§ 1.Geometrical Elements.We consider space as filled with points, lines and planes, and these we call the elements out of which our figures are to be formed, calling any combination of these elements a “figure.”By a line we mean a straight line in its entirety, extending both ways to infinity; and by a plane, a plane surface, extending in all directions to infinity.We accept the three-dimensional space of experience—the space assumed by Euclid—which has for its properties (among others):—Through any two points in space one and only one line may be drawn;Through any three points which are not in a line, one and only one plane may be placed;The intersection of two planes is a line;A line which has two points in common with a plane lies in the plane, hence the intersection of a line and a plane is a single point; andThree planes which do not meet in a line have one single point in common.These results may be stated differently in the following form:—I. A plane is determined—A point is determined—1. By three points which do not lie in a line;2. By two intersecting lines;3. By a line and a point which does not lie in it.1. By three planes which do not pass through a line;2. By two intersecting lines3. By a plane and a line which does not lie in it.A line is determined—1. By two points;2. By two planes.It will be observed that not only are planes determined by points, but also points by planes; that therefore the planes may be considered as elements, like points; and also that in any one of the above statements we may interchange the words point and plane, and we obtain again a correct statement, provided that these statements themselves are true. As they stand, we ought, in several cases, to add “if they are not parallel,” or some such words, parallel lines and planes being evidently left altogether out of consideration. To correct this we have to reconsider the theory of parallels.Fig. 1.§ 2.Parallels. Point at Infinity.—Let us take in a plane a line p (fig. 1), a point S not in this line, and a line q drawn through S. Then this line q will meet the line p in a point A. If we turn the line q about S towards q’, its point of intersection with p will move along p towards B, passing, on continued turning, to a greater and greater distance, until it is moved out of our reach. If we turn q still farther, its continuation will meet p, but now at the other side of A. The point of intersection has disappeared to the right and reappeared to the left. There is one intermediate position where q is parallel to p—that is where it does not cut p. In every other position it cuts p in some finite point. If, on the other hand, we move the point A to an infinite distance in p, then the line q which passes through A will be a line which does not cut p at any finite point. Thus we are led to say:Everyline through S which joins it to any point at an infinite distance in p is parallel to p. But by Euclid’s 12th axiom there is but one line parallel to p through S. The difficulty in which we are thus involved is due to the fact that we try to reason about infinity as if we, with our finite capabilities, could comprehend the infinite. To overcome this difficulty, we may say that all points at infinity in a lineappearto us as one, and may be replaced by a single “ideal” point.We may therefore now give the following definitions and axiom:—Definition.—Lines which meet at infinity are called parallel.Axiom.—All points at an infinite distance in a line may be considered as one single point.Definition.—This ideal point is called thepoint at infinityin the line.The axiom is equivalent to Euclid’s Axiom 12, for it follows from either that through any point only one line may be drawn parallel to a given line.This point at infinity in a line is reached whether we move a point in the one or in the opposite direction of a line to infinity. A line thus appears closed by this point, and we speak as if we could move a point along the line from one position A to another B in two ways, either through the point at infinity or through finite points only.It must never be forgotten that this point at infinity is ideal; in fact, the whole notion of “infinity” is only a mathematical conception, and owes its introduction (as a method of research) to the working generalizations which it permits.§ 3.Line and Plane at Infinity.—Having arrived at the notion of replacing all points at infinity in a line by one ideal point, there is no difficulty in replacing all points at infinity in a plane by one ideal line.To make this clear, let us suppose that a line p, which cuts two fixed lines a and b in the points A and B, moves parallel to itself to a greater and greater distance. It will at last cut both a and b at their points at infinity, so that a line which joins the two points at infinity in two intersecting lines lies altogether at infinity. Every other line in the plane will meet it therefore at infinity, and thus it contains all points at infinity in the plane.All points at infinity in a plane lie in a line, which is called theline at infinityin the plane.It follows that parallel planes must be considered as planes having a common line at infinity, for any other plane cuts them in parallel lines which have a point at infinity in common.If we next take two intersecting planes, then the point at infinity in their line of intersection lies in both planes, so that their lines at infinity meet. Hence every line at infinity meets every other line at infinity, and they are therefore all in one plane.All points at infinity in space may be considered as lying in one ideal plane, which is called theplane at infinity.§ 4.Parallelism.—We have now the following definitions:—Parallel lines are lines which meet at infinity;Parallel planes are planes which meet at infinity;A line is parallel to a plane if it meets it at infinity.Theorems like this—Lines (or planes) which are parallel to a third are parallel to each other—follow at once.This view of parallels leads therefore to no contradiction of Euclid’sElements.As immediate consequences we get the propositions:—Every line meets a plane in one point, or it lies in it;Every plane meets every other plane in a line;Any two lines in the same plane meet.§ 5.Aggregates of Geometrical Elements.—We have called points, lines and planes the elements of geometrical figures. We also say that an element of one kind contains one of the other if it lies in it or passes through it.All the elements of one kind which are contained in one or two elements of a different kind form aggregates which have to be enumerated. They are the following:—I. Of one dimension.1. Therow, or range,of pointsformed by all points in a line, which is called its base.2. Theflat pencilformed by all the lines through a point in a plane. Its base is the point in the plane.3. Theaxial pencilformed by all planes through a line which is called its base or axis.II. Of two dimensions.1. The field of points and lines—that is, a plane with all its points and all its lines.2. The pencil of lines and planes—that is, a point in space with all lines and all planes through it.III. Of three dimensions.The space of points—that is, all points in space.The space of planes—that is, all planes in space.IV. Of four dimensions.The space of lines, or all lines in space.§ 6.Meaning of “Dimensions.”—The word dimension in the above needs explanation. If in a plane we take a row p and a pencil with centre Q, then through every point in p one line in the pencil will pass, and every ray in Q will cut p in one point, so that we are entitled to say a row contains as many points as a flat pencil lines, and, we may add, as an axial pencil planes, because an axial pencil is cut by a plane in a flat pencil.The number of elements in the row, in the flat pencil, and in the axial pencil is, of course, infinite and indefinite too, but the same in all. This number may be denoted by ∞. Then a plane contains ∞² points and as many lines. To see this, take a flat pencil in a plane. It contains ∞ lines, and each line contains ∞ points, whilst each point in the plane lies on one of these lines. Similarly, in a plane each line cuts a fixed line in a point. But this line is cut at each point by ∞ lines and contains ∞ points; hence there are ∞² lines in a plane.A pencil in space contains as many lines as a plane contains points and as many planes as a plane contains lines, for any plane cuts the pencil in a field of points and lines. Hence a pencil contains ∞² lines and ∞² planes.The field and the pencil are of two dimensions.To count the number of points in space we observe that each point lies on some line in a pencil. But the pencil contains ∞² lines, and each line ∞ points; hence space contains ∞³ points. Each plane cuts any fixed plane in a line. But a plane contains ∞² lines, and through each pass ∞ planes; therefore space contains ∞³ planes.Hence space contains as many planes as points, but it contains an infinite number of times more lines than points or planes. To count them, notice that every line cuts a fixed plane in one point. But ∞² lines pass through each point, and there are ∞² points in the plane. Hence there are ∞4lines in space.The space of points and planes is of three dimensions, but the space of lines is of four dimensions.A field of points or lines contains an infinite number of rows and flat pencils; a pencil contains an infinite number of flat pencils and of axial pencils; space contains a triple infinite number of pencils and of fields, ∞4rows and axial pencils and ∞5flat pencils—or, in other words, each point is a centre of ∞² flat pencils.§ 7. The above enumeration allows a classification of figures. Figures in a row consist of groups of points only, and figures in the flat or axial pencil consist of groups of lines or planes. In the plane we may draw polygons; and in the pencil or in the point, solid angles, and so on.We may also distinguish the different measurements We have—In the row, length of segment;In the flat pencil, angles;In the axial pencil, dihedral angles between two planes;In the plane, areas;In the pencil, solid angles;In the space of points or planes, volumes.Segments of a Line§ 8. Any two points A and B in space determine on the line through them a finite part, which may be considered as being described by a point moving from A to B. This we shall denote by AB, and distinguish it from BA, which is supposed as being described by a point moving from B to A, and hence in a direction or in a “sense” opposite to AB. Such a finite line, which has a definite sense, we shall call a “segment,” so that AB and BA denote different segments, which are said to be equal in length but of opposite sense. The one sense is often called positive and the other negative.In introducing the word “sense” for direction in a line, we have the word direction reserved for direction of the line itself, so that different lines have different directions, unless they be parallel, whilst in each line we have a positive and negative sense.We may also say, with Clifford, that AB denotes the “step” of going from A to B.Fig. 2.§ 9. If we have three points A, B, C in a line (fig. 2), the step AB will bring us from A to B, and the step BC from B to C. Hence both steps are equivalent to the one step AC. This is expressed by saying that AC is the “sum” of AB and BC; in symbols—AB + BC = AC,where account is to be taken of the sense.This equation is true whatever be the position of the three points on the line. As a special case we haveAB + BA = 0,(1)and similarlyAB + BC + CA = 0,(2)which again is true for any three points in a line.We further writeAB = −BA.where − denotes negative sense.We can then, just as in algebra, change subtraction of segments into addition by changing the sense, so that AB − CB is the same as AB + (−CB) or AB + BC. A figure will at once show the truth of this. The sense is, in fact, in every respect equivalent to the “sign” of a number in algebra.§ 10. Of the many formulae which exist between points in a line we shall have to use only one more, which connects the segments between any four points A, B, C, D in a line. We haveBC = BD + DC, CA = CD + DA, AB = AD + DB;or multiplying these by AD, BD, CD respectively, we getBC · AD = BD · AD + DC · AD = BD · AD − CD · ADCA · BD = CD · BD + DA · BD = CD · BD − AD · BDAB · CD = AD · CD + DB · CD = AD · CD − BD · CD.It will be seen that the sum of the right-hand sides vanishes, hence thatBC · AD + CA · BD + AB · CD = 0(3)for any four points on a line.Fig. 3.§ 11. If C is any point in the line AB, then we say that C divides the segment AB in the ratio AC/CB, account being taken of the sense of the two segments AC and CB. If C lies between A and B the ratio is positive, as AC and CB have the same sense. But if C lies without the segment AB,i.e.if C divides AB externally, then the ratio is negative. To see how the value of this ratio changes with C, we will move C along the whole line (fig. 3), whilst A and B remain fixed. If C lies at the point A, then AC = 0, hence the ratio AC : CB vanishes. As C moves towards B, AC increases and CB decreases, so that our ratio increases. At the middle point M of AB it assumes the value +1, and then increases till it reaches an infinitely large value, when C arrives at B. On passing beyond B the ratio becomes negative. If C is at P we have AC = AP = AB + BP, henceAC=AB+BP= −AB− 1.CBPBPBBPIn the last expression the ratio AB : BP is positive, has its greatest value ∞ when C coincides with B, and vanishes when BC becomes infinite. Hence, as C moves from B to the right to the point at infinity, the ratio AC : CB varies from −∞ to −1.If, on the other hand, C is to the left of A, say at Q, we have AC = AQ = AB + BQ = AB − QB, hence AC/CB = AB/QB − 1.Here AB < QB, hence the ratio AB : QB is positive and always less than one, so that the whole is negative and < 1. If C is at the point at infinity it is −1, and then increases as C moves to the right, till for C at A we get the ratio = 0. Hence—“As C moves along the line from an infinite distance to the left to an infinite distance at the right, the ratio always increases; it starts with the value −1, reaches 0 at A, +1 at M, ∞ at B, now changes sign to −∞, and increases till at an infinite distance it reaches again the value −1.It assumes therefore all possible values from -∞ to +∞, and each value only once, so that not only does every position ofCdetermine a definite value of the ratioAC : CB,but also, conversely, to every positive or negative value of this ratio belongs one single point in the lineAB.