MATH.18:5 - Archetypal Grounding

MATH.18:5.1 - Recover an order from an operation, then compare allowed maps

One account starts with a set S and a binary operation a∨b satisfying associativity, commutativity and idempotence:

(a∨b)∨c=a∨(b∨c); a∨b=b∨a; a∨a=a.

Another starts with a partial order in which every pair has a least upper bound. An upper bound of a and b is an element z with a<=z and b<=z. The least upper bound lies below every such z. Antisymmetry makes it unique. We want to move constructions between these descriptions.

From the operation, define a<=b to mean a∨b=b. Idempotence gives reflexivity. If a∨b=b and b∨a=a, commutativity gives a=b, establishing antisymmetry. If a∨b=b and b∨c=c, associativity gives:

a∨c=a∨(b∨c)=(a∨b)∨c=b∨c=c.

Thus the relation is transitive. Also a∨b is an upper bound of a and b. If z is any upper bound, then (a∨b)∨z=a∨(b∨z)=a∨z=z. So a∨b<=z and the original operation supplies the least upper bound.

Conversely, define a∨b to be the least upper bound in the ordered account. Uniqueness makes the operation commutative and idempotent. Both (a∨b)∨c and a∨(b∨c) are the least upper bound of the same three elements, giving associativity.

The round trips recover both primitives. Starting from the operation returns it as the least upper bound just proved. Starting from the order returns it because a<=b holds exactly when b is the least upper bound of a and b.

Now the intended use expands: can every order-preserving map be used as a map preserving the binary operation? A join-preserving map is monotone: apply it to a∨b=b. The converse fails. Take subsets of {u,v} ordered by inclusion, with union as join, and target {0,1} with its usual order. Define h to return 1 only on the full set and 0 on all other subsets. It is monotone, yet:

h({u} union {v})=1, while max(h({u}),h({v}))=0.

For constructions that combine joins, select join-preserving maps in both accounts. If all monotone maps are needed, retain that broader ordered account and its different transformation class. The objects matched; the expanded map claim required and received its own answer.

MATH.18:5.2 - An enlarged domain changes the available solution

The natural numbers N={0,1,2,...} embed in the integers Z. Inclusion preserves 0, addition and equality of expressions evaluated on natural-number inputs. It also reflects those equalities: two included natural numbers are equal in Z precisely when they were equal in N.

Suppose the next task is to solve x+1=0. The integer solution x=-1 is outside N. The interpretation useful for additive calculations therefore has not supplied a natural-number solution.

To express the original existential question in Z, retain its range: “there exists an integer x with x>=0 and x+1=0.” That statement remains false. To use Z as a calculation space for another natural-number equation, compute there and test the recovered candidates against the source domain.

The revised use keeps the beneficial integer calculations and makes their return condition explicit. A demand to identify all natural-number and integer solution questions would instead fail this comparison.

MATH.18:5.3 - Compare linear maps with matrices, then add a length question

Additional structure – finite-dimensional real vector spaces. In one account, each space comes with a specified ordered basis, and maps between spaces are all linear maps. In the other, an object is a nonnegative integer n and a map n to m is an m-by-n real matrix. Maps compose by matrix multiplication.

For a space V with basis (b1,...,bn), let c_V:V -> R^n send a vector to its coefficients in that basis. Interpret f:V -> W by the matrix of c_W∘f∘c_V^(-1). Composition agrees because the adjacent coordinate conversion and its inverse cancel. Identity maps become identity matrices.

The return takes n to R^n with its standard basis and a matrix to its linear map. The round trip on matrices is literal. The round trip on V returns its coordinate space, with recovery supplied by c_V. For each f, the equation matrix(f)∘c_V=c_W∘f establishes the required compatibility. This compares the entire selected system of linear maps, including their compositions, rather than only matching vector values.

Now ask for lengths in Euclidean space. With basis b1=(1,0), b2=(0,2), coordinates (0,1) represent a vector of length 2; their ordinary coordinate length is 1. The linear comparison did not include the inner product. Carry it as a Gram matrix: here G=diag(1,4) and squared length is c^T*G*c. Retaining G lets us calculate the same vector’s length after changing coordinates.

A different question asks which linear maps preserve lengths. For a map from V to W represented by matrix M, with Gram matrices G_V and G_W, require M^T*G_W*M=G_V. This follows by comparing the squared length c^T*G_V*c of every input with (M*c)^T*G_W*(M*c) of its output. Use that condition to select the length-preserving maps. The earlier interpretation of arbitrary linear maps remains available when length preservation is not required.

MATH.18:5.4 - Recover an operation using a chosen origin

On the integers, one account supplies the ternary operation t(x,y,z)=x-y+z and a marked origin 0. It recovers addition as t(x,0,y) and negation as t(0,x,0). Conversely, addition and negation recover t. Substitution verifies both returns for every integer input.

Now remove the marked origin. Choosing any integer c gives an addition x+_c y=t(x,c,y)=x-c+y, identity c and inverse 2c-x. Combining these operations again gives the same t(x,y,z). The ternary operation alone therefore permits several choices of origin and corresponding binary operations.

For example, c=0 makes the sum of 2 and 3 equal 5; c=1 makes it equal 4 under +_1. Both choices recover the original ternary operation. A question using only t can continue without selecting an origin; recovering the former addition needs its marked origin again. This distinguishes a usable one-way interpretation from a return that silently supplies extra structure.