RRemarks on the the circumcenter of mass
Serge Tabachnikov ∗ Emmanuel Tsukerman † Given a homogeneous polygonal lamina P , one way to find its center of massis as follows: triangulate P , assign to each triangle its centroid, taken withthe weight equal to the area of the triangle, and find the center of massof the resulting system of point masses. That the resulting point, CM ( P ),does not depend on the triangulation, is a consequence of the ArchimedesLemma: if an object is divided into smaller objects, then the center of massof the compound object is the weighted average of the centers of mass of theparts, with the weights equal to the respective areas .Replace, in the above construction, the centroids of the triangles by theircircumcenters. The resulting weighted average is called the circumcenter ofmass of the polygon P , denoted by CCM ( P ). This point is well defined, thatis, does not depend on the triangulation (assuming that degenerate trianglesare avoided), see Figure 1.This construction is mentioned in the 19th century book [7], where itis attributed to the Italian algebraic geometer G. Bellavitis. We learnedabout this reference from B. Gr¨unbaum who, together with G. C. Shephard,studied this construction in the early 1990s [5]. Independently, and at aboutthe same time, the circumcenter of mass was rediscovered by V. Adler [1, 2]as an integral of a discrete dynamical system called recutting of polygons.The explicit formulas are as follows. Let the coordinates of the verticesof the polygon P , taken in the cyclic order, be ( x i , y i ) , i = 1 , . . . , n . Then ∗ Department of Mathematics, Pennsylvania State University, University Park,PA 16802, and ICERM, Brown University, Box 1995, Providence, RI 02912;[email protected] † Department of Mathematics, University of California, Berkeley, CA 94720-3840;[email protected] a r X i v : . [ m a t h . M G ] O c t igure 1: Circumcenter of mass CCM ( P ) =14 A ( P ) (cid:32) n − (cid:88) i =0 y i ( x i − + y i − − x i +1 − y i +1 ) , n − (cid:88) i =0 − x i ( x i − + y i − − x i +1 − y i +1 ) (cid:33) , where A ( P ) is the signed area of P . For comparison, CM ( P ) =16 A ( P ) (cid:32) n − (cid:88) i =0 ( x i + x i +1 )( x i y i +1 − x i +1 y i ) , n − (cid:88) i =0 ( y i + y i +1 )( x i y i +1 − x i +1 y i ) (cid:33) . The construction of the circumcenter of mass extends to higher dimen-sions, and to the elliptic and hyperbolic geometries. We studied it in [10] inrelation with the so-called discrete bicycle transformation [9]. See also thepaper by A. Akopyan [3].The construction in R n is similar. Given a simplicial polytope P , considerits non-degenerate triangulation. Assign the circumcenter CC (∆ i ) to eachsimplex ∆ i of the triangulation, and take the center of mass of these pointswith weights equal to the oriented volumes of the respective simplices: CCM ( P ) = 1Vol( P ) (cid:88) i Vol(∆ i ) CC (∆ i ) . (1)The result does not depend on the triangulation.The explicit formula is as follows. Let F = ( V , . . . , V n ) be a face of P , where V i are vectors in R n . Let A ( F ) be the n × n matrix made of2ectors V i , and let A i ( F ) be obtained form A ( F ) by replacing i th row with( | V i | , . . . , | V n | ). Then the i th component of the circumcenter of mass isgiven by CCM ( P ) i = 12( n !)Vol( P ) (cid:88) F ⊂ ∂P det A i ( F ) . One can take affine combinations tCM + (1 − t ) CCM, t ∈ R , resultingin a line, called the generalized Euler line of the polytope P (for a triangle,the Euler line is the line through the centroid and the circumcenter; it passesthrough