IIntersecting the sides of a polygon
Anton Izosimov ∗ Abstract
Consider the map S which sends a planar polygon P to a new polygon S ( P ) whose vertices are theintersection points of second nearest sides of P . This map is the inverse of the famous pentagram map.In this paper we investigate the dynamics of the map S . Namely, we address the question of whethera convex polygon stays convex under iterations of S . Computer experiments suggest that this almostnever happens. We prove that indeed the set of polygons which remain convex under iterations of S has measure zero, and moreover it is an algebraic subvariety of codimension two. We also discuss theequations cutting out this subvariety, as well as their geometric meaning in the case of pentagons. Let P be a planar polygon, and let S ( P ) be the polygon whose vertices are the intersection points ofsecond nearest sides of P , see Figure 1. The map S is the inverse of the celebrated pentagram map ,defined by R. Schwartz [6] and studied by many others. Indeed, as can be seen in Figure 1, one canrecover P from S ( P ) by intersecting consecutive shortest diagonals (i.e. diagonals connecting second-nearest vertices), which is precisely the definition of the pentagram map. In what follows, we denotethe pentagram map by D , so that S = D − and D = S − .As can be seen from Figure 1, the image of a convex polygon under the inverse pentagram map S does not have to be convex. Moreover, computer experiments show that a randomly chosen polygonbecomes non-convex after several applications of S , see Figure 2. The goal of the present paper is todetermine necessary and sufficient conditions on a convex polygon P which guarantee that all successiveimages of P under S , i.e. S ( P ) , S ( S ( P )) , . . . , are convex polygons. Our main result is that the setof polygons with this property has zero measure and moreover is a codimension two algebraic surface.Furthermore, we present explicit equations describing this surface, i.e. explicit conditions on P whichensure that all its successive images under the map S are convex.We will state our main result in two different forms: in terms of a certain vector d P associated withthe polygon P , and in terms of a certain operator G P also associated with P .To define d P , consider a convex planar n -gon P in the affine plane. Assume that the vertices of P are labeled in cyclic order by residues modulo n . Denote by d i the vector connecting the vertices i − i + 1, and let A i be the area of the triangle cut out by d i , see Figure 3. Then d P is defined by S ( P ) P (a) S ( P ) P (b) Figure 1: The inverse pentagram map. ∗ Department of Mathematics, University of Arizona, e-mail: [email protected] a r X i v : . [ m a t h . M G ] D ec igure 2: The orbit of a polygon under the inverse pentagram map. Each iteration is nor-malized by means of an appropriate affine transformation. d P := n (cid:88) i =1 d i A i . (1)The first way to formulate our main result is as follows: for a convex polygon P , all successive imagesof P under the inverse pentagram map S are convex if and only if d P = 0. Since this is equivalentto two algebraic equations on the coordinates of vertices, it follows that the set of convex n -gons P such that S k ( P ) is convex for every k ≥ n -dimensionalspace of all convex n -gons.Another way to state our result is in terms of a certain operator introduced by M. Glick [2]. Thisoperator is defined as follows. Assume that we are given a polygon P in the projective plane P . Let v i ∈ R be the vector of homogeneous coordinates of the i ’th vertex of P . Then Glick’s operator G P : R → R is defined by G P ( v ) := nv − n (cid:88) i =1 v i − ∧ v ∧ v i +1 v i − ∧ v i ∧ v i +1 v i . (2)Observe that the right-hand side does not change under rescaling of v i ’s, so this is indeed a well-defined operator. Furthermore, as any operator in the 3-space, Glick’s operator can be interpreted asa projective mapping P → P . Theorem 1.1.
Consider a convex planar polygon P in the affine plane with at least five vertices. Thenthe following conditions are equivalent:1. All successive images of the polygon P under the inverse pentagram map S are convex.2. The associated vector d P defined by (1) is zero.3. The associated Glick’s operator G P defined by (1) , regarded as a projective mapping, is affine. We prove this theorem by establishing equivalences 2 ⇔ ⇔
3. The proof of 2 ⇔ ⇔
3, that part is based on more subtleproperties of Glick’s operator and its connection with pentagram dynamics. We discuss these propertiesin Section 3, after which we complete the proof in Section 4.In the case of pentagons, there is one more condition equivalent to the above three:
Theorem 1.2.
