In these notes we will outline a proof of Lobachevskiĭ s main equation for the angle of parallelism, namely

Transcription

1 Mathematics 30 Spring Semester, 003 Notes on the angle of parallelism c W. Taylor, 003 In these notes we will outline a proof of Lobachevskiĭ s main equation for the angle of parallelism, namely ( ) Π(y) () tan e Ky. (See Theorem on page 4 below.) The proof will invoke the following assumptions, which we probably will not have the time to prove.. The function Π is differentiable at zero. Thus Π (0) is a well-defined number, clearly < 0. We define K Π (0); for any given plane, K is a (non-negative) constant for that plane. So for the rest of our deduction, we think of K as a constant. This is the K that appears in Equation ().. There are constants R > 0 and D > 0 such that, if ABC is a right triangle with hypotenuse of length c D, then (a) ABC has defect δ < Rc. (b) BC c sin( A) < Rc. 3. lim y 0 Π(y) π/. (This is the first thing proved in Theorem 3. on page 8. It essentially requires Theorem 3.0 on the previous page.) Geometric derivation of a differential equation. In order to derive a differential equation for the angle of parallelism Π(y), we will evaluate its derivative Π (y) through consideration of a certain construction (which takes place in neutral geometry). Let us be given a directed line m and a point F on m. On the perpendicular to m choose points P and P as indicated, () P α E y γ β P α y m F l l Ω

2 so that F P y, F P y + y. Let directed lines l and l be asymptotic to P and P through P and P, respectively. (The common asymptotic direction of l, l and m is denoted Ω in Diagram ().) Thus (3) Π(y) α and Π(y + y) α. Let E be the foot of P on l. It will be left as an exercise to show that E is interior to P P l, and hence that (4) α + β + γ π, where β and γ are as indicated in the diagram. We also know that (5) α + γ + π/ π δ, where δ is the defect of the right triangle P P E. Subtracting Equation (4) from Equation (5), we easily obtain (6) Π(y + y) Π(y) α α β π/ δ. By the transitivity of the asymptotic relation, l is asymptotic to l at P. Since E is the foot of P on l, β must be the angle of parallelism for the segment P E; thus, by Assumption b, for sufficiently small y, we have (7) β Π( P E ) Π( y sin α + Q( y)), where Q( y) < R y. Since Π is differentiable at zero (Assumption ), we have Π(x) Π(0) + xπ (0) + xe(x) (8) π Kx + xe(x), where E(x) is a quantity that approaches zero with x. Applying (8) to the right-hand side of (7), we have β π K( y sin α + Q( y)) + ( y sin α + Q( y))e( y sin α + Q( y)) (9) π K sin α y + y G( y), where G(x) is a quantity that approaches zero with x. Combining Equations (6) and (9) we obtain and so (0) Π(y + y) Π(y) K sin α y + y G( y) δ, Π(y + y) Π(y) y K sin α + G( y) δ y. We now consider the limit of Equation (0) as y approaches 0. By continuity, α Π(y + y) approaches α, and so sin α approaches sin α. We

3 already know that G( y) approaches 0, and it is immediate from Assumption a that δ/ y approaches 0. Therefore () Π (y) We have thus proved Π(y + y) Π(y) lim y 0 y K sin α. Theorem. For any plane P, the angle-of-parallelism function Π is differentiable, and there is a non-negative constant K such that Π satisfies the differential equation 3 () Π (y) K sin(π(y)). It is now our objective to solve this differential equation. More precisely, we will show that Lobachevskiĭ s formula () yields the unique solution of () that satisfies Π(0) π/. Solution of the differential equation (). The trigonometric identity sin θ sin θ cos θ, with θ Π(y)/, converts Equation () to Π (y) K sin Π(y) ( cos Π(y) K cos Π(y) sin Π(y) cos Π(y) Multiplying both sides of this equation by sec Π(y), we see that ( ) ( ) Π(y) Π(y) (3) sec Π (y) K tan. We now make the substitution u(y) tan(π(y)/). It is not hard to see that the left-hand side of the differential equation (3) is du/dy. Therefore, we may proceed to integrate (3) as follows: (4) ) du dy Ku d dy (eky u) e Ky ( du dy + Ku) 0 e Ky u A u A e Ky ( ) Π(y) tan A e Ky, where A in the third line is a constant of integration. Now the value of A is easily determined by the fact (Assumption 3) that we must have Π(0) π/. If we substitute the initial conditions y 0, Π(y) π/ into.