the orthocenter as well).In this note we are interested in the uniqueness of this construction.Suppose that to every non-degenerate simplex ∆ ⊂ R n a ‘center’ C (∆) ∈ R n is assigned so that the following assumptions hold:1. The map ∆ (cid:55)→ C (∆) commutes with similarities (both orientation-preserving and orientation-reversing);2. The map ∆ (cid:55)→ C (∆) is invariant under the permutations of the verticesof the simplex ∆;3. The map ϕ : ∆ (cid:55)→ Vol(∆) C (∆) is polynomial in the coordinates of thevertices of the simplex ∆; Theorem 1
Under these assumptions, C (∆) is an affine combination of thecenter of mass and the circumcenter: C (∆) = tCM (∆) + (1 − t ) CC (∆) , where the constant t ∈ R depends on the map ∆ (cid:55)→ C (∆) (and does notdepend on the simplex ∆ ). Let x , . . . , x n be Cartesian coordinates in R n . Let ∆ = ( V , . . . , V n ) be asimplex, and let V j = ( x j , . . . , x jn ) , j = 0 , . . . , n , be the coordinates of its3ertices (where j is a superscript, not an exponent). Let V = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x x · · · x n x x · · · x n ... ... . . . ... x n x n · · · x nn · · · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , a multiple of the oriented volume of ∆, and X i,jk = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x x · · · x n x x · · · x n ... ... . . . ... x i − x i − · · · x ni − x j x k x j x k · · · x nj x nk x i +1 x i +1 · · · x ni +1 ... ... . . . ... x n x n · · · x nn · · · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Skew( x x · · · (cid:100) x i − i · · · x n − n x i − j x i − k ) , ≤ i, j, k ≤ n. where Skew is skew-symmetrization over superscripts. Evidently, X i,jk = X i,kj , and the number of such polynomials equals n ( n + 1) /
2. Both deter-minants, V and X i,jk , are skew-symmetric under permutations of the verticesof the simplex. Lemma 2.1
The polynomials X i,jk constitute a linear basis of the space S of homogeneous polynomials of degree n + 1 in the variables x , x , . . . , x nn ,skew-symmetric under permutations of the superscripts. Proof.
Since X i,jk = Skew( x x · · · (cid:100) x i − i · · · x n − n x i − j x i − k ) , a monomial x x · · · (cid:100) x i − i · · · x n − n x i − j x i − k determines X i,jk . We will showthat:1. There exists no nonidentity permutation acting on superscripts thatmaps this monomial to itself. 4. These monomials lie in different orbits under this action.The former shows that this monomial does not cancel in the expression of X i,jk . The latter will then imply that different monomials give rise to differentdeterminants.Suppose that there exists a permutation σ such that x x · · · (cid:100) x i − i · · · x n − n x i − j x i − k = x σ (0)1 x σ (1)2 · · · (cid:92) x σ ( i (cid:48) − i (cid:48) · · · x σ ( n − n x σ ( i (cid:48) − j (cid:48) x σ ( i (cid:48) − k (cid:48) . The superscripts i − σ ( i (cid:48) −
1) are the unique ones which occur twice.Therefore σ ( i (cid:48) −
1) = i −
1. The corresponding subscripts are j, k and j (cid:48) , k (cid:48) ,so that { j, k } = { j (cid:48) , k (cid:48) } . Dividing both sides by x i − j x i − k , we get x x · · · (cid:100) x i − i · · · x n − n = x σ (0)1 x σ (1)2 · · · (cid:92) x σ ( i (cid:48) − i (cid:48) · · · x σ ( n − n , so that i = i (cid:48) and σ is the identity.Let f ∈ S . Then f is equal to its skew-symmetrization. Write f in itsmonomial basis: f ( x , . . . , x nn ) = (cid:88) c αβ x αβ , | α | = | β | = n + 1 . Consider the skew-symmetrization of a monomial x αβ . If some number ap-pears in α with multiplicity 3 or greater, then α must be