Lecture 35: Hilbert’s Third Problem

Lecture 35: Hilbert’s Third Problem 35 35.1 Hilbert’s Third Problem Polygons in the Plane Defnition 35.1 Given polygons P and Q on the plane, P is scissors-congruent to Q (denoted P ∼ Q) if we can divide P , using fnitely many straight cuts, into a set of polygons R1 through Rn ; and we can divide Q into the same collection R1 , . . . , Rn . Example 35.2 This triangle and quadrilateral are scissors-congruent: Guiding Question Given two polygons P and Q, when is P ∼ Q? There’s one obvious obstruction: we need P and Q to have the same area, since the area is preserved by rearranging pieces. This is actually an if and only if condition: Theorem 35.3 If P and Q have the same area, then P ∼ Q. Proof Outline. The idea is to rearrange both P and Q to a rectangle with dimensions 1 × area(P ), where A is the common area – if P and Q are scissors-congruent to the same rectangle, then they’re scissors-congruent to each other. First, cut P into triangles T1 , . . . , Ts . Then each triangle T is scissors-congruent to some rectangle: cut the triangle as shown, and move the red pieces to the blue triangles. Then we can show that any rectangle R is scissors-congruent to another rectangle with dimensions 1 × area(R). This part is a bit fnicky and involves a bunch of cases, so we’ll skip it. Finally, we have a bunch of triangles, which are each scissors-congruent to a rectangle of height 1. So we can concatenate all these rectangles, to get a bigger rectangle of height 1. This means P is scissors-congruent to a 1 × area(P ) rectangle, and by performing the same argument for Q, then P ∼ Q. 35.2 The Question We can now consider what happens in 3 dimensions. Defnition 35.4 If P and Q are polytopes (meaning they have fnitely many vertices, edges, and faces, and each face is a polygon), then P ∼ Q if you can use fnitely many straight cuts to decompose P and Q into the same polytope pieces. Lecture 35: Hilbert’s Third Problem Again, there’s an obvious condition: if P ∼ Q, then they must have the same volume. Guiding Question (Hilbert’s Third Problem) If two polytopes have the same volume, are they scissors-congruent? In 1900, David Hilbert made a list of around twenty problems, which he considered the most important problems in modern mathematics. These problems were very infuential. This question was on that list, and he was pretty sure that the answer was no. In fact, this was the frst one of his questions to be answered (in 1901), by his student Max Dehn. He showed more precisely that a cube and a tetrahedron of the same volume are not scissors-congruent. 35.3 Some Algebra At the heart of this problem is a certain algebraic construction. Defnition 35.5 Given two abelian groups G and H, their tensor product G⊗H is the abelian group generated by elements of the form g ⊗ h for g ∈ G and h ∈ H, satisfying the relations (g + g ′ ) ⊗ h = g ⊗ h + g ′ ⊗ h, and similarly g ⊗ (h + h′ ) = g ⊗ h + g ⊗ h′ . You can think of G ⊗ H as taking M Z(g ⊗ h) g,h (which gives an integer for every pair (g, h)), and then quotienting out by the subgroup generated by ((g + g ′ ) ⊗ h) − (g ⊗ h) − (g ′ ⊗ h) and so on. This gives a group; and because we’ve quotiented out by the subgroup, now our generators satisfy the given relations. L Student Question. Is the direct sum Z(g ⊗ h) the free group? Answer. Not exactly – it’s a free abelian group, since everything commutes. But if you’d like, you can instead start with the free group, and throw the condition that everything commutes into the quotient as well. We can think of elements of G ⊗ H as expressions combining the terms g ⊗ h, which we can simplify using the given relations. Proposition 35.6 The defnition has a few consequences: • 0 ⊗ h = g ⊗ 0 = 0. • If a ∈ Z, then (ag) ⊗ h = a(g ⊗ h) = g ⊗ ah (by using linearity). • If we take lists of generators G = ⟨g1 , . . . , gr ⟩ and H = ⟨h1 , . . . , hs ⟩, then G ⊗ H = ⟨gi ⊗ hj ⟩. Example 35.7 We have Z ⊗ G ∼ = G. Proof. We must have (a ⊗ g) 7→ ag = 1 ⊗ ag. (So the isomorphism sends (a ⊗ g) to ag, and its inverse sends g to 1 ⊗ g.) Example 35.8 We have (Z2 ) ⊗ G ∼ = G × G. Lecture 35: Hilbert’s Third Problem Proof. We must have (a, b) ⊗ g 7→ (ag, bg). Note that in general, not everything in G ⊗ H is of the form g ⊗ h: you can have expressions g1 ⊗ h1 + g2 ⊗ h2 which you can’t simplify any further using these relations. So this isn’t the same as looking at ordered pairs. Example 35.9 We have C2 ⊗ C3 = 0. Proof. Take x ⊗ y for any x ∈ C2 and y ∈ C3 . We have 3x = x, so then x ⊗ y = 3x ⊗ y = x ⊗ 3y = x ⊗ 0 = 0, since 3y = 0 in C3 . So you can start with two nontrivial groups and take their tensor product, and get the 0 group. If we just took the product, that would be a nontrivial group of size 6. So the tensor product is somewhat subtle. Student Question. Does this happen with 2 and 3 replaced by any distinct positive integers? Answer. Not necessarily distinct, but it does happen if the integers are relatively prime. 