[ notation ]

The volume of a cuboid

In a linear space (such as the space of infinitesimal displacements, or tangents, at a point on a smooth manifold), E, one can construct a notion of volume for suitable cuboids. This depends on a notion of distance obtained from a metric on E. The metric is formalized as a linear map (E| g :G), with G the linear space of gradients on E, G = {linear (E| :{scalars})}, so that a displacement r in E gets mapped to g(r), a linear map from displacements in E to scalars: the inner product of r with another displacement, s, in E is then g(r,s). Linearity means that, for any scalar h, g(h.r, s) = h.g(r,s) = g(r, h.s) and, for any displacement q in E, g(r+q, s) = g(r,s) + g(q,s) and g(r, s+q) = g(r,s) + g(r,q).

The length of a displacement r in E is the square root of g(r,r). This is the minimal intelligible cuboid, of dimension 1. This has two corners and one edge: it may help to think of one corner as the base and the edge as the diagonal to the opposite corner. A cuboid of higher dimension is intelligible in terms of a base corner and several edges, emanating from that corner: the edges are described by the displacements, in E, along them; the diagonal of the cuboid is the sum of these displacements. The volume depends on the lengths of the edges and the angles between them. When E is of dimension dim, the volume of such a cuboid depends on the metric, g, at the base-point and the displacements, (dim| p :E), which I'll give stereotypically as [h,i,j,k] (with a suspicion of reading as a change in co-ordinates [t,x,y,z]).

The area of a rhombus (the 2-dimensional cuboid) with edges [j,k] is generally given as the outer product of j and k. To compute this we need J = sqrt(g(j,j)), K = sqrt(g(k,k)) and a θ for which J.K.cos(θ) is g(j,k), giving an area of J.K.sin(θ). There's an orthodox understanding of that in 3 dimensions, as (the length of) a vector, j^k, at right angles to both j and k (with some chiral reading of what negative means). I need to make sense of that …

Geometrically, what we do is fix k (say) and shift j by some change parallel to k after which g(j',k) is 0. The adjustment is easy enough to compute, with j'= j+a.k for some scalar a: we want g(j+a.k,k) = 0, so -g(j,k) = a.g(k,k) so we're replacing j with j'= j- k.g(j,k)/g(k.k). Furthermore the vector j^k is perpendicular to j and k: so g(j^k) is a gradient which annihilates j and k. Now j^k is the same after this change as before (because k^k = 0, so (j-a.k)^k = j^k), but the new form has g(j,k)=0 hence cos(θ) = 0, so sin(θ) is ±1, with the sign telling us whether [j,k] is left- or right-handed. This gives the area as K.J' with J'.J' = g(j',j') and j'= j- k.g(j,k)/g(k,k).

Now there is a natural antisymmetric product of [j,k]: it is ^([j,k]) or (j×k - k×j)/2 with × denoting the tensor product multiplier (E| e-> (E| f-> e×f = (G| a-> a(f).e :E) :{linear (E|:G)}) :). The measure can give us a linear map which will turn this into a vector, which we can combine with a gradient field (possibly obtained, using the metric, from a vector field) so as to integrate the latter (or the vector field) over our cuboid.

So a crucial issue is the relationship between antisymmetry, the metric and the measure. The last is an antisymmetric combiner on lists of displacements, giving us the volumes of cuboids.

So the measure gives us the length of a line ? In which case the metric is a subsidiary entity, because we can get it by integrating along lines and looking at derivatives.

How does the measure give us the length of a line ? We have, for a trajectory 0<a<1, a-> r(a) with tangent r'(a), a length-rate s given by 0<a<1, a-> sqrt(g(r'(a), r'(a))), using the metric g at r(a), and a total length T= integral(s). On my 1-dimensional cuboid with side k, this is just sqrt(g(k,k)). Use of sqrt is bugging me, OK ?

So let's leave aside the lengths of lines and volumes of cuboids and go back to what I know I can do with integration. If I've got (manifold: f :{scalars}) and a trajectory ({a: 0<a<1}| p :manifold) I can integrate up the gradient, df, of the former along the latter as f(p(1)) - f(p(0)) = integral({a: 0<a<1}: df(p(a))*p'(a) :{scalars}), which knows how to integrate in a chart, p, of the 1-manifold (:p|). But I want to be integrating over the 1-manifold itself, not over the chart, describing this as the integral of df along (:p|), because the change doesn't depend on re-parameterisations of p.

Now formally p'(a) is actually a linear map from tangents of your parameter space (which happens to be 1-dimensional) to those of the manifold. Tangents to a 1-dimensional space look just like scalars and any linear map from scalars to tangents is entirely characterized by its value at scalar 1, since its value at any other scalar is simply obtained by scaling this one answer.

Reparameterising our interval 0<a<1 introduces a derivative of the reparameterisation, which is a linear isomorphism (assuming the reparameterisation is sensible) between the tangent bundles of the original parameter space and the new one. This converts p'(a) into a linear map from the new parameter-rates to tangents of the manifold. The process of integration with respect to the new parameters effectively applies a matching re-parameterisation to how it integrates so as to yield the same result regardless of parameterisation.

The right thing to integrate along a 1-submanifold is thus a gradient (that's what df is). The process of integration I want to be using is what one gets by choosing a chart and integrating using that, which happens to be chart-independent when the quantity integrated is a gradient. Furthermore, the integral only depends on the gradient's action on tangents parallel to the 1-manifold.

When integrating over an area, or 2-manifold, model the process in terms of a cuboid with sides [j,k] as the stereotypical area. This wants to consume a gradient×gradient to which it pays attention only to: the antisymmetric part and, within that; its restriction to the 2-manifold's tensor bundle, specifically its action on ^([j,k]) which is antiSymmetric(2,{tangents}). So we can integrate the anti-symmetric part of a gradient×gradient field. Note that this the same kind of thing as the metric of space-time, whose anti-symmetric part is usually taken to be zero and/or ignorable.

Likewise we can integrate up ×(n| :{gradients}) for any integer n, attending only to the anti-symmetric part, over an n-manifold. This is a natural process of integration having no dependence at all on the metric or any particular chart or variety of chart.

So what did I integrate up along a trajectory to get its length ? It was the rate of traversal of length, as seen by the metric, along the trajectory. That's a g(tangent). Using a parameterisation (range| p :{manifold}) of some trajectory (:p|), the length is integral(range| a-> sqrt(g(p'(a),p'(a))) :{scalars}). Consider doubling the scaling: q(a) = p(2.a) gives q'(a) = 2.p'(2.a) but the range of integration is half as long, so the integral is unchanged. Likewise, consider shifting the scaling so as to make g(q'(a),q'(a)) always 1: use q = p&on;r for some reparameterisation r and find q'(a) = r'(a).p'(r(a)), so we need 1 = g(q',q') = square(r').g((p'&on;r),(p'&on;r)).

Now really p' is linear ({parameter-rates}| p' :{tangents}) so we really get g&on;p' linear ({parameter-rates}| :{gradients}). When I form g(p'(a),p'(a)), I need to mean ({rates}| u-> ({rates}| v-> g(p'(a,u), p'(a,u)) :{scalars}) :) and I expect it to have a square root ({rates}| s :{scalars}) for which this is s×s = (: u-> (: v-> s(u).s(v) :) :), give or take a possible sign.

and the process of integration ensures that the rate you use doesn't matter, because going faster (so having a bigger tangent) covers a shorter parameter interval to cover the same terrain.