[Relations between segments of lines are interesting as showing an application of algebra to geometry. The genesis of such relations from algebraic identities is very simple. For example, if a, b, c, x be any four quantities, thena+b+c=x;(a − b)(a − c)(x − a)(b − c)(b − a)(x − b)(c − a)(c − b)(x − c)(x − a)(x − b)(x − c)this may be proved, cumbrously, by multiplying up, or, simply, by decomposing the right-hand member of the identity into partial fractions. Now take a line ABCDX, and let AB = a, AC = b, AD = c, AX = x. Then obviously (a − b) = AB − AC = −BC, paying regard to signs; (a − c) = AB − AD = DB, and so on. Substituting these values in the identity we obtain the following relation connecting the segments formed by five points on a line:—AB+AC+AD=AX.BC · BD · BXCD · CB · CXDB · DC · DXBX · CX · DXConversely, if a metrical relation be given, its validity may be tested by reducing to an algebraic equation, which is an identity if the relation be true. For example, if ABCDX be five collinear points, proveAD · AX+BD · BX+CD · CX= 1.AB · ACBC · BACA · CBClearing of fractions by multiplying throughout by AB · BC · CA, we have to prove−AD · AX · BC − BD · BX · CA − CD · CX · AB = AB · BC · CA.Take A as origin and let AB = a, AC = b, AD = c, AX = x. Substituting for the segments in terms of a, b, c, x, we obtain on simplificationa²b − ab² = −ab² + a²b, an obvious identity.An alternative method of testing a relation is illustrated in thefollowing example:—If A, B, C, D, E, F be six collinear points, thenAE · AF+BE · BF+CE · CF+DE · DF= 0.AB · AC · ADBC · BD · BACD · CA · CBDA · DB · DCClearing of fractions by multiplying throughout by AB · BC · CD · DA, and reducing to a common origin O (calling OA = a, OB = b, &c.), an equation containing the second and lower powers of OA ( = a), &c., is obtained. Calling OA = x, it is found that x = b, x = c, x = d are solutions. Hence the quadratic has three roots; consequently it is an identity.The relations connecting five points which we have instanced above may be readily deduced from the six-point relation; the first by taking D at infinity, and the second by taking F at infinity, and then making the obvious permutations of the points.]Projection and Cross-ratios§ 12. If we join a point A to a point S, then the point where the line SA cuts a fixed plane π is called the projection of A on the plane π from S as centre of projection. If we have two planes π and π′ and a point S, we may project every point A in π to the other plane. If A′ is the projection of A, then A is also the projection of A′, so that the relations are reciprocal. To every figure in π we get as its projection a corresponding figure in π′.We shall determine such properties of figures as remain true for the projection, and which are called projective properties. For this purpose it will be sufficient to consider at first only constructions in one plane.Fig. 4.Fig. 5.Let us suppose we have given in a plane two lines p and p′ and a centre S (fig. 4); we may then project the points in p from S to p′. Let A′, B′ ... be the projections of A, B ..., the point at infinity in p which we shall denote by I will be projected into a finite pointI′ in p′, viz. into the point where the parallel to p through S cuts p′. Similarly one point J in p will be projected into the point J′ at infinity in p′. This point J is of course the point where the parallel to p′ through S cuts p. We thus see that every point in p is projected into a single point in p′.Fig. 5 shows that a segment AB will be projected into a segment A′B′ which is not equal to it, at least not as a rule; and also that the ratio AC : CB is not equal to the ratio A′C′ : C′B′ formed by the projections. These ratios will become equal only if p and p′ are parallel, for in this case the triangle SAB is similar to the triangle SA′B′. Between three points in a line and their projections there exists therefore in general no relation. But between four points a relation does exist.§ 13. Let A, B, C, D be four points in p, A′, B′, C, D′ their projections in p′, then the ratio of the two ratios AC : CB and AD : DB into which C and D divide the segment AB is equal to the corresponding expression between A′, B′, C′, D′. In symbols we haveAC:AD=A′C′:A′D′.CBDBC′B′D′B′This is easily proved by aid of similar triangles.Fig. 6.Through the points A and B on p draw parallels to p′, which cut the projecting rays in C2, D2, B2and A1, C1, D1, as indicated in fig. 6. The two triangles ACC2and BCC1will be similar, as will also be the triangles ADD2and BDD1.The proof is left to the reader.This result is of fundamental importance.The expression AC/CB : AD/DB has been called by Chasles the “anharmonic ratio of the four points A, B, C, D.” Professor Clifford proposed the shorter name of “cross-ratio.” We shall adopt the latter. We have then theFundamental Theorem.—The cross-ratio of four points in a line is equal to the cross-ratio of their projections on any other line which lies in the same plane with it.§ 14. Before we draw conclusions from this result, we must investigate the meaning of a cross-ratio somewhat more fully.If four points A, B, C, D are given, and we wish to form their cross-ratio, we have first to divide them into two groups of two, the points in each group being taken in a definite order. Thus, let A, B be the first, C, D the second pair, A and C being the first points in each pair. The cross-ratio is then the ratio AC : CB divided by AD : DB. This will be denoted by (AB, CD), so that(AB, CD) =AC:AD.CBDBThis is easily remembered. In order to write it out, make first the two lines for the fractions, and put above and below these the letters A and B in their places, thus, A/*B : A/*B; and then fill up, crosswise, the first by C and the other by D.§ 15. If we take the points in a different order, the value of the cross-ratio will change. We can do this in twenty-four different ways by forming all permutations of the letters. But of these twenty-four cross-ratios groups of four are equal, so that there are really only six different ones, and these six are reciprocals in pairs.We have the following rules:—I. If in a cross-ratio the two groups be interchanged, its value remains unaltered,i.e.(AB, CD) = (CD, AB) = (BA, DC) = (DC, BA).II. If in a cross-ratio the two points belonging to one of the two groups be interchanged, the cross-ratio changes into its reciprocal,i.e.(AB, CD) = 1/(AB, DC) = 1/(BA, CD) = 1/(CD, BA) = 1/(DC, AB).From I. and II. we see that eight cross-ratios are associated with (AB, CD).III. If in a cross-ratio the two middle letters be interchanged, the cross-ratio α changes into its complement 1 − α,i.e.(AB, CD) = 1 − (AC, BD).[§ 16. If λ = (AB, CD), μ = (AC, DB), ν = (AD, BC), then λ, μ, ν and their reciprocals 1/λ, 1/μ, 1/ν are the values of the total number of twenty-four cross-ratios. Moreover, λ, μ, ν are connected by the relationsλ + 1/μ = μ + 1/ν = ν + 1/λ = −λμν = 1;this proposition may be proved by substituting for λ, μ, ν and reducing to a common origin. There are therefore four equations between three unknowns; hence if one cross-ratio be given, the remaining twenty-three are determinate. Moreover, two of the quantities λ, μ, ν are positive, and the remaining one negative.The following scheme shows the twenty-four cross-ratios expressed in terms of λ, μ, ν.](AB, CD)(BA, DC)(CD, AB)(DC, BA)λ1 − μ1/(1 − ν)(AD, BC)(BC, AD)(CB, DA)(DA, CB)(λ − 1)/λμ/(μ − 1)ν(AC, DB)(BD, CA)(CA, BD)(DB, AC)1/(1 − λ)1/μ(ν − 1)/ν(AC, BD)(BD, AC)(CA, DB)(DB, CA)1 − λμν/(ν − 1)(AB, DC)(BA, CD)(CD, BA)(DC, AB)1/λ1/(1 − μ)1 − ν(AD, CB)(BC, DA)(CB, AD)(DA, BC)λ/(λ − 1)(μ − 1)/μ1/ν§ 17. If one of the points of which a cross-ratio is formed is the point at infinity in the line, the cross-ratio changes into a simple ratio. It is convenient to let the point at infinity occupy the last place in the symbolic expression for the cross-ratio. Thus if I is a point at infinity, we have (AB, CI) = −AC/CB, because AI : IB = −1.Every common ratio of three points in a line may thus be expressed as a cross-ratio, by adding the point at infinity to the group of points.Harmonic Ranges§ 18. If the points have special positions, the cross-ratios may have such a value that, of the six different ones, two and two become equal. If the first two shall be equal, we get λ = 1/λ, or λ² = 1, λ = ±1.If we take λ = +1, we have (AB, CD) = 1, or AC/CB = AD/DB; that is, the points C and D coincide, provided that A and B are different.If we take λ = −1, so that (AB, CD) = −1, we have AC/CB = −AD/DB.Hence C and D divide AB internally and externally in the same ratio.The four points are in this case said to beharmonic points, and CandDare said to be harmonic conjugates with regard toAandB.But we have also (CD, AB) = −1, so that A and B are harmonic conjugates with regard to C and D.The principal property of harmonic points is that their cross-ratio remains unaltered if we interchange the two points belonging to one pair, viz.(AB, CD) = (AB, DC) = (BA, CD).For four harmonic points the six cross-ratios become equal two and two:λ = −1, 1 − λ = 2,λ= ½,1= −1,1= ½,λ − 1= 2.λ − 1λ1 − λλHence if we get four points whose cross-ratio is 2 or ½, then they are harmonic, but not arranged so that conjugates are paired. If this is the case the cross-ratio = −1.§ 19. If we equate any two of the above six values of the cross-ratios, we get either λ = 1, 0, ∞, or λ = −1, 2, ½, or else λ becomes a root of the equation λ² − λ + 1 = 0, that is, an imaginary cube root of −1. In this case the six values become three and three equal, so that only two different values remain. This case, though important in the theory of cubic curves, is for our purposes of no interest, whilst harmonic points are all-important.§ 20. From the definition of harmonic points, and by aid of § 11, the following properties are easily deduced.If C and D are harmonic conjugates with regard to A and B, then one of them lies in, the other without AB; it is impossible to move from A to B without passing either through C or through D; the one blocks the finite way, the other the way through infinity. This is expressed by saying A and B are “separated” by C and D.For every position of C there will be one and only one point D which is its harmonic conjugate with regard to any point pair A, B.If A and B are different points, and if C coincides with A or B, D does. But if A and B coincide, one of the points C or D, lying between them, coincides with them, and the other may be anywhere in the line. It follows that, “if of four harmonic conjugates two coincide, then a third coincides with them, and the fourth may be any point in the line.”If C is the middle point between A and B, then D is the point at infinity; for AC : CB = +1, hence AD : DB must be equal to −1.The harmonic conjugate of the point at infinity in a line with regard to two pointsA, Bis the middle point ofAB.This important property gives a first example how metric properties are connected with projective ones.[§ 21.Harmonic properties of the complete quadrilateral and quadrangle.Fig. 7.Fig. 8.A figure formed by four lines in a plane is called acomplete quadrilateral, or, shorter, afour-side. The four sides meet in six points, named the “vertices,” which may be joined by three lines (other than the sides), named the “diagonals” or “harmonic lines.” The diagonals enclose the “harmonic triangle of the quadrilateral.” In fig. 7, A′B′C′, B′AC, C′AB, CBA′ are the sides, A, A′, B, B′, C, C′ the vertices, AA′, BB′, CC′ the harmonic lines, and αβγ the harmonic triangle of the quadrilateral. A figure formed by four coplanar points is named acomplete quadrangle, or, shorter, afour-point. The four points may be joined by six lines, named the “sides,” which intersect in three other points, termed the “diagonal or harmonic points.” The harmonic points are the vertices of the “harmonic triangle of the complete quadrangle.” In fig. 8, AA′, BB′ are the points, AA′, BB′, A′B′, B′A, AB, BA′ are the sides, L, M, N are the diagonal points, and LMN is the harmonic triangle of the quadrangle.The harmonic property of the complete quadrilateral is: Any diagonal or harmonic line is harmonically divided by the other two; and of a complete quadrangle: The angle at any harmonic point is divided harmonically by the joins to the other harmonic points. To prove the first theorem, we have to prove (AA′, βγ), (BB′, γα), (CC′, βα) are harmonic. Consider the cross-ratio (CC′, αβ). Then projecting from A on BB′ we have A(CC′, αβ) = A(B′B, αγ). Projecting from A′ on BB′, A′(CC′, αβ) = A′(BB′, αγ). Hence (B′B, αγ) = (BB′, αγ),i.e.the cross-ratio (BB′, αγ) equals that of its reciprocal; hence the range is harmonic.The second theorem states that the pencils L(BA, NM), M(B′A, LN), N(BA, LM) are harmonic. Deferring the subject of harmonic pencils to the next section, it will suffice to state here that any transversal intersects an harmonic pencil in an harmonic range. Consider the pencil L(BA, NM), then it is sufficient to prove (BA′, NM′) is harmonic. This follows from the previous theorem by considering A′B as a diagonal of the quadrilateral ALB′M.]This property of the complete quadrilateral allows the solution of the problem:To construct the harmonic conjugateDto a pointCwith regard to two given pointsAandB.Through A draw any two lines, and through C one cutting the former two in G and H. Join these points to B, cutting the former two lines in E and F. The point D where EF cuts AB will be the harmonic conjugate required.This remarkable construction requires nothing but the drawing of lines, and is therefore independent of measurement. In a similar manner the harmonic conjugate of the line VA for two lines VC, VD is constructed with the aid of the property of the complete quadrangle.