Why the S combinator is curvature

Continues from Why approximate sum is composition.

With the same conventions as in that post, we just translate the S combinator

$S = \lambda w.\lambda y.\lambda x.((w x) (y x))$

into

$S = \delta_{z}^{w} \delta_{z}^{y} \delta_{z}^{x} \delta_{1/z}^{\delta_{1/z}^{x} y} \delta_{1/z}^{x} w$

With the notation from arXiv:2110.08178, page 4, “Curvature as deviation from linearity”, we see that

$S = \delta_{z}^{w} LIN_{1/z,1/z}(x,y,w)$

therefore S is the deviation from linearity, aka curvature.

Another interpretation of the S combinator comes from $(S A) B$ , which in translation has the form

$(S A) B = \delta_{1/z}^{B} \delta_{1/z}^{A} \delta_{z}^{w} \delta_{z}^{y} \delta_{z}^{x} \delta_{1/z}^{\delta_{1/z}^{x} y} \delta_{1/z}^{x} w$

From two beta reductions like (the first one)

$\delta_{1/z}^{A} \delta_{z}^{w} C \rightarrow C[w=A]$

we arrive to

$(S A) B \rightarrow \delta_{z}^{x} \delta_{1/z}^{\delta_{1/z}^{x} B} \delta_{1/z}^{x} A$

which is a relative dilation, of coefficient $1/z$ , based at w and the coefficient $1/z$ . Let’s use a notation which I can write on this page:

$\left(\delta, w, 1/z\right)_{1/z}^{B} A = \delta_{z}^{x} \delta_{1/z}^{\delta_{1/z}^{x} B} \delta_{1/z}^{x} A$

$(S A) B \rightarrow \left(\delta, w, 1/z\right)_{1/z}^{B} A$

It appears in the abstraction elimination, or in the process of turning lambda terms into SKI combinators. Say $T[A]$ is the conversion of the term A. Part of the definition of T (the one which is the abstraction elimination) is

$T[\lambda w. (A B)] = ((S T[\lambda w. A]) T[\lambda w. B]$

(which is a very awkward and so human based way to use the same w in two lambda operations… intuitively clear though)

In translation this would read as

$T\left[ \delta_{z}^{w} \delta_{1/z}^{B} A \right] = \left(\delta, w, 1/z\right)_{1/z}^{T\left[ \delta^{w}_{z} B\right]} T\left[ \delta^{w}_{z} A \right]$

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chorasimilarity

Why the S combinator is curvature

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