Canad.Math. Bull. Vol. þ’ (Ë), z§Ë@ pp. Ëzh–Ëhþ http://dx.doi.org/˧. Ëþh/CMB-z§Ëþ-§@@- ©Canadian Mathematical Society z§Ëþ

Discrete Space-time and Lorentz Transformations

Gerd Jensen and Christian Pommerenke

Abstract. Alfred Schild established conditions where Lorentz transformations map world-vectors (ct, x, y, z) with integer coordinates onto vectors of the same kind. _e problem was dealt with in the context of tensor and spinor calculus. Due to Schild’s number-theoretic arguments, the subject is also interesting when isolated from its physical background. Schild’s paper is not easy to understand. _erefore, we ûrst present a streamlined version of his proof which is based on the use of null vectors. _en we present a purely algebraic proof that is somewhat shorter. Both proofs rely on the properties of Gaussian integers.

1 Introduction _e points of the Minkowski space-time and the Lorentz transformations are usually ⊺ ′ written in the form X⃗ = (t, x, y, z) ∈ R and X⃗ = LX⃗, where L is a real × matrix ′ ′ ′ ′ and t z − x z − y z − z z = tz − xz − yz − zz. _e speed of light is here assumed to be Ë. An equivalent representation is obtained by the vector space ⎧ ⎫ ⎪ t + z x + iy ⎪ (Ë.Ë) M ∶= ⎨ X = Œ ‘ ∶ t, x, y, z ∈ R⎬ ⎪ x − iy t − z ⎪ ⎩⎪ ⎭⎪ of Hermitian matrices X. _ey satisfy det X = tz −xz −yz −zz. Given such a Hermitian ⊺ matrix X, write X⃗ for the associated Lorentz vector (t, x, y, z)⊺ .If ⎧ ⎫ ⎪ a b ⎪ (Ë.z) Γ ∶= ⎨ A = Œ ‘ ∶ a, b, c, d ∈ C, S det AS = Ë⎬ ⎪ c d ⎪ ⎩⎪ ⎭⎪ ′ ∗ and A ∈ Γ, then X ∶= AXA is again Hermitian and has a form analogous to X in (Ë.Ë) ′ ′ ′ ′ ′ with uniquely determined t , x , y , z . Because det X = det X, the linear mapping ′ ∗ (Ë.h) X = AXA is a Lorentz transformation. It uniquely determines a real matrix such that × LA (Ë. ) ′ . X⃗ = LAX⃗ In [6], A. Schild found conditions that all components of are integers; an equiv- LA alent formulation is that maps the lattice onto itself. His main result is the LA Z following theorem.

Received by the editors January ˒, z§Ë . Published electronically October zË, z§Ëþ. AMS subject classiûcation: zzE h, z§H’’,ÇhA§þ. Keywords: Lorentz transformation, integer lattice, Gaussian integers. Ëzh

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_eorem Ë.Ë (Schild) (i) Let A ∈ Γ, det A ∈ {Ë, i, −Ë, −i}, and either (a) a, b, c, d ∈ Z + iZ and SaSz + SbSz + ScSz + SdSz even or (b) a, b, c, d i Ë Ë i . ∈ Z + Z + z ( + ) (Ë. ) _en LA deûned by has integer components. (ii) Γ Ë For every B ∈ such that the components of LB are integers, there is u ∈ C, SuS = such that A ∶= uB fulûlls the conditions in (i).