4 4 Equation (4), we easily obtain tan π/ A e 0 A; hence A. Now Lobachevskiĭ s Equation () is immediate from (4): Theorem. For any plane P, there is a non-negative constant K such that the angle of parallelism Π(y) satisfies Lobachevskiĭ s Equation ( ) Π(y) ( ) tan e Ky. The first mention of Equation () in our textbook is in Equation (6. ) on page 9. It has a proof there, based on a long theory of co-ordinates on horocycles. It is worth noting the general shape of Π(x) that is implied by Theorem. Clearly y 0 corresponds to Π(y)/ an angle of tangent, i.e. Π(y)/ π/4; thus Π(0) π/. Then as y, as long as K > 0, the negative exponential function approaches 0, and so Π(y)/ is an angle whose tangent is approaching 0. In other words, in hyperbolic geometry, Π(y) 0 as y. (This is one of the conclusions of Theorem 3. on page 8.) Corollary 3. (Hyperbolic geometry, i.e. K > 0.) Let l be a directed line and m a line. There exists a directed line l that meets m at right angles and is asymptotic to l. Proof. Let A be a point on m, and let l be the unique directed line through A that is parallel to l. If l is perpendicular to m at A, then we may take l to be l and be done. Otherwise, there is a point Q on m such that AQ makes an acute angle α with l. Now let a be such that e Ka tan(α/), i.e., let a (/K) log tan(α/). It follows from Theorem that α is the angle of parallelism for a. Let B be the point on AQ with AB a, and let l be the line perpendicular to m at B. Since α is the angle of parallelism for a, we must l asymptotic to l through B. By transitivity, l is asymptotic to l, and therefore is as desired. The value of K. The theory does not tell us a value of K. Clearly, if K 0 then Π(y) π/ for every y, and the plane is Euclidean. If K 0, then Π(y) < π/ for all y > 0, and the plane is hyperbolic. In other words, neutral geometry does not imply a value of K. In fact, any value of K 0 leads to a consistent theory. (This was the great discovery of Bolyai, Gauss, Lobachevskiĭ and Beltrami.) In fact, concerning planes in the world where we live, there is no experiment that can distinguish the Euclidean case (with K 0) from the non-euclidean case with very small K 0. If, for instance, K is so small that K ( parsec) is less than e.g..000, then no instrument or measurement can distinguish our geometry from Euclidean geometry.

5 5 In Equation () (with K > 0) we encounter a feature that plays no role in Euclidean geometry, namely the dependence of K on the unit of length. This dependence is inherent in the simple fact that one can change the unit of length without changing any angles. (Axiomatically, this means that if all distance are multiplied by a fixed factor, and angles are left alone, the axioms will still hold.) Thus the left-hand side of Equation () which involves only the angle α Π(y) is invariant under any change of unit length. Since () is an equation, the quantity Ka appearing on the righthand side must also be invariant under such a change. A change in unit length will of course change a, and this change must obviously result in an inversely proportional change in K. If one likes, one can think of K as implying a new unit of length, which we might (temporarily) call the Lob after Lobachevskiĭ K Lobs equal one ordinary unit of length. Now, if L is the length in Lobs of segment P F of diagram, then its length a in ordinary units is a L/K; thus tan α e Ka e L, and K is not apparent in the relationship between L and α. For the rest of these notes, we assume that we are working in a particular plane, and hence that K can be considered fixed. Most of our conclusions will involve the constant K. A slight generalization of Lobachevskiĭ s Equation. Theorem was the special case of Theorem 4 (just below) that has ψ π/. Here we give the easy derivation for the general case. A φ (5) a B Q ψ n l m Ω Theorem 4. Let l and m be asymptotic directed lines, A a point on l and B a point on m. Let ψ be the measure of BA m, let φ be the measure of AB l, and let a be the length of segment AB (as diagrammed by the solid