§ 22.Harmonic Pencils.—The theory of cross-ratios may be extended from points in a row to lines in a flat pencil and to planes in an axial pencil. We have seen (§ 13) that if the lines which join four points A, B, C, D to any point S be cut by any other line in A′, B′, C′, D′, then (AB, CD) = (A′B′, C′D′). In other words, four lines in a flat pencil are cut by every other line in four points whose cross-ratio is constant.Definition.—By the cross-ratio of four rays in a flat pencil is meant the cross-ratio of the four points in which the rays are cut by any line. If a, b, c, d be the lines, then this cross-ratio is denoted by (ab, cd).Definition.—By the cross-ratio of four planes in an axial pencil is understood the cross-ratio of the four points in which any line cuts the planes, or, what is the same thing, the cross-ratio of the four rays in which any plane cuts the four planes.In order that this definition may have a meaning, it has to be proved that all lines cut the pencil in points which have the same cross-ratio. This is seen at once for two intersecting lines, as their plane cuts the axial pencil in a flat pencil, which is itself cut by the two lines. The cross-ratio of the four points on one line is therefore equal to that on the other, and equal to that of the four rays in the flat pencil.If two non-intersecting lines p and q cut the four planes in A, B, C, D and A′, B′, C′, D′, draw a line r to meet both p and q, and let this line cut the planes in A″, B″, C″, D″. Then (AB, CD) = (A′B′, C′D′), for each is equal to (A″B″, C″D″).§ 23. We may now also extend the notion of harmonic elements, viz.Definition.—Four rays in a flat pencil and four planes in an axial pencil are said to be harmonic if their cross-ratio equals -1, that is, if they are cut by a line in four harmonic points.If we understand by a “median line” of a triangle a line which joins a vertex to the middle point of the opposite side, and by a “median line” of a parallelogram a line joining middle points of opposite sides, we get as special cases of the last theorem:The diagonals and median lines of a parallelogram form an harmonic pencil; andAt a vertex of any triangle, the two sides, the median line, and the line parallel to the base form an harmonic pencil.Taking the parallelogram a rectangle, or the triangle isosceles, we get:Any two lines and the bisections of their angles form an harmonic pencil.Or:In an harmonic pencil, if two conjugate rays are perpendicular, then the other two are equally inclined to them; and, conversely,if one ray bisects the angle between conjugate rays, it is perpendicular to its conjugate.This connects perpendicularity and bisection of angles with projective properties.§ 24. We add a few theorems and problems which are easily proved or solved by aid of harmonics.An harmonic pencil is cut by a line parallel to one of its rays in three equidistant points.Through a given point to draw a line such that the segment determined on it by a given angle is bisected at that point.Having given two parallel lines, to bisect on either any given segment without using a pair of compasses.Having given in a line a segment and its middle point, to draw through any given point in the plane a line parallel to the given line.To draw a line which joins a given point to the intersection of two given lines which meet off the drawing paper (by aid of § 21).Correspondence. Homographic and Perspective Ranges§ 25. Two rows, p and p′, which are one the projection of the other (as in fig. 5), stand in a definite relation to each other, characterized by the following properties.1.To each point in either corresponds one point in the other; that is, those points are said to correspond which are projections of one another.2.The cross-ratio of any four points in one equals that of the corresponding points in the other.3.The lines joining corresponding points all pass through the same point.If we suppose corresponding points marked, and the rows brought into any other position, then the lines joining corresponding points will no longer meet in a common point, and hence the third of the above properties will not hold any longer; but we have still a correspondence between the points in the two rows possessing the first two properties. Such a correspondence has been called aone-one correspondence, whilst the two rows between which such correspondence has been established are said to beprojectiveorhomographic. Two rows which are each the projection of the other are thereforeprojective. We shall presently see, also, that any two projective rows may always be placed in such a position that one appears as the projection of the other. If they are in such a position the rows are said to be inperspective position, or simply to be inperspective.§ 26. The notion of a one-one correspondence between rows may be extended to flat and axial pencils, viz. a flat pencil will be said to be projective to a flat pencil if to each ray in the first corresponds one ray in the second, and if the cross-ratio of four rays in one equals that of the corresponding rays in the second.Similarly an axial pencil may be projective to an axial pencil. But a flat pencil may also be projective to an axial pencil, or either pencil may be projective to a row. The definition is the same in each case: there is a one-one correspondence between the elements, and four elements have the same cross-ratio as the corresponding ones.§ 27. There is also in each case a special position which is calledperspective, viz.1. Two projective rows are perspective if they lie in the same plane, and if the one row is a projection of the other.2. Two projective flat pencils are perspective—(1) if they lie in the same plane, and have a row as a common section; (2) if they lie in the same pencil (in space), and are both sections of the same axial pencil; (3) if they are in space and have a row as common section, or are both sections of the same axial pencil, one of the conditions involving the other.3. Two projective axial pencils, if their axes meet, and if they have a flat pencil as a common section.4. A row and a projective flat pencil, if the row is a section of the pencil, each point lying in its corresponding line.5. A row and a projective axial pencil, if the row is a section of the pencil, each point lying in its corresponding line.6. A flat and a projective axial pencil, if the former is a section of the other, each ray lying in its corresponding plane.That in each case the correspondence established by the position indicated is such as has been called projective follows at once from the definition. It is not so evident that the perspective position may always be obtained. We shall show in § 30 this for the first threecases. First, however, we shall give a few theorems which relate to the general correspondence, not to the perspective position.§ 28.Two rows or pencils, flat or axial, which are projective to a third are projective to each other; this follows at once from the definitions.§ 29.If two rows, or two pencils, either flat or axial, or a row and a pencil, be projective, we may assume to any three elements in the one the three corresponding elements in the other, and then the correspondence is uniquely determined.For if in two projective rows we assume that the points A, B, C in the first correspond to the given points A′, B′, C′ in the second, then to any fourth point D in the first will correspond a point D′ in the second, so that(AB, CD) = (A′B′, C′D′).But there is only one point, D′, which makes the cross-ratio (A′B′, C′D′) equal to the given number (AB, CD).The same reasoning holds in the other cases.§ 30. If two rows are perspective, then the lines joining corresponding points all meet in a point, the centre of projection; and the point in which the two bases of the rows intersect as a point in the first row coincides with its corresponding point in the second.This follows from the definition. The converse also holds, viz.If two projective rows have such a position that one point in the one coincides with its corresponding point in the other, then they are perspective, that is, the lines joining corresponding points all pass through a common point, and form a flat pencil.For let A, B, C, D ... be points in the one, and A′, B′, C′, D′ ... the corresponding points in the other row, and let A be made to coincide with its corresponding point A′. Let S be the point where the lines BB′ and CC′ meet, and let us join S to the point D in the first row. This line will cut the second row in a point D″, so that A, B, C, D are projected from S into the points A, B′, C′, D″. The cross-ratio (AB, CD) is therefore equal to (AB′, C′D″), and by hypothesis it is equal to (A′B′, C′D′). Hence (A′B′, C′D″) = (A′B′, C′D′), that is, D″ is the same point as D′.§ 31. If two projected flat pencils in the same plane are in perspective, then the intersections of corresponding lines form a row, and the line joining the two centres as a line in the first pencil corresponds to the same line as a line in the second. And conversely,If two projective pencils in the same plane, but with different centres, have one line in the one coincident with its corresponding line in the other, then the two pencils are perspective, that is, the intersection of corresponding lines lie in a line.The proof is the same as in § 30.§ 32. If two projective flat pencils in the same point (pencil in space), but not in the same plane, are perspective, then the planes joining corresponding rays all pass through a line (they form an axial pencil), and the line common to the two pencils (in which their planes intersect) corresponds to itself. And conversely:—If two flat pencils which have a common centre, but do not lie in a common plane, are placed so that one ray in the one coincides with its corresponding ray in the other, then they are perspective, that is, the planes joining corresponding lines all pass through a line.§ 33. If two projective axial pencils are perspective, then the intersection of corresponding planes lie in a plane, and the plane common to the two pencils (in which the two axes lie) corresponds to itself. And conversely:—If two projective axial pencils are placed in such a position that a plane in the one coincides with its corresponding plane, then the two pencils are perspective, that is, corresponding planes meet in lines which lie in a plane.The proof again is the same as in § 30.§ 34. These theorems relating to perspective position become illusory if the projective rows of pencils have a common base. We then have:—In two projective rows on the same line—and also in two projective and concentric flat pencils in the same plane, or in two projective axial pencils with a common axis—every element in the one coincides with its corresponding element in the other as soon as three elements in the one coincide with their corresponding elements in the other.Proof(in case of two rows).—Between four elements A, B, C, D and their corresponding elements A′, B′, C′, D′ exists the relation (ABCD) = (A′B′C′D′). If now A′, B′, C′ coincide respectively with A, B, C, we get (AB, CD) = (AB, CD′), hence D and D′ coincide.The last theorem may also be stated thus:—In two projective rows or pencils, which have a common base but are not identical, not more than two elements in the one can coincide with their corresponding elements in the other.Thus two projective rows on the same line cannot have more than two pairs of coincident points unless every point coincides with its corresponding point.Fig. 9.Fig. 10.Fig. 11.It is easy to construct two projective rows on the same line, which have two pairs of corresponding points coincident. Let the points A, B, C as points belonging to the one row correspond to A, B, and C′ as points in the second. Then A and B coincide with their corresponding points, but C does not. It is, however, not necessary that two such rows have twice a point coincident with its corresponding point; it is possible that this happens only once or not at all. Of this we shall see examples later.§ 35. If two projective rows or pencils are in perspective position, we know at once which element in one corresponds to any given element in the other. If p and q (fig. 9) are two projective rows, so that K corresponds to itself, and if we know that to A and B in p correspond A′ and B′ in q, then the point S, where AA′ meets BB′, is the centre of projection, and hence, in order to find the point C′ corresponding to C, we have only to join C to S; the point C′, where this line cuts q, is the point required.If two flat pencils, S1and S2, in a plane are perspective (fig. 10), we need only to know two pairs, a, a′ and b, b′, of corresponding rays in order to find the axis s of projection. This being known, a ray c′ in S2, corresponding to a given ray c in S1, is found by joining S2to the point where c cuts the axis s.A similar construction holds in the other cases of perspective figures.On this depends the solution of the following general problem.