If η ∈ {Ë, i, −Ë, −i}, and A satisûes the conditions in (i), then so does ηA. Because det(ηA) = ηz det A = ± det A, one can assume that det(ηA) = Ë or det(ηA) = i. _erefore the conclusion det A Ë, i, Ë, i in (ii) can be replaced with det A Ë, i . ∈ { − − } ∈ { √} _is is no loss of generality for multiplication with a suitable power of (Ë + i)~ z converts A ∈ Γ into an element of SL(z, C). For Schild the physical aspects were essential, so the problem was dealt with in the context of tensor and spinor calculus. However, mainly because of Schild’s num- ber theoretic arguments, the subject is also interesting when isolated from its physical background. _erefore, a streamlined presentation, which is reduced to the quintes- sential mathematical features, might be useful. In the following, Schild’s proof is presented using only simple matrix calculus. His method is geometrical in nature; the problem is translated to mapping properties of the spin transformations A. Additionally, a more direct, algebraic proof is given. It must be emphasized, however, that Schild’s method points to aspects and problems that could hardly be found in the course of a diòerent reasoning. Some of these, e.g., the question of the equivalence of timelike vectors with integer components under Lorentz transformations with integer components, are dealt with in [6]; further, there are outlined unpublished results of H. S. M. Coxeter concerning the generators of the discrete group; see also [ ]. Both proofs rest upon the unique factorization in the ring of Gaussian integers. _e crucial step in revealing the structure of the transformation matrices is in each case the comparison of the factorizations of a number that is both a square and a product of two diòerent numbers (in the proofs of _eorem h. and Lemma .h). At the outset, integral Lorentz matrices are deûned as those elements of Z × that satisfy the ˧ orthogonality conditions for the columns. _eorem Ë.Ë disentangles the complexity signiûcantly and brings it down to Ç integer parameters and z constraints; what is more, the smaller set of parameters allows deeper insight into the group struc- ture. Another interesting and explicit method was presented by J. D. Louck in [þ], based on the biquaternionic parametrization of SL(z, C).

2 Preliminaries 2.1 Four-vectors _e connection between (Ë.h) and (Ë. ) can be made explicit by means of the Kro- necker product and the vectorization of matrices ([h,Chap. ]). For z × z matrices A aËË aËz and B, the Kronecker product A B is the block matrix aËË B aËz B , = ‰ azË azz Ž ⊗ × ‰ azË B azz B Ž and vec is the vector , , , ⊺ obtained by stacking the columns of on A (aËË azË aËz azz) A

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top of one another.If C, D, X are also z × z matrices, then ⊺ (z.Ë) vec(AXB ) = (B ⊗ A) vec X, (z.z) (A ⊗ B)(C ⊗ D) = AC ⊗ BD, z.h) det(A ⊗ B) = (det A)z(det B)z.( _e points of space-time have the equivalent representations X, X⃗, and vec X = ⊺ (t + z, x − iy, x + iy, t − z)⊺ . _e vectors X⃗ and vec X are connected by a unitary matrix T: Ë § § Ë vec Ë ⎛§ Ë i § ⎞ X ⎜ − ⎟ (z. ) √ = TX⃗ with T ∶= √ ⎜ ⎟ . z z ⎜§ Ë i § ⎟ ⎝Ë § § −Ë⎠ Using (z.Ë), relation (Ë.h) becomes

′ vec X = (A ⊗ A) vec X, and with (z. ), it follows that the matrix in (Ë. ) has the representation LA (z.þ) ∗ . LA = T (A ⊗ A)T From (z.þ), (z.h), and (Ë.z) one obtains (z.@) det Ë for Γ. LA = A ∈ With (z.z) it follows from (z.þ) that , −Ë . LBA = LB LA LA = LA−Ë By substituting special matrices X in (Ë.h), one easily shows the following lemma. Ë Lemma z.Ë LA = LB holds if and only if B = uA with u ∈ C, SuS = . From (z.þ) (in its expanded form ( .h)) and (z.@) it is seen that belongs to the LA group SO+ Ë, h of restricted Lorentz transformations. Actually the mapping ( ) A → LA + produces a double cover of SO (Ë, h) by SL(z, C) (see, for instance, [Ë, Sec. @.h]), + so _eorem Ë.Ë characterizes all members of SO (Ë, h) with integer components. For technical reasons, following Schild, we deal with Γ instead of SL(z, C), because this allows for a smaller number of case distinctions. Moreover, multiplication with a suit- √ able power of (Ë + i)~ z converts A ∈ Γ into an element of SL(z, C).

