EEXTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE
DIRK SIERSMA
Abstract.
We determine all critical configurations for the Area function on polygonswith vertices on a circle or an ellipse. For isolated critical points we compute their Morseindex, resp index of the gradient vector field. We relate the computation at an isolateddegenerate point to an eigenvalue question about combinations. In the even dimensionalcase non-isolated singularities occur as ‘zigzag trains’ . Introduction
A traditional problem for the maximal area of a polygon is the Isoperimetric Problem.It dates back to the antiquity. Many mathematicians where involved in this problem, e.g.Steiner; see e.g. the historical overview of Bl˚asj¨o [1].Several problems of maximal area with other constraints have been studied; many ofthem concern computer science, e.g. as maximal triangle in a convex polygon.Recently there became more attention for the other critical points of Area as a functionon a configuration space. A goal is a Morse-theoretic approach: determine all criticalpoints, check if they are Morse and if so compute the Morse index. In specific problemscritical points and indices can be successfully described by nice geometric properties.This has been carried out e.g. for the signed Area function on linkages with given edgelength [7], [9] and more recently in the context of the isoperimetric problem [8].We continue this approach in a very elementary case: A polygon with n vertices ona circle. We allow self-intersections and coinciding vertices. We use the signed Areafunction sA . This has the advantage and disadvantage that due to orientation we canhave negative area and in case of multiple covered regions we count with multiplicities.Our main theorem 1 gives geometric criteria for the critical points and determines alsothe Hesse matrix at those points. Most of the critical points are of Morse type and look as aregular star, but several of them have zigzag behaviour. The Morse index is determined bycombinatorial data. The remaining cases are degenerate: All points coincide (degeneratestar) which has an isolated critical point in case n is odd; and zigzag-trains, which arenon-isolated in case n is even.We compute also the index of the gradient vector field at the degenerate star by Euler-characteristic arguments. In the last section we discuss the Eisenbud-Levine-Khimshiashvilimethod to calculate this index. This relates nicely to a combinatorial question.Note, that the problem of extremal area polygons in an ellipse is also solved due to theexistence of an area preserving affine map. Date : October 9, 2020.2000
Mathematics Subject Classification.
Key words and phrases.
Area, polygon, ellipse, critical point, Morse index. a r X i v : . [ m a t h . M G ] O c t DIRK SIERSMA
This paper is a by-product of a study (in progress) of polygons with vertices slidingalong a set of curves. This was triggered by the work of Wilson [12] in 1917.Part of the work was done at Luminy. The author thanks CIRM for the excellent work-ing conditions, Wilberd van der Kallen for his advice around Mathematica computations,George Khimshiashvili and Gaiane Panina for useful discussions.2.
Polygons in a circle
In this section we give a complete description of critical polygons and compute indicesat isolated zero’s of the gradient vector field.We first state the formula for signed Area sA . Take unit circle with center O . We fix thepoint P of the polygon with vertices P , · · · , P n and work with the reduced configurationspace ( S ) n − . We use α i = ∠ P i OP i +1 as coordinates ( i = 1 , · · · , n − π , we choose values in the interval ( − π, π ]. α n = ∠ P n OP = 2 ωπ − (cid:80) n − i =1 α i .The signed area function is sA = n (cid:88) i =1 sin α i , We next describe some special types of polygons (see also Figure 1):
Definition 1.
