]> Smooth images of a simplex

Smooth images of a simplex

The isomorphism between an n-simplex and S(n) delivers 1+n scalar fields on the simplex, whose sum is 1. Any n of those scalar fields are independent variables on S(n), hence – via the isomorphism – on any n-simplex. This makes them suitable for use as a chart on a smooth manifold, when the n-simplex is a neighbourhood (i.e. subsumes an open patch of the smooth manifold); their gradients form a basis of the gradient bundle, yielding a dual basis of the tangent bundle. However, because we have 1+n variables with just one dependency, we can exploit the situation to obtain 1+n vector directions corresponding to our 1+n scalar fields – namely: for given i in 1+n we have the associated scalar field x(i) = ({positive scalars}: p(i) ←p :S(n)); the gradients of these fields, dx(i) for i in 1+n, sum to zero everywhere (because sum(x) = 1 is constant, so sum(dx) is zero); for each i in 1+n and position p in S(n)'s interior, there is a unique direction at p for which, along a path in this direction at p, all the scalar fields x(j), for j other than i, vary at equal rates; x(i) then varies at −n times the rate of the other n scalar fields. [Construction: use the dx(j) for j not equal to i as a basis for gradients, construct the dual basis of tangents, add the members of it; this is a tangent parallel to which the x(j) with j not equal to i all vary at rate 1.] We thus obtain ({tangents}: v :1+n) for which v(i)·dx(i) is n, v(i)·dx(j) = −1 for j, i distinct; furthermore, sum(v)·dx(i) is zero for each i, as the sum(: v(j)·dx(i) ←j |1+n) = zero since it comprises n, from v(i)·dx(i), plus n copies of −1 from the other v(j)·dx(j); making a total of zero. Since (: dx |1+n) spans {gradients}, sum(v)·u is zero for every gradient u, so sum(v) is zero.

Again, for each i, the v(j) with j not equal to i form a basis of {tangents} at each point in the simplex's interior, so can be used to express any tangent as sum(t.v) for some ({scalars}: t :1+n) with i not in (:t|). Now, sum(v) is zero, so adding an arbitrary constant ({scalars}:|1+n) to t won't change sum(t.v), though it will clearly change sum(t), and t(i) will no longer be zero. This will let us force sum(t) to be any particular value we like (e.g. zero or one), but it will equally allow us (by adding constant ({zero}:|1+n) to t, to make it defined everywhere, hence simply a list of length 1+n scalars, finding its minimum – if t has negative entries, this delivers the biggest of them, otherwise zero is the minimum – subtract this minimum from each entry in the list to obtain a non-negative list with some zero entries, then ignore the zero entries) to find a replacement for t, giving the same value for sum(t.v), with (:t|) still not being all of 1+n but now with (|t:) subsumed by {positives}. This constraint on t (not defined on all of 1+n, but positive where defined) provides a unique choice of ({positives}: t :1+n) to represent each tangent as sum(t.v): uniqueness follows from the fact that the only linear dependence among v's outputs is sum(v) = zero, and adding any constant to each member of t gives t values on all of 1+n, unless the constant were negative, in which case those i ignored by t would have negative values in place of t(i).

So any tangent in the interior of the simplex is a positive sum of the vectors associated with our scalar fields.

We can stitch together any finite-dimensional smooth manifold as a union of overlapping simplices, optionally only overlapping in their boundaries (if they overlap otherwise, cut them up into smaller simplices which don't). In the overlap, even if it is only the boundary, each simplex supplies its own family of co-ordinates: we must ask how the co-ordinates of one relate to those of the other. The portion of the smooth manifold in which two simplices overlap is equipped with two isomorphisms between the overlap and, for each isomorphism, a sub-set of a canonical simplex. By composing one isomorphism with the reverse of the other, we can obtain a mapping from one subset of a canonical simplex to another subset of a canonical simplex (not necessarily the same one).