2.2 Gaussian Integers Let Z[i] = Z + iZ denote the ring of Gaussian integers and E = {Ë, i, −Ë, −i} its group of units. _e associates of x ∈ Z[i] are x, ix, −x, and −ix. _e prime elements of Z[i] are called Gaussian primes. _e Gaussian integers form a unique factorization domain ([z, Sec.Ëz.Ç]), which will be simply referred to by UFD; two factorizations of an integer are associates of each other. A natural number that is prime in Z is a Gaussian prime if and only if it is congruent to −Ë (mod ) (see [z, Sec.Ëþ.Ë]); otherwise, it is the product of two complex conjugate Gaussian primes. If the greatest common divisor of real integers r, s, . . . is determined in Z and in Z[i],

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the results are associates; the positive value will be denoted by gcd(r, s, ... ). For X as in (Ë.Ë) with t, x, y, z ∈ Z, let (z.6) δ(X) ∶= gcd(t, x, y, z). Because z = (Ë+i)(Ë−i), all real Gaussian primes are odd. For w ∈ Z[i] the product ρ(w) of its positive Gaussian prime factors is odd and has a factorization (z.Ç) ρ(w) = σ(w)r(w)z with square-free σ(w). All factors of ρ(w) are congruent to −Ë (mod ). For a ∈ Z[i] the parity π(a) ∈ {§, Ë} is deûned by π(a) ≡ Re a + Im a (mod z). If π(a) = §, then a is called even; if π(a) = Ë, then a is called odd. _at extends the terminology for Z, for Im a = § if a ∈ Z, hence π(a) = § for even and π(a) = Ë for odd . Furthermore, is a ring homomorphism i . _e set a π Z[ ] → Zz (z.’) i z , ⊺ Z[ ]§ ∶= {(ξ§ ξË) ∶ π(ξ§) = π(ξË)} is a submodule of the Z[i]-module Z[i]z.It contains the vectors Ë i § Ë (z.˧) + , , ; e§ ∶= Œ ‘ eË ∶= Œ ‘ ez ∶= Œ ‘ § Ë + i Ë , is a basis of i z. Each i has a representation Ë with {e§ ez} Z[ ]§ w ∈ Z[ ] w = ( + i)ŵ + ε uniquely determined ŵ ∈ Z[i] and ε ∈ {§, Ë}. _en ε = π(w), hence π(w) = § if and only if (Ë + i)S w. Moreover, (z.ËË) , ⊺ i z Ë ̂ , Ë ̂ . (ξ§ ξË) ∈ Z[ ]§ ⇐⇒ ξ§ = ( + i)ξ§ + ε ξË = ( + i)ξË + ε In _eorem Ë.Ë, the parity requirement in (a) is automatically fulûlled in (b).

Let a, b, c, d i Ë Ë i and ad bc . _en π a b c d §. Lemma z.z ∈ Z[ ]+ z ( + ) − ∈ E ( + + + ) =

Proof Write a a′ Ë Ë i with a′ i and likewise for b, c, d. _en = + z ( + ) ∈ Z[ ] ad bc a′d′ b′c′ Ë Ë i a′ d′ b′ c′ , − = − + z ( + )( + − − ) ′ ′ ′ ′ a + b + c + d = a + b + c + d + z(Ë + i) ′ ′ ′ ′ ′ ′ = (Ë − i)‰ ad − bc − (a d − b c )Ž + z(b + c ) + z(Ë + i). Since ad −bc ∈ E, this is in Z[i] and is divisible by Ë+i, hence π(a +b +c +d) = §.

2.3 Integrity

If X ∈ M has integer components, then the components of X⃗ are integers if and only if the diagonal elements of X are both even or both odd.Let (z.Ëz) , M§ ∶= {X ∈ M∶ X⃗ ∈ Z }

z.Ëh) Γ Γ .( § ∶ = {A ∈ ∶ LA(Z ) ⊂ Z } _e next lemma follows immediately from the deûnition (z.Ëh).