Let all | α i | be equal.A regular star is a polygon with vertices on a circle and all angles α i are equal and differentfrom 0 and π . In case all α i = π we call it a complete fold .Regular stars are determined by the winding number ω of the polygon with respect tothe center of the circle. All angles are equal to πωn .A zigzag is a polygon with vertices on a circle and all | α i | are equal and different from zero,but not all α i have the same sign. Let f be the number of forwards edges (0 < α i < π )and b the number of backward edges ( α i <
0) in the polygon. NB. f + b = n . We definea number m = [ n − ]; so n = 2 m + 1 if n is odd and n = 2 m + 2 if n is even.In case f (cid:54) = b and | α i | = ωπ | f − b | , ω (cid:54) = 0 we call them zigzag stars . Zigzag stars are determinedby a sequence of signs of α i . Their realization in the plane is a | f − b | -regular star withsome multiple edges.In case f = b gives rise to a 1-dimensional singularities, since | α i | can be arbitrary; wecall them zigzag trains . They only occur if the number of vertices is even. In that casethey include the degenerate star and complete fold.A degenerate star is the polygon with all α i equal to zero. Theorem 1.
The signed area function for polygons on a circle (defined on the reducedconfiguration space) has critical points iff all | α i | are equal. These critical points areisolated or (if the number of vertices is even) contain also a 1-dimensional singular set.Moreover The isolated singularity types are regular stars, zigzag stars and if n = odd alsodegenerate stars, All regular and zigzag stars are Morse critical points, (i) If f > b , the number of critical points with b backward edges is (cid:0) nb (cid:1) ( m − b ) ,each with Morse index f − , XTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE 3
Figure 1.
Some critical configurations(ii) If f < b , the number of critical points with f forward edges is (cid:0) nf (cid:1) ( m − f ) ,each with Morse index f . Degenerate stars are degenerate isolated critical points if n is odd. Their (complex)Milnor number is n − and the real gradient map has local degree ι = 2( − m (cid:0) n − m − (cid:1) . The non-isolated case only occurs if n = even and includes the complete fold, zigzagtrains and degenerate stars. The non-isolated part of the critical set contains (cid:0) n n (cid:1) branches, which meet only at the complete fold and the degenerate stars.Proof. The conditions for critical points arecos α i = cos α n ( i = 1 , · · · n − | α i | are equal. If f (cid:54) = b thereare finitely many solutions; if f = b infinitely many.1. This follows from the fact that all | α i | are equal.2. First we count the critical points with b edges. Since the polygon has n edges,we have (cid:0) nb (cid:1) different possibilities. We also have to take into account the windingnumber ω of the polygon, which determines the embedding. This winding numberfor f > b is between 1 and m − b . This explains the factor in the statement. Thecase f < b is similar.Second, the Hessian matrix has diagonal term − sin α i − sin α n and subdiagonalterms p := sin α n . We first consider positive regular stars; i.e. f = n and p > α = α = · · · = α n = πωn , where ω (cid:54) = 0 , n . The matrix is: − p − p − p · · · − p − p − p − p · · · − p − p − p − p · · · − p · · · · · · · · · · · ·− p − p − p · · · − p DIRK SIERSMA
The eigenvalues are − np, − p, − p, · · · , − p . It follows that the regular star (with ω >
0) has a maximum if 1 ≤ ω < n and a minimum if n < ω ≤ n − ω = n we have a complete fold and a vanishing Hessian.Next the zigzag case:Assume α n >
0. If not, change the cyclic ordering such that this is the case. TheHesse matrix can change, but not the index. The Hesse matrix has as elements h ij = − p if i (cid:54) = j and h ii = 0 if the i th edge is backwards and h ii = − p if the i th edge is forward (this happens, resp b and f − b = 2 is: − p − p · · · − p − p − p · · · − p − p − p − p · · · − p · · · · · · · · · · · ·− p − p − p · · · − p A direct computation shows now for any b that the characteristic equation is:( λ + ( b + f − pλ + ( b − f ) p ) · ( λ − p ) b − · ( λ + p ) f − = 0This explains the index formula of the theorem. Note that the index is independentof the places of the zigzags.We have excluded the case where b = f . This case corresponds to the noniso-lated critical points and the Hessian determinant is zero (follows also from theabove formula).3. Degenerate star: ( n = 2 m + 1)In case all α i = 0 it is clear that the 2-jet vanishes at the critical point. The 3-jetis ( α + · · · + α n − − ( α + · · · + α n − ) ). Therefore the Milnor number is 2 n − ( n odd).Next we will use that the sum of all indices of zero’s of the gradient vector fieldis equal to the Euler characteristic of the configuration space; which is zero in ourcase. We know the Morse indices of all non-degenerate critical points. The index ι of the degenerate critical point is therefore equal to: ι = − m − (cid:88) b =0 ( − b (cid:0) nb (cid:1) ( m − b ) = 2( − m (cid:0) n − m − (cid:1) . NB. This reduction of the first summation to a single binomial coefficient is dueto repeated use of the formula (cid:0) nk (cid:1) = (cid:0) n − k − (cid:1) + (cid:0) n − k (cid:1) .4. Non isolated singularities:These occur only in case n is even. Besides the degenerate star and the completefold, where all | α i | are 0 or π we get exactly the case b = f , which gives rise to zigzagtrains. They can be distinguished by the positions of forward and backward edges.There are exactly (cid:0) n n (cid:1) possibilities. They correspond to branches, parametrized via | α | in the open interval (0 , π ).The complete fold and degenerate star are limitingcases. (cid:3) XTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE 5
Example 1.
When n = 7 we find the following critical polygons: (pictures in Figure 1).b=0: 3 maxima (index 6): regular stars with resp. ω = 1 , , ω = 1 and 7 zigzag starswith ω = 2 (pentagonal shape),b=2: 21 saddle points with index 4 : zigzag stars with ω = 1 (triangular shape),next: a degenerate star with gradient-index -20 ,b=5: 21 saddle points with index 2 : zigzag stars with ω = − ω = − ω = − ω = − , − , − −
14 + 21 −
20 + 21 −
14 + 3 = 0
Proposition 1.
The critical values of sA are • for regular stars and zigzag star: ( n − b ) sin πωn − b , • for degenerate stars, complete folds and zigzag trains: . In figure 2 we show the critical values (in case n = 7, positive ω ), together with thecommon value | α i | . Negative ω results to adding a minus sign. Figure 2.
Extremal values, listed with | α i | Note that the absolute maximum does not occur for the regular polygon, but for aregular star with 2 πω/n nearest to π/
2. This is because we measure Area with overlaps.All the marked points on the upper sin-curve: y = n sin x correspond to maxima; on theother sin-curves y = ( n − b ) sin x (with b >
0) one finds only saddle points. Drawingsimilar figures for n even show several critical points with different b but the same value.The absolute maximum is n in this case.3. Eisenbud-Levine-Khimshiashvili index computations
In case n = 2 m + 1 we know that sA has an isolated critical point at a degenerate starand the Eisenbud-Levine-Khimshiashvili (ELK) signature formula is available ([3],[6]) to DIRK SIERSMA compute the gradient index. This can serve as a check for correctness of our computations,but also relates to a nice combinatorial question. For computational reasons we replaceArea sA by its 3-jet: f = j ( sA ) and compute its index. This does not change the index,see [2].3.1. Milnor Algebra.
The general case.
We first give a sketch of the ELK method for the gradient vectorfield. Let f : R n , → R be a real analytic function, such that the complexification has anisolated zero. Let J ( f ) = ( f , · · · , f n ) be the ideal generated by the partial derivatives of f in the ring O n,O of real analytic function germs. Let M f = O n, /J ( f ) , this is the (real)Milnor Algebra of f . Its dimension is finite if and only if the complexification of f has anisolated singularity. Let h f be the Hessian determinant of the germ f and [ h f ] ∈ M f itsequivalence class modulo J ( f ).Let l be any linear function such that l ([ h f ]) >
0. Define a nondegenerate bilinear form β on M f by β ( g, h ) = l ( g ∗ h ), where ∗ is the multiplication in O n, . The ELK signatureformula states that the index of the gradient vector field at the origin is equal to thesignature of the bilinear form β on M f , i.e the number of positive eigenvalues minus thenumber of negative eigenvalues.In order to compute the signature we work as follows- choose a set of function germs, which induce a basis of M f ,- write down the multiplication table (modulo J ( f )) on this basis,- compute [ h f ] and choose the linear function l as above,- substitute the l -values in the multiplication table; this gives a scalar valued sym-metric matrix, compute its signature.3.1.2. The case of our 3-jet.
