Commentarii Mathematici Helvetici

A Poncelet theorem in space.

through Po, we let Lx be the other tangent line through Po meeting C in Pu L2 the other tangent line through Pi meeting C" in P2, and so forth.Let /7(po,Lo) dénote the configuration of lines {L,}l=0, i, obtained in this way.It is clear that any closed polygon inscribed in C and circumscribed about C which contains Po as vertex must be the configuration 17(Pn 10>.Consequently, our question is équivalent to asking when n(PotLo) is finitei.e., when does Ln Lo for some ni Although it seems difficult to write down an explicit gênerai answer to this problem,t Poncelet did succeed in proving the beautiful and remarkable Figure 1 THEOREM.For any pair of ellipses as above, there exists a closed polygon inscribed in C and circumscribed about C if, and only if, there exist infinitely many such polygons, one with a vertex at any given point Pe C".
In other words, The condition that the configuration n(PofLo) be finite is independent of the starting point (Po, Lo).
As mentioned before, the original proof by Poncelet was synthetic, and we shall follow the lead of Jacobi and give an analytic discussion of the question based on elliptic functions.Consequently, we consider the problem in the projective plane P2 over the complex numbers.Let x [x0, Xi,x2] be homogeneous coordinates and consider a plane conic C given by a quadratic équation where the coefficient matrix Q (qtJ) is symmetric.The condition that the algebraic curve C be smooth is that the matrix Q be nonsingular.In this case C is a complex submanifold of P2, and therefore defines a Riemann surface.
Since C has degree 2, every line L in P2 meets this conic in two points counting multiplicities.Taking a fixed point PoeC and line Lo not containing Po, the stereographic projection mapping P » Q establishes a biholomorphic map- ping C~> P1 between C and the line Lo (cf.Fig. 2).
At a point PeC with coordinates x the projective tangent line Tp(C) to C at Figure 2 P is given by the équation Consequently, in terms of the dual coordinates £ [£0, eu £2] on the dual projective space P2* of Unes in P2, the Gauss mapping P » TP(C) is given by £= Qx.It follows that the dwa/ curve C* c P2* of tangent lines to C is again a smooth conic with équation (Q"1^, £) 0, where Q"1 is the inverse matrix to Q. Dualizing Fig. 2, we find that through each point PeP2 there pass two tangent lines to C, again Figure 3 counting multiplicities, and C* may be rationally parametrized by the mapping P-* T, where P varies on a fixed tangent Une To and T is the other tangent to C passing through P as depicted in Fig. 3. Now let C be another conic in the plane P2 and assume that C and C are nowhere tangent, so that they meet transversely at four points.Then the same will be true of the dual curves C* and C* in P2*, and in the product C'x C^-P^P1 we consider the incidence correspondence E ={(P, L):PeL) of points PeC and tangent lines L to C such that PeL.It is clear that E is the basic variety underlying the Poncelet construction.E is an algebraic curve in PxxPx; the transversality of C and C assures that for every point (P, T)eE at least one of the coordinate axes {P}xC* and C'x{T\ through (P, T) meets E in two distinct points, from which it follows that E is nonsingular.The adjunction formula for curves on an algebraic surface then gives that the genus of E is one.
Thèse facts will also corne out directly during the proof of the Poncelet theorem.Namely, the construction in Fig. 1 suggests considering the pair of involutions i.E-^E and ï\E-*E defined respectively by i(P,L) (P,L') and i'(P,L) (P',L) where P' is the residual intersection of L with C and V is the other tangent to C through P as depicted in Fig. 4. In terms of the notation in Fig. 1, i(Pl9 U) (Pl9 Ll+1), i'(Pn Ll+1) (Pl+1, L,+1), so that the composition j=ï°i satisfies /(P,, L,) (P,+i, Ll+1).Consequently, /n(P0, Lo) (Pn, Ln), from which we conclude that: The configuration n(PojLo) will be finite exactly when jn(P0, Lo) (POy Lo) for some integer n.
Figure 4 Evidently now our problem is équivalent to looking for flxed points of the automorphism jn on the curve E, to which we now turn.Taking the quotient of E by the first involution i is the same as considering the map E -» C* given by (P,L)->L.This realizes E as a 2-sheeted covering of the Riemann sphère P1.