Downloaded from https://www.cambridge.org/core. 27 Sep 2021 at 22:51:20, subject to the Cambridge Core terms of use. Discrete Space-time and Lorentz Transformations Ëz6 Γ Lemma z.h LA has integer components if and only if A ∈ §.

− − Clearly Γ is a semigroup. Because − Ë and (z.@), we have Ë Γ if Γ . § LA Ë = LA A ∈ § A ∈ § Hence, Γ is a group. §

3 Schild’s Proof _e fundamental entities in this proof are null vectors (see [@,Chap. I]). _ey span the Minkowski space-time and can be represented by vectors in Cz. Lorentz transfor- mations are already determined by their action on the null vectors and correspond to linear homogeneous transformations of Cz (spin transformations).

3.1 Null Vectors ∗ X ∈ M is called a null vector if tz − xz − yz − zz = §. From det(AXA ) = det X for A ∈ Γ, we have the following lemma.

′ ∗ Lemma h.Ë For A ∈ Γ and X a null vector, X = AXA is a null vector.

For ξ ∈ Cz, ξ ξ ξ ξ ξ (h.Ë) ξξ∗ § ξ ξ § § § Ë = Œξ ‘ ‰ § ˎ = Œ ‘ Ë ξË ξ§ ξË ξË is a null vector.If X is future-pointing (t ≥ §), the converse also holds.

∗ ∗ Lemma h.z A future-pointing null vector X has the form (h.Ë), and ̂ξ̂ξ = ξξ holds if and only if there is u ∈ C with SuS = Ë and ̂ξ = uξ.

If , then ∗ with √ and i √ ; if z z, Proof t > SzS X = ξξ ξ§ = t + z ξË = (x − y)~ t + z t = z then § and ∗ with √ and √ . _e simple proof of x = y = X = ξξ ξ§ = t + z ξË = t − z the second assertion is omitted.

Let be the set of future-pointing null vectors in (see (z.Ëz)), Deûnition h.h V§ M§ i.e., z z z z §, § . V§ ∶= {X ∈ M§∶ t − x − y − z = t ≥ }

If ξ ∈ Z[i]z, then ∗ i z . ξξ ∈ M§ ⇐⇒ ξ ∈ Z[ ]§

(i) i z ∗ _eorem h. If ξ ∈ Z[ ]§ and r ∈ N, then rξξ ∈ V§. (ii) i z Conversely, for each X ∈ V§ there is ξ ∈ Z[ ]§ such that ∗ (h.z) X = sξξ and s is a square-free product of positive Gaussian primes. _en ξ is uniquely determined up to a factor u ∈ C with SuS = Ë. Further, s = σ(δ(X)) (with δ and σ as deûned in (z.6) and (z.Ç)).

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Proof (i) is trivial. (ii) is proved in three steps. ′ ′ Ë. We suppose at ûrst that δ(X) = Ë. Let ρ(t + z) = σrz and ρ(t − z) = σ r z be the factorizations according to (z.Ç). _en