The Milnor algebra M f is graded by degree and has di-mension 2 n − . The partial derivatives are homogeneous of degree 2 and give rise to therelations: x = x = · · · = x n − = − n − (cid:88) ≤ i 1. We first look to the effect in this degree of the relations x = x = · · · = x n − mentioned above. It follows that the class [ x I ] = [ x i x i · · · x i n − n − ] is completely determinedby the unordered ( n − i , i , · · · , i n − ) modulo 2. E.g. I k consisting of 2 k tuples equal 1 and the others equal 0. Let the corresponding class of monomials be w k = [ x I k ] , ( k = 0 , · · · m ). Note w m = [ x x · · · x m ] . Next we add the remaining relation x p = − n − (cid:80) ≤ i 1) : (cid:18) p (cid:19) w p − + { n − + 2 p ( n − − p ) } w p + (cid:18) n − − p (cid:19) w p +1 = 0 . XTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE 7 The solution space is 1-dimensional and determined by the following recurrence relations: w p − = − m + 2 − p p − w p ( p = 1 , · · · , m ) , which determines w k = ( − m − k (2 m − k )!!(2 k − m − w m ( k = 0 , · · · , m ) . Next: h f = x x · · · x n − − (cid:80) ≤ i,j ≤ n − x i x x ··· x n − x j . In M f we have[ h f ] = − ( n − w m − ( n − n − w m − = − ( n − w m + 2( n − w m = n w m . We fix l ( w m ) = ω . This determines al the other l ( w k ). All other basis elements of M f have lower degree. We set l = 0 for all of them. Examples m = 1 : w = − ω , w = ωm = 2 : w = ω , w = − ω , w = ωm = 3 : w = − ω , w = ω , w = − ω , w = ω The multiplication matrix. Next consider the matrix B of l ( β ) in block form ( B i,j )where B i,j is the submatrix constructed by multiplying monomials of degree i with thoseof degree j . Due to symmetry we have B i,j = B j,i . Note that B i,j = 0 if i + j ≥ n . Dueto the choice of l we also have that B i,j = 0 if i + j < n − B has now the following anti-diagonal block form: (e..g. n = 5) B , B , 00 0 B , B , B , Next observe that due to the symmetry: signature B = signature B m,m .3.2. A combinatorial question. The computation of the signature of B m,m is related toan eigenvalue question of combinatorial type. This question was solved in a manuscriptof the paper for m=1,2,3 only. Due to the suggestions of an anonymous referee andthe assistance of Wilberd van der Kallen it was possible to give the full solution. Wewill publish the combinatorial result (in a more general case) in a separate paper [5]and summarize below the results. Relevant notions are the representation theory of thesymmetric group and the relation between invariant subspaces and the eigenspaces of theintertwining matrices. As references we mention [4] and [10].Consider combinations C mm of m elements out of a set of 2 m elements. Take an arbitrarytuple of real numbers b , · · · , b m . We constitute a matrix, where the rows and columnsare indexed (in lexicografic order) by elements of C mm . The matrix elements are definedas follows: < σ, τ > = b p if σ ∩ τ has p elements ( 0 ≤ p ≤ m ). DIRK SIERSMA Figure 3. B , and B , ; coloured by b p . Proposition 2. The matrix B m,m has the