§ 36. Three pairs of corresponding elements in two projective rows or pencils being given, to determine for any element in one the corresponding element in the other.We solve this in the two cases of two projective rows and of two projective flat pencils in a plane.ProblemI.—Let A, B, C be three points in a row s, A′, B′, C′ the corresponding points in a projective row s′, both being in a plane; it is required to find for any point D in s the corresponding point D′ in s′.ProblemII.—Let a, b, c be three rays in a pencil S, a′, b′, c′ the corresponding rays in a projective pencil S′, both being in the same plane; it is required to find for any ray d in S the corresponding ray d′ in S′.The solution is made to depend on the construction of an auxiliary row or pencil which is perspective to both the given ones. This is found as follows:—Solution of ProblemI.—On the line joining two corresponding points, say AA′ (fig. 11), take any two points, S and S′, as centres of auxiliary pencils. Join the intersection B1of SB and S′B′ to the intersection C1of SC and S′C′ by the line s1. Then a row on s1will be perspective to s with S as centre of projection, and to s′ with S′ as centre. To find now the point D′ on s′ corresponding to a point D on s we have only to determine the point D1, where the line SD cuts s1, and to draw S′D1; the point where this line cuts s′ will be the required point D′.Proof.—The rows s and s′ are both perspective to the row s1, hence they are projective to one another. To A, B, C, D on s correspond A1, B1, C1, D1on s1, and to these correspond A′, B′, C′, D′ on s′; so that D and D′ are corresponding points as required.Fig. 12.Fig. 13.Solution of ProblemII.—Through the intersection A of two corresponding rays a and a′ (fig. 12), take two lines, s and s′, as bases of auxiliary rows. Let S1be the point where the line b1, which joins B and B′, cuts the line c1, which joins C and C′. Then a pencil S1will be perspective to S with s as axis of projection. To find the ray d′ in S′ corresponding to a given ray d in S, cut d by s at D; project this point from S1to D′ on s′ and join D′ to S′. This will be the required ray.Proof.—That the pencil S1is perspective to S and also to S′ follows from construction. To the lines a1, b1, c1, d1in S1correspond the lines a, b, c, d in S and the lines a′, b′, c′, d′ in S′, so that d and d′ are corresponding rays.In the first solution the two centres, S, S′, areanytwo points on a line joining any two corresponding points, so that the solution of the problem allows of a great many different constructions.But whatever construction be used, the pointD′,corresponding toD,must be always the same, according to the theorem in § 29. This gives rise to a number of theorems, into which, however, we shall not enter. The same remarks hold for the second problem.§ 37.Homological Triangles.—As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem.Theorem.—If ABC and A′B′C′ (fig. 13) be two triangles, such that the lines AA′, BB′, CC′ meet in a point S, then the intersections of BC and B′C′, of CA and C′A′, and of AB and A′B′ will lie in a line. Such triangles are said to be homological, or in perspective. The triangles are “co-axial” in virtue of the property that the meets of corresponding sides are collinear and copolar, since the lines joining corresponding vertices are concurrent.Proof.—Let a, b, c denote the lines AA′, BB′, CC′, which meet at S. Then these may be taken as bases of projective rows, so that A, A′, S on a correspond to B, B′, S on b, and to C, C′, S on c. As the point S is common to all, any two of these rows will be perspective.IfS1be the centre of projection of rowsb and c,S2” ” ”c and a,S3” ” ”a and b,and if the line S1S2cuts a in A1, and b in B1, and c in C1, then A1, B1will be corresponding points in a and b, both corresponding to C1in c. But a and b are perspective, therefore the line A1B1, that is S1S2, joining corresponding points must pass through the centre of projection S3of a and b. In other words, S1, S2, S3lie in a line. This is Desargues’ celebrated theorem if we state it thus:—Theorem of Desargues.—If each of two triangles has one vertex on each of three concurrent lines, then the intersections of corresponding sides lie in a line, those sides being called corresponding which are opposite to vertices on the same line.The converse theorem holds also, viz.Theorem.—If the sides of one triangle meet those of another in three points which lie in a line, then the vertices lie on three lines which meet in a point.The proof is almost the same as before.§ 38.Metrical Relations between Projective Rows.—Every row contains one point which is distinguished from all others, viz. the point at infinity. In two projective rows, to the point I at infinity in one corresponds a point I′ in the other, and to the point J′ at infinity in the second corresponds a point J in the first. The points I′ and J are in general finite. If now A and B are any two points in the one, A′, B′ the corresponding points in the other row, then(AB, JI) = (A′B′, J′I′),orAJ/JB : AI/IB = A′J′/J′B′ : A′I′/I′B′.But, by § 17,AI/IB = A′J′/J′B′ = −1;therefore the last equation changes intoAJ · A′I′ = BJ · B′I′,that is to say—Theorem.—The product of the distances of any two corresponding points in two projective rows from the points which correspond to the points at infinity in the other is constant, viz. AJ · A′I′ = k. Steiner has called this number k thePower of the correspondence.[The relation AJ · A′I′ = k shows that if J, I′ be given then the point A′ corresponding to a specified point A is readily found; hence A, A′ generate homographic ranges of which I and J′ correspond to the points at infinity on the ranges. If we take any two origins O, O′, on the ranges and reduce the expression AJ · A′I′ = k to its algebraic equivalent, we derive an equation of the form αxx′ + βx + γx′ + δ = 0. Conversely, if a relation of this nature holds, then points corresponding to solutions in x, x′ form homographic ranges.]§ 39.Similar Rows.—If the points at infinity in two projective rows correspond so that I′ and J are at infinity, this result loses its meaning. But if A, B, C be any three points in one, A′, B′, C′ the corresponding ones on the other row, we have(AB, CI) = (A′B′, C′I′),which reduces toAC/CB = A′C′/C′B′ or AC/A′C′ = BC/B′C′,that is, corresponding segments are proportional. Conversely, if corresponding segments are proportional, then to the point at infinity in one corresponds the point at infinity in the other. If we call such rowssimilar, we may state the result thus—Theorem.—Two projective rows are similar if to the point at infinity in one corresponds the point at infinity in the other, and conversely, if two rows are similar then they are projective, and the points at infinity are corresponding points.From this the well-known propositions follow:—Two lines are cut proportionally (in similar rows) by a series of parallels. The rows are perspective, with centre of projection at infinity.If two similar rows are placed parallel, then the lines joining homologous points pass through a common point.§ 40. If two flat pencils be projective, then there exists in either, one single pair of lines at right angles to one another, such that the corresponding lines in the other pencil are again at right angles.Fig. 14.To prove this, we place the pencils in perspective position (fig. 14) by making one ray coincident with its corresponding ray. Corresponding rays meet then on a line p. And now we draw the circle which has its centre O on p, and which passes through the centres S and S′ of the two pencils. This circle cuts p in two points H and K. The two pairs of rays, h, k, and h′, k′, joining these points to S and S′ will be pairs of corresponding rays at right angles. The construction gives in general but one circle, but if the line p is the perpendicular bisector of SS′, there exists an infinite number, andto every right angle in the one pencil corresponds a right angle in the other.Principle of Duality§ 41. It has been stated in § 1 that not only points, but also planes and lines, are taken as elements out of which figures are built up. We shall now see that the construction of one figure which possesses certain properties gives rise in many cases to the construction of another figure, by replacing, according to definite rules, elements of one kind by those of another. The new figure thus obtained will then possess properties which may be stated as soon as those of the original figure are known.We obtain thus a principle, known as theprinciple of dualityor ofreciprocity, which enables us to construct to any figure not containing any measurement in its construction areciprocalfigure, as it is called, and to deduce from any theorem areciprocaltheorem, for which no further proof is needed.It is convenient to print reciprocal propositions on opposite sides of a page broken into two columns, and this plan will occasionally be adopted.We begin by repeating in this form a few of our former statements:—Two points determine a line.Two planes determine a line.Three points which are not in a line determine a plane.Three planes which do not pass through a line determine a point.A line and a point without it determine a plane.A line and a plane not through it determine a point.Two lines in a plane determine a point.Two lines through a point determine a plane.These propositions show that it will be possible, when any figure is given, to construct a second figure by taking planes instead of points, and points instead of planes, but lines where we had lines.For instance, if in the first figure we take a plane and three points in it, we have to take in the second figure a point and three planes through it. The three points in the first, together with the three lines joining them two and two, form a triangle; the three planes in the second and their three lines of intersection form a trihedral angle. A triangle and a trihedral angle are therefore reciprocal figures.Similarly, to any figure in a plane consisting of points and lines will correspond a figure consisting of planes and lines passing through a point S, and hence belonging to the pencil which has S as centre.The figure reciprocal to four points in space which do not lie in a plane will consist of four planes which do not meet in a point. In this case each figure forms a tetrahedron.§ 42. As other examples we have the following:—To a rowis reciprocalan axial pencil,to a flat pencil”a flat pencil,to a field of points and lines”a pencil of planes and lines,to the space of points”the space of planes.For the row consists of a line and all the points in it, reciprocal to it therefore will be a line with all planes through it, that is, an axial pencil; and so for the other cases.This correspondence of reciprocity breaks down, however, if we take figures which contain measurement in their construction. For instance, there is no figure reciprocal to two planes atright angles, because there is no segment in a row which has a magnitude as definite as a right angle.We add a few examples of reciprocal propositions which are easily proved.Theorem.—If A, B, C, D are any four points in space, and if the lines AB and CD meet, then all four points lie in a plane, hence also AC and BD, as well as AD and BC, meet.Theorem.—If α, β, γ, δ are four planes in space, and if the lines αβ and γδ meet, then all four planes lie in a point (pencil), hence also αγ and βδ, as well as αδ and βγ, meet.Theorem.—If of any number of lines every one meets every other, whilst all do notlie in a point, then all lie in a plane.lie in a plane, then all lie in a point(pencil).§ 43. Reciprocal figures as explained lie both in space of three dimensions. If the one is confined to a plane (is formed of elements which lie in a plane), then the reciprocal figure is confined to a pencil (is formed of elements which pass through a point).But there is also a more special principle of duality, according to which figures are reciprocal which lie both in a plane or both in a pencil. In the plane we take points and lines as reciprocal elements, for they have this fundamental property in common, that two elements of one kind determine one of the other. In the pencil, on the other hand, lines and planes have to be taken as reciprocal, and here it holds again that two lines or planes determine one plane or line.Thus, to one plane figure we can construct one reciprocal figure in the plane, and to each one reciprocal figure in a pencil. We mention a few of these. At first we explain a few names:—A figure consisting of n points in a plane will be called an n-point.A figure consisting of n lines in a plane will be called an n-side.A figure consisting of n planes in a pencil will be called an n-flat.A figure consisting of n lines in a pencil will be called an n-edge.It will be understood that an n-side is different from a polygon of n sides. The latter has sides of finite length and n vertices, the former has sides all of infinite extension, and every point where two of the sides meet will be a vertex. A similar difference exists between a solid angle and an n-edge or an n-flat. We notice particularly—A four-point has six sides, of which two and two are opposite, and three diagonal points, which are intersections of opposite sides.A four-side has six vertices, of which two and two are opposite, and three diagonals, which join opposite vertices.A four-flat has six edges, of which two and two are opposite, and three diagonal planes, which pass through opposite edges.A four-edge has six faces, of which two and two are opposite, and three diagonal edges, which are intersections of opposite faces.