For pentagons, the three conditions of Theorem 1.1 are equivalent to the following one:4. The conic inscribed in P and the conic circumscribed about P are concentric. i +1 ii − d i A i d P := n (cid:88) i =1 d i A i Figure 3: To the construction of the vector d P .2igure 4: Polygons dual with respect to the line at infinity. Lines of the same color and styleare parallel to each other. Apparently, there should be an elementary way of proving this by establishing equivalence 2 ⇔ d P = 0. However, we are not awareof such a proof. What we do instead is directly prove the equivalence 1 ⇔
4. The main ingredient ofthe proof is E. Kasner’s theorem on pentagons.
Remark 1.3.
S. Tabachnikov [7] proved that Kasner’s theorem holds for all Poncelet polygons (i.e.polygons inscribed in conic and circumscribed about a conic). Therefore, Theorem 1.2 should be truefor such polygons too. We do not consider this case here so as to not encumber the exposition.
Remark 1.4.
Glick (personal communication) suggested the following geometric interpretation of thecondition d P = 0, valid regardless of the number of vertices. Say that two polygons P in Q are dualwith respect to the line at infinity if sides of P are parallel to shortest diagonals of Q , while sides of Q are parallel to shortest diagonals of P , see Figure 4. Then it turns out that a polygon P admits sucha dual if and only if d P = 0. Acknowledgments.
The author is grateful to Niklas Affolter, Max Glick, Boris Khesin, RichardSchwartz, and Sergei Tabachnikov for fruitful discussions. Figure 2 was created with help of an appletwritten by Richard Schwartz. Figures 7, 8, 9, and 10 were created with help of software packageCinderella. This work was supported by NSF grant DMS-2008021.
Wr start the proof of Theorem 1.1 by showing the equivalence 2 ⇔
3: Glick’s operator’s G P is affineif and only d P = 0. Given a polygon in the affine plane, denote by ( x i , y i ) the Cartesian coordinatesof its vertices. We assume that the vertices are labeled in the counter-clockwise order. As in theintroduction, denote by d i the i ’th shortest diagonal, and by A i the area cut out by that diagonal, asshown in Figure 3. Then a straightforward calculation shows that Glick’s operator G P is given by G P xyz = n xyz − n (cid:88) i =1 A i (cid:18) det (cid:18) xy d i (cid:19) − z · det (cid:18) x i − x i +1 y i − y i +1 (cid:19)(cid:19) x i y i From this formula it follows that the preimage of the line at infinity under G P is given in homogeneouscoordinates x , y , z bydet (cid:18) xy d P (cid:19) = z (cid:32) n + n (cid:88) i =1 A i det (cid:18) x i − x i +1 y i − y i +1 (cid:19)(cid:33) . Now assume that G P is affine. Then the line at infinity is mapped to itself, so the above equationimplies d P = 0. Conversely, assume that d P = 0. Then the coefficient of z in the above equationis easily seen to be invariant under coordinate transformations, and by choosing coordinates in sucha way that the origin is in the interior of the polygon, one shows that the coefficient of z is alwayspositive and in particular does not vanish. At the same time, since d P = 0, the coefficients of x and y vanish, so the preimage of the line at infinity is the line at infinity, as desired. Y LW Z
Figure 5: To the proof of Proposition 3.1.
In this section we discuss properties of Glick’s operator G P and its connection with the dynamics ofthe pentagram map D . This is mainly an overview of [2], but for some of the results we present arefined version. Proposition 3.1.
Let P be a convex polygon in the affine plane with at least five vertices. Then thefollowing is true:1. For the dual polygon P ∗ , one has G P ∗ = G ∗ P (recall that the dual polygon is the polygon in thedual projective plane whose vertices are the sides of initial polygon).2. Let conv( P ) be the convex hull of vertices of P , and let int(conv( P )) be its interior. Then G P (conv( P )) ⊂ int(conv( P )) . In particular, the projective mapping G P is well-defined at allpoints of conv( P ) (i.e. none of those points belong to the kernel of G P when the latter is regardedas an operator in the -space).3. The intersection (cid:84) k ≥ conv( D k ( P )) = (cid:84) k ≥ int(conv( D k ( P ))) is a single point, known as thelimit point of the pentagram orbit D k ( P ) . This point is a fixed point of G P . Moreover, it is theonly fixed point of G P in conv( P ) . Proof.
