When a sphere (globe or ball) is turned about its centre a number of times, there is at least one point that starts and ends at the same place. This a consequence of the algebraic statment that a 3x3 orthogonal matrix with determinant has 1 as an eigenvector with eigenvalue 1. One of my students recently asked me for a geometric proof and I came up with the following construction. (Of which I am evidently quite proud!)

We begin by making a more precise statement of the above
proposition. Let `a`

, `b`

, `c`

,
..., `z`

be a succession of points on the sphere and
`A`

, `B`

, `C`

, ...,
`Z`

be a sucession of rotations; where `A`

is a rotation about the axis through `a`

,
`B`

a rotation about the axis through `b`

and so on. We will show how to find a point on the sphere so that
the combined effect of these rotations is a rotation about the
axis through that point.

It is enough to show how this can be done for two successive rotations since the rest of the argument/construction follows by induction/recursion.

So we restrict ourselves to the simpler case of two points
`a`

and `b`

on the sphere and two rotations
`A`

and `B`

respectively about the axes
through these points.

Let me introduce some elementary notions and results. The term
"great circle" is used for the circle obtained by cutting a
sphere with a plane passing through the centre of the sphere; any
two points on the sphere lie on a (common) great circle. The term
"antipode of a point `p`

" denotes a point
`q`

that is diametrically opposite to the given point
`p`

; the axis through a point also passes through its
antipode. Two points on the sphere that are not antipodal lie on
a unique great circle, whereas the antipode of `p`

lies on every great circle through `p`

.

Consider the point `c`

which goes to the point
`b`

under the rotation `A`

and
`cb`

be the great circle containing these two points.
Let `d`

denote a point on `bc`

that lies
halfway between `b`

and `c`

. There are two
such points which are antipodes; pick any one. Let
`ad`

be the great circle containing `a`

and
`d`

.

Similarly, let `e`

be the point which
`a`

goes to under the rotation `B`

and
`ae`

be the great circle containing these two points.
Let `f`

denote a point on `ae`

that lies
halfway between `a`

and `e`

; let
`bf`

denote the great circle which contains
`b`

and `f`

.

A point `g`

(and its antipode) where the great
circle `ad`

meets the great circle `bf`

is
fixed under the combination of the two rotations. In fact, the
result of the rotation `A`

followed by the rotation
`B`

is a rotation `G`

about the axis
through the point `g`

.

The proof that this construction works relies on the following
description of a rotation `X`

of a sphere about the
axis through a point `x`

on the sphere. Let
`y`

be any other point and `z`

be its image
under the rotation. As above let `t`

be the point
halfway between `y`

and `z`

on the great
circle containing these two points. The rotation `X`

is obtained as a succession of two reflections; first a
reflection in the plane containing the `x`

,
`t`

and the centre of the sphere and second a
reflection in the plane containing `x`

, `z`

and the centre of the sphere. One can see this by noting that the
composite of two reflections is indeed a rotation; moreover, a
rotation that fixes `x`

is uniquely determined by what
it does to `y`

.

We use the above description to write the composite of
`A`

and `B`

as succession of four
reflections through planes as follows:

the plane containing `d`

, `a`

and the
centre of the sphere; the plane containing `b`

,
`a`

and the centre of the sphere; the plane containing
`a`

, `b`

and the centre of the sphere; the
plane containing `f`

, `b`

and the centre of
the sphere

The two reflections in the middle cancel each other out. We
are then left with the composite of two reflections. Clearly, the
points that lie on the fixed plane of each of these reflections
are fixed by this composite rotation. This intersection is an
axis of the sphere through the point `g`

.

## Barak A. Pearlmutter, Tue Oct 21 14:46:18 2008