The branch points correspond to the fîxed points of i, and thèse occur for the Unes L which are tangent to C as well as to C, i.e., to the four distinct points in C'*nC*.It is now clear that E is a nonsingular Riemann surface with Euler characteristic x(JE) 2x(P1)-4 0, and, consequently, E has genus 1.
The basic deep fact underlying Poncelet is that there exists on E a nonvanishing holomorphic differential <o such that the map Jpo gives an analytic isomorphism of Riemann surfaces E -> CI A where A is the period lattice of o).Now the involutions i and i' on E are induced by automorphisms ï and V on the universal covering C of E ; we can write ï{u) Then, since i2 0, we hâve r2(M) a2M + (a + l)T M + A for some À 6 A. Thus a ±l, and in fact a -l since otherwise i would hâve no fixed points in contradiction to Fig. 5.It follows that Jn(u)= ii + n(r'-T), and hence jn(P,L) (P,L) if, and only if, n(r'-r)eA.This condition is clearly independent of (P, L) E and so the Poncelet theorem is proved.We want to discuss this condition in more détail.Given our plane conics C and C, by linear algebra we may change coordinates so that c= where /3 [j30> Pi, £2] is a point in P2^).The resulting map P2^)-^{pairs of conics} is generally finite, and those pairs of conics for which n{r'-T)eA détermine an algebraic curve Vn^Pf&).In fact, it is clear that Vn is locally determined by one analytic condition on j8, and also that Vn is globally of an algebraic character.

Consequently:
The Poncelet condition holds for pairs of conics corresponding to a countable set of curves in the parameter space Pfp).
2. Poncelet theorem for quadrics in space Following the Poncelet theorem in the plane it seems natural to look for polyhedra in R3 which are inscribed in one surface and circumscribed about another.At first glance this doesn't seem to work, since there are in gênerai oo1 tangent planes to an algebraic surface passing through a point in space.However, upon closer inspection it is possible to generate polyhedra from a pair of quadric surfaces, and the question of whether or not we obtain a finite figure turns out to again repose on elementary properties of elliptic curves.We now describe how this goes.
Let x [x0, JCi, x2, x3] be homogeneous coordinates in complex projective space P3.A quadric surface S is defined by (Qx, x) ».J where Q (ql]) is a symmetric matrix.S is nonsingular exactly when detQ^O, and in this case S is a complex submanifold in P3.As before, for PeS with coordinates x the projective tangent plane Tp(S) to S at P is given by the équation (Qx, y) 0. Denoting by £ [£o> £1, &> £3] the dual coordinates in the dual projective space P3* of planes {£3=o ê*t 0} in P3, the Gauss mapping of S to îts dual surface S* of tangent planes to S is given by £ Qx.Consequently, S* is a nonsingular quadric surface with équation (Q1^, £) 0.
If S and S' are two quadric surfaces meeting transversely, then the same is true of the dual surfaces S* and S'*. (5)In this case the intersections c=snsf, E s*nsf* are nonsingular algebraic curves.Geometrically, E is the set of bitangent planes to the pair of surfaces S and Sr.
Especially noteworthy about a smooth quadric surface S are the lines contained in it.Thèse may be described in the following three steps: (i) Since S has degree 2, any line meeting S in three or more points must lie entirely in the surface.Therefore, if 0 is any point in the intersection S H TP(S) of S with its tangent plane at P, the line PQ will lie in this intersection.It follows that: The intersection S H Tp(S) consists of two straight Unes; any line lying in S and passing through P must be one of these.(6) Figure 6 (ii) To describe ail Unes lying in S, we pick one such Lo, and call any line L an A-line if it is either equal to or disjoint from Lo and a B-line if L meets Lo in one point.In this case L and Lo span a 2-plane LaL0.
If two Unes L, VV Lo meet in a point, then the plane they span must meet Lo in a point, and so one of the two is an A-line and the other a B-line.(7)Conversely, if L is an A-line and V a B-line, then the plane LoaL' will meet L in a point which must therefore be a point on L H L'. In summary, The Unes on S fall into two disjoint families, the A-lines and B-lines.Each A-line L meets each B-line L' once at a point P LC\Lf whose tangent plane Tp(S) is LaL'.Two distinct A-lines or B-lines fail to meet.