1. See [2, Proposition 3.3].2. See [2, Proposition 4.1]. The statement of that proposition is that G P (conv( P )) ⊂ conv( P ), butit is actually proved that G P (conv( P )) ⊂ int(conv( P )).3. The existence of the limit point follows from [6, Theorem 3.1]. Denote that point by X . Then,by [2, Proposition 1.2], we have G P ( X ) = X . So it remains to show that there are no otherfixed points in conv( P ). Assume that Y (cid:54) = X is another fixed point, G P ( Y ) = Y . Note that Y must be in the interior of P , since G P (conv( P )) ⊂ int(conv( P )). Let L be the line in R P through X and Y . Then G P restricts to and defines a Moebius transformation of L . Furthermore, G P preserves each of the two connected components of L \ { X, Y } . Now, let W and Z be theintersection points of L with the boundary of the polygon, as shown in Figure 5. Then, since G P (conv( P )) ⊂ int(conv( P )), it follows that G P maps W inside the interval W X and Z insidethe interval Y Z . At the same time, this Moebius transformation preserves X and Y . ButMoebius transformations with these properties do not exist. Indeed, for any non-trivial Moebiustransformation which has two fixed points and preserves both of the intervals between thosepoints, one of the fixed points must be attractive and the other one repelling. So, if W is movedby G P towards X , then Z should be moved away from Y . Thus, it is indeed not possible that G P has two fixed points in conv( P ). We now prove the equivalence 1 ⇔
3: for a convex polygon P , all its successive images under theinverse pentagram map S are convex if and only if the projective mapping G P is affine. To that end,we will reformulate the former condition in terms of the dual polygon. The following is well-known. Proposition 4.1.
Projective duality intertwines the inverse pentagram map S with the direct pentagrammap D = S − . In other words, S ( P ) ∗ = D ( P ∗ ) . O Figure 6: To the proof of Lemma 4.2.
Proof.
The sides of S ( P ) ∗ are vertices of S ( P ), which, by definition of S , are intersections of second-nearest sides of P . At the same time, the sides of D ( P ∗ ) are shortest diagonals of P ∗ , i.e. diagonalsconnecting second nearest vertices. But second nearest vertices of P ∗ are second nearest sides of P , sothe diagonals connecting them are precisely the intersections of second-nearest sides of P . So, S ( P ) ∗ and D ( P ∗ ) have the same sides and hence coincide, as claimed.We now reformulate the convexity condition for the polygon S k ( P ) in terms of its dual D k ( P ∗ ).Note that it is in general not true that the dual of a convex polygon is convex. In fact, this statementdoes not even make sense, since the definitions of convexity in the initial and dual planes requiredifferent structures. To define convexity in the initial plane, one needs to pick an affine chart, which isdetermined by choosing a line . Likewise, convexity in the dual plane also becomes well-defined uponchoice of a line, i.e. a point in the initial plane. So, in order to be able to talk about convexity in boththe initial and dual planes, in the initial plane one needs to pick a line (which defines an affine chart)and a point (which defines an affine chart in the dual plane). The following is a folklore result. Lemma 4.2.
Consider a projective plane P with a fixed line L and point O . Assume that P is apolygon in that plane which is convex in the affine chart P \ L and contains the point O in its interior.Then the dual polygon P ∗ is convex in the affine chart ( P ) ∗ \ O and contains the point L in ins interior. Remark 4.3.
The choice of a point O turns P \ L into a vector space. In that setting, the lemmacan be reformulated by saying that the dual of a convex polygon containing the origin is also a convexpolygon containing the origin. This is in fact true for any convex set, with an appropriate definition ofduality. Proof of Lemma 4.2.
Figure 6 shows a polygon P in the affine chart P \ L . Let A be an arbitraryvertex of P . Consider the straight line interval connecting O to A . Then none of the sides of P non-adjacent to A intersect that interval, and neither does the line at infinity L . Therefore, all the sidesof P non-adjacent to A , as well as the line L belong to the same connected component in the space oflines in P \ { O, A } . But this means that all vertices of P ∗ not adjacent to the side A , as well as thepoint L , belong to the same connected component of ( P ) ∗ \ ( O ∪ A ). In other words, these points lieon the same side of the line A in the affine plane ( P ) ∗ \ O , and since this holds for every side A of P ∗ ,this precisely means that P ∗ is convex and contains the point L in its interior.Now, consider a convex polygon P in the affine plane, and let L be the line at infinity. Fix a point O in the interior of P . Then, by Lemma 4.2, the dual polygon P ∗ is convex in the affine chart ( P ) ∗ \ O and contains the point L in its interior. Proposition 4.4.
For any given k > , the polygon S k ( P ) is convex if and only its dual D k ( P ∗ ) contains the point L ∈ ( P ) ∗ in its interior. Proof.
