On Unfoldings of Stretched Polyhedra
Gözde Sert Sergio Zamora



On Unfoldings of Stretched Polyhedra

Gözde Sert, Sergio Zamora

Penn State University






We give a short proof of a result obtained by Mohammad Ghomi concerning existence of nets of a convex polyhedron after a suitable linear transformation.






§ INTRODUCTION



A net of a polyhedron is an arrangement of edge-joined polygons in the plane which can be folded along its edges to become the faces of the polyhedron. The first known record of this procedure is the renaissance book Instructions for Measuring with Compass and Ruler by Albrecht Dürer <cit.>. In this book, Dürer shows how to cut and develop some figures, including all five regular polyhedra.  In 1975 Geoffrey Shephard <cit.> posed the problem to determine whether all convex polyhedra have a net.













Recently Mohammad Ghomi <cit.> proved the existence of a net for any polyhedron after a suitable stretching in one direction (theorem <ref> below). In particular, every polyhedron is affinely equivalent to one with a net, and having a net does not depend on the combinatorial structure of the polyhedron. In this note we give an elementary proof of that result based on the arm lemma, which allows us to considerably shorten it and make it less technical.



Let P be a convex polyhedron, then there is a linear stretching L, such that L(P) has a net.




















§ PRELIMINARIES





Let a, b, c be vertices of a convex polyhedron such that ab and ac are edges. We will denote by ∡ bac the intrinsic angle at a from ab to ac measured counterclockwise. For distinct x,y,z ∈^2, arg (x) will denote the argument from -π to π of x as a complex number, and ∠ yxz will denote the angle at x measured counterclockwise.



Since the polyhedron is convex, for any a,b,c vertices such that ab and ac are edges, we have

    ∡ bac + ∡  cab  < 2π.
 

 

 
(Arm Lemma) Suppose { u_0, u_1, … , u_m     }, { v_0, v_1, … , v_m      }⊂^2 satisfy


  * u_0 = v_0.

  * | u_j - u_j-1| = | v_j - v_j-1| for j=1,2, …, m.

  * u_j - u_j-1 and v_j - v_j-1 are almost horizontal for j = 1, 2, … , m (their argument is in ( -π/10,π/10) ).

  * arg (v_i - v_i-1 )≥ arg (u_i - u_i-1) for i ∈{ 1, 2, …, m }.


Then the broken lines determined by { u_0, u_1, … , u_m     } and { v_0, v_1, … , v_m      } do not cross (see figure <ref>) and v_m - u_m is almost vertical (its argument lies in the interval (  π/2 - π/10 , π/2 + π/10) ) 














We will prove this lemma by induction on m. The case m=1 is elementary plane geometry.


Suppose it holds for m=ν and take m = ν +1. Construct another sequence { w_1, w_2, …, w_m   } such that w_1 = v_1 and u_j+1u_jw_jw_j+1 is a parallelogram for j ∈{ 1, 2, … , m-1}. Applying the induction hypothesis to { w_1, w_2, … w_m   } and {v_1, v_2, …, v_m  }, we see that they do not cross and arg( v_m - w_m) ∈(  4 π/10 , 6π/10). Also, arg(w_m - u_m )= arg (w_1 - u_1) = arg (v_1 - u_1) ∈(  4 π/10 , 6π/10), which implies arg(v_m - u_m)∈(  4 π/10 , 6π/10).










Let S be a flat surface homeomorphic to a closed disc. Consider a map f  S →ℝ^2 such that restricted to the interior of S is an isometric immersion. Then f is injective if and only if f(∂ S) does not self intersect. 




One implication is trivial. Since local isometries are conformal maps, the other one follows from the fact that for conformal maps f  S →ℝ^2, the number of preimages f^-1(x) equals the winding number I(f(∂ S), x ) for all x ∈^2  \ f(∂ S), which is a standard fact in complex analysis (<cit.>, p. 384). If f(∂ S) does not self intersect, by the Jordan Curve Theorem <cit.> the winding number I(f(∂ S), x ) equals 0 or 1, then the function is injective.






§ THE CONSTRUCTION





To obtain a net one has to cut a convex polyhedron P along a spanning tree T of the graph of its edges. This way we obtain a flat surface P_T, which is homeomorphic to a closed disc. The surface P_T can be mapped isometrically face by face into the plane in an essentially unique way; denote this map by f_T P_T→ℝ^2. 