Since any B-line meets Lo once, it follows that the A-lines and B-lines are each parametrized by P1.In this way S is a doubly ruled surface, Le., it is the surface traced out by oo1 Unes in two distinct ways.This also shows that, as a complex manifold, S =* P1 x P1.
(iii) For any line L c P3 the set of ail planes containing L is a line L* in the dual projective space P3*.The line L lies on S if, and only if, L* lies on 5*.If this is not the case, then L* will meet S* in two points.Thus: Through a Une LcP3 not lying on S there pass exactly two tangent planes to S. Any plane T containing a Une Le s is tangent to S somewhere along L. Now we can describe the analogue of the Poncelet construction relative to a pair of smooth quadric surfaces S and S'intersecting transversely in p\( 8) Letting E S*H S'* be the curve of bitangent planes, each plane Te E will meet SUS' in a figure of the sort consisting of the pair of Unes LA, LB on S, together with a pair of lines LA, Lb on S'.This is by (i) above.The four lines are distinct since S and S' meet transversely, and P LAniB, P' L'AnL'B are the points where T is tangent to S, S', respectively.According to (iii) there will be one plane f other than T passing through LA and tangent to S', and by (iii) again this plane will be tangent to S somewhere along LA.The picture of T, together with T, is something like that shown in Fig. 8.There we are letting It is important to note that by (ii) the line L'B meets LA LA in the same point as LA, etc., so that when flattened out the shaded régions form a configuration as shown in Fig. 9.We set f=iA(T), and in this way define four involutions 1 We will comment on the construction in IR3 at the end of the paper.Figure 8 ïaî te, i'a, *b on the curve E We may thmk of thèse involutions as reflections in the four sides of the shaded quadrilatéral m Fig 7 If we begm with a fixed bitangent plane To e E and successively apply ail thèse reflections we generate a polyhedral figure I7(T0) What makes this configuration so remarkable îs that the polyhedron îs both inscribed in and circumscnbed about the pair of quadnc surfaces S and S' More precisely, The planes Ten(T0) are ail tangent to both S and S\ and thc tertices of II(T{)) are ail points lying on S H S'(9) The edges o//7(T0) are Unes lying alternately on S and S' Figure 9 Naturally we now ask whether or not our configuration may be finite.By analogy with Poncelet in the plane we shall prove the THEOREM.For S and S' smooth quadric surfaces meeting transversely in P3, there exists a finite polyhedron both inscribing and circumscribing S and S' if, and only if, there exist infinitely many such.
In other words, the question of whether or not /7(T0) is finite is independent of the initial bitangent plane To.Contrary to the plane case, it is not immediately apparent that there exist S and S' generating finite polyhedra, especially since at first glance this seems to impose three conditions on the 3-parameter famjly of pairs S, S'.This important point will be discussed following the proof of the theorem.
The proof is essentially the same as before.The four involutions generate a subgroup G <= Aut (E) of the automorphism group of the Riemann surface E, and we are asking whether or not the orbit G {To} is finite.If we apply the adjunction formula(10) to each of the two inclusions Ec S*c:p3*, we fincj that the canonical bundle of E is trivial.Hence the genus of E is 1 and the argument proceeds as in the plane case.
For later use, it is of interest to prove this in a more elementary fashion.
Instead of E we consider the curve C=SDS' and the mapping ttaiC-^P1 sending a point Pe C to the A-line L on S which contains P. Since L meets S'in two points, 7rA is a 2-sheeted covering, and branching occurs exactly over those A-Unes La S which are tangent to S'.To show that the genus of C is 1, it will suffice to prove that there are four such branch points.Dualizing, we then obtain that the genus of E is 1.Now the other projection irB : C-» P1 given by sending Pe C to the J3-line on S containing it is again a 2-sheeted covering, having by symmetry the same number of branch points as tta.Thus, it will suffice to prove that there are exactly eight lines on S which are tangent to S'.
If a Une LcSis tangent to S' at P, then PeSDS' and Tp(Sf)eS* by the second statement in (iii) above.Conversely, if PeSDS' and Tp(S') TP>(S) for some P'eS, then the line WP lies in S and is tangent to S'.If PeSDS' has coordinates x, then Tp(Sr)eS* if, and only if, Q'x Qy for some y satisfying (Oy, y) 0. This is the case if, and only if, (Q'x, CTlQ'x) 0. It follows that the points PeSnS' for which Tp(S')gS* are defined by (Qx, x) 0, (Q'x, x) 0, {Q'x, QlQ'x) 0.
