Introduction.

In 1 we investigate a new series of line involutions in a projective space of three dimensions over the field of complex numbers.
These are defined by a simple involutorial transformation of the points in which a general line meets a nonsingular quadric surface bearing a curve of symbol Af.
Then in 2 we show that any line involution with the properties that (A) It has no complex of invariant lines, and (B) Its singular lines form a complex consisting exclusively of the lines which meet a twisted curve, is necessarily of the type discussed in 1.
No generalization of these results to spaces of more than three dimensions has so far been found possible.

1.

Let Q be a nonsingular quadric surface bearing reguli Af and Af, and let **zg be a Af curve of order K on Q.
A general line L meets Q in two points, Af and Af, through each of which passes a unique generator of the regulus, Af, whose lines are simple secants of Aj.
On these generators let Af and Af be, respectively, the harmonic conjugates of Af and Af with respect to the two points in which the corresponding generator meets Aj.
The line Af is the image of L.
Clearly, the transformation is involutorial.

We observe first that no line, l, can meet its image except at one of its intersections with Q.
For if it did, the plane of L and l' would contain two generators of Af, which is impossible.
Moreover, from the definitive transformation of intercepts on the generators of Af, it is clear that the only points of Q at which a line can meet its image are the points of Aj.
Hence the totality of singular lines is the T order complex of lines which meet Aj.

The invariant lines are the lines of the congruence of secants of **zg, since each of these meets Q in two points which are invariant.
The order of this congruence is Af, since Af secants of a curve of symbol (B) on a quadric surface pass through an arbitrary point.
The class of the congruence is Af, since an arbitrary plane meets **zg in K points.

Since the complex of singular lines is of order K and since there is no complex of invariant lines, it follows from the formula Af that the order of the involution is Af.

There are various sets of exceptional lines, or lines whose images are not unique.
The most obvious of these is the quadratic complex of tangents to Q, each line of which is transformed into the entire pencil of lines tangent to Q at the image of the point of tangency of the given line.
Thus pencils of tangents to Q are transformed into pencils of tangents.
It is interesting that a 1: 1 correspondence can be established between the lines of two such pencils, so that in a sense a unique image can actually be assigned to each tangent.
For the lines of any plane, **yp, meeting Q in a conic C, are transformed into the congruence of secants of the curve C' into which C is transformed in the point involution on Q.
In particular, tangents to C are transformed into tangents to C'.
Moreover, if Af and Af are two planes intersecting in a line l, tangent to Q at a point P, the two free intersections of the image curves Af and Af must coincide at P', the image of P, and at this point Af and Af must have a common tangent l'.
Hence, thought of as a line in a particular plane **yp, any tangent to Q has a unique image and moreover this image is the same for all planes through L.

Each generator, **yl, of Af is also exceptional, for each is transformed into the entire congruence of secants of the curve into which that generator is transformed by the point involution on Q.
This curve is of symbol Af since it meets **yl, and hence every line of Af in the Af invariant points on **yl and since it obviously meets every line of Af in a single point.
The congruence of its secants is therefore of order Af and class Af.

A final class of exceptional lines is identifiable from the following considerations: Since no two generators of Af can intersect, it follows that their image curves can have no free intersections.
In other words, these curves have only fixed intersections common to them all.
Now the only way in which all curves of the image family of Af can pass through a fixed point is to have a generator of Af which is not a secant but a tangent of **zg, for then any point on such a generator will be transformed into the point of tangency.
Since two curves of symbol Af on Q intersect in Af points, it follows that there are Af lines of Af which are tangent to Aj.
Clearly, any line, l, of any bundle having one of these points of tangency, T, as vertex will be transformed into the entire pencil having the image of the second intersection of L and Q as vertex and lying in the plane determined by the image point and the generator of Af which is tangent to **zg at T.
A line through two of these points, Af and Af, will be transformed into the entire bilinear congruence having the tangents to **zg at Af and Af as directrices.

A conic, C, being a (1, 1) curve on Q, meets the image of any line of Af, which we have already found to be a Af curve on Q, in Af points.
Hence its image, C', meets any line of Af in Af points.
Moreover, C' obviously meets any line Af in a single point.
Hence C' is a Af curve on Q.
Therefore, the congruence of its secants, that is the image of a general plane field of lines, is of order Af and class Af.
Finally, the image of a general bundle of lines is a congruence whose order is the order of the congruence of invariant lines, namely Af and whose class is the order of the image congruence of a general plane field of lines, namely Af.