missing some twodistinct numbers i, j ∈ { , , . . . , n } . Each permutation σ has a counterpart σ ( i j ) of opposite sign which maps x αβ to the same monomial. Therefore theskew-symmetrization of x αβ in this case is zero.Now suppose that α contains n + 1 different elements of { , , . . . , n } .Since the entries of β are elements of { , , . . . , n } , there exist some β i and β j with i (cid:54) = j such that β i = β j . Each permutation σ has a counterpart σ ( α i α j )of opposite sign which maps x αβ to the same element.It follows that the only monomials appearing in f are those for which α is a permutation of (0 , , . . . , (cid:91) i − , . . . , n − , i − , i − α is of this form. Let α = ( α , α , . . . , α n , α n +1 ) and β = ( β , β , . . . , β n , β n +1 ). Suppose that β i = β i = . . . = β i k . For the mono-mial to not vanish under skew-symmetrization, the corresponding multiset α i , α i , . . . , α i k cannot be invariant under any transpositions. Knowing thestructure of α , we see that this implies that k = 1 , k = 2, then( α i , α i ) = ( i − , i − k = 3 then ( α i , α i , α i ) = ( i − , i − , q ), with q (cid:54) = i −
1. This proves the claim. (cid:50) ϕ : ∆ (cid:55)→ Vol(∆) C (∆), and let ( y , . . . , y n ) be its com-ponents. Our assumption 3 implies that each y (cid:96) , (cid:96) = 1 , . . . , n , is a polynomialin the variables x , x , . . . , x nn . The assumption 1, applied to scaling, impliesthat these polynomials are homogeneous of degree n + 1, and the assump-tion 2 that they are skew-symmetric under permutations of the superscripts.Lemma 2.1 implies that y (cid:96) = (cid:88) A (cid:96)i,jk X i,jk , (cid:96) = 1 , . . . , n, where the coefficients A (cid:96)i,jk satisfy A (cid:96)i,jk = A (cid:96)i,kj . We always assume thatsummation is over repeated indices. Example 2.2
The center of mass and the circumcenter of mass correspondto the functions y (cid:96) = 1 n + 1 (cid:88) X i,i(cid:96) and y (cid:96) = 12 (cid:88) X (cid:96),ii , respectively. In terms of the coefficients, one has: A (cid:96)i,jk = 12( n + 1) ( δ ij δ lk + δ ik δ lj ) and A (cid:96)i,jk = 12 δ i(cid:96) δ jk , (2)where δ is the Kronecker symbol.We now show that Archimedes Lemma is automatically satisfied for anychoice of coefficients A (cid:96)i,jk . Let ∆ = ( V , . . . , V n ) be a simplex and O a point.Consider the simplices∆ i = ( V , . . . , V i − , O, V i +1 , . . . , V n ) , i = 0 , . . . , n. Lemma 2.3
For every choice of i, j, k , one has: X i,jk (∆) = n (cid:88) i =0 X i,jk (∆ i ) . roof. Let f : R n → R n be a mapping. Consider the ( n + 2) × ( n + 2)-determinant0 = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( O ) f ( V ) f ( V ) · · · f ( V n )1 1 1 · · ·
11 1 1 · · · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) f ( V ) f ( V ) · · · f ( V n )1 1 · · · (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) f ( O ) f ( V ) · · · f ( V n )1 1 · · · (cid:12)(cid:12)(cid:12)(cid:12) + . . . + ( − n +1 (cid:12)(cid:12)(cid:12)(cid:12) f ( O ) f ( V ) · · · f ( V n − )1 1 · · · (cid:12)(cid:12)(cid:12)(cid:12) . Let f ( x , . . . , x n ) = ( x , . . . , x i − , x j x k , x i +1 , . . . , x n ) . Then, taking the orientations of the simplices into account, the above equalityfor determinants yields the result. (cid:50)
In the next section, we shall use assumption 1, namely, the fact thatthe map ∆ (cid:55)→ C (∆) commutes with parallel translations, rotations, andtranspositions of coordinates, to conclude that the coefficients A (cid:96)i,jk must beaffine combinations of the ones in (2). Introduce infinitesimal parallel translations in r th direction and rotations inthe p, q -plane: ξ r = ∂∂x r , η pq = x p ∂∂x q − x q ∂∂x r , p, q, r = 1 , . . . , n. Let σ denote a transposition of coordinates. The next lemma describes theaction of these transformations on the polynomials X i,jk . Lemma 3.1
