Consider a piece of open flat country-side, such as is common in East Anglia, where I lived for many years. Two common geographical features on the plain are the embankment (often with things moving along on top) and the ditch (usually with water at its bottom). When they meet, if the embankment passes over a bridge, we see a topologically non-trivial 2-dimensional surface (toroidal in character). Since folk generally find it a struggle describing non-trivial surfaces, I'll take this form as a start-point for illustrating the kinds of things that can happen.

Some oik with a spray-can might walk about the vicinity of the bridge
marking lines – on the walls and ceiling of the under-pass as well as on
the ground. One such line runs from the right bank of the brook in the bottom
of the ditch, across the ground, up the side of the embankment, over the
bridge, down the other side of the embankment and across some more ground to
the left hand of the brook. The oik then jumped the brook and sprayed a new
line: along the ground back along the brook to under the bridge, up the wall
arching over on the under-side of the bridge (the oik hops the brook again) to
come down the wall on the other side of the brook and continue along the side
of the brook to rejoin it opposite the point where the oik started. These two
lines and the brook all lie in the surface

, they share common
end-points, meet no-where in between, yet change sides of one another in
between.

Come back in summer, when the ditch has dried up, with a strip of (highly
flexible and easily stretched) sticky tape which has been used to catch a
large number of ants: it is stuck to their backs, but their legs are free to
wiggle and to grip. Lie this along one of our lines or along the bottom of
the ditch, so that the ants can walk about on the surface (clinging tight to
the ceiling where necessary). Nail down the ends of the tape: the ants can
carry the middle

portion about, in the surface, to re-position the
tape. All our lines have the same end-points, but the ants can't move their
tape onto a different one of the three lines from the one on which they
started – at least, not without some ants losing their grip for a
while.

For the sake of definiteness, it helps to have a description of a surface:
in the present case a natural form of this would be to describe the terrain,
in three dimensions – this describes the two-dimensional surface in
terms of an embedding in three dimensions. Since the surface has several
points vertically one above another, at least in its middle

region, we
can't simply describe it in terms of height as a function of
position

. If we could, the surface would be topologically trivial.

None the less, our three-dimensional space does admit of description in
terms of height and position: imagine the horizontal surface whose height is
half-way between the top of the embankment and the bottom of the ditch; any
vertical line meets this surface in exactly one point; through any point in
our three dimensional space there is a unique vertical line, joining the point
to the one directly above or below it in the horizontal surface. There is
also a unique vertical line which meets both the ridge of the embankment and
the bottom of the ditch: call this the axis

; imagine we have put a
compass where this meets our horizontal surface. Using the compass
as origin

we can now describe any point in three dimensions as an
altitude (its distance from the horizontal surface, measured positive if the
point is above the surface, negative if below), distance from the axis and
direction from the axis, as given by the compass. (The compass point and
radius constitute polar coordinates

in the horizontal plane; with
altitude, these form cylindrical

polar coordinates in three
dimensions.)

We would ideally describe our surface in terms of some parameterisation of the surface by giving the three-dimensional position of the point indicated by the values of the parameters; but I'm trying to get a grip on what parameterisation to use.

A second way of describing our surface in three dimensions is to concoct a
function of position which is zero on the surface, negative in
the ground

(and perhaps concrete, if the bridge involves any) and
positive in air. Chose one compass direction (of the two opposites available)
going along the top of the ridge and use it as reference direction: this can
be used to define a scalar parameter φ which ranges between −π
and +π and represents compass direction – the chosen ridge direction
is φ = 0, its opposite is the bounding values φ = ±π –
by rotating the chosen direction E-N-W-S-wards through angle φ (which
amounts to a S-W-N-E-wards rotation if φ is negative) to obtain the
compass direction identified by φ. Define a function of 3-position by the
following formula in φ, altitude h and cylindrical radius R (given here
with a few of its synonyms):

- h.(1−h.h−R.R) +R.R.cos(2.φ)
- h.(1−h.h) +R.R.(cos(2.φ) −h)
- h.(1−h).(1+h) +R.R.(cos(2.φ) −h)

The surface on which this is zero has the desired form. Notice that the function is unchanged under reflections in the vertical planes through the ridge atop the embankment and the bottom-line of the ditch, consequently also under a half-turn about our axis: and that it is exactly negated (which doesn't change where the zero surface is) if we combine (: −h ←h :) with a reflection in either of the planes given by cos(2.φ) = 0.