§ 1.Geometrical Elements.We consider space as filled with points, lines and planes, and these we call the elements out of which our figures are to be formed, calling any combination of these elements a “figure.”
By a line we mean a straight line in its entirety, extending both ways to infinity; and by a plane, a plane surface, extending in all directions to infinity.
We accept the three-dimensional space of experience—the space assumed by Euclid—which has for its properties (among others):—
Through any two points in space one and only one line may be drawn;
Through any three points which are not in a line, one and only one plane may be placed;
The intersection of two planes is a line;
A line which has two points in common with a plane lies in the plane, hence the intersection of a line and a plane is a single point; and
Three planes which do not meet in a line have one single point in common.
These results may be stated differently in the following form:—
I. A plane is determined—
A point is determined—
1. By three points which do not lie in a line;
2. By two intersecting lines;
3. By a line and a point which does not lie in it.
1. By three planes which do not pass through a line;
2. By two intersecting lines
3. By a plane and a line which does not lie in it.
A line is determined—
1. By two points;
2. By two planes.
It will be observed that not only are planes determined by points, but also points by planes; that therefore the planes may be considered as elements, like points; and also that in any one of the above statements we may interchange the words point and plane, and we obtain again a correct statement, provided that these statements themselves are true. As they stand, we ought, in several cases, to add “if they are not parallel,” or some such words, parallel lines and planes being evidently left altogether out of consideration. To correct this we have to reconsider the theory of parallels.
§ 2.Parallels. Point at Infinity.—Let us take in a plane a line p (fig. 1), a point S not in this line, and a line q drawn through S. Then this line q will meet the line p in a point A. If we turn the line q about S towards q’, its point of intersection with p will move along p towards B, passing, on continued turning, to a greater and greater distance, until it is moved out of our reach. If we turn q still farther, its continuation will meet p, but now at the other side of A. The point of intersection has disappeared to the right and reappeared to the left. There is one intermediate position where q is parallel to p—that is where it does not cut p. In every other position it cuts p in some finite point. If, on the other hand, we move the point A to an infinite distance in p, then the line q which passes through A will be a line which does not cut p at any finite point. Thus we are led to say:Everyline through S which joins it to any point at an infinite distance in p is parallel to p. But by Euclid’s 12th axiom there is but one line parallel to p through S. The difficulty in which we are thus involved is due to the fact that we try to reason about infinity as if we, with our finite capabilities, could comprehend the infinite. To overcome this difficulty, we may say that all points at infinity in a lineappearto us as one, and may be replaced by a single “ideal” point.
We may therefore now give the following definitions and axiom:—
Definition.—Lines which meet at infinity are called parallel.
Axiom.—All points at an infinite distance in a line may be considered as one single point.
Definition.—This ideal point is called thepoint at infinityin the line.
The axiom is equivalent to Euclid’s Axiom 12, for it follows from either that through any point only one line may be drawn parallel to a given line.
This point at infinity in a line is reached whether we move a point in the one or in the opposite direction of a line to infinity. A line thus appears closed by this point, and we speak as if we could move a point along the line from one position A to another B in two ways, either through the point at infinity or through finite points only.
It must never be forgotten that this point at infinity is ideal; in fact, the whole notion of “infinity” is only a mathematical conception, and owes its introduction (as a method of research) to the working generalizations which it permits.
§ 3.Line and Plane at Infinity.—Having arrived at the notion of replacing all points at infinity in a line by one ideal point, there is no difficulty in replacing all points at infinity in a plane by one ideal line.
To make this clear, let us suppose that a line p, which cuts two fixed lines a and b in the points A and B, moves parallel to itself to a greater and greater distance. It will at last cut both a and b at their points at infinity, so that a line which joins the two points at infinity in two intersecting lines lies altogether at infinity. Every other line in the plane will meet it therefore at infinity, and thus it contains all points at infinity in the plane.
All points at infinity in a plane lie in a line, which is called theline at infinityin the plane.
It follows that parallel planes must be considered as planes having a common line at infinity, for any other plane cuts them in parallel lines which have a point at infinity in common.
If we next take two intersecting planes, then the point at infinity in their line of intersection lies in both planes, so that their lines at infinity meet. Hence every line at infinity meets every other line at infinity, and they are therefore all in one plane.
All points at infinity in space may be considered as lying in one ideal plane, which is called theplane at infinity.
§ 4.Parallelism.—We have now the following definitions:—
Parallel lines are lines which meet at infinity;
Parallel planes are planes which meet at infinity;
A line is parallel to a plane if it meets it at infinity.
Theorems like this—Lines (or planes) which are parallel to a third are parallel to each other—follow at once.
This view of parallels leads therefore to no contradiction of Euclid’sElements.
As immediate consequences we get the propositions:—
Every line meets a plane in one point, or it lies in it;
Every plane meets every other plane in a line;
Any two lines in the same plane meet.
§ 5.Aggregates of Geometrical Elements.—We have called points, lines and planes the elements of geometrical figures. We also say that an element of one kind contains one of the other if it lies in it or passes through it.
All the elements of one kind which are contained in one or two elements of a different kind form aggregates which have to be enumerated. They are the following:—
I. Of one dimension.
1. Therow, or range,of pointsformed by all points in a line, which is called its base.2. Theflat pencilformed by all the lines through a point in a plane. Its base is the point in the plane.3. Theaxial pencilformed by all planes through a line which is called its base or axis.
1. Therow, or range,of pointsformed by all points in a line, which is called its base.
2. Theflat pencilformed by all the lines through a point in a plane. Its base is the point in the plane.
3. Theaxial pencilformed by all planes through a line which is called its base or axis.
II. Of two dimensions.
1. The field of points and lines—that is, a plane with all its points and all its lines.2. The pencil of lines and planes—that is, a point in space with all lines and all planes through it.
1. The field of points and lines—that is, a plane with all its points and all its lines.
2. The pencil of lines and planes—that is, a point in space with all lines and all planes through it.
III. Of three dimensions.
The space of points—that is, all points in space.The space of planes—that is, all planes in space.
The space of points—that is, all points in space.
The space of planes—that is, all planes in space.
IV. Of four dimensions.
The space of lines, or all lines in space.
The space of lines, or all lines in space.
§ 6.Meaning of “Dimensions.”—The word dimension in the above needs explanation. If in a plane we take a row p and a pencil with centre Q, then through every point in p one line in the pencil will pass, and every ray in Q will cut p in one point, so that we are entitled to say a row contains as many points as a flat pencil lines, and, we may add, as an axial pencil planes, because an axial pencil is cut by a plane in a flat pencil.
The number of elements in the row, in the flat pencil, and in the axial pencil is, of course, infinite and indefinite too, but the same in all. This number may be denoted by ∞. Then a plane contains ∞² points and as many lines. To see this, take a flat pencil in a plane. It contains ∞ lines, and each line contains ∞ points, whilst each point in the plane lies on one of these lines. Similarly, in a plane each line cuts a fixed line in a point. But this line is cut at each point by ∞ lines and contains ∞ points; hence there are ∞² lines in a plane.
A pencil in space contains as many lines as a plane contains points and as many planes as a plane contains lines, for any plane cuts the pencil in a field of points and lines. Hence a pencil contains ∞² lines and ∞² planes.The field and the pencil are of two dimensions.
To count the number of points in space we observe that each point lies on some line in a pencil. But the pencil contains ∞² lines, and each line ∞ points; hence space contains ∞³ points. Each plane cuts any fixed plane in a line. But a plane contains ∞² lines, and through each pass ∞ planes; therefore space contains ∞³ planes.
Hence space contains as many planes as points, but it contains an infinite number of times more lines than points or planes. To count them, notice that every line cuts a fixed plane in one point. But ∞² lines pass through each point, and there are ∞² points in the plane. Hence there are ∞4lines in space.The space of points and planes is of three dimensions, but the space of lines is of four dimensions.
A field of points or lines contains an infinite number of rows and flat pencils; a pencil contains an infinite number of flat pencils and of axial pencils; space contains a triple infinite number of pencils and of fields, ∞4rows and axial pencils and ∞5flat pencils—or, in other words, each point is a centre of ∞² flat pencils.
§ 7. The above enumeration allows a classification of figures. Figures in a row consist of groups of points only, and figures in the flat or axial pencil consist of groups of lines or planes. In the plane we may draw polygons; and in the pencil or in the point, solid angles, and so on.
We may also distinguish the different measurements We have—
In the row, length of segment;In the flat pencil, angles;In the axial pencil, dihedral angles between two planes;In the plane, areas;In the pencil, solid angles;In the space of points or planes, volumes.
In the row, length of segment;
In the flat pencil, angles;
In the axial pencil, dihedral angles between two planes;
In the plane, areas;
In the pencil, solid angles;
In the space of points or planes, volumes.
Segments of a Line
§ 8. Any two points A and B in space determine on the line through them a finite part, which may be considered as being described by a point moving from A to B. This we shall denote by AB, and distinguish it from BA, which is supposed as being described by a point moving from B to A, and hence in a direction or in a “sense” opposite to AB. Such a finite line, which has a definite sense, we shall call a “segment,” so that AB and BA denote different segments, which are said to be equal in length but of opposite sense. The one sense is often called positive and the other negative.
In introducing the word “sense” for direction in a line, we have the word direction reserved for direction of the line itself, so that different lines have different directions, unless they be parallel, whilst in each line we have a positive and negative sense.
We may also say, with Clifford, that AB denotes the “step” of going from A to B.
§ 9. If we have three points A, B, C in a line (fig. 2), the step AB will bring us from A to B, and the step BC from B to C. Hence both steps are equivalent to the one step AC. This is expressed by saying that AC is the “sum” of AB and BC; in symbols—
AB + BC = AC,
where account is to be taken of the sense.
This equation is true whatever be the position of the three points on the line. As a special case we have
AB + BA = 0,
(1)
and similarly
AB + BC + CA = 0,
(2)
which again is true for any three points in a line.
We further write
AB = −BA.
where − denotes negative sense.
We can then, just as in algebra, change subtraction of segments into addition by changing the sense, so that AB − CB is the same as AB + (−CB) or AB + BC. A figure will at once show the truth of this. The sense is, in fact, in every respect equivalent to the “sign” of a number in algebra.
§ 10. Of the many formulae which exist between points in a line we shall have to use only one more, which connects the segments between any four points A, B, C, D in a line. We have
BC = BD + DC, CA = CD + DA, AB = AD + DB;
or multiplying these by AD, BD, CD respectively, we get
BC · AD = BD · AD + DC · AD = BD · AD − CD · ADCA · BD = CD · BD + DA · BD = CD · BD − AD · BDAB · CD = AD · CD + DB · CD = AD · CD − BD · CD.
BC · AD = BD · AD + DC · AD = BD · AD − CD · AD
CA · BD = CD · BD + DA · BD = CD · BD − AD · BD
AB · CD = AD · CD + DB · CD = AD · CD − BD · CD.
It will be seen that the sum of the right-hand sides vanishes, hence that
BC · AD + CA · BD + AB · CD = 0
(3)
for any four points on a line.