So consider a continuum isomorphism between a subset of S(n) and a subset of S(m); with no loss of generality, presume that n is no greater than m. We should be able to describe the derivatives of S(n)'s co-ordinates with respect to those of S(m); the notion smooth manifold should then come down to these derivatives being smooth functions of position in S(n), possibly joined with a continuity constraint. Now, unless n is greater than m, S(n) appears as a sub-set of S(m), comprising those f in S(m) for which (:f:n) = f – i.e., n subsumes (:f|); consequently, so does S(n)'s subset. There must needs be a continuous deformation of S(n)'s subset, within S(m), into its image under the smooth isomorphism, via the intersection on the manifold of the two simplices. Whether that can already be inferred or needs to be taken as an axiom isn't yet clear to me.

Consider the easy case first. , e.g. that there should be a continuous deformation, within S(m), transforming the given subset of S(n) continuously, within S(m), onto its image i position to positive measures on finite collections: either or both might map position I presume a context that comprehends continuity, in which each position is surrounded by at least one open neighbourhood within the context. We can define mappings from such a context to scalar values, or to lists of scalar values; we can use notions of continuity for mappings to and from scalars, and (inferred therefrom) to and from lists of scalars, to distinguish which mappings from our context are continuous.

Smooth deformation of a simplex

Now consider a smooth transformation ({continuum isomorphisms}: (S(n)| f(t) |S(n)) ← t |{scalars between zero and one}) of S(n) onto itself – i.e. f(0) is the identity on S(n), f(1) is some smoothly-equivalent continuum isomorphism of S(n). (Where S(n) is seen as a face of S(i+n) for some positive natural i, this is like smoothly transforming one of the n-faces of S(i+n) within S(i+n) onto another n-face, then comparing the results via each face's isomorphism with S(n).) For each p in S(n), (S(n): f(t,p) ← t :) is a smooth trajectory starting at p; it may also end at p, but need not. Suppose, furthermore, that f preserves the faces of S(n) – that is, for each face F of S(n), (: (F: f(t) :F) ←t :) is in fact a smooth transformation of F, i.e. is a smooth mapping ({continuum isomorphisms}: (F|f(t)|F) ←t |{scalars between zero and one}).

So what I'm really describing here is a trajectory, in {simplicial isomorphism (S(n)||S(n))}, where a simplicial isomorphism of S(n) is a mapping (S(n)|f|S(n)) for which, for each face F of S(n), (:f:F) and (:reverse(f):F) are continuum isomorphisms (F||F). [Note: (:f:F) = (:f(p)←p:F), (:reverse(f):F) =(:p←f(p):F).] The collection of simplicial isomorphisms acquires its notions of continuity from those of S(n); it is closed under composition; composing f before or after the identity on S(n), which is a simplicial isomorphism of S(n), yields f, making S(n) a natural identity for the binary operator composition. Our trajectory in R(n) = {simplicial isomorphisms of S(n)} starts at the identity and can explore a path-connected neighbourhood of the identity.

What can a simplicial isomorphism do ? By definition, it preserves each face, so let's start with the vertices: they can't move. Between any two vertices, there's an edge: it has to be transformed onto itself without moving the ends. That's easy enough; the scalar fields associated with the two vertices add up to one at each point on the edge, each being one at one end, zero at the other end and continuously varying in between; this gives us two mutually complementary continuum isomorphisms, natural to the simplex, between the edge and the scalars between zero and one; either equips us with an isomorphism between the collection of simplicial isomorphisms on the edge and the collection of simplicial isomorphisms on the scalars from zero to one (which is a 1-simplex). The latter are easy to describe: each is a continuous function, f, on the unit interval (i.e. the scalars from zero to one) of which both zero and one are fixed points (i.e. f(one) = one, f(zero) = zero), which is differentiable throughout the interior of the interval, with positive gradient throughout (and it may need to vary continuously, but I'm not sure …) If the line were a rubber string stretched between fixed end-points, this would amount to pulling along the string to stretch it (some more) in some places and allow it to relax a bit in others, thereby moving each point along the line to a continuously varying degree.

So an edge can stretch and shrink along its length, keeping its ends fixed. A face has three edges, each doing its own stretching and shrinking; picture a triangular rubber sheet, clamped all along each edge to a triangular frame which holds each edge stretched-and-shrunk along its length; the sheet in between is the rest of the face; we can stretch and squeeze it, we can even twist it (still keeping within the plane of the triangle).

Valid CSSValid XHTML 1.1 Written by Eddy.