One has σ ( X i,jk ) = − X σ ( i ) ,σ ( j ) σ ( k ) , ξ r ( X i,jk ) = ( δ ij δ rk + δ ik δ rj ) V,η pq ( X i,jk ) = δ qj X i,pk + δ qk X i,pj − δ pi X q,jk − δ pj X i,qk − δ pk X i,qj + δ qi X p,jk . roof. The first equality follows from the fact that, along with the transpo-sition of indices, exactly two rows of the determinant X i,jk are interchanged.For the second equality, notice that X i,jk = x j x k ∂∂x i ( V ) , ∂∂x r ( V ) = 0 , x j ∂∂x i ( V ) = δ ij V. It follows that ξ r ( X i,jk ) = (cid:20) ∂∂x r , x j x k ∂∂x i (cid:21) ( V ) = δ rj x k ∂∂x i ( V )+ δ rk x j ∂∂x i ( V ) = ( δ ij δ rk + δ ik δ rj ) V. The third equality is proved similarly. (cid:50)
The covariance of the map ∆ (cid:55)→ C (∆) with respect to rigid motions isexpressed by the next equations on the coefficients A (cid:96)i,jk . Proposition 3.2
For every transposition σ of the indices , . . . , n , one has A (cid:96)i,jk = A σ ( (cid:96) ) σ ( i ) ,σ ( j ) σ ( k ) . (3) The covariance with respect to infinitesimal translations is given by (cid:88) A (cid:96)i,ir = 12 δ (cid:96)r for all (cid:96), r, (4) and with respect to infinitesimal rotations by A (cid:96)a,qc δ pb + A (cid:96)a,qb δ pc − A (cid:96)p,bc δ qa − A (cid:96)a,pc δ qb − A (cid:96)a,pb δ qc + A (cid:96)q,bc δ pa − A pa,bc δ (cid:96)q + A qa,bc δ (cid:96)p = 0 , (5) for all a, b, c, p, q, (cid:96) . Proof.
A transposition of coordinates is reflection in a hyperplane, and itchanges the sign of V . Hence the covariance of the map ∆ (cid:55)→ C (∆) withrespect to σ implies the equality (cid:88) A (cid:96)i,jk X σ ( i ) ,σ ( j ) σ ( k ) = (cid:88) A σ ( (cid:96) ) i,jk X i,jk for all (cid:96) . Since the polynomials X i,jk form a basis, for each term on the right,there is a matching term on the left: A σ ( (cid:96) ) i,jk = A (cid:96)σ ( i ) ,σ ( j ) σ ( k ) . σ ( (cid:96) ) by (cid:96) , we obtain (3.2).To establish (4), we use Lemma 3.1 to calculate: ξ r (cid:16) y (cid:96) V (cid:17) = (cid:88) A (cid:96)i,jk ( δ ij δ rk + δ ik δ rj ) = (cid:88) A (cid:96)i,ir . On the other hand, y (cid:96) /V is the (cid:96) th component of the map ∆ (cid:55)→ C (∆), andthe infinitesimal translation in the r th direction sends it to δ (cid:96)r . By translationcovariance, the above sum equals δ (cid:96)r , as claimed.Likewise, the infinitesimal rotation in the p, q -plane annihilates y (cid:96) for (cid:96) distinct from p, q , and sends y q to y p , and y p to − y q ; in short, y (cid:96) (cid:55)→ y p δ (cid:96)q − y q δ (cid:96)q . On the other hand, by Lemma 3.1, η pq ( y (cid:96) ) = (cid:88) A (cid:96)i,jk ( δ qj X i,pk + δ qk X i,pj − δ pi X q,jk − δ pj X i,qk − δ pk X i,qj + δ qi X p,jk )= 2 A (cid:96)i,qj X i,pj − A (cid:96)p,jk X q,jk − A (cid:96)i,pj X i,qj + A (cid:96)q,jk X p,jk . Equate this to y p δ (cid:96)q − y q δ (cid:96)q = (cid:88) ( A pi,jk δ (cid:96)q − A qi,jk δ (cid:96)p ) X i,jk , and then, for fixed a, b, c , equate the coefficients in front of X a,bc in bothexpressions to obtain (5). (cid:50) Now we need to solve the system of linear equations (3)–(5) on the un-knowns A (cid:96)i,jk . We use (3) to reduce the number of variables.Consider the following four cases. If |{ i, j, k, (cid:96) }| = 4 then, applying anappropriate sequence of transpositions, we obtain: A (cid:96)i,jk = A , =: t . If |{ i, j, k, (cid:96) }| = 3, then one has four sub-cases, and A (cid:96)i,jk is equal to A , =: u, or A , =: v, or A , =: w, or A , =: s. Likewise, if |{ i, j, k, (cid:96) }| = 2, then one has five sub-cases, and A (cid:96)i,jk is equal to A , =: φ, or A , =: ψ, or A , =: α, or A , =: β, or A , =: γ. Finally, if |{ i, j, k, (cid:96) }| = 1, then A (cid:96)i,jk = A , =: ν . Thus we have 11 un-knowns. 