Thèse are three quadrics meeting transversely(11) in 2 2 2 8 points, which is what we wished to prove.
Returning to our curve JE, the genus is 1 and the quotient by any of the four involutions is P1.Representing E-C/A as before, we can write It is then clear that the orbit G {To} is finite exactly when the différences T.-îjeQ-il.
(1) Since this condition is independent of To, we conclude the proof of our theorem as before.Q.E.D.
We now discuss the conditions on S and S' which will insure that our polyhedron is finite.Although (1) seems to impose three conditions on the pair S, S', we shall prove the relation î'b °Îb °ïa °U identity, (2) or, equivalently, Ti + T2 -T3 -T4 G A, (3) which shows that (1) contains only two conditions.In fact, using (3)  That the intersection is transverse follows from the équations for S and S' given below.
To prove (2) we enlarge Fig. 9 to contain four quadrilatéral: LAi LA, Each quadrilatéral represents a différent plané, ail flattened out as in Fig. 9 to aid in visualization.The diagram of involutions is and (2) is équivalent to LBl LB4.This, in turn, follows from the observation that if an A-line LAl and B-line LB4 meet in a point Pe S H S\ then there is a unique B-line LBl LB4 on S' passing through P.
To show that the condition ( 4) is nonvacuous, we use the following argument shown to us by Robert Steinberg.First, by simultaneously diagonalizing the pair of quadratic forms Q and Q', we may choose coordinates such that and The transversality condition is PJPjïl (i?6;).Take P [0,0, 0,1], H to be the hyperplane x3 0, and tt : P3 -{P} -» H given by ir(x) [xo/x3, xjx3, x2/x3] to be the linear projection.Set Ç SHH, so that C {xo+x? + X2 0}.Let C" tt(S H Sr), which is a curve with équation It remains to check that the conditions (4) and (4') are not mutually exclusive, Le., the hypersurfaces they define meet somewhere in a pair of nonsingular quadrics in gênerai position.For this we call S and S' symmetric if there exists a linear transformation of P3 carrying S to S' and S' to S. If S and Sf are given by the above équations, then the transformation y, Vft x, takes S' to S and S to the quadric £?«o /3T1yf 0. Consequently, S and Sr are symmetric in case the sets of complex numbers {/3l} {Aj8J~1} f°r some A#0, i.e., if j3, fiftklfii for some permutation (i, /, k, /) of (0,1, 2, 3).In particular, the symmetric pairs of quadrics form a hypersurface in the P3(/8) of ail pairs of quadrics, and moreover, there exist symmetric pairs in gênerai position.Now let P, H, and tt:P3-{P}-» H be as above.For a gênerai pair of plane conics in H, we can find a symmetric pair S and S' of quadric surfaces such that C SnH and C -7r(SDSf).Namely, the conditions which must be met are -j83 Aal i 0,l,2 Writing the first équation as j3, Aa, + j83 and substituting in the second yields À2aoa2 + A(aoi33 + a2i33) + j33 Àai!33 + j33.Canceling the pi and writing A yp3y both sides are divisible by pi, and the équation becomes y2aoa2 + -cx1) 0, which has the solution y (ao + a2-ai)/a0a2-Now we are done.Choose C and C for which the plane Poncelet condition is satisfied, and then choose a symmetric pair of quadrics S and S' associated to C and C as above.Then (4') is satisfied for S, S', and by symmetry (4) is also satisfied.Summarizing: The pairs of smooth quadric surfaces meeting transversely in P3 and which satisfy the Poncelet conditions of having a finite polyhedron both inscribed in and circumscribed about the pair of surfaces form a dense countable union of algebraic curves among ail pairs of quadrics.
Finally, we would like to discuss the second Poncelet theorem over thç real numbers.First, note that the remarks (i) and (ii) above about Unes on a quadric hold for a real quadric S^Pl if, and only if S is of hyperbolic type, as in Fig. 6, that is, given by a real quadratic form Q with two négative and two positive eigenvalues.For such a quadric, moreover, and any line L c pj, there will be two tangent planes to S containing L if, and only if L meets S.