If f_T is injective then the image f_T(P_T) is a net of P. However f_T might not be injective as figure <ref> shows. Therefore Shephard's problem is asking if for any convex polyhedron P there is a spanning tree T such that f_T is injective.




Since the number of edges of P is finite, we can rotate it so that none of its edges is orthogonal to the x axis. Then, after a suitable stretching in the direction of the x axis, all the edges of the obtained polyhedron Q  form a sufficiently small angle with the x axis (less than π/20 N will do, where N is the number of edges of P).


We now define an ordering on the set of vertices of Q. We say that v ≤ v^' if the first coordinate of v is less or equal than the first coordinate of v^'. We will denote by y and z the minimal and maximal vertices respectively.

We define an ascending sequence as a set of vertices { p_0, p_1, … , p_n   } such that p_i ≤ p_i+1 and p_i p_i+1 are connected by an edge for all i ∈{ 0,1, …, n-1 }. We say that a spanning tree T with root z is increasing if for any vertex v∈ Q there is a (unique) ascending sequence from v to z contained in T. 

Note that all terminal edges of an increasing tree form an angle close to π with e_1. Otherwise, the path in T joining the corresponding leaf to z wouldn't be an ascending sequence (see figure <ref>). Also, all the ends of the surface Q \ T point in the direction of the x axis. 








The following theorem clearly implies theorem <ref>.






(Ghomi) If Q is cut along any increasing tree T, then the unfolding map f_T is injective.







First, choose an unfolding f_T of Q_T in which the images of all the edges are almost horizontal (they form an angle less than π/10 with e_1 ∈ ^2). Next, we are going to prove that f_T(∂ Q_T) does not self intersect.

Consider y^' = f_T(y) ∈ ^2. Then, starting at y^', the counterclockwise image of ∂ Q_T is a sequence of piecewise linear curves (broken lines) going almost horizontally rightwards and leftwards, alternately. We are going to denote the first broken line going right as R_1, the first one going left as L_1, and so on. Since T is ascending, when we switch from going rightwards to leftwards, we rotate counterclockwise and when we switch from going leftwards to rightwards, we rotate clockwise (see figure <ref>). 









As defined above, if the sequence of broken lines R_1L_1R_2… 
L_n doesn't self intersect, then R_1L_1 R_2 … L_nR_n+1 doesn't self intersect either.  





The proof goes by induction on the number n of times it has gone leftwards (in figure <ref>, n=3). For the case n=1, observe that the condition arg (v_i - v_i-1 )≥ arg (u_i - u_i-1) for i =1, 2, …, m in the arm lemma is implied by arg (v_1 - v_0 )≥ arg (u_1 - u_0) and ∠ v_i+1v_iv_i-1+ ∠ u_i-1u_iu_i+1≤ 2π for i∈{ 1,2, …, m-1}. Therefore the arm lemma completes the base of induction.

Suppose the assertion is true for n ≤ k and consider the case n=k+1.  Note that the edges of ∂ Q_T are paired in such a way that we glue paired edges together to obtain Q from Q_T. Each one will be called the dual of the other.

Observe that when we start R_2, we are traveling the dual edges of the leftmost part of L_1. If the length of R_2 is greater than or equal to the length of L_1, we can apply the induction hypothesis to R_2L_2R_3 … L_nR_n+1 and the result will follow. 

If the length of R_2 is less than the length of L_1, we extend R_2 with a broken line S parallel to the portion of L_1 without the dual of R_2 (see figure <ref>). By the arm lemma, S will be above L_1, and by the induction hypothesis S does not touch L_2R_3 … L_nR_n+1.










Now, the image of ∂  Q_T will self intersect for the first time in a point y^'' while going leftwards. Because T is increasing, f_T(∂ Q_T) contains a simple closed curve γ starting at y^'' consisting of a sequence of broken lines turning clockwise when changing from going leftwards to rightwards and counterclockwise in the other case.







Since all ends point rightwards, we can contract the curve γ to the point  y^'' moving leftwards all the time. Such contraction can be performed in the same way in Q_T. Therefore only one point of ∂ Q _T is sent to y^''. This is only possible if y^' = y^'' and f_T restricted to ∂ Q_T is injective.


Applying lemma <ref> with S=Q_T we get that f_T is a net, completing the proof.


As an obvious consequence of the proof we get the following result.



(Ghomi) For any convex polyhedron, a sufficiently long stretch in almost any direction yields a polyhedron with a net.






9

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