6 6 lines in (5)). Then a, φ and ψ are related by (6) tan φ tan ψ e Ka. Proof. The Euclidean version is easily proved (e.g. using co-ordinate geometry), and so we will present only the hyperbolic version. Let Q denote a point on AB, and n the directed line perpendicular to AB at Q and directed to the same side of AB as are l and m. By Corollary 3, we may choose Q so that n is asymptotic to l and m. There are now three cases: (i) Q between A and B, (ii) A between B and Q, and (iii) B between A and Q. We prove only case (i), leaving the other two to the reader. (In fact cases (ii) and (iii) have similar proofs, so the reader need only check case (ii).) In diagram (5), one may make two applications of Theorem : tan φ e K AQ ; tan ψ e K QB. Multiplying these equations together yields tan φ tan ψ e K( AQ + QB ) e Ka, where the last equation used the fact that Q is between A and B. A development of hyperbolic trigonometry from Lobachevskiĭ s Equation. Consider a right triangle ABC, with right angle at C, angles of measure α and β at A and B, respectively, and side-lengths a, b and c, as indicated in Diagram (7) below. (7) B a C β c b α A We shall prove that the various quantities in this right triangle satisfy four laws: the hyperbolic counterpart to the cosine as a ratio: (8) tanh Ka cos β tanh Kc, the hyperbolic counterpart of the law of sines, (9) sin α sinh Ka sinh Kc,

7 7 the hyperbolic counterpart of the law of Pythagoras, (0) cosh Kc cosh Ka cosh Kb, and and a new law which has no serious counterpart in Euclidean geometry: () tan α tan β cosh Kc. In Equations (8 ) cosh x and sinh x are the hyperbolic functions cosh x ex + e x ; sinh x ex e x. In fact, we prove (8 ) in reverse order, so we begin with (). The law tan α tan β / cosh Kc. As depicted in Diagram () just below, let us augment our picture of ABC with an auxiliary directed line l that is asymptotic to the ray CA through B; we let θ be the measure of the angle BA l. B β θ l () a c C b α π α A Applying Equation (6) to the right side of this diagram, we immediately have which is easily transformed to tan θ tan π α e Kc, (3) tan θ e Kc tan α. For a second equation involving θ and α, we augment the previous picture by reflecting segments AC and AB in the vertical line BC, as indicated in

8 8 the following diagram: B (4) c β β θ l A α C A An application of Equation (6) to Diagram (4) yields tan α ( tan β + θ ) e Kc, which immediately leads to ( tan β + θ ) (5) e Kc tan(α/). At this point the reader is asked to recall the addition formula for the tangent function: tan u + tan v tan(u + v) tan u tan v. It will be used twice in the next calculation. We eliminate θ from Equations (3) and (5) with the following calculation: tan(β + θ/) tan(θ/) tan β + tan(θ/) tan(β + θ/) e Kc tan(α/) e Kc tan(α/) + e Kc e Kc [tan(α/)] tan(α/) + e Kc e Kc tan (α/) + e Kc tan(α/) cosh Kc tan α. We now immediately obtain ( ) tan α tan β cosh Kc.

9 The law of Pythagoras, hyperbolic version. Let F denote the foot of C on AB. We know (e.g. by the Exterior Angle Theorem) that F is between A and B, as in the following diagram. 9 B a C β F c δ γ α b A Thus the angle measures denoted γ and δ in the diagram add to a right angle. Therefore, by well-known properties of the tangent function, (6) tan γ tan δ. Three applications of () and one application of (6) yield (0 ) cosh Ka cosh Kb tan β tan δ tan β tan α cosh Kc. tan γ tan α Note that (0) is derived in our textbook as (6.3-6) on page 97. Connections with the original theorem of Pythagoras. The power series for the hyperbolic cosine is cosh x + x + 4 x4 + + (n)! xn +. To see what (0) says in the Euclidean case, we apply this power series to each instance of the cosh that appears in (0), thereby obtaining + K c + ( + K a + ) ( + K b + ) + K a + K b +, where the dots represent terms that have fourth and higher powers of K. Thus c a + b +, where the dots now represent terms that are quadratic or higher in K. Thus the ordinary law of Pythagoras, c a + b, may be seen as a limiting form of (0) as K 0 (the Euclidean case).