Consider the function's value for h>1 or h<−1: in such a case, (1−h.h) is negative; cos(2.φ) −h is dominated by h and has opposite sign to it; so the two terms in the second form of the formula have equal sign; their sum is non-zero (even at R.R = 0) so our function is non-zero for h >1 or <−1: our surface does not reach up or down by more than unit distance from the horizontal plane.

For h=0 we have R.R.cos(2.φ), which is zero (at the origin and)
wherever cos(2.φ) = 0; that is, on the two diagonal

lines in the
horizontal plane, which bisect the quadrants into which our ditch and
embankment quarter the plain. I shall return to the axial planes through
these diagonal lines, but for now note that these are the planes in which to
reflect, while negating h, if one wishes to negate our scalar function:
indeed, the combination amounts to a half-turn about one of the two diagonal
lines, on which cos(2.φ) = 0.

Far from the bridge, for large R, we have h.(1−h.h) +R.R(cos(2.φ) −h) dominated by R.R, at least on the surface (since there −1<h<1 so h.(1−h.h) is between ±2/sqrt(3) which is small compared to R.R), so we may expect h to be cos(2.φ) to within of order 1/R/R (from above, the result is exact when cos(2.φ) is 0 – even for small R). Crucially, at large radius, we only get one solution for h, the surface is reducible to a flat plane, albeit with some minor geometric distortion: reality looks normal far from the bridge.

Now let us examine the surface's cross-sections in planes through the axis, which we can specify in terms of the value of φ for a horizontal direction in the plane, with the direction φ±π also in the plane – all that is going to matter is the value of cos(2.φ), which is the same in these directions. The axial half-turn symmetry means that the surface's intersection with an axial plane is necessarily symmetric under reflection in the axis. I'll start with the axial planes along the embankment and ditch, then go on to the diagonals (the only directions on which h is 0) and finally fill in the gaps between these four sample planes.

At φ=0, π or −π, along the embankment, we have cos(2.φ)=1 so the formula reduces to (1−h).(h.(1+h) +R.R) = (1−h).(square(h+1/2) +R.R −1/4) giving zero on the line h=1 (the top of the embankment) and on the circle whose centre is at altitude −1/2 on the axis, with radius 1/2 – so this passes through the origin at its top and touches h=−1 at its bottom: the downwards unit vector is a diameter of the circle. In the vertical plane through the ridge of the embankment, the circle is the walls of the tunnel under the bridge: the region above h=1 is in air, as is the interior of the circle; the region below h=1 outside the circle is the soil of the embankment and the concrete of the bridge. The function is positive in the ground and negative in the air – the reverse of what I originally sought, but I don't mind a bit.

For φ = ±π/2, along the ditch, we have cos(2.φ)=−1 so the formula is (1+h).(h.(1−h) −R.R) = (1+h).(1/4 −square(h−1/2) −R.R) giving us zero on the line h=−1 (the bottom of the ditch) and on the circle which has the upwards unit vector as a diameter; the circle is the cross-section of our bridge, the line is the bottom of our ditch. In the circle's interior and for h < −1, our function is positive (ground/concrete); above h=−1, the exterior of the circle yields negative value (air).

We thus see two orthogonal vertical planes, meeting in the axis, in which we have, starting at the top and working down:

- a horizontal line along the top of the embankment
- a circle, whose top cuts this line horizontally at right angles, whose centre is on the axis,
- a circle with the same diameter, centred one diameter below the first, on the axis, but aligned with the first horizontal line; the top of this second circle crosses the bottom of the first at right angles in the horizontal plane, meeting at the origin;
- a horizontal line along the bottom of the ditch, in the plane of the first circle, meeting the second circle at right-angles at its bottom.