§ 11. If C is any point in the line AB, then we say that C divides the segment AB in the ratio AC/CB, account being taken of the sense of the two segments AC and CB. If C lies between A and B the ratio is positive, as AC and CB have the same sense. But if C lies without the segment AB,i.e.if C divides AB externally, then the ratio is negative. To see how the value of this ratio changes with C, we will move C along the whole line (fig. 3), whilst A and B remain fixed. If C lies at the point A, then AC = 0, hence the ratio AC : CB vanishes. As C moves towards B, AC increases and CB decreases, so that our ratio increases. At the middle point M of AB it assumes the value +1, and then increases till it reaches an infinitely large value, when C arrives at B. On passing beyond B the ratio becomes negative. If C is at P we have AC = AP = AB + BP, hence
In the last expression the ratio AB : BP is positive, has its greatest value ∞ when C coincides with B, and vanishes when BC becomes infinite. Hence, as C moves from B to the right to the point at infinity, the ratio AC : CB varies from −∞ to −1.
If, on the other hand, C is to the left of A, say at Q, we have AC = AQ = AB + BQ = AB − QB, hence AC/CB = AB/QB − 1.
Here AB < QB, hence the ratio AB : QB is positive and always less than one, so that the whole is negative and < 1. If C is at the point at infinity it is −1, and then increases as C moves to the right, till for C at A we get the ratio = 0. Hence—
“As C moves along the line from an infinite distance to the left to an infinite distance at the right, the ratio always increases; it starts with the value −1, reaches 0 at A, +1 at M, ∞ at B, now changes sign to −∞, and increases till at an infinite distance it reaches again the value −1.It assumes therefore all possible values from -∞ to +∞, and each value only once, so that not only does every position ofCdetermine a definite value of the ratioAC : CB,but also, conversely, to every positive or negative value of this ratio belongs one single point in the lineAB.
[Relations between segments of lines are interesting as showing an application of algebra to geometry. The genesis of such relations from algebraic identities is very simple. For example, if a, b, c, x be any four quantities, then
this may be proved, cumbrously, by multiplying up, or, simply, by decomposing the right-hand member of the identity into partial fractions. Now take a line ABCDX, and let AB = a, AC = b, AD = c, AX = x. Then obviously (a − b) = AB − AC = −BC, paying regard to signs; (a − c) = AB − AD = DB, and so on. Substituting these values in the identity we obtain the following relation connecting the segments formed by five points on a line:—
Conversely, if a metrical relation be given, its validity may be tested by reducing to an algebraic equation, which is an identity if the relation be true. For example, if ABCDX be five collinear points, prove
Clearing of fractions by multiplying throughout by AB · BC · CA, we have to prove
−AD · AX · BC − BD · BX · CA − CD · CX · AB = AB · BC · CA.
Take A as origin and let AB = a, AC = b, AD = c, AX = x. Substituting for the segments in terms of a, b, c, x, we obtain on simplification
a²b − ab² = −ab² + a²b, an obvious identity.
An alternative method of testing a relation is illustrated in thefollowing example:—If A, B, C, D, E, F be six collinear points, then
Clearing of fractions by multiplying throughout by AB · BC · CD · DA, and reducing to a common origin O (calling OA = a, OB = b, &c.), an equation containing the second and lower powers of OA ( = a), &c., is obtained. Calling OA = x, it is found that x = b, x = c, x = d are solutions. Hence the quadratic has three roots; consequently it is an identity.
The relations connecting five points which we have instanced above may be readily deduced from the six-point relation; the first by taking D at infinity, and the second by taking F at infinity, and then making the obvious permutations of the points.]
Projection and Cross-ratios
§ 12. If we join a point A to a point S, then the point where the line SA cuts a fixed plane π is called the projection of A on the plane π from S as centre of projection. If we have two planes π and π′ and a point S, we may project every point A in π to the other plane. If A′ is the projection of A, then A is also the projection of A′, so that the relations are reciprocal. To every figure in π we get as its projection a corresponding figure in π′.
We shall determine such properties of figures as remain true for the projection, and which are called projective properties. For this purpose it will be sufficient to consider at first only constructions in one plane.
Let us suppose we have given in a plane two lines p and p′ and a centre S (fig. 4); we may then project the points in p from S to p′. Let A′, B′ ... be the projections of A, B ..., the point at infinity in p which we shall denote by I will be projected into a finite pointI′ in p′, viz. into the point where the parallel to p through S cuts p′. Similarly one point J in p will be projected into the point J′ at infinity in p′. This point J is of course the point where the parallel to p′ through S cuts p. We thus see that every point in p is projected into a single point in p′.
Fig. 5 shows that a segment AB will be projected into a segment A′B′ which is not equal to it, at least not as a rule; and also that the ratio AC : CB is not equal to the ratio A′C′ : C′B′ formed by the projections. These ratios will become equal only if p and p′ are parallel, for in this case the triangle SAB is similar to the triangle SA′B′. Between three points in a line and their projections there exists therefore in general no relation. But between four points a relation does exist.
§ 13. Let A, B, C, D be four points in p, A′, B′, C, D′ their projections in p′, then the ratio of the two ratios AC : CB and AD : DB into which C and D divide the segment AB is equal to the corresponding expression between A′, B′, C′, D′. In symbols we have
This is easily proved by aid of similar triangles.
Through the points A and B on p draw parallels to p′, which cut the projecting rays in C2, D2, B2and A1, C1, D1, as indicated in fig. 6. The two triangles ACC2and BCC1will be similar, as will also be the triangles ADD2and BDD1.
The proof is left to the reader.
This result is of fundamental importance.
The expression AC/CB : AD/DB has been called by Chasles the “anharmonic ratio of the four points A, B, C, D.” Professor Clifford proposed the shorter name of “cross-ratio.” We shall adopt the latter. We have then the
Fundamental Theorem.—The cross-ratio of four points in a line is equal to the cross-ratio of their projections on any other line which lies in the same plane with it.
§ 14. Before we draw conclusions from this result, we must investigate the meaning of a cross-ratio somewhat more fully.
If four points A, B, C, D are given, and we wish to form their cross-ratio, we have first to divide them into two groups of two, the points in each group being taken in a definite order. Thus, let A, B be the first, C, D the second pair, A and C being the first points in each pair. The cross-ratio is then the ratio AC : CB divided by AD : DB. This will be denoted by (AB, CD), so that
This is easily remembered. In order to write it out, make first the two lines for the fractions, and put above and below these the letters A and B in their places, thus, A/*B : A/*B; and then fill up, crosswise, the first by C and the other by D.
§ 15. If we take the points in a different order, the value of the cross-ratio will change. We can do this in twenty-four different ways by forming all permutations of the letters. But of these twenty-four cross-ratios groups of four are equal, so that there are really only six different ones, and these six are reciprocals in pairs.
We have the following rules:—
I. If in a cross-ratio the two groups be interchanged, its value remains unaltered,i.e.
(AB, CD) = (CD, AB) = (BA, DC) = (DC, BA).
II. If in a cross-ratio the two points belonging to one of the two groups be interchanged, the cross-ratio changes into its reciprocal,i.e.
(AB, CD) = 1/(AB, DC) = 1/(BA, CD) = 1/(CD, BA) = 1/(DC, AB).
From I. and II. we see that eight cross-ratios are associated with (AB, CD).
III. If in a cross-ratio the two middle letters be interchanged, the cross-ratio α changes into its complement 1 − α,i.e.(AB, CD) = 1 − (AC, BD).
[§ 16. If λ = (AB, CD), μ = (AC, DB), ν = (AD, BC), then λ, μ, ν and their reciprocals 1/λ, 1/μ, 1/ν are the values of the total number of twenty-four cross-ratios. Moreover, λ, μ, ν are connected by the relations
λ + 1/μ = μ + 1/ν = ν + 1/λ = −λμν = 1;
this proposition may be proved by substituting for λ, μ, ν and reducing to a common origin. There are therefore four equations between three unknowns; hence if one cross-ratio be given, the remaining twenty-three are determinate. Moreover, two of the quantities λ, μ, ν are positive, and the remaining one negative.
The following scheme shows the twenty-four cross-ratios expressed in terms of λ, μ, ν.]
§ 17. If one of the points of which a cross-ratio is formed is the point at infinity in the line, the cross-ratio changes into a simple ratio. It is convenient to let the point at infinity occupy the last place in the symbolic expression for the cross-ratio. Thus if I is a point at infinity, we have (AB, CI) = −AC/CB, because AI : IB = −1.
Every common ratio of three points in a line may thus be expressed as a cross-ratio, by adding the point at infinity to the group of points.
Harmonic Ranges
§ 18. If the points have special positions, the cross-ratios may have such a value that, of the six different ones, two and two become equal. If the first two shall be equal, we get λ = 1/λ, or λ² = 1, λ = ±1.
If we take λ = +1, we have (AB, CD) = 1, or AC/CB = AD/DB; that is, the points C and D coincide, provided that A and B are different.
If we take λ = −1, so that (AB, CD) = −1, we have AC/CB = −AD/DB.Hence C and D divide AB internally and externally in the same ratio.
The four points are in this case said to beharmonic points, and CandDare said to be harmonic conjugates with regard toAandB.
But we have also (CD, AB) = −1, so that A and B are harmonic conjugates with regard to C and D.
The principal property of harmonic points is that their cross-ratio remains unaltered if we interchange the two points belonging to one pair, viz.
(AB, CD) = (AB, DC) = (BA, CD).
For four harmonic points the six cross-ratios become equal two and two:
Hence if we get four points whose cross-ratio is 2 or ½, then they are harmonic, but not arranged so that conjugates are paired. If this is the case the cross-ratio = −1.
§ 19. If we equate any two of the above six values of the cross-ratios, we get either λ = 1, 0, ∞, or λ = −1, 2, ½, or else λ becomes a root of the equation λ² − λ + 1 = 0, that is, an imaginary cube root of −1. In this case the six values become three and three equal, so that only two different values remain. This case, though important in the theory of cubic curves, is for our purposes of no interest, whilst harmonic points are all-important.
§ 20. From the definition of harmonic points, and by aid of § 11, the following properties are easily deduced.
If C and D are harmonic conjugates with regard to A and B, then one of them lies in, the other without AB; it is impossible to move from A to B without passing either through C or through D; the one blocks the finite way, the other the way through infinity. This is expressed by saying A and B are “separated” by C and D.
For every position of C there will be one and only one point D which is its harmonic conjugate with regard to any point pair A, B.
If A and B are different points, and if C coincides with A or B, D does. But if A and B coincide, one of the points C or D, lying between them, coincides with them, and the other may be anywhere in the line. It follows that, “if of four harmonic conjugates two coincide, then a third coincides with them, and the fourth may be any point in the line.”
If C is the middle point between A and B, then D is the point at infinity; for AC : CB = +1, hence AD : DB must be equal to −1.The harmonic conjugate of the point at infinity in a line with regard to two pointsA, Bis the middle point ofAB.
This important property gives a first example how metric properties are connected with projective ones.
[§ 21.Harmonic properties of the complete quadrilateral and quadrangle.
A figure formed by four lines in a plane is called acomplete quadrilateral, or, shorter, afour-side. The four sides meet in six points, named the “vertices,” which may be joined by three lines (other than the sides), named the “diagonals” or “harmonic lines.” The diagonals enclose the “harmonic triangle of the quadrilateral.” In fig. 7, A′B′C′, B′AC, C′AB, CBA′ are the sides, A, A′, B, B′, C, C′ the vertices, AA′, BB′, CC′ the harmonic lines, and αβγ the harmonic triangle of the quadrilateral. A figure formed by four coplanar points is named acomplete quadrangle, or, shorter, afour-point. The four points may be joined by six lines, named the “sides,” which intersect in three other points, termed the “diagonal or harmonic points.” The harmonic points are the vertices of the “harmonic triangle of the complete quadrangle.” In fig. 8, AA′, BB′ are the points, AA′, BB′, A′B′, B′A, AB, BA′ are the sides, L, M, N are the diagonal points, and LMN is the harmonic triangle of the quadrangle.
