On the Critical Lattices of a Sphere and Four‐Dimensional Hypersphere

Mr. E. S. Barnes of Trinity College, Cambridge, has pointed out to me that in finding the critical lattices of the three‐† and four‐‡ dimensional spheres I have assumed that the independent lattice points on the boundary form bases. My proof, (I), for the sphere S : x 1 2 + x 2 2 + x 2 3 ⩽ is valid since no use is made of the assumption that P 3 is on the surface of S . The proof holds without change, provided every critical lattice Λ 3 contains two points P 1 , P 2 on the surface of S which together with some other point P 3 form a basis of Λ 3 ; i.e. provided P 1 , P 2 form a basis of the two‐dimensional sub‐lattice Λ 1 of Λ 3 in the plane OP 1 P 2 . This is clearly true since, as is well known, (A) every two‐dimensional lattice Λ 2 which contains two independent points on the circumference of the unit circle with centre 0, and is admissible for this circle, is necessarily generated by these points. In passing we notice that, in (A), d (Λ 2 ) ⩽ 1. Similarly my proof for the hypersphere requires only that three of the four points generating the (four‐dimensional) critical lattice should be on the boundary. Thus it suffices to show that (B) every admissible three‐dimensional lattice, λ 3 , say, of the unit 3‐spherc, which contains three independent points on the surface of the sphere, is generated by these points. (We notice here also that, necessarily, d (Λ 3 ) ⩽ 1.) As in (I), choose the axes so that the plane x 3 = 0 contains two of the three independent lattice points on the surface of the sphere and denote by Λ 2 the two‐dimensional sub‐lattice of Λ 3 in this plane. Then, for some k with 0 < k < 1, the lattice Λ 3 is built up of an infinity of two‐dimensional lattices Λ 3 , say, in the planes x 3 = nk ( n = 0, ±1, ±2, …). Now Λ 2 contains two independent lattice points on the circumference of x 1 2 + x 2 2 ⩽ 1, x 3 = 0. Thus d (Λ 2 ) ⩽ 1, and, by (A), these points generate Λ 2 , so that d (Λ 3 ) = kd (Λ 2 ). If k > ½, the only points of Λ 3 on the surface of the sphere lie in the planes x 3 = 0 or ± k , and so (B) holds. If k ⩽ ½, then d (Λ 2 ) ⩽ ½ d (Λ 2 ) ⩽ ½ < √½; whereas it was shown in (I) that, if Λ 3 is admissible, d (Λ 3 ) ⩾ √½. This establishes (B) and completes the proof given in (II) for the fourdimensional hypersphere. The process can be carried a stage further. Using the results of (II) we can show in the same way that (C) if Λ 4 is an admissible lattice of the unit 4‐sphere, containing a set of four independent points on the boundary, then, with one exception, the set forms a basis of Λ 4 . Let Λ 3 be the sub‐lattice of Λ 4 in x 4 = 0 (the axes having been suitably chosen). Then, by (B), if d (Λ 4 ) = kd (Λ 2 ), (C) is certainly true if k > ½. But d (Λ 3 ) ⩽ 1. Hence, if k ⩽ ½, d (Λ 4 ) ⩽ ½, which, from the results of (II), can hold if and only if k = ½ and Λ 4 is the critical lattice x 1 = ζ 1 +½ζ 4 , x 2 = ζ 2 +½ζ 4 , x 3 = ζ 3 +½ζ 4 , x 4 = ½ζ 4 , defined in II(36). This lattice contains the points of intersection with the boundary of the set of four mutually perpendicular axes of coordinates, and these points do not form a basis. This gives the single exception. Even in this case, however, there are other sets of four independent lattice points on the boundary which do form a basis, for example the points (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (½, ½, ½, ½). I am grateful to the referee for suggestions as to the wording of this note.