10 0 Review of some ordinary trigonometry: (7) sin λ tan λ + tan λ ; (8) (9) cos λ tan λ + tan λ ; tan λ tan λ tan λ. Notice now, for instance, that Lobachevskiĭ s Equation for the angle of parallelism can now be presented as tan Π(a) tan Π(a)/ tan Π(a)/ (30) e Ka e Ka e Ka e Ka sinh Ka. Hyperbolic version of the law of sines (9). In Diagram (), we note that β + θ is the angle of parallelism for the length a. Thus by (30), we have (3) tan(β + θ) sinh Ka. From (3) and (9) we immediately obtain (3) tan θ tan(θ/) tan (θ/) e Kc tan(α/) e Kc tan (α/). Now comes a long calculation; to simplify, in this calculation we will abbreviate tan(α/) by t and e Kc by f. Our strategy is this: in (3) we already have one calculation of tan(β+θ). Thus we will make a second calculation of tan(β +θ) using () and (3), and then equate the two values of tan(β +θ).

11 Here goes: tan(β + θ) tan β + tan θ tan β tan θ cosh Kc tan α + ft f t ft cosh Kc tan α f t f t + f t + f t + f t ft f t ft f t f( t )( f t ) + ft ( + f ) t( + f )( f t ) f( t )t f f t f t + f t + t + f t 4 + f t f f 4 t ( + t )( + f t ) ( f )( + f t ) f t + t f (33) + t t f f sin α sinh Kc (where the last line used (7)). Equating (3) and (33), we then have (9 ) sinh Ka sin α sin α sinh Ka sinh Kc. sinh Kc Hyperbolic version of the ratio for the cosine (8). immediate from (9) that tan sin α α sin α sinh Ka sinh Kc sinh Ka. From this equation and Equation (), we calculate tan β tan α cosh Kc sinh Kc sinh Ka sinh Ka It is almost cosh Kc

12 Therefore sec β tan β + sinh Kc sinh Ka + sinh Ka cosh Kc sinh Ka cosh Kc sinh Kc sinh Ka + sinh Ka ( + sinh Kc) sinh Ka cosh Kc (sinh Ka + ) sinh Kc sinh Ka cosh Kc cosh KA sinh Kc sinh Ka cosh Kc tanh Kc tanh Ka. From here it follows almost immediately that (8 ) cos β tanh Ka tanh Kc. Comment on Equations (8) and (9). Since sinh(x) and tanh(x) have power series that begin sinh(x) x + 6 x3 + tanh(x) x 3 x3 +, it follows immediately from Equations (8) and (9) that for small K, (8-Euclid) cos β Kb Kc b c ; (9-Euclid) sin α Ka Kc a c. Thus in the Euclidean limit, our formulas yield the well known expression for the trigonometric functions as the appropriate side-ratios in a right triangle. Moreover, from (9) we may derive a formula that is more traditionally known as the law of sines. Suppose that we have an arbitrary triangle ABC, as diagrammed in (34): (34) B β F a h α C b A