Next consider the planes bisecting the angle between the two planes
studied thus far, φ = odd.π/4, whereon cos(2.φ) = 0: we are left
with h.(1−h.h−R.R), giving us the line h=0 and the unit circle
h.h+R.R=1 in each plane. The horizontal line h=0 bisects this circle. Within
the circle, our function is positive (ground) above the line, negative (air)
below; outside the circle, negative above, positive below. Each plane's line
is horizontal at zero altitude, through the origin (where the two circles in
the earlier planes met): these two lines are the diagonals of the quadrants
into which the ditch and embankment break up the plain. The two circles
are polar

great circles of the unit sphere about our origin – in
each case, the upper half circle is a diagonal cross-section of the bridge,
the lower half circle is the corresponding cross-section of the tunnel. Each
semi-circle meets

the exterior region of the same sign as it at the
corners where the horizontal diameter emerges from the interior: these two
diagonal planes are the transition

where the surface's appearance
shifts between distorted forms of the cross-sections parallel to the
embankment and to the ditch.

For φ between −π/4 and π/4, cos(2φ) > 0, consider
the axial plane with direction φ: intersect this with the surface. At
zero, we have a line at h=1 and a circle below the origin; to either side, the
line's outer reaches drop, approaching h=cos(2.φ) asymptotically at large
radius, with a hump

centered on the axis, which the line still meets at
h=1. Below the line the circle has been deformed: at its two crossings of the
axis, it is still tangent to h=0 at the origin and to
h=−1 opposite

it; but as φ increases, its middle

grows
wider, particularly near h=0; as φ approaches ±π/4, the
deformation turns the circle into the diameter and lower half of the unit
circle centred on the origin.

Meanwhile, as this limit approaches, the line over the top has deformed downwards: its outer reaches' horizontal asymptote approaches h=0 from above and its hump in the middle lowers onto the upper half of the unit circle; at the diagonal planes the two fragments (line and loop) meet at the ends of the unit circle's diameter, fuse, and come apart again in a new quadrant as a loop above and a line below; the loop is initially the upper half-circle and diameter, the line being now the lower half-circle and the rest of h=0. Thus, between π/4 and 3π/4, we have a loop above a line; the line is asymptotic to h=cos(2.φ) < 0 at large radius and dips downwards, at small radius, to form a trough and pass horizontally through the axis at h=−1; the loop lies within the upper half of the unit circle but outside the little circle which has the axis between h=0 and h=−1 as diameter. As φ passes through π/2, the familiar straight line above a circle appears, after which the entire sequence runs in reverse.

Using x = R.cosφ, y = R.sinφ, we can re-write our function as
h.(1−h.h−x.x−y.y) +x.x−y.y and examine the non-axial
planes and their intersection with the surface. (The x and y used here are
the Cartesian

co-ordinates associated with R and φ as polar

co-ordinates.)

Remember that the surface meets the two diagonal

planes in h=0 and
the unit circle in each of these planes: the tangent to the latter is vertical
where it meets h=0. Considering both diagonal planes, we have four vertical
tangents to the surface: these happen at the four points with x, y in
{±1/sqrt(2)}. So let us consider the plane x = 1/sqrt(2), wherein our
function is h.(1/2 −h.h −y.y) +1/2 −y.y and h, y are
orthonormal co-ordinates. Use s = 1/2 −y.y to simplify this as
h.(s−h.h) +s or s.(1+h)−h.h.h. Our vertical tangents happen at
s=0, with h=0; nearer the axis (well, strictly, the centre-line of our plane
– the projection of the axis onto the plane), s is positive; on the
surface, h.h.h is positive but less than one; at y=0, s=1/2 and h=1. Further
from the axis, s is negative (and grows rapidly), 0>h>−1 so that
h.h.h has the right sign; for large y, s(1+h) dominates h.h.h, so h approaches
1, with error 1+h = h.h.h/s behaving as something slightly less than
−1/s (which is positive).

