NOTE ON A THEOREM RELATING TO SURFACES.
[From the Philosophical Magazine, vol. xxv. (1863), pp. 61, 62.]
The following apparently self-evident geometrical theorem requires, I think, a
proof; viz. the theorem is—“ If every plane section of a surface of the order m + n
break up into two curves of the orders m and n respectively, then the surface breaks
up into two surfaces of the orders m, n respectively.”
To fix the ideas, suppose n = 2. Imagine any line meeting the surface in m + 2
points, the section includes a conic which meets the line in two of the m+ 2 points,
say the points A, A'Q). Suppose that the plane revolves round the line AA', the
section will always include a conic which passes through these same two points A, A';
and it is to be shown that the sheet, the locus of this conic, is a surface of the
second order. In fact the conic in question, say APA', by its intersection with an
arbitrary plane traces out a branch of the intersection of the given surface with
the arbitrary plane. And if ABA'B' be the conic in any particular plane through
A, A', and if the arbitrary plane meet this conic in the points B, B', then the branch
passes through these points B, B'. Imagine the plane ABA’B' revolving round BB'
until it coincides with the arbitrary plane; the section includes a conic passing through
the points B, B', and the before-mentioned branch is this conic; that is, the conic
APA' by its intersection with an arbitrary plane traces out a conic; or, what is the
same thing, the sheet, the locus of the conic APA', is met by an arbitrary plane in
a conic, that is, the sheet is a surface of the second order; and the given surface thus
includes a surface of the second order, and is therefore made up of two surfaces of
the orders m and 2 respectively. The demonstration seems to me to add at least
1 The figure referred to will be at once understood by considering A, A' as the poles of an ellipsoid, or
say of a sphere, ABA'B' the meridian of projection, APA' any other meridian, BPB' the equator or any other
great circle.