9ow the strategy is to consider particular cases of (5) and (4).To start with, consider (5) with (cid:96) = q = b = c (cid:54) = p = a . One obtains − A qp,pq + A qq,qq − A pp,qq = 0 , or 2 ψ + φ = ν. Likewise, (4) with (cid:96) = r yields ( n − ψ + ν = 1 / . It followsthat ν = 12 − ( n − ψ, φ = 12 − ( n + 1) ψ. (6)We pause to check against Example 2.2. For the center of mass, ψ = 12( n + 1) , ν = 1 n + 1 , and the rest of variables vanish; for the circumcenter of mass, φ = ν = 1 / , and the rest vanishes. In both cases, (6) holds.To finish the proof of Theorem 1, we need to show that all variables,except φ, ψ, ν , vanish. We proceed in a similar fashion: (5) with (cid:96) = q = a = b = c (cid:54) = p yields α + 2 β + γ = 0 , (5) with (cid:96) = q = c (cid:54) = p = a = b yields γ = α . Hence β = − α . Next, (4) with (cid:96) (cid:54) = r yields ( n − s + α + β = 0 , hence s = 0.Next, (5) with (cid:96) = q = c , but distinct from pairwise distinct p, a, b ,yields t = 0. It remains to eliminate u, v, w and α . Toward this, (5) with (cid:96) = q = c (cid:54) = p = b and distinct from a yields γ = v + w, hence α = v + w. Likewise, (5) with (cid:96) = q = c (cid:54) = p = a and distinct from b , yields − s + β − u = 0 , hence α + u = 0 . Next, (5) with (cid:96) = q = c = a , but distinct from pairwise distinct p, b , yields u + v + s = 0 , hence u + v = 0 , and (5) (cid:96) = q = c = b , but distinct from pairwise distinct p, a , yields 2 v + w =0 . We have obtained four linear equations on u, v, w, α , and the only solutionof this system is zero. This completes the proof.10
Final remarks (i) Degenerate simplices can be safely ignored when calculating the centerof mass: such a simplex has a finite centroid and zero volume, making nocontribution to the total sum. Not so for the circumcenter of mass: althoughthe volume of a nearly degenerate simplex tends to zero, its circumcenter maygo to infinity, and the contribution to the sum (1) may be non-negligible. Themap ϕ : ∆ (cid:55)→ Vol(∆) CC (∆), being polynomial in the coordinates of thevertices, is continuous.For example, consider an isosceles right triangle ABC . Its circumcenteris the midpoint M of the hypothenuse AC . Consider the triangulation infigure 2 consisting of three triangles, one of which, AM C , is degenerate. Ifone ignored this triangle, then, by the Archimedes Lemma, the circumcenterof mass of (cid:52)
ABC would be the midpoint of the segment connecting themidpoints of the hypothenuses AB and BC of the triangles ABM and
BCM .The latter point is the circumcenter of the quadrilateral
ABCM , not the triangle
ABC . MA B C
Figure 2: Contribution of a degenerate triangle(ii) One may wish to extend the notion of the circumcenter of mass tomore general sets. For example, let γ ( t ) be a parameterized smooth curve,star-shaped with respect to point O , see figure 3. It is natural to define theCircumcenter of Mass by continuity as (cid:82) C ( t ) dA (cid:82) dA , (7)where C ( t ) denotes the limiting ε → O to thecircumcenter of the infinitesimal triangle Oγ ( t ) γ ( t + ε ), and dA is the areaof this infinitesimal triangle. However, this does not give anything new: theintegral (7) is the center of mass of the lamina bounded by the curve [10].11 C(t) (cid:97) (t)