properties: • The eigenspaces are independent of a (generic) choice of b , · · · b m , • The eigenvalues are Z -linear combinations of b , · · · b m . This results is clear from Schur’s lemma and Young’s rule as the referee suggested. Hisremark about Specht modules made it possible to compute the eigenvalues. That workwas carried out in [5]. Proposition 3. The matrix B m,m has eigenvalues λ k = k (cid:88) j =0 m − k (cid:88) p =0 ( − k − j (cid:18) kj (cid:19)(cid:18) m − jp (cid:19)(cid:18) j − k − m + n − k + m − p (cid:19) b j + p each of multiplicity 1 with dimension of eigenspaces: µ k = (2 m )!(2 m − k +1) k !(2 m − k +1)! . Example 2. B , has eigenvalues: • b + b with multiplicity 1, and • b − b with multiplicity 1. B , has eigenvalues: • b + 4 b + b with multiplicity 1, • b − b + b with multiplicity 2 and • b − b with multiplicity 3. B , has eigenvalues: • b + 9 b + 9 b + b with multiplicity 1, • b − b − b + b with multiplicity 9, • − b + 3 b − b + b with multiplicity 5 and • − b − b + 3 b + b with multiplicity 5.3.3. Final step; computation of the index.Proposition 4. The restriction of B m,m to the 1-dimensional degree m part of the Milnoralgebra has the eigenvalues λ k = ( − m m + 12 m − k + 1 ω ( k = 0 , · · · , m ) XTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE 9 Proof. Take l ( w p ) = b m − p . Note b p = ( − p ω (2 p )!!(2 m − p − m − ( p = 0 , · · · , m ) . We insertthis in the formula for λ k in Proposition 4. The formula in the present Proposition, whichis in fact an enormous reduction, was expected by numerical results. We discovered akey recurrence using the package [11]. See Appendix. The proof of the formula becomesstraight forward after checking the initial cases. (cid:3) Example 3. Set ω = 1; m = 1 eigenvalues − , − m = 2 eigenvalues 1 , , m = 3 eigenvalues − , − , − , − m = 8 eigenvalues 1 , , , , , , , , . Set ω = 1 and conclude that the eigenvalues of B m,m have the same sign and the indexis equal to ι = 2( − m (cid:0) m − m − (cid:1) . This coincides with the result in Theorem 1. References [1] V. Bl˚asj¨o, The isoperimetric problem , American Mathematical Monthly, 112, 6 (2005), 525-566.[2] A. Cima, A. Gasull, J. Torregrosa, On the relation between index and multiplicity J.London Math.Soc. (2) 57 (1998), no. 3, 757-768.[3] Eisenbud, D , Levine, H.I, An algebraic formula for the degree of a c ∞ map germ, Ann. of Math. (2)106 (1977), no. 1, 19–44.[4] W. Fulton, Young Tableaux, London Mathematical Society Student Texts 35, Cambridge UniversityPress, 1997.[5] Kallen, W. van der, Siersma, D, Subset Permutations and Eigenvalues of the Universal IntertwiningMatrix , manuscript in preparation.[6] Khimshiashvili, G, The local degree of a smooth mapping. (Russian), Sakharth. SSR Mecn. Akad.Moambe 85 (1977), no. 2, 309–312.[7] Khimshiashvili G., Panina G., Cyclic polygons are critical points of area . Zap. Nauchn. Sem. S.-Petersburg. Otdel. Mat. Inst. Steklov. (POMI), 2008, 360, 8, 238–245.[8] Khimshiashvili, G., Panina, G., Siersma, D., Extremal Area’s of Polygons with Fixed Perimeter . Zap.Nauchn. Sem. S.