′ ′ ρ‰(t + z)(t − z)Ž = σσ rzr z. From UFD, it follows that ρ(x + iy) = ρ(x − iy), and because (h.h) (t + z)(t − z) = (x + iy)(x − iy), we must have ′ ′ σσ rzr z = ρ‰(t + z)(t − z)Ž = ρ(x + iy)z. ′ ′ As a consequence of UFD it follows that σ = σ and σrr S(x + iy). Furthermore, ′ ′ ′ σrr S(x − iy), so σrr S zx and σrr S zy. Now σ is odd, so σ S x and σ S y. From σ S t + z and σ S t − z it follows that σ S zt and σ S zz, hence σ S t and σ S z. But then σ = Ë, since δ(X) = Ë. ′ _e non-real prime factors of and of form pairs , and ′ , of t + z t − z (ζ j ζ j) (ζk ζ k ) complex conjugate numbers. With and ′ ′ then ζ = ∏ j ζ j ζ = ∏k ζk ′ ′ ′ ′ t + z = rzζζ = (rζ)(rζ) and t − z = r zζ ζ′ = (r ζ )(r′ζ′). From (h.h) and UFD, we may suppose that i , ′ i and i , ζ j S(x + y) ζk S(x + y) ζ j S(x − y) ′ i . From (h.h) follows further that i i ′. _en ζ k S(x − y) ρ(x + y) = ρ(x − y) = rr ′ ′ x +iy = u(rζ)(r ζ ) and x −iy = u(rζ)(r′ζ′), where u is a unit of Z[i]. Since δ(X) = Ë ′ ′ ⊺ and therefore s = σ(δ(X)) = Ë, (h.z) holds with ξ ∶= (ruζ, r ζ ) . From π(ruζ) = z ′z ′ ′ ′ ′ , it follows that i z. π(r ζζ) = π(t + z) = π(t − z) = π(r ζ ζ ) = π(r ζ ) ξ ∈ Z[ ]§ Now let be arbitrary and , hence Ë. _erefore, z. X ∈ V§ d = δ(X) δ(X~d) = ∗ for suitable , ⊺ i z.If z then ∗. X~d = ξξ ξ = (ξ§ ξË) ∈ Z[ ]§ ρ(d) = σ(d)r X = σ(d)rξ(rξ) Suppose also that ∗ with a square-free product of positive Gaussian h. X = sË ξË ξË sË primes. Since the highest power of a positive Gaussian prime that divides all compo- nents of ∗ or all components of ∗ is even, it must be . Now the uniqueness ξξ ξË ξË sË = s assertion follows from Lemma h.z.

Γ −Ë _eorem h.þ A ∈ § if and only if LA(V⃗ §) ⊂ V⃗ § and LA (V⃗ §) ⊂ V⃗ §.

If Γ , then and −Ë by (z.Ëh) and Lemma h.Ë. For Proof A ∈ § LA(V⃗ §) ⊂ V⃗ § LA (V⃗ §) ⊂ V⃗ § the converse we show that maps a basis of into . LA Z Z Now Ë, Ë, §, § ⊺ , Ë, Ë, §, § ⊺ , Ë, §, Ë, § ⊺ , and Ë, §, §, Ë ⊺ X⃗§ ∶= ( − ) X⃗Ë ∶= ( ) X⃗z ∶= ( ) X⃗h ∶= ( ) belong to . For §, Ë, z, h at least one component of , , , ⊺ V§ i = (ti xi yi zi ) ∶= LA(X⃗i ) must be odd, for otherwise z −Ë z z by hypothesis, which is X⃗i = LA (LA(X⃗i )~ ) ∈ Z wrong. Since the square of a real integer leaves a remainder of Ë or § on division by , it follows from z z z z that and exactly one of , , are odd. ti = xi + yi + zi ti xi yi zi Because of the invariance of the bilinear form associated with tz − xz − yz − zz, it follows that z. Since is odd, at least one of , , t§ tË − x§xË − y§ yË − z§zË = t§ tË x§xË y§ yË , say , is odd. _en , , , are odd and , , , must be even. Hence, z§zË x§xË t§ tË x§ xË y§ yË z§ zË , , and are all even, and t§ + tË x§ + xË y§ + yË z§ + zË Ë Ë Ë, §, §, § ⃗ ⃗ ⃗ ⃗ . LA( ) = LA‰ z(X§ + XË)Ž = z‰ LA(X§) + LA(XË)Ž ∈ Z

Downloaded from https://www.cambridge.org/core. 27 Sep 2021 at 22:51:20, subject to the Cambridge Core terms of use. Discrete Space-time and Lorentz Transformations Ëz’ _en §, Ë, §, § Ë, §, §, § , LA( ) = LA(X⃗Ë) − LA( ) §, §, Ë, § Ë, §, §, § , LA( ) = LA(X⃗z) − LA( ) §, §, §, Ë Ë, §, §, § LA( ) = LA(X⃗h) − LA( ) are also in Z .