13 Let F be the foot of the perpendicular from C to AB, and let h be the length of segment CF, as noted in Diagram (34). Applying (9) to the two small triangles in Diagram (34), we see that sin α sinh Ka sinh Kh sin β sinh Kb. The same pattern applies also to c and γ as one might check in detail and so in general, for any triangle, we have (35) sin α sinh Ka sin β sinh Kb sin γ sinh Kc. Equation (35) is known as the general law of sines. In our textbook, Equations (8) and (9) appear as Equations (6.3 4b) and (6.3 4a), respectively, on page 96. The general law of sines (35) may be seen on page Side-Angle-Side reconsidered. As we easily recall, the SAS axiom states, for triangles A B C and A B C, that if segments A B and B C are congruent, respectively, to A B and B C, and if A B C is congruent to A B C, then all six corresponding parts (three segments and three angles) of the two triangles are congruent. In stating this axiom, it is implicitly understood that we are speaking of triangles A B C and A B C that lie in the same plane P. This is because, in fact, the axioms refer to what happens within a single plane. This is a point that we almost always ignore mostly because for much of our work the issue does not arise, but also because (historically) we were stuck in the idea that geometry is merely a description of the one and only space in which we live. In discussing hyperbolic geometry, however, we are forced to consider the possibility that more than one sort of plane might make sense. If we, however cautiously, entertain the idea that hyperbolic geometry is consistent (something we shall later prove), then we are tacitly assuming that there are at least two planes possible, a Euclidean one and a non-euclidean one. Now, faced with the existence of more than one plane, we might ask, Does SAS hold between planes? More precisely, suppose that A B C and A B C satisfy the aforementioned hypotheses of SAS, but with the points taken from two different planes P and P A, B, C P and A, B, C P ; can we then conclude that the two triangles are congruent? In fact, we already know enough about geometry to give an easy answer of No! to this question. If P is Euclidean and P is hyperbolic, then no triangle of P is congruent to any triangle of P. The simplest reason here is simply that all triangles in P have defect zero, and all triangles in P have non-zero defect. With slightly more work, we may also see that the answer is no if P and P are hyperbolic planes with distinct values of the constant K. Suppose for instance that K is the value of K for P, and K is the value of K for

14 4 P, with K K. Suppose we had congruent right triangles A B C and A B C (right angles at C and C ), lying in the two planes P and P. From Equation () it is immediate that cosh K c tan α tan β cosh K c, where c is the common length of segments A B and A B, α is the common measure of the angles at A and A, and β is the common measure of the angles at B and B. The absurdity of this equation for K K shows that no right triangle of P can be congruent to a triangle of P. The result for general triangles is a short step away; we omit the details. With that background, we now state and prove Theorem 5. (SAS for triangles in planes of the same K). Suppose that P and P are hyperbolic planes with the same value K valid in both. If A B C P and A B C P satisfy the hypotheses of SAS, namely that A B and B C are congruent, respectively, to A B and B C, and that A B C is congruent to A B C, then the two triangles are congruent. Proof. Suppose without loss of generality that A B B C. Let F be the foot of the perpendicular from C to A B, and perform a similar construction of F in P. Clearly F lies between A and B and ray C F is interior to the angle at C. We first prove that F B C F B C. Applying Equation (8) to F B C and F B C, and using the given congruences, yields tanh K B F cos( A B C ) tanh K B C cos( A B C ) tanh K B C tanh K B F. Since the tanh function is one-one, we now have (36) B F B F. A similar calculation using (9) yields (37) C F C F. To complete the congruence of F B C and F B C, we apply (9) to yield sin( B C F ) sinh K B F sinh K B C sinh K B F sinh K B C sin( B C F ). Since the sine function is one-one on the interval [0, π], we now have (38) B C F B C F.

15 We shall now prove that the right triangles A F C and A F C are congruent. We have congruence for the leg C F in Equation (37). Congruence for the leg A F may be found by subtracting (36) from the given congruence A B A B to yield (39) A F A F. The parts of A F C for which we still need to prove a congruence are the hypotenuse C A and the two acute angles F A C and F C A. For the hypotenuse, we apply (0) to the right triangles A F C and A F C, using (37) and (39), to yield cosh K C A cosh K C F cosh K F A cosh K C F cosh K F A cosh K C A. Since cosh x is one-to-one for positive x, we have (40) C A C A. Now that we have all the corresponding sides of A F C and A F C congruent, their two acute angle can be handled with Equation (9), much as we did in proving (37); we skip the details. Finally, since we began with F interior to the angle at C (and correspondingly in the plane P ), we have (4) and (4) B A B F + F A B F + F A B A C rad B C F rad + F C A rad B C F rad + F C A rad C rad. Now the congruence of F B C and F B C is contained in Equations (40), (4) and (4). One should especially notice that although we made use of the important Equations (8), (9) and (0), the exact form of these equations was immaterial to the argument. In this context, the significant feature of these equations is simply that there is some definite formulas of this type (i.e. relating various angles and sides) are true generally in the geometry, not depending on which particular plane is under consideration. Another way to say it is this: on page 5 our text defines a function φ(a, b, γ) which tells the length of the third side of a triangle, given sides a and b and the angle γ between those sides. It is pointed out there that the SAS axiom guarantees the existence of a φ(a, b, γ) that is valid for all triangles. Theorem 5 says that, moreover, one φ(a, b, γ) is valid for all triangles in all planes that have a fixed K-value. 5