-Petersburg. Otdel. Mat. Inst. Steklov. (POMI), 2019, 841, 136-145.[9] Panina G., Zhukova A., Morse index of a cyclic polygon , Cent. Eur. J. Math., 9(2) (2011), 364-377.[10] Prasad, A., Representation Theory: A Combinatorial Viewpoint , Cambridge Studies in AdvancedMathematics, Cambridge University Press, 2015.[11] Wegschaider, K, P ackage MultiSum version 2.3 written by Kurt Wegschaider, enhanced by AxelRiese and Burkhard Zimmermann, Copyright Research Institute for Symbolic Computation (RISC),Johannes Kepler University, Linz, Austria.[12] Wilson, E.B, Relating to Infinitesimal Methods in Geometry , The American Mathematical Monthly,Vol 23, No.5 (May 1917), p. 241-243. Appendix We discovered the key identity (1) below usingPackage MultiSum version 2.3 written by Kurt Wegschaiderenhanced by Axel Riese and Burkhard ZimmermannCopyright Research Institute for Symbolic Computation (RISC),Johannes Kepler University, Linz, Austria. Recall that ( − n !! = n (( n − n (cid:62) (cid:54) p (cid:54) m − k , 0 (cid:54) j (cid:54) k , put F ( m, k, j, p ) = ( − k + p (cid:0) kj (cid:1)(cid:0) m − jp (cid:1)(cid:0) j − k + mj + p (cid:1) (2( j + p ))!!( − j + 2 m − p − m − . Otherwise put F ( m, k, j, p ) = 0.Put F j ( m, k, j, p ) = (2 m − k − − k − k + m − F ( m − , k − , j, p + 1)+ 4( m − k − m ) F ( m − , k − , j, p ) − m − k − m ) F ( m − , k − , j, p + 1) − (2 m − k + 2 m − F ( m − , k − , j, p + 1) − (2 k − m − F ( m − , k − , j + 1 , p + 1)+ 2(2 m − k − m ) F ( m − , k − , j, p )+ 2(2 m − m − k ) F ( m − , k − , j, p + 1)+ 2(2 m − k − m ) F ( m − , k − , j + 1 , p )+ 2(2 m − m − k ) F ( m − , k − , j + 1 , p + 1) − (2 m − m − F ( m − , k − , j, p + 1) − (2 m − m − F ( m − , k − , j + 1 , p + 1))and F p ( m, k, j, p ) =( k − − k − k − m − F ( m − , k − , j, p ) − k − m − k + 2 m − F ( m − , k − , j, p )+ 4( m − m − m − k ) F ( m − , k − , j, p ) − (2 m − m − k + 2 m − F ( m − , k − , j, p )+ 2(2 m − m − m − k ) F ( m − , k − , j, p ) − (2 m − m − m − F ( m − , k − , j, p )) . XTREMAL AREA OF POLYGONS, SLIDING ALONG A CIRCLE 11 Then we have the key identity4( k − k − k − m − F ( m − , k − , j, p )+ 4( k − k − m − k + 2 m − F ( m − , k − , j, p )(1) + (2 m − m − k − k + 2 m − F ( m − , k − , j, p )+ (2 m − m − m − k − F ( m − , k − , j, p )= F j ( m, k, j + 1 , p ) − F j ( m, k, j, p ) + F p ( m, k, j, p + 1) − F p ( m, k, j, p ) . It yields a recurrence for the (cid:80) j,p F ( m, k, j, p ).Similarly, put g ( m, p ) = 4 m F ( m − , , , p − − (2 m − m + 1) F ( m, , , p − m − m + 1) F ( m, , , p − − (2 m − m + 1) F ( m + 1 , , , p − − (2 m − m + 1) F ( m + 1 , , , p − . Then (2 m − m + 1) F ( m, p − 2) + (2 m − m + 1) F ( m + 1 , p − 2) = g ( m, p + 1) − g ( m, p ),which yields a recurrence for the (cid:80) p F ( m, , , p ).One concludes that (cid:80) p F ( m, , , p ) = ( − m for m (cid:62) (cid:88) j,p F ( m, k, j, p ) = ( − m (2 m + 1) / (2( m − k ) + 1)for m (cid:62) k (cid:62) Utrecht University, Department of Mathematics, P.O.Box 80.010, 3508TA, Utrecht,The Netherlands Email address ::