~~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 . ~~