35.4 Back to Polytopes If we have two polytopes of the same volume, we need another way to see they aren’t scissors-congruent. Given a polytope P , each edge has a length ℓ, and a dihedral angle θ – where we take the perpendiculars to the edge on each face, and look at the angle between them. θ ℓ Then ℓ and θ are real numbers, so we can take the tensor product ℓ ⊗ θ ∈ R ⊗ R/2πZ. (Here R and R/2πZ are both uncountably generated infnite groups.) Then every edge gives an element in this tensor product, so we can sum over the edges: Defnition 35.10 The Dehn invariant of a polytope P is d(P ) = X ℓi ⊗ θi ∈ R ⊗ R/2πZ, i where the sum is taken over all edges of P . We’ll see that the Dehn invariant is preserved by scissors-congruence – then we can use it to tell whether two polytopes are scissors-congruent, similarly to how we can use volume. And it turns out that we can actually calculate it for some examples – a cube and regular tetrahedron – and see that they give diferent results. Theorem 35.11 The Dehn invariant is preserved by scissors-congruence: if P ∼ Q, then d(P ) = d(Q). Lecture 35: Hilbert’s Third Problem Proof. Consider what happens to the polytope when we cut it. First, we’ll look at what happens to the original edges. One possibility is that the edge is cut into two pieces, while the dihedral angle on each side remains the same: ℓ2 θ ℓ 1 Then we have ℓ = ℓ1 + ℓ2 , so ℓ ⊗ θ = ℓ1 ⊗ θ + ℓ2 ⊗ θ. Another possibility is that we cut by a plane that contains this edge. Then the entire edge remains intact, but the angles change: now we have two polytopes each with an edge of length ℓ, and angles θ1 and θ2 respectively. θ1 θ2 In this case, we have θ = θ1 + θ2 , so ℓ ⊗ θ = ℓ ⊗ θ1 + ℓ ⊗ θ2 . In both cases, we used the linear property of the tensor product. Meanwhile, we can also create new edges: when we make a cut along a face, this produces an edge on each of the two pieces. Lecture 35: Hilbert’s Third Problem θ1 θ2 ℓ Then we had a contribution of 0 previously, and a contribution of ℓ ⊗ θ1 + ℓ ⊗ θ2 after the cut. But we have θ1 + θ2 = π, so ℓ ℓ ⊗ θ1 + ℓ ⊗ θ2 = ℓ ⊗ π = ⊗ 2π = 0 2 (since the second group is R/2πZ). The fnal case is if we have a new edge in the interior of the polytope. Then we have a bunch of angles which add up to 2π: So the same thing happens: when we add up its contributions, we have ℓ ⊗ θ1 + · · · + ℓ ⊗ θk = ℓ ⊗ 2π = 0. So in all cases, the Dehn invariant is preserved by cutting the polytope. Student Question. Does the argument used to show ℓ ⊗ π = 0 imply that any ℓ ⊗ θ is 0? Answer. No – we can only scale by integers. So this shows that if θ is a rational multiple of π, then ℓ ⊗ θ = 0. But we’ll see later that if θ isn’t a rational multiple of π, we get something nonzero. But this is something to watch out for – it happens a scary number of times that someone defnes a complicated invariant, but it turns out the invariant is always 0. Note 35.12 We haven’t thoroughly gone through all possibilities, but this gives a favor of why the theorem is true – in every case, if we compare the old and new contributations, then the relations we quotiented out by imply that the invariant doesn’t change. Now we’ll show that the invariant actually does something interesting. Theorem 35.13 If C is a cube and T is a regular tetrahedron, then d(C) = d(T ). Proof. A cube has 12 edges, each of which has dihedral angle d(C) = 12ℓ ⊗ π 2. So we get π = ℓ ⊗ 6π = 0. 2 Lecture 35: Hilbert’s Third Problem So the Dehn invariant of a cube is always 0. For a regular tetrahedron, we have 6 edges which are symmetric, so we get d(T ) = 6ℓ′ ⊗ α for some angle α. α To fnd α, we can drop an altitude. This altitude ends at the center of the base: So the short side of the right triangle is a third of the height of the base. h α 1 h 3 Then, since the two triangles have the same height, we have cos α = 1 . 3 So we’ve found α, and now we want to show that for this value of α, the Dehn invariant is nonzero. Claim 35.14. α is not a rational multiple of π. Proof Outline. One way to prove this is by trig identities – suppose α = identities to write cos bα = (cos α)b 2b + · · · . 2aπ b , and then use trigonometric Then there’s an enormous power of 3 in the denominator, since cos α = 13 , so this can’t be 1. Claim 35.15. For any α ̸∈ Qπ, and any ℓ ∈ R, ℓ ⊗ α is nonzero. Lecture 35: Hilbert’s Third Problem Proof. Think of R as a vector space over Q (of uncountable dimension). Then π and α are linearly independent, so we can fll them out into a basis: we can write R = Qα ⊕ Qπ ⊕ W, where W comes from the uncountably many remaining basis elements. Then we can defne a linear map f : R → Q (a map of Q-vector spaces) sending α to 1, and every other basis element to 0 – we can do this because to defne a linear map, it sufces to defne it on each basis element. Then we have a group homomorphism g : R ⊗ R/2πZ → R, where z ⊗ x 7→ zf (x̃), where x̃ is x mod 2π – since 2π ∈ ker(f ), this is well-defned. (We can also check that this is compatible with the relations of the tensor product.) Now we have g(ℓ ⊗ α) = ℓ ̸= 0. And since g(ℓ ⊗ α) is nonzero, then ℓ ⊗ α must be nonzero as well. So d(T ) = ℓ′ ⊗ α is nonzero, which means d(C) = ̸ d(T ). The Dehn invariant lies in some enormous group. In this proof, in order to show it was nonzero, we mapped it to a much simpler group – the real numbers – and showed instead that its image is nonzero. Note 35.16 This was proven in 1901. In 1968 Sydler showed the converse – that if P and Q have the same volume and Dehn invariants, then they’re scissors-congruent. This is known in dimension 4 as well. But in higher dimensions, it’s not actually known how to characterize when two polytopes are scissors-congruent.