2.

The preceding observations make it clear that there exist line involutions of all orders greater than 1 with no complex of invariant lines and with a complex of singular lines consisting exclusively of the lines which meet a twisted curve Aj.
We now shall show that any involution with these characteristics is necessarily of the type we have just described.

To do this we must first show that every line which meets **zg in a point P meets its image at P.
To see this, consider a general pencil of lines containing a general secant of Aj.
By (1), the image of this pencil is a ruled surface of order Af which is met by the plane of the pencil in a curve, C, of order Af.
On C there is a Af correspondence in which the Af points cut from C by a general line, l, of the pencil correspond to the point of intersection of the image of L and the plane of the pencil.
Since C is rational, this correspondence has K coincidences, each of which implies a line of the pencil which meets its image.
However, since the pencil contains a secant of **zg it actually contains only Af singular lines.
To avoid this contradiction it is necessary that C be composite, with the secant of **zg and a curve of order Af as components.
Thus it follows that the secants of **zg are all invariant.
But if this is the case, then an arbitrary pencil of lines having a point, P, of **zg as vertex is transformed into a ruled surface of order Af having Af generators concurrent at P.
Since a ruled surface of order N with N concurrent generators is necessarily a cone, it follows finally that every line through a point, P, of **zg meets its image at P, as asserted.

Now consider the transformation of the lines of a bundle with vertex, P, on **zg which is effected by the involution as a whole.
From the preceding remarks, it is clear that such a bundle is transformed into itself in an involutorial fashion.
Moreover, in this involution there is a cone of invariant lines of order Af, namely the cone of secants of **zg which pass through P.
Hence it follows that the involution within the bundle must be a perspective De Jonquieres involution of order Af and the invariant locus must have a multiple line of multiplicity either Af or Af.
The first possibility requires that there be a line through P which meets **zg in Af points;
the second requires that there be a line through P which meets **zg in Af points.
In each case, lines of the bundles are transformed by involutions within the pencils they determine with the multiple secant.
In the first case the fixed elements within each pencil are the multiple secant and the line joining the vertex, P, to the intersection of **zg and the plane of the pencil which does not lie on the multiple secant.
In the second, the fixed elements are the lines which join the vertex, P, to the two intersections of **zg and the plane of the pencil which do not lie on the multiple secant.
The multiple secants, of course, are exceptional and in each case are transformed into cones of order Af.

Observations similar to these can be made at each point of Aj.
Hence **zg must have either a regulus of Af-fold secants or a regulus of Af-fold secants.
Moreover, if Af, no two of the multiple secants can intersect.
For if such were the case, either the plane of the two lines would meet **zg in more than K points or, alternatively, the order of the image regulus of the pencil determined by the two lines would be too high.
But if no two lines of the regulus of multiple secants of **zg can intersect, then the regulus must be quadratic, or in other words, **zg must be either a Af or a Af curve on a nonsingular quadric surface.

We now observe that the case in which **zg is a Af curve on a quadric is impossible if the complex of singular lines consists exclusively of the lines which meet Aj.
For any pencil in a plane containing a Af-fold secant of **zg has an image regulus which meets the plane of the pencil in Af lines, namely the images of the lines of the pencil which pass through the intersection of **zg and the multiple secant, plus an additional component to account for the intersections of the images of the general lines of the pencil.
However, if there is no additional complex of singular lines, the order of the image regulus of a pencil is precisely Af.
This contradicts the preceding observations, and so, under the assumption of this paper, we must reject the possibility that **zg is a Af curve on a quadric surface.

Continuing with the case in which **zg is a Af curve on a quadric Q, we first observe that the second regulus of Q consists precisely of the lines which join the two free intersections of **zg and the planes through any one of the multiple secants.
For each of these lines meets Q in three points, namely two points on **zg and one point on one of the multiple secants.

Now consider an arbitrary line, l, meeting Q in two points, Af and Af.
If **ya is the multiple secant of **zg which passes through Af and **yb is the simple secant of **zg which passes through Af, and if Af are the points in which **ya meets **zg, and if Af is the image of Af on the generator **yb, it follows that the image of the line Af is Af.