The harmonic property of the complete quadrilateral is: Any diagonal or harmonic line is harmonically divided by the other two; and of a complete quadrangle: The angle at any harmonic point is divided harmonically by the joins to the other harmonic points. To prove the first theorem, we have to prove (AA′, βγ), (BB′, γα), (CC′, βα) are harmonic. Consider the cross-ratio (CC′, αβ). Then projecting from A on BB′ we have A(CC′, αβ) = A(B′B, αγ). Projecting from A′ on BB′, A′(CC′, αβ) = A′(BB′, αγ). Hence (B′B, αγ) = (BB′, αγ),i.e.the cross-ratio (BB′, αγ) equals that of its reciprocal; hence the range is harmonic.
The second theorem states that the pencils L(BA, NM), M(B′A, LN), N(BA, LM) are harmonic. Deferring the subject of harmonic pencils to the next section, it will suffice to state here that any transversal intersects an harmonic pencil in an harmonic range. Consider the pencil L(BA, NM), then it is sufficient to prove (BA′, NM′) is harmonic. This follows from the previous theorem by considering A′B as a diagonal of the quadrilateral ALB′M.]
This property of the complete quadrilateral allows the solution of the problem:
To construct the harmonic conjugateDto a pointCwith regard to two given pointsAandB.
Through A draw any two lines, and through C one cutting the former two in G and H. Join these points to B, cutting the former two lines in E and F. The point D where EF cuts AB will be the harmonic conjugate required.
This remarkable construction requires nothing but the drawing of lines, and is therefore independent of measurement. In a similar manner the harmonic conjugate of the line VA for two lines VC, VD is constructed with the aid of the property of the complete quadrangle.
§ 22.Harmonic Pencils.—The theory of cross-ratios may be extended from points in a row to lines in a flat pencil and to planes in an axial pencil. We have seen (§ 13) that if the lines which join four points A, B, C, D to any point S be cut by any other line in A′, B′, C′, D′, then (AB, CD) = (A′B′, C′D′). In other words, four lines in a flat pencil are cut by every other line in four points whose cross-ratio is constant.
Definition.—By the cross-ratio of four rays in a flat pencil is meant the cross-ratio of the four points in which the rays are cut by any line. If a, b, c, d be the lines, then this cross-ratio is denoted by (ab, cd).
Definition.—By the cross-ratio of four planes in an axial pencil is understood the cross-ratio of the four points in which any line cuts the planes, or, what is the same thing, the cross-ratio of the four rays in which any plane cuts the four planes.
In order that this definition may have a meaning, it has to be proved that all lines cut the pencil in points which have the same cross-ratio. This is seen at once for two intersecting lines, as their plane cuts the axial pencil in a flat pencil, which is itself cut by the two lines. The cross-ratio of the four points on one line is therefore equal to that on the other, and equal to that of the four rays in the flat pencil.
If two non-intersecting lines p and q cut the four planes in A, B, C, D and A′, B′, C′, D′, draw a line r to meet both p and q, and let this line cut the planes in A″, B″, C″, D″. Then (AB, CD) = (A′B′, C′D′), for each is equal to (A″B″, C″D″).
§ 23. We may now also extend the notion of harmonic elements, viz.
Definition.—Four rays in a flat pencil and four planes in an axial pencil are said to be harmonic if their cross-ratio equals -1, that is, if they are cut by a line in four harmonic points.
If we understand by a “median line” of a triangle a line which joins a vertex to the middle point of the opposite side, and by a “median line” of a parallelogram a line joining middle points of opposite sides, we get as special cases of the last theorem:
The diagonals and median lines of a parallelogram form an harmonic pencil; and
At a vertex of any triangle, the two sides, the median line, and the line parallel to the base form an harmonic pencil.
Taking the parallelogram a rectangle, or the triangle isosceles, we get:
Any two lines and the bisections of their angles form an harmonic pencil.Or:
In an harmonic pencil, if two conjugate rays are perpendicular, then the other two are equally inclined to them; and, conversely,if one ray bisects the angle between conjugate rays, it is perpendicular to its conjugate.
This connects perpendicularity and bisection of angles with projective properties.
§ 24. We add a few theorems and problems which are easily proved or solved by aid of harmonics.
An harmonic pencil is cut by a line parallel to one of its rays in three equidistant points.
Through a given point to draw a line such that the segment determined on it by a given angle is bisected at that point.
Having given two parallel lines, to bisect on either any given segment without using a pair of compasses.
Having given in a line a segment and its middle point, to draw through any given point in the plane a line parallel to the given line.
To draw a line which joins a given point to the intersection of two given lines which meet off the drawing paper (by aid of § 21).
Correspondence. Homographic and Perspective Ranges
§ 25. Two rows, p and p′, which are one the projection of the other (as in fig. 5), stand in a definite relation to each other, characterized by the following properties.
1.To each point in either corresponds one point in the other; that is, those points are said to correspond which are projections of one another.
2.The cross-ratio of any four points in one equals that of the corresponding points in the other.
3.The lines joining corresponding points all pass through the same point.
If we suppose corresponding points marked, and the rows brought into any other position, then the lines joining corresponding points will no longer meet in a common point, and hence the third of the above properties will not hold any longer; but we have still a correspondence between the points in the two rows possessing the first two properties. Such a correspondence has been called aone-one correspondence, whilst the two rows between which such correspondence has been established are said to beprojectiveorhomographic. Two rows which are each the projection of the other are thereforeprojective. We shall presently see, also, that any two projective rows may always be placed in such a position that one appears as the projection of the other. If they are in such a position the rows are said to be inperspective position, or simply to be inperspective.
§ 26. The notion of a one-one correspondence between rows may be extended to flat and axial pencils, viz. a flat pencil will be said to be projective to a flat pencil if to each ray in the first corresponds one ray in the second, and if the cross-ratio of four rays in one equals that of the corresponding rays in the second.
Similarly an axial pencil may be projective to an axial pencil. But a flat pencil may also be projective to an axial pencil, or either pencil may be projective to a row. The definition is the same in each case: there is a one-one correspondence between the elements, and four elements have the same cross-ratio as the corresponding ones.
§ 27. There is also in each case a special position which is calledperspective, viz.
1. Two projective rows are perspective if they lie in the same plane, and if the one row is a projection of the other.
2. Two projective flat pencils are perspective—(1) if they lie in the same plane, and have a row as a common section; (2) if they lie in the same pencil (in space), and are both sections of the same axial pencil; (3) if they are in space and have a row as common section, or are both sections of the same axial pencil, one of the conditions involving the other.
3. Two projective axial pencils, if their axes meet, and if they have a flat pencil as a common section.
4. A row and a projective flat pencil, if the row is a section of the pencil, each point lying in its corresponding line.
5. A row and a projective axial pencil, if the row is a section of the pencil, each point lying in its corresponding line.
6. A flat and a projective axial pencil, if the former is a section of the other, each ray lying in its corresponding plane.
That in each case the correspondence established by the position indicated is such as has been called projective follows at once from the definition. It is not so evident that the perspective position may always be obtained. We shall show in § 30 this for the first threecases. First, however, we shall give a few theorems which relate to the general correspondence, not to the perspective position.
§ 28.Two rows or pencils, flat or axial, which are projective to a third are projective to each other; this follows at once from the definitions.
§ 29.If two rows, or two pencils, either flat or axial, or a row and a pencil, be projective, we may assume to any three elements in the one the three corresponding elements in the other, and then the correspondence is uniquely determined.
For if in two projective rows we assume that the points A, B, C in the first correspond to the given points A′, B′, C′ in the second, then to any fourth point D in the first will correspond a point D′ in the second, so that
(AB, CD) = (A′B′, C′D′).
But there is only one point, D′, which makes the cross-ratio (A′B′, C′D′) equal to the given number (AB, CD).
The same reasoning holds in the other cases.
§ 30. If two rows are perspective, then the lines joining corresponding points all meet in a point, the centre of projection; and the point in which the two bases of the rows intersect as a point in the first row coincides with its corresponding point in the second.
This follows from the definition. The converse also holds, viz.
If two projective rows have such a position that one point in the one coincides with its corresponding point in the other, then they are perspective, that is, the lines joining corresponding points all pass through a common point, and form a flat pencil.
For let A, B, C, D ... be points in the one, and A′, B′, C′, D′ ... the corresponding points in the other row, and let A be made to coincide with its corresponding point A′. Let S be the point where the lines BB′ and CC′ meet, and let us join S to the point D in the first row. This line will cut the second row in a point D″, so that A, B, C, D are projected from S into the points A, B′, C′, D″. The cross-ratio (AB, CD) is therefore equal to (AB′, C′D″), and by hypothesis it is equal to (A′B′, C′D′). Hence (A′B′, C′D″) = (A′B′, C′D′), that is, D″ is the same point as D′.
§ 31. If two projected flat pencils in the same plane are in perspective, then the intersections of corresponding lines form a row, and the line joining the two centres as a line in the first pencil corresponds to the same line as a line in the second. And conversely,
If two projective pencils in the same plane, but with different centres, have one line in the one coincident with its corresponding line in the other, then the two pencils are perspective, that is, the intersection of corresponding lines lie in a line.
The proof is the same as in § 30.
§ 32. If two projective flat pencils in the same point (pencil in space), but not in the same plane, are perspective, then the planes joining corresponding rays all pass through a line (they form an axial pencil), and the line common to the two pencils (in which their planes intersect) corresponds to itself. And conversely:—
If two flat pencils which have a common centre, but do not lie in a common plane, are placed so that one ray in the one coincides with its corresponding ray in the other, then they are perspective, that is, the planes joining corresponding lines all pass through a line.
§ 33. If two projective axial pencils are perspective, then the intersection of corresponding planes lie in a plane, and the plane common to the two pencils (in which the two axes lie) corresponds to itself. And conversely:—
If two projective axial pencils are placed in such a position that a plane in the one coincides with its corresponding plane, then the two pencils are perspective, that is, corresponding planes meet in lines which lie in a plane.
The proof again is the same as in § 30.
§ 34. These theorems relating to perspective position become illusory if the projective rows of pencils have a common base. We then have:—
In two projective rows on the same line—and also in two projective and concentric flat pencils in the same plane, or in two projective axial pencils with a common axis—every element in the one coincides with its corresponding element in the other as soon as three elements in the one coincide with their corresponding elements in the other.
Proof(in case of two rows).—Between four elements A, B, C, D and their corresponding elements A′, B′, C′, D′ exists the relation (ABCD) = (A′B′C′D′). If now A′, B′, C′ coincide respectively with A, B, C, we get (AB, CD) = (AB, CD′), hence D and D′ coincide.
The last theorem may also be stated thus:—
In two projective rows or pencils, which have a common base but are not identical, not more than two elements in the one can coincide with their corresponding elements in the other.
Thus two projective rows on the same line cannot have more than two pairs of coincident points unless every point coincides with its corresponding point.
It is easy to construct two projective rows on the same line, which have two pairs of corresponding points coincident. Let the points A, B, C as points belonging to the one row correspond to A, B, and C′ as points in the second. Then A and B coincide with their corresponding points, but C does not. It is, however, not necessary that two such rows have twice a point coincident with its corresponding point; it is possible that this happens only once or not at all. Of this we shall see examples later.
§ 35. If two projective rows or pencils are in perspective position, we know at once which element in one corresponds to any given element in the other. If p and q (fig. 9) are two projective rows, so that K corresponds to itself, and if we know that to A and B in p correspond A′ and B′ in q, then the point S, where AA′ meets BB′, is the centre of projection, and hence, in order to find the point C′ corresponding to C, we have only to join C to S; the point C′, where this line cuts q, is the point required.