Assume that the polygon S k ( P ) is convex. Observe that D k ( S k ( P )) = P , so S k ( P ) containsthe polygon P and hence the point O in its interior. Therefore, since S k ( P ) is convex and contains thepoint O , by Lemma 4.2 the polygon D k ( P ∗ ) = S k ( P ) ∗ contains the point L . Conversely, assume that D k ( P ∗ ) contains the point L . Note that since the pentagram map D preserves convexity, and P ∗ isconvex, we also have that D k ( P ∗ ) is convex. Therefore, since D k ( P ∗ ) is convex and contains the point L , by Lemma 4.2 we have that S k ( P ) = D k ( P ∗ ) ∗ is convex too, as desired. Corollary 4.5.
Each of the polygons S k ( P ) , where k > , is convex if and only if the limit point ofthe sequence D k ( P ∗ ) is the line at infinity L . R P ( A ) := B . Proof.
The limit point is the unique point in the intersection (cid:84) k> conv( D k ( P ∗ )), so L ∈ conv( D k ( P ∗ ))for every k > L .Now, to complete the proof of Theorem 1.1, it suffices to show that the limit point of the pentagramorbit D k ( P ∗ ) is L if and only if the projective mapping G P is affine. First, assume that L is thelimit point for D k ( P ∗ ). Then, L , viewed as a point in the dual plane, is fixed by the mapping G P ∗ (Proposition 3.1, Item 3). But in view of the equality G P ∗ = G ∗ P (Proposition 3.1, Item 1), this isequivalent to saying that L , viewed as the line in the initial plane, is invariant under G P . So, G P preserves the line at infinity and hence is affine. Conversely, assume that the mapping G P is affine.Then the line at infinity L , viewed as a point in the dual plane, is a fixed point of G P ∗ . And since L isinside P ∗ , it follows from Proposition 3.1, Item 3 that L must be the limit point of D k ( P ∗ ), as desired.Thus, Theorem 1.1 is proved. We now prove the equivalence 1 ⇔ P , all its successiveimages under the inverse pentagram map S are convex if and only if the conic inscribed in P andthe conic circumscribed about P are concentric. The strategy of the proof is as follows. Consider apentagon P , and let its inscribed and circumscribed conics be given by homogeneous quadratic forms Q I and Q C respectively. Let R P := Q − I Q C . Note that since the forms Q I and Q C are defined up to aconstant factor, R P is well-defined as a projective map P → P . The geometric meaning of that mapis the following: it takes a point A ∈ P to point B ∈ P such that the polar line of A with respectto the circumscribed conic is the same as the polar line of B with respect to the inscribed conic, seeFigure 7. We will show that the operator R P has all the same properties as Glick’s operator G P . Fromthat it follows that persistence of convexity is equivalent to R P being affine. But R P is affine preciselywhen the inscribed and circumscribed conics are concentric.To describe the properties of R P , we use the following classical theorem of Kasner. Let D , as before,be the pentagram map, and let I be the map which sends a pentagon P to a new pentagon whosevertices are the tangency points of the sides of P and the inscribed conic, see Figure 8. Figure 8: To the definition of the operation I .6igure 9: Kasner’s theorem. Theorem 5.1 (Kasner [3]) . The operations D and I on pentagons commute: DI = ID , see Figure 9. Corollary 5.2.
For any pentagon P , one has R D ( P ) = R P . Proof.
Consider Figure 10. Observe that the sides of P are polar to the vertices of I ( P ) with respectto the inscribed conic and to the vertices of I − ( P ) with respect to the circumscribed conic. Therefore, R P ( I − ( P )) = I ( P ). But since the map I commutes with projective transformations, this is the sameas to say that R P ( P ) = I ( P ) . At the same time, we have R D ( P ) ( P ) = D − ( R D ( P ) ( D ( P ))) = D − ( I ( D ( P ))) = I ( P ) , where the last equality follows from Kasner’s theorem. Thus, the mappings R P and R D ( P ) both map P to I ( P ), while R − D ( P ) ◦ R P maps P to itself. But the only projective automorphism of a genericpentagon is the identity, so we must have R D ( P ) = R P for generic, and hence, by continuity, for allpolygons. Remark 5.3.
One can also use a more precise version of Kasner’s theorem which takes into accountlabeling of vertices to show that R P and R D ( P ) map every vertex of P to the same vertex of I ( P ) andhence coincide. Proposition 5.4.
Let P be a convex pentagon in the affine plane. Then the associated operator R P has all the properties listed in Proposition 3.1. Namely, the following is true:1. For the dual pentagon P ∗ , one has R P ∗ = R ∗ P .2. One has R P (conv( P )) ⊂ int(conv( P )) .3. The limit point (cid:84) k ≥ conv( D k ( P )) of the pentagram orbit D k ( P ) is a fixed point of R P . Moreover,it is the only fixed point of R P in conv( P ) . Figure 10: To the proof of Corollary 5.2.7 roof.