3.2 Reformulation of the Problem Let ∆ Γ i z i z , −Ë i z i z , where Deûnition h.@ § ∶= ™ A ∈ ∶ A‰Z[ ]§Ž ⊂ Z[ ]§ A ‰Z[ ]§Ž ⊂ Z[ ]§ž i z is deûned in (z.’). Z[ ]§ ∆ Γ Γ Ë ∆ _eorem h.6 § ⊂ §. Conversely, for B ∈ § there is u ∈ C with SuS = and uB ∈ §.

Let ∆ and . By _eorem h. (ii) there exists i z such that Proof A ∈ § X ∈ V§ ξ ∈ Z[ ]§ ∗ X = σ(δ(X))ξξ and ′ ∗ ∗ ∗ ∗ X = AXA = σ(δ(X))A(ξξ )A = σ(δ(X))(Aξ)(Aξ) . Since i z by hypothesis, this implies by _eorem h. (i) that ∗ . Aξ ∈ Z[ ]§ AXA ∈ V§ Similarly, −Ë −Ë ∗ .Hence, Γ by _eorem h.þ. A X(A ) ∈ V§ A ∈ § _e converse is proved in three steps. At ûrst it is shown that for each i z there is with Ë and Ë. ξ ∈ Z[ ]§ u ∈ C SuS = (h. ) i z . uBξ ∈ Z[ ]§ Let ∗. _en , so ′ by _eorem h.þ. From X ∶= ξξ X ∈ V§ X ∶= LB(X) ∈ V§ ′ ∗ ∗ ∗ (h.þ) X = B(ξξ )B = (Bξ)(Bξ) and _eorem h. (ii), it follows that Ë and that there is i z such that σ(δ(X)) = η ∈ Z[ ]§ ′ ′ ∗ h.@) X = σ(δ(X ))ηη .( Since and −Ë have integer components, we see that ′ , and therefore LB LB δ(X ) = δ(X) ′ σ(δ(X )) = σ(δ(X)) = Ë. Now (h. ) follows from (h.þ), (h.@), and Lemma h.z. Next we show that can be chosen independent of , i z i z. z. u ξ i.e., uB‰Z[ ]§Ž ⊂ Z[ ]§ For the vectors and (see (z.˧)) there exist, by step Ë, , such that e§ ez u§ uË ∈ C (h.6) ′ i z , ′ i z u§Be§ = e§ ∈ Z[ ]§ uËBez = ez ∈ Z[ ]§ and Ë. _en Su§S = SuËS = −Ë ′ ′ . u§ e§ + uË ez = B (e§ + ez) Since −Ë Γ , there is, by step Ë, with Ë and i z. B ∈ § u ∈ C SuS = u(u§ e§ + uË ez) ∈ Z[ ]§ Hence, there are p, q ∈ Z[i] with π(p) = π(q) and Ë i , . ( + )uu§ + uuË = p uuË = q So uu i and uu Ë Ë i p q i , the latter because of π p q §. _is Ë ∈ Z[ ] § = z ( − )( − ) ∈ Z[ ] ( − ) = is only possible if , . From (h.6) it now follows that ′ i z uu§ uuË ∈ E uBe§ = uu§ e§ ∈ Z[ ]§