If two flat pencils, S1and S2, in a plane are perspective (fig. 10), we need only to know two pairs, a, a′ and b, b′, of corresponding rays in order to find the axis s of projection. This being known, a ray c′ in S2, corresponding to a given ray c in S1, is found by joining S2to the point where c cuts the axis s.
A similar construction holds in the other cases of perspective figures.
On this depends the solution of the following general problem.
§ 36. Three pairs of corresponding elements in two projective rows or pencils being given, to determine for any element in one the corresponding element in the other.
We solve this in the two cases of two projective rows and of two projective flat pencils in a plane.
ProblemI.—Let A, B, C be three points in a row s, A′, B′, C′ the corresponding points in a projective row s′, both being in a plane; it is required to find for any point D in s the corresponding point D′ in s′.
ProblemII.—Let a, b, c be three rays in a pencil S, a′, b′, c′ the corresponding rays in a projective pencil S′, both being in the same plane; it is required to find for any ray d in S the corresponding ray d′ in S′.
The solution is made to depend on the construction of an auxiliary row or pencil which is perspective to both the given ones. This is found as follows:—
Solution of ProblemI.—On the line joining two corresponding points, say AA′ (fig. 11), take any two points, S and S′, as centres of auxiliary pencils. Join the intersection B1of SB and S′B′ to the intersection C1of SC and S′C′ by the line s1. Then a row on s1will be perspective to s with S as centre of projection, and to s′ with S′ as centre. To find now the point D′ on s′ corresponding to a point D on s we have only to determine the point D1, where the line SD cuts s1, and to draw S′D1; the point where this line cuts s′ will be the required point D′.
Proof.—The rows s and s′ are both perspective to the row s1, hence they are projective to one another. To A, B, C, D on s correspond A1, B1, C1, D1on s1, and to these correspond A′, B′, C′, D′ on s′; so that D and D′ are corresponding points as required.
Solution of ProblemII.—Through the intersection A of two corresponding rays a and a′ (fig. 12), take two lines, s and s′, as bases of auxiliary rows. Let S1be the point where the line b1, which joins B and B′, cuts the line c1, which joins C and C′. Then a pencil S1will be perspective to S with s as axis of projection. To find the ray d′ in S′ corresponding to a given ray d in S, cut d by s at D; project this point from S1to D′ on s′ and join D′ to S′. This will be the required ray.
Proof.—That the pencil S1is perspective to S and also to S′ follows from construction. To the lines a1, b1, c1, d1in S1correspond the lines a, b, c, d in S and the lines a′, b′, c′, d′ in S′, so that d and d′ are corresponding rays.
In the first solution the two centres, S, S′, areanytwo points on a line joining any two corresponding points, so that the solution of the problem allows of a great many different constructions.But whatever construction be used, the pointD′,corresponding toD,must be always the same, according to the theorem in § 29. This gives rise to a number of theorems, into which, however, we shall not enter. The same remarks hold for the second problem.
§ 37.Homological Triangles.—As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem.
Theorem.—If ABC and A′B′C′ (fig. 13) be two triangles, such that the lines AA′, BB′, CC′ meet in a point S, then the intersections of BC and B′C′, of CA and C′A′, and of AB and A′B′ will lie in a line. Such triangles are said to be homological, or in perspective. The triangles are “co-axial” in virtue of the property that the meets of corresponding sides are collinear and copolar, since the lines joining corresponding vertices are concurrent.
Proof.—Let a, b, c denote the lines AA′, BB′, CC′, which meet at S. Then these may be taken as bases of projective rows, so that A, A′, S on a correspond to B, B′, S on b, and to C, C′, S on c. As the point S is common to all, any two of these rows will be perspective.
and if the line S1S2cuts a in A1, and b in B1, and c in C1, then A1, B1will be corresponding points in a and b, both corresponding to C1in c. But a and b are perspective, therefore the line A1B1, that is S1S2, joining corresponding points must pass through the centre of projection S3of a and b. In other words, S1, S2, S3lie in a line. This is Desargues’ celebrated theorem if we state it thus:—
Theorem of Desargues.—If each of two triangles has one vertex on each of three concurrent lines, then the intersections of corresponding sides lie in a line, those sides being called corresponding which are opposite to vertices on the same line.
The converse theorem holds also, viz.
Theorem.—If the sides of one triangle meet those of another in three points which lie in a line, then the vertices lie on three lines which meet in a point.
The proof is almost the same as before.
§ 38.Metrical Relations between Projective Rows.—Every row contains one point which is distinguished from all others, viz. the point at infinity. In two projective rows, to the point I at infinity in one corresponds a point I′ in the other, and to the point J′ at infinity in the second corresponds a point J in the first. The points I′ and J are in general finite. If now A and B are any two points in the one, A′, B′ the corresponding points in the other row, then
(AB, JI) = (A′B′, J′I′),
or
AJ/JB : AI/IB = A′J′/J′B′ : A′I′/I′B′.
But, by § 17,
AI/IB = A′J′/J′B′ = −1;
therefore the last equation changes into
AJ · A′I′ = BJ · B′I′,
that is to say—
Theorem.—The product of the distances of any two corresponding points in two projective rows from the points which correspond to the points at infinity in the other is constant, viz. AJ · A′I′ = k. Steiner has called this number k thePower of the correspondence.
[The relation AJ · A′I′ = k shows that if J, I′ be given then the point A′ corresponding to a specified point A is readily found; hence A, A′ generate homographic ranges of which I and J′ correspond to the points at infinity on the ranges. If we take any two origins O, O′, on the ranges and reduce the expression AJ · A′I′ = k to its algebraic equivalent, we derive an equation of the form αxx′ + βx + γx′ + δ = 0. Conversely, if a relation of this nature holds, then points corresponding to solutions in x, x′ form homographic ranges.]
§ 39.Similar Rows.—If the points at infinity in two projective rows correspond so that I′ and J are at infinity, this result loses its meaning. But if A, B, C be any three points in one, A′, B′, C′ the corresponding ones on the other row, we have
(AB, CI) = (A′B′, C′I′),
which reduces to
AC/CB = A′C′/C′B′ or AC/A′C′ = BC/B′C′,
that is, corresponding segments are proportional. Conversely, if corresponding segments are proportional, then to the point at infinity in one corresponds the point at infinity in the other. If we call such rowssimilar, we may state the result thus—
Theorem.—Two projective rows are similar if to the point at infinity in one corresponds the point at infinity in the other, and conversely, if two rows are similar then they are projective, and the points at infinity are corresponding points.
From this the well-known propositions follow:—
Two lines are cut proportionally (in similar rows) by a series of parallels. The rows are perspective, with centre of projection at infinity.
If two similar rows are placed parallel, then the lines joining homologous points pass through a common point.
§ 40. If two flat pencils be projective, then there exists in either, one single pair of lines at right angles to one another, such that the corresponding lines in the other pencil are again at right angles.
To prove this, we place the pencils in perspective position (fig. 14) by making one ray coincident with its corresponding ray. Corresponding rays meet then on a line p. And now we draw the circle which has its centre O on p, and which passes through the centres S and S′ of the two pencils. This circle cuts p in two points H and K. The two pairs of rays, h, k, and h′, k′, joining these points to S and S′ will be pairs of corresponding rays at right angles. The construction gives in general but one circle, but if the line p is the perpendicular bisector of SS′, there exists an infinite number, andto every right angle in the one pencil corresponds a right angle in the other.
Principle of Duality
§ 41. It has been stated in § 1 that not only points, but also planes and lines, are taken as elements out of which figures are built up. We shall now see that the construction of one figure which possesses certain properties gives rise in many cases to the construction of another figure, by replacing, according to definite rules, elements of one kind by those of another. The new figure thus obtained will then possess properties which may be stated as soon as those of the original figure are known.
We obtain thus a principle, known as theprinciple of dualityor ofreciprocity, which enables us to construct to any figure not containing any measurement in its construction areciprocalfigure, as it is called, and to deduce from any theorem areciprocaltheorem, for which no further proof is needed.
It is convenient to print reciprocal propositions on opposite sides of a page broken into two columns, and this plan will occasionally be adopted.
We begin by repeating in this form a few of our former statements:—
Two points determine a line.
Two planes determine a line.
Three points which are not in a line determine a plane.
Three planes which do not pass through a line determine a point.
A line and a point without it determine a plane.
A line and a plane not through it determine a point.
Two lines in a plane determine a point.
Two lines through a point determine a plane.
These propositions show that it will be possible, when any figure is given, to construct a second figure by taking planes instead of points, and points instead of planes, but lines where we had lines.
For instance, if in the first figure we take a plane and three points in it, we have to take in the second figure a point and three planes through it. The three points in the first, together with the three lines joining them two and two, form a triangle; the three planes in the second and their three lines of intersection form a trihedral angle. A triangle and a trihedral angle are therefore reciprocal figures.
Similarly, to any figure in a plane consisting of points and lines will correspond a figure consisting of planes and lines passing through a point S, and hence belonging to the pencil which has S as centre.
The figure reciprocal to four points in space which do not lie in a plane will consist of four planes which do not meet in a point. In this case each figure forms a tetrahedron.
§ 42. As other examples we have the following:—
For the row consists of a line and all the points in it, reciprocal to it therefore will be a line with all planes through it, that is, an axial pencil; and so for the other cases.
This correspondence of reciprocity breaks down, however, if we take figures which contain measurement in their construction. For instance, there is no figure reciprocal to two planes atright angles, because there is no segment in a row which has a magnitude as definite as a right angle.
We add a few examples of reciprocal propositions which are easily proved.
Theorem.—If A, B, C, D are any four points in space, and if the lines AB and CD meet, then all four points lie in a plane, hence also AC and BD, as well as AD and BC, meet.
Theorem.—If α, β, γ, δ are four planes in space, and if the lines αβ and γδ meet, then all four planes lie in a point (pencil), hence also αγ and βδ, as well as αδ and βγ, meet.
Theorem.—If of any number of lines every one meets every other, whilst all do not
lie in a point, then all lie in a plane.
lie in a plane, then all lie in a point(pencil).
§ 43. Reciprocal figures as explained lie both in space of three dimensions. If the one is confined to a plane (is formed of elements which lie in a plane), then the reciprocal figure is confined to a pencil (is formed of elements which pass through a point).
But there is also a more special principle of duality, according to which figures are reciprocal which lie both in a plane or both in a pencil. In the plane we take points and lines as reciprocal elements, for they have this fundamental property in common, that two elements of one kind determine one of the other. In the pencil, on the other hand, lines and planes have to be taken as reciprocal, and here it holds again that two lines or planes determine one plane or line.
Thus, to one plane figure we can construct one reciprocal figure in the plane, and to each one reciprocal figure in a pencil. We mention a few of these. At first we explain a few names:—
A figure consisting of n points in a plane will be called an n-point.
A figure consisting of n lines in a plane will be called an n-side.
A figure consisting of n planes in a pencil will be called an n-flat.
A figure consisting of n lines in a pencil will be called an n-edge.
It will be understood that an n-side is different from a polygon of n sides. The latter has sides of finite length and n vertices, the former has sides all of infinite extension, and every point where two of the sides meet will be a vertex. A similar difference exists between a solid angle and an n-edge or an n-flat. We notice particularly—
A four-point has six sides, of which two and two are opposite, and three diagonal points, which are intersections of opposite sides.
A four-side has six vertices, of which two and two are opposite, and three diagonals, which join opposite vertices.
A four-flat has six edges, of which two and two are opposite, and three diagonal planes, which pass through opposite edges.
A four-edge has six faces, of which two and two are opposite, and three diagonal edges, which are intersections of opposite faces.