1. The conic inscribed in P ∗ is the dual of the conic circumscribed about P , i.e. it is givenby the quadratic form Q − C . Likewise, the conic circumscribed about P ∗ is given by Q − I . So, R P ∗ = ( Q − C ) − Q − I = Q C Q − I = ( Q − I Q C ) ∗ = R ∗ P .
2. Any point in the convex hull of P is inside the circumscribed conic (i.e. in the contractiblecomponent of the complement to that conic in P ), or belongs to that conic. Therefore, the polarline of that point with respect to the circumscribed conic is either outside or tangent to thatconic, and the point polar to that line with respect to the inscribed conic is inside the inscribedconic and hence inside P . So, R P (conv( P )) ⊂ int(conv( P )), as claimed.3. By Corollary 5.2, we have R P = R D k ( P ) , so R P (conv( D k ( P ))) = R D k ( P ) (conv( D k ( P )) , which,by above, is a subset of conv( D k ( P )). Therefore, R P (cid:16)(cid:92) k ≥ conv( D k ( P )) (cid:17) ⊂ (cid:92) k ≥ R p (conv( D k ( P ))) ⊂ (cid:92) k ≥ conv( D k ( P )) , meaning that the limit point of the pentagram orbit D k ( P ) is indeed fixed by R P . As for theuniqueness of the fixed point, it follows from other properties in the same way as for G P .From this proposition it follows that for a convex pentagon P , all its successive images under themap S are convex if and only if the associated projective mapping R P is affine. Indeed, the proof ofthe corresponding statement for G P is based on Proposition 3.1, and since that proposition also holdsfor R P , the result follows. Now, it remains to show that R P is affine if and only if the conic inscribedin P and the conic circumscribed about P are concentric. To that end, recall that the center of theconic is, by definition, the point polar to the line at infinity with respect to that conic. Therefore, theimage of the line at infinity under the map R P is the line polar with respect to the inscribed conic tothe center of the circumscribed conic. Thus, the centers coincide precisely when R P preserves the lineat infinity, i.e. is affine, as desired. Remark 5.5.
Implication 4 ⇒ P is a convex pentagon with concentric inscribed and circumscribedconics ⇒ all pentagons S k ( P ) are convex) can also be established as follows. First assume that theinscribed and circumscribed conics of P are confocal . Then, as shown in [4], there exists an affinetransformation D P taking P to its pentagram image D ( P ). Therefore, we have S k ( P ) = D − kP ( P ),and S k ( P ) is convex for every k . Now assume that the inscribed and circumscribed conics of P areconcentric, but not necessarily confocal. Then there exists a (generally speaking, defined over C )affine transformation Q which makes those concentric conics confocal. Then, since Q ( P ) has confocalinscribed and circumscribed conics, all the pentagons S k ( Q ( P )) are convex, and the same holds for S k ( P ) = Q − ( S k ( Q ( P ))), as claimed. Remark 5.6.
Instead of proving the equivalence 1 ⇔
4, we could have proved 3 ⇔
4, i.e. G P is affine ⇔ R P is affine, as follows. First, recall that by Clebsch’s theorem for any pentagon P there exists aprojective transformation D P such that D P ( P ) = D ( P ), see e.g. [6, Theorem 2.1]. Furthermore, anypentagon is projectively equivalent to its dual, see e.g. [1, Proposition 5]. So, since I ( P ) is polar to P and hence projective to P ∗ , there exists a projective transformation such that I ( P ) = I P ( P ). It is thena direct corollary of Kasner’s theorem that transformations D P and I P commute, cf. [5, Corollary 7].Therefore, D P commutes with R P = I P . Further, as observed in [2], D P coincides, as a linear operatorin 3-space, with G P − · Id, so R P commutes with G P . Now, assume that R P is affine. Then the lineat infinity L is a fixed point of the dual operator R ∗ P . Moreover, by Proposition 5.4, Item 3, it is theonly fixed point of R ∗ P in the interior of P ∗ . Then, since G ∗ P commutes with R ∗ P and preserves theinterior of P ∗ , it must preserve L , which means that G P is affine. Likewise, if G P is affine, then R P isaffine too, as desired. eferences [1] D. Fuchs and S. Tabachnikov. Self-dual polygons and self-dual curves. Functional analysis andother mathematics , 2:203–220, 2009.[2] M. Glick. The limit point of the pentagram map.
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