Figure 3: Continuous limit of the circumcenter of mass(iii) Although the rational map ∆ (cid:55)→ CC (∆) is discontinuous, the poly-nomial map ϕ : P (cid:55)→ Vol( P ) CCM ( P ), defined on simplicial polytopes in R n , is continuous and is a valuation: ϕ ( P ∪ P ) + ϕ ( P ∩ P ) = ϕ ( P ) + ϕ ( P ) . This valuation is isometry covariant; see, e.g., [8] for the theory of valuations. R n -valued continuous isometry covariant valuations on convex compactsubsets of R n were classified in [6] as linear combinations of intrinsic moments.Namely, given a convex set K , let K ε be its ε -neighborhood. Then themoment vector (cid:90) K ε x dx (8)is a polynomial in ε , and its coefficients span the space V n of continuousisometry covariant valuations (the free term being Vol( K ) CM ( K )). Onehas: dim V n = n + 1.The next proposition states that the circumcenter of mass is not a linearcombination of the intrinsic moments. Proposition 4.1
The map ϕ : P (cid:55)→ Vol( P ) CCM ( P ) is not an element ofthe space V n . Proof.
We argue in dimension 2; the general case is similar.Consider an isosceles triangle K ( α ) whose base is aligned with the x -axisand has length 2, whose axis of symmetry is the y -axis, and whose base angleis α . If α is close to zero then the moment vector (8) is close to the zerovector, and so are the coefficients of the powers of ε in (8) that constitute abasis in V . 12ssume that ϕ is a linear combination of the three basic vectors of thespace V (with constant coefficients, independent of K ). Then ϕ ( K ( α )) → α →
0. However, a straightforward computation shows thatlim α → ϕ ( K ( α )) = (cid:18) , − (cid:19) (without computation, this can be seen in figure 2). This is a contradiction. (cid:50) (iv) We finish with a conjecture and two problems.Assume that one assigns a “center” to every simplicial polytope in R n so that the center depends analytically on the polytope, commutes withdilations, and satisfies the Archimedes Lemma (with the weights equal tothe respective volumes). Conjecture 4.2
The space of such centers is 1-dimensional: they are affinecombinations of the centers of mass and the circumcenters of mass.
For n = 2, this is proved in [10].As we mentioned, the construction of the circumcenter of mass extendsto the spherical and hyperbolic geometries [3, 10]. Thus one has versions ofConjecture 4.2 for S n and H n as well.Next, we pose the following problem. Problem 4.3
Describe R n -valued continuous isometry covariant valuationson simplicial polytopes in R n . Finally, it is interesting to find an axiomatic description of the centersfor simplicial polygons and polytopes, discussed in this note, in the threegeometries of constant curvature (for the center of mass, see [4]). Thesecenters should be isometry covariant and satisfy some additivity condition(Archimedes Lemma or valuation-like). Such a description should includethe valuations from Problem 4.3. At the moment of writing, we do not knowsuch an axiomatic description.
Acknowledgments . This work is an extension of the project that origi-nated in the Summer@ICERM 2012 program; it is a pleasure to acknowledgethe inspiring atmosphere and hospitality of the institute. We are grateful to13. Adler, A. Akopyan, Yu. Baryshnikov and B. Gr¨unbaum for their interestand help. The first author was supported by the NSF grant DMS-1105442,and the second author was supported by a NSF Graduate Research Fellow-ship under Grant No. DGE 1106400
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