We clearly get the same bell-shaped

curve at x = −1/sqrt(2);
and the planes y = ±1/sqrt(2) must yield similar curves turned
upside-down. Clearly, further from the origin, planes perpendicular to the
embankment or the ditch will show smoothed

versions of the same
curves. Between 1/sqrt(2) and 1/2, such planes must show versions of the
curve in which the top of the bell widens, the waist narrows and the tangent
at z=0 points backwards

– that is, the curve (starting at
h=−1 far away, with h rising towards zero as we approach) passes below
its h=0 point, keeps coming towards the axis and upwards, stops moving towards
the axis and loops back to pass upwards through h=0 while moving *away
from* the axis, keeps moving upwards but slows its radial movement,
looping back once more to approach the axis (still ascending as it
approaches), eventually crossing it horizontally at h=1 before repeating the
entire process in reverse again. The planar sections given by fixed x between
1/2 and 1/sqrt(2) are Ω-shaped.

At x=1/2, we're going to touch a little circle in the y=0 plane, whose
diameter is the portion of axis from h=0 to h=−1: our plane meets this
at y=0, x=1/2, h=−1/2, and the surface has a vertical tangent at this
point: our Ω

has pinched off

at its waist, like a
meandering river pinching off an ox-bow lake. For x smaller than 1/2 we shall
clearly see a line below, with h=−1 as distant asymptote but belling
upwards to a hump in the middle, and a loop above (the cross-section of the
bridge's span) which contracts down onto the little circle we're expecting at
x=1/2.

So let me now examine x=1/2 in detail. Our scalar function is now h.(3/4 −h.h −y.y) +1/4 −y.y; which we can re-write as (1/4 −y.y).(1+h) +h.(1/2 −h.h). This gives h(h.h −1/2) positive but at most 1/2 for −1/2 < y < 1/2, hence either 1>h>1/sqrt(2) or 0>h>−1/sqrt(2); the latter case allows h.(h.h−1/2) to be at most 1/3/sqrt(6) but yields two solutions where this holds; h.(3/4 −h.h −y.y) +1/4 −y.y known sample points: (y, h) in (0,1), (0,−1/2) known asymptote h=−1 for large y: h roughly 1 −2/(4.y.y+9)

Next, consider the intersection of our surface with planes through the horizonal axes parallel to ditch and ridge but passing through the origin. The horizontal plane through either of these lines is h=0; the vertical planes are the first axial planes I considered, with their straight line and circle cross-sections.

An intermediate inclined plane has one half above h=0, the other below;
the former cuts the embankment, the latter the ditch. To describe such a
plane, the Cartesian co-ordinates are appropriate: a typical plane is given by
h.cos(r) = z.sin(r) for some angle r, with z being either x (for planes whose
horizontal direction is y-wards, parallel to the ditch) or y (for planes
rising or falling along the direction of the ditch but horizontal parallel to
the ridge). The angle r is the angle between the plane and the other

horizontal direction.

On a plane h.cos(r) = x.sin(r), our scalar function becomes x.tan(r).(1−y.y−x.x(1+tan(r).tan(r))) +x.x−y.y, in which 1+tan(r).tan(r) is sec(r).sec(r) and is constant on the plane. Rearranging, we obtain x.tan(r) + x.x.(1 −x.tan(r).sec(r).sec(r)) −y.y(1+x.tan(r)) which will be zero exactly when y.y = (x.tan(r) +x.x.(1−x.tan(r).sec(r).sec(r))) / (1+x.tan(r)) which may be rearranged as (tan(r)+x.sec(r).sec(r)).(1−x.tan(r)).x/(1+x.tan(r)). We know −1<h<1 on the surface; and h is x.tan(r), so 1−x.tan(r) is positive and x is at most 1/tan(r); likewise 1+x.tan(r) is positive and x is at least −1/tan(r). Since y.y must be positive, we obtain x.(tan(r)+x.sec(r).sec(r)) >0 or, on multiplying by cos(r).cos(r), x.(sin(2.r)/2 +x) >0 so that either 0<x with 2.x>−sin(2.r) or 0>x with 2.x<−sin(2.r).