16 6 The isomorphism of two hyperbolic planes. Suppose that H and H are hyperbolic planes that share one value of K. We will use Theorem 5 to prove that H and H are isomorphic. A function F :H H is called an isomorphism of planes if (i) F is a bijection. (ii) F preserves distances; i.e. if A, B H, then AB F (A)F (B). (iii) F preserves angles; i.e. if A, B and C are non-collinear points of H, then ABC rad F (A)F (B)F (C) rad It can be proved that under these hypotheses, we also have (iv) F preserves halfplanes. (v) F preserves rays. (vi) F preserves segments. (vii) F preserves asymptotically parallel rays. If there exists an isomorphism F : H H, then H and H are said to be isomorphic. Remarks like those prior to Theorem 5 make it clear that hyperbolic planes with different K-values are not isomorphic. In the remaining cases, we do have isomorphism: Theorem 6. (N. I. Lobachevskiĭ) Any two hyperbolic planes with the same value of K are isomorphic. Proof. Let H and H be such planes. We proceed to construct the desired F. Let S be a half-plane of H, let r be a ray in H that lies in the edge of S, and let A be the vertex of r. Similarly let S be a half-plane of H, let r be a ray in H that lies in the edge of S, and let A be the vertex of r. To begin the definition of F, we let F (A ) A. Now suppose X A ; we will define F (X) using a sort of polar coordinate system like those that are found in our textbook. Consider the ray A X. Case A X lies in the half-plane S. By Axiom 5, there is a unique ray s emanating from A in the half-plane S such that angles r A X and r s have the same measure. There is a unique point Y on s such that A Y A X. We then define F (X) to be this Y. Case A X lies in the half-plane opposite to S. Similar to Case, except that we must locate s in the half-plane opposite to S. Case 3 A X lies in the edge of S. (I.e. A X is either the ray r or its opposite ray.) We leave it to the reader to see how F (X) should be

17 defined in this case. We may now consider that F (X) has been defined for every X. We leave the proof that F is a bijection as an exercise. It remains to prove that F preserves both distances and angles. We first observe that F preserves angles at A ; this is almost immediate from the definition, and we omit the details. Next, it is not hard to see that F preserves distances along any line that passes through A ; we omit the details. We next prove that F preserves all distances, i.e. that B C B C whenever B, C H and B, C are defined as B F (B ) and C F (C ). By the previous paragraph, we need only prove this in the case that A, B and C are non-collinear, i.e. that they form the vertices of a triangle. It is now obvious that A B C and A B C satisfy the hypotheses of SAS: A B A B and A C A C because F preserves distances along lines through A. Moreover, as we said before, angles at A are preserved by F, and so for these triangles, A A. Therefore by Theorem 5, we have B C B C. Thus all distances are preserved. The proof that all angles are preserved will be left as an exercise for the reader. Here I should like to point out and acknowledge that Theorem 6 illustrates a trend in twentieth-century mathematics, that of classifying mathematical objects. Here the (potentially wild) collection of all planes is tamed by being classified according to a single real parameter, namely K. It s like classifying shoes up to size, but much more thorough. Shoes of the same size may still have interesting differences between them. But planes of the same K are the same according to all geometrically interesting properties. In fact the classification can be made even simpler, as follows. In fact, if K is the relevant constant for a hyperbolic plane P, then P can be changed to a plane with constant K, simply by changing all lengths by a factor K /K. Therefore, in fact, all hyperbolic planes are isomorphic to one another if we are allowed to re-scale lengths. Now here is where the real mystery or challenge begins. If there is really only one hyperbolic plane H, then anything we discover (such as any equation in this report) can be viewed as part of a description of H. If our description is sufficiently complete, we might expect to be able to actually describe H to the point of saying what H is. Slowly this realization dawned on the early workers in the field. It turns out for a full construction of H it is more efficient to start over in a slightly different way, but these facts will not be forgotten. 7