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and ′ i z. Since and are a basis of i z, we have i z uBez = uuË ez ∈ Z[ ]§ e§ ez Z[ ]§ uB‰Z[ ]§Ž ⊂ i z. Z[ ]§ Likewise there is with Ë and −Ë i z i z. From h. v ∈ C SvS = vB ‰Z[ ]§Ž ⊂ Z[ ]§ i z −Ë i z i z i z uv‰Z[ ]§Ž = uB‰vB ‰Z[ ]§ŽŽ ⊂ uB‰Z[ ]§Ž ⊂ Z[ ]§

it follows that i z, hence and uvez ∈ Z[ ]§ uv ∈ E −Ë −Ë i z −Ë −Ë i z i z . u B ‰Z[ ]§Ž = (uv) vB ‰Z[ ]§Ž ⊂ Z[ ]§ _eorem h.6s hows that ∆ is big enough to substantially represent all elements § of Γ . _e matrices with integer components can therefore be characterized by § LA mapping properties of the matrices ∆ .It follows immediately from the deûnition A ∈ § that ∆ is a group. Further properties are given in the following lemma. § ∆ Lemma h.Ç If A ∈ §, then (i) det A ∈ E, (ii) ∆ uA ∈ § if and only if u ∈ E.

(i) _e columns of Ë+i Ë are in i z, hence Ë i det det Proof M ∶= A‰ § Ë Ž Z[ ]§ ( + ) A = M ∈ Z[i]. From S det MSz = z, it follows that det M ∈ (Ë + i)E, so (Ë + i) det A ∈ (Ë + i)E. (ii)If , then clearly ∆ . Conversely, let ∆ . _en −Ë i z, u ∈ E uA ∈ § uA ∈ § A ez ∈ Z[ ]§ hence −Ë i z and therefore . uez = uA(A ez) ∈ Z[ ]§ u ∈ E _e lemma shows that each , Γ , can be represented by exactly four matrices LB B ∈ § in ∆ : , i , , and i ; then . § A = uB A −A − A LB = LA = LiA = L−A = L−iA

3.3 Proof of Theorem 1.1(i) In case (a) both components of are even (see Section z.z). Because det , the Ae§ A ∈ E same holds for −Ë . _erefore, , −Ë i z. A e§ Ae§ A e§ ∈ Z[ ]§ In case (b) both components of and of −Ë are odd. _erefore , −Ë Ae§ A e§ Ae§ A e§ ∈ i z in this case , too. Z[ ]§ In both cases the sum of the components of is , and § Aez a+b+c+d π(a+b+c+d) = (see Lemma z.z in case (b)), so both components of have the same parity, hence Aez i z. _e sum of the components of −Ë is in both cases det Aez ∈ Z[ ]§ A ez (d−b−c+a)~ A = (a + b + c + d)~ det A − z(b + c)~ det A; since in both cases b + c ∈ Z[i] and det A ∈ E, this expression is in i and has parity §. _erefore, −Ë i z. Since , Z[ ] A ez ∈ Z[ ]§ {e§ ez} is a basis of i z, it follows that ∆ . _e assertion now follows from the ûrst Z[ ]§ A ∈ § statement in _eorem h.6.

3.4 Proof of Theorem 1.1(ii) By Lemma z.ht he hypothesis stands for Γ . By _eorem h.6t here is such B ∈ § u that ∆ .Let , , be the i z-vectors from (z.˧). _en we have A ∶= uB ∈ § e§ eË ez Z[ ]§ , , i z, since ∆ . According to (z.ËË), i z and i z Ae§ AeË Aez ∈ Z[ ]§ A ∈ § Ae§ ∈ Z[ ]§ AeË ∈ Z[ ]§