The vertical tangents to our surface happen when the gradient of our scalar field annihilates vertical vectors: which is as much as to say that the co-efficient of dh in the gradient is zero at the points on the surface where the tangents touch it. This coefficient is 1−R.R−3.h.h, which is zero on an ellipse with R-wards major axis of length 1 and h-wards minor axis of length 1/sqrt(3); this is independent of φ, so we need to spin this ellipse about our axis to obtain an ellipsoid, wide and squat like a squashed ball, on whose surface the gradient has zero vertical component. Where this meets our surface, on which the scalar field itself is zero, we shall have vertical tangents to the surface.

In each axial plane, we see the same ellipse, inside the box

1>R>−1, 1>h.sqrt(3)>−1, aka R.R<1, 3.h.h<1. The
zero-surface meets each axial plane in a line – which passes through
h=±1 on the axis so might miss the ellipse – and a loop. The
latter passes through the origin and R=0, h=±1 (the opposite ±,
this time) so must necessarily meet the ellipse, once either side of the
axis.

On the ellipsoid, R.R = 1 −3.h.h so (1 −h.h −R.R) is 2.h.h, so our scalar function is just 2.h.h.h +R.R.cos(2.φ), or h.h.(2.h −3.cos(2.φ)) + cos(2.φ).

In the embankment's axial plane, cos(2.φ) = 1, our function is 2.h.h.h −3.h.h +1, which is (2.h +1).(h.h −2.h +1) = (2.h+1).(h−1).(h−1): at h=1 we know we have only horizontal tangents (we aren't on the ellipse); at h = −1/2 we need R.R = 1 −3/4 = 1/4 so R = ±1/2, which is indeed on both the surface and the ellipsoid – it's where our little circle has its sides. Likewise in the ditch's axial plane we have 2.h.h.h +3.h.h −1 = (2.h−1).(h+1).(h+1) with vertical tangents at h=1/2, R=±1/2. On either diagonal plane, we have simply 2.h.h.h, giving h=0 hence R=1, where, indeed, we have the vertical tangents of the unit circle as it passes through its horizontal diameter.

We can parameterise the ellipse R.R +3.h.h = 1 using an angle r with R=cos(r), h = −sin(r)/sqrt(3): this r lies between −π/2 and π/2 and I am confident that, where ellipse and surface meet, r does not attain the bounds (so, for instance, tan(r) is always finite). In these terms, our function is now −2.power(3, sin(r))/3/sqrt(3) +cos(r).cos(r).cos(2.φ) and the vertical tangents to its zero-surface happen when 3.sqrt(3).cos(2.φ) = 2.sin(r).tan(r).tan(r).

Our function is h.(1−h).(1+h) +R.R.(cos(2.φ) −h) so its gradient in 3-space is (1−3.h.h−R.R).dh +2.R.(cos(2.φ)−h).dR −2.R.R.sin(2.φ).dφ. Any 3-vector which this maps to 0 is a tangent to the surface – i.e. a direction in the surface. At each point there are two dimensions worth of such tangents.

Clearly any scalar function of 3-position restricts to a scalar function
on our surface, though some scalar functions on our surface might not be
susceptible of continuation out to the rest of our 3-space. If we are to
build physical models of our universe, in particular of propagating
wave-fronts (e.g. of light), we need to be able to construct scalar fields
whose gradients vary smoothly, are never zero and look like

wave-fronts.

To describe gradients on our surface, we need to do so in terms of

So, can I construct a global parameterisation of the surface ? It should be possible using two angular co-ordinates going from −turn/2 to +turn/2 each, modulo turn, with the [turn/2, turn/2] point being mapped to infinity, the {[turn/2, a]: angle a} line being the top of the ridge, across the bridge, the {[a, turn/2]: angle a} line the bottom of the ditch, under the tunnel, {[zero, a]: angle a} looping round the centre of the bridge's span and {[a, zero]: angle a} looping round the tunnel's walls half way under, so that [zero, zero] is the top of the arch, under the span of the bridge.

Written by Eddy.