[math-fun] A pentatope (not Heronian --- relax!)
Consider the 4-space simplex PQRST having edge-lengths QP = 1, RP = sqrt(2), RQ = sqrt(2), SP = sqrt(3), SR = sqrt(3), SQ = sqrt(3), TP = 2, TQ = sqrt(2), TR = 1, TS = 1. Let lattice vertices be labelled even / odd according to the sum of their Cartesian coordinate components. The triangle PQR has sides 1,sqrt(2),sqrt(2): if it is embedded in the lattice with P is even then Q is odd, so R is odd, so P is odd. By contradiction, it must be impossible to embed PQRST in the lattice |Z^4. However, PQRST does have the rational embedding in |Q^4 P = [0,0,0,0], Q = [0,0,0,1], R = [1,0,1,0], S = [1,1,0,0], T = [5,1,1,0]/3. Why am I telling you this? The proofs that a Heronian simplex is lattice embeddable in 2 and 3 dimensions require only that the simplex be rationally embedded, and have edge lengths squaring to integers. This example shows that any proof of Heronian embeddability in |Z^4 would have to somehow involve extra properties besides these two. [That's always assuming, of course, that somebody can manage to find any to embed: Sascha Kurz has recently calculated that no Heronian pentatope can exist with diameter < 600,000.] Incidentally, I have prepared a proof of the construction of lattice embeddings in 2- and 3-space via complex-number and quaternion GCD. I'll try to upload this somewhere public shortly --- in the meantime, anyone interested is welcome to request a copy (don't all rush at once!). Fred Lunnon
Sorry everybody --- my edge order got transposed --- it's all complete twaddle. Back to the drawing board! WFL On 2/5/12, Fred lunnon <fred.lunnon@gmail.com> wrote:
Consider the 4-space simplex PQRST having edge-lengths QP = 1, RP = sqrt(2), RQ = sqrt(2), SP = sqrt(3), SR = sqrt(3), SQ = sqrt(3), TP = 2, TQ = sqrt(2), TR = 1, TS = 1.
Let lattice vertices be labelled even / odd according to the sum of their Cartesian coordinate components. The triangle PQR has sides 1,sqrt(2),sqrt(2): if it is embedded in the lattice with P is even then Q is odd, so R is odd, so P is odd. By contradiction, it must be impossible to embed PQRST in the lattice |Z^4.
However, PQRST does have the rational embedding in |Q^4 P = [0,0,0,0], Q = [0,0,0,1], R = [1,0,1,0], S = [1,1,0,0], T = [5,1,1,0]/3.
Why am I telling you this? The proofs that a Heronian simplex is lattice embeddable in 2 and 3 dimensions require only that the simplex be rationally embedded, and have edge lengths squaring to integers. This example shows that any proof of Heronian embeddability in |Z^4 would have to somehow involve extra properties besides these two.
[That's always assuming, of course, that somebody can manage to find any to embed: Sascha Kurz has recently calculated that no Heronian pentatope can exist with diameter < 600,000.]
Incidentally, I have prepared a proof of the construction of lattice embeddings in 2- and 3-space via complex-number and quaternion GCD. I'll try to upload this somewhere public shortly --- in the meantime, anyone interested is welcome to request a copy (don't all rush at once!).
Fred Lunnon
Deep breath and start again ... The good news is that in the meantime I found an even more compact example. As previously, each triangle is separately embeddable; not sure about solids. Squared edge-lengths in order (this time!) QP,RP,RQ,SP,SQ,SR,TP,TQ,TR,TS; 1, 2, 3, 2, 3, 2, 1, 2, 1, 1. Cartesian coordinates of a rational embedding of vertices: P,Q,R,S,T -> (0,0,0,0), (1,0,0,0), (0,1,1,0), (0,1,0,1), (0,1,2,2)/3. Proof that there is no lattice embedding: By parallelepiped determinant, the volume equals 1/24. We may choose lattice axes so that T = (0,0,0,0). Now TS^2 = TR^2 = TP^2 = 1, therefore choose S = (0,0,0,1), R = (0,0,1,0), P = (1,0,0,0). Also QP^2 = 1, and the volume is nonzero; but col 2 remains vacant, therefore must choose Q = (0,1,0,0). This implies RQ^2 = 2, contradicting RQ^2 = 3 above. QED (mopping brow) Finally, the counterexample stands up, and lattice embedding in 4-space falls over --- or at least, looks pretty wobbly. Fred Lunnon On 2/6/12, Fred lunnon <fred.lunnon@gmail.com> wrote:
Sorry everybody --- my edge order got transposed --- it's all complete twaddle.
Back to the drawing board! WFL
On 2/5/12, Fred lunnon <fred.lunnon@gmail.com> wrote:
Consider the 4-space simplex PQRST having edge-lengths QP = 1, RP = sqrt(2), RQ = sqrt(2), SP = sqrt(3), SR = sqrt(3), SQ = sqrt(3), TP = 2, TQ = sqrt(2), TR = 1, TS = 1.
Let lattice vertices be labelled even / odd according to the sum of their Cartesian coordinate components. The triangle PQR has sides 1,sqrt(2),sqrt(2): if it is embedded in the lattice with P is even then Q is odd, so R is odd, so P is odd. By contradiction, it must be impossible to embed PQRST in the lattice |Z^4.
However, PQRST does have the rational embedding in |Q^4 P = [0,0,0,0], Q = [0,0,0,1], R = [1,0,1,0], S = [1,1,0,0], T = [5,1,1,0]/3.
Why am I telling you this? The proofs that a Heronian simplex is lattice embeddable in 2 and 3 dimensions require only that the simplex be rationally embedded, and have edge lengths squaring to integers. This example shows that any proof of Heronian embeddability in |Z^4 would have to somehow involve extra properties besides these two.
[That's always assuming, of course, that somebody can manage to find any to embed: Sascha Kurz has recently calculated that no Heronian pentatope can exist with diameter < 600,000.]
Incidentally, I have prepared a proof of the construction of lattice embeddings in 2- and 3-space via complex-number and quaternion GCD. I'll try to upload this somewhere public shortly --- in the meantime, anyone interested is welcome to request a copy (don't all rush at once!).
Fred Lunnon
participants (1)
-
Fred lunnon