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(h.Ça) (Ë + i)a = (Ë + i)̂a + ε, (Ë + i)c = (Ë + i)̂c + ε, ′ (h.Çb) (Ë + i)b = (Ë + i)̂b + ε , ′ (Ë + i)d = (Ë + i)d̂+ ε with , , , i and , ′ §, Ë . Further, i z means that ̂a ̂c ̂b d̂∈ Z[ ] ε ε ∈ { } Aez ∈ Z[ ]§ (h.’) Ë i ′′, a + b = ( + )gË + ε (h.˧) Ë i ′′ c + d = ( + )gz + ε with , i and ′′ §, Ë . From (h.’) and (h.Ça), it follows that gË gz ∈ Z[ ] ε ∈ { } Ë i Ë i Ë i ′′ Ë i , ( + )b = ( + )(( + )gË + ε − ̂a − ( − )ε) + ε ′ and comparing that with (h.Çb), one obtains ε = ε . Furthermore, addition of (h.’) and (h.˧) shows that π(a + b + c + d) = §. Finally, we note that det A ∈ E by Lemma h.Ç(i). If ε = §, then a, b, c, d ∈ Z[i] and all requirements in (a) hold.If ε = Ë, then (h.Ça) means that Ë i Ë+i , and likewise for the other relations (h.Ç). Hence, a = ̂a + + = (̂a − )+ all requirements ofË(bi) are satisûedz.

4 Alternative Proof 4.1 Reformulation of the Problem At the outset, the condition that the components of are integers is expressed LA through the components of A. Γ Γ Lemma .Ë Let A ∈ . _en A ∈ § if and only if ( .Ë) i aË az ± ah a ∈ Z[ ] , , , , , , for all permutations (aË az ah a ) of (a b c d), and ( .z) ±SaSz ± SbSz ± ScSz ± SdSz ∈ zZ whenever the number of plus signs is even. It is Γ if and only if the components of are integers. We need (z.þ) to Proof A ∈ § LA be expanded at full length:

Ë Ë (SaSz+SbSz+ScSz+SdSz) Re(ab+cd) − Im(ab+cd) (SaSz−SbSz+ScSz−SdSz) ⎛ z z ⎞ h) ⎜ Re(ac+bd) Re(ad+bc) − Im(ad−bc) Re(ac−bd) ⎟ .( . LA = ⎜ ⎟ ⎜ Im(ac+bd) Im(ad+bc) Re(ad−bc) Im(ac−bd) ⎟ Ë Ë (SaSz+SbSz−ScSz−SdSz) Re(ab−cd) − Im(ab−cd) (SaSz−SbSz−ScSz+SdSz) ⎝ z z ⎠ Obviously ( .Ë) and ( .z) are suõcient for Γ . Conversely, if has integer com- A ∈ § LA ponents, then from ( .h) one obtains ( .z) and ab ± cd, ac ± bd, ad ± bc ∈ Z[i]. From ab + cd = g and ab − cd = p with g, p ∈ Z[i] follows that zcd = g − p and zcd − zdc = g − p − (g − p) = zi Im(g − p) ∈ zZ[i], so cd − dc ∈ Z[i], hence

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ab ± dc ∈ Z[i]. Similarly, ac ± db ∈ Z[i] and ad ± cb ∈ Z[i]. _is proves ( .Ë) for all permutations with ûrst term a. By forming complex conjugates and negatives, one obtains all other expressions ( .Ë).

4.2 Proof of Theorem1.1(i) If (a) holds, ( .Ë) and ( .z) are obvious. In case (b),

′ ′ ( . ) (Ë + i)a = (Ë + i)a + Ë (a ∈ Z[i]) and likewise for b, c, d.It follows that

′ ′ ′ ′ ′ ′ ′ ′ zi[(ad − bc) − (a d − b c )] = (Ë + i)(a − c − b + d ). ′ ′ ′ ′ Because det A ∈ E, this implies that all expressions (Ë+i)(±a ±b ±c ±d ) are divisible by z, the signs being arbitrary. From ( . ) and the analogs for b, c, d it follows that

′ ′ ′ ab Ë Ë i a′ Ë Ë i b Ë a′b Ë Ë i a′ Ë i b Ë , = z [( + ) + ][( − ) + ] = + z (( + ) + ( − ) + ) ′ ′ ′ cd Ë Ë i c′ Ë Ë i d Ë c′d Ë Ë i c′ Ë i d Ë , = z [( + ) + ][( − ) + ] = + z (( + ) + ( − ) + )

′ ′ ab cd a′b c′