intersection of two triangles joined side to side is not a polygon.

Another example is given on the next page: for $n \ge 0$, let the $n$-simplices be arbitrary $(n+1)$-element subsets of an infinite set $X$, and congruence is equinumerosity. If $n = 0$ then the polyhedra are just finite sets and interior disjointness is set-theoretic disjointness. Sah then states that if $n>0$ then no $n$-simplices are interior disjoint (because D2 never holds) and moreover he states that '(S) is clear and cardinality is a complete invariant'. As far as I can tell (S) does not actually hold in this case either, as the intersection of two $(n+1)$-element sets may have fewer elements but not be empty.

Keeping the Euclidean simplices in mind, the most obvious way to fix the statement of axiom (S) would be to require $A$ and $B$ to either be interior disjoint or else satisfy the current definition of (S). This still does not fix the finite subset case, though, as no two simplices are interior disjoint.

Presumably the entire book is not built on nonsense, so what is the proper definition used by Sah and other authors that refer to this book? Another definition of abstract scissors congruence in terms of category theory (specifically the idea of double categories) is given by Inna Zakharevich in her thesis Scissors Congruence and K-Theory, but it strikes me as a bit more complicated, and I was hoping for a more elementary definition requiring less heavy machinery.

Answer 1

I agree that Sah's definition is not quite correct, but it's not difficult to fix. The key point is this: when you have an intersection that contains $0$ simplices, you declare it to be empty. If you look at the definition of an assembler (which was partially designed to fix Sah's definition) it uses the categorical structure of the assembler to keep track of this. The point of axiom (S) here is that "you can always refine a subdivision." If you have two different subdivisions of a polytope, you can find a common refinement; if you have a subpolytope of a polytope and a subdivision of the large one, then you can refine the subdivision so that it restricts to a subdivision of the smaller one.

To fix the definition you need to interpret intersection as "take the union of all simplices contained in the intersection." So in both cases that you're worried about, the intersection is actually empty and everything holds automatically. This is actually the point of the definition of an assembler: since you take the "intersection" inside the assembler (the categorical definition of pullback) you automatically disregard any non-valid sets which are contained in the intersection.

Also: quick explanation of what an assembler is. It's the partial order on valid simplices. The "covering families" inside the Grothendieck topology are exactly Sah's families of covers. Axiom (R) says any two covers have a common subcover. Axiom (E) says that the empty set can be covered by nothing. Disjointness says that there are no valid simplices contained in the intersection of two simplices. That's pretty much it. The goal is to fix Sah's definition and to also allow it to be analyzed by $K$-theoretic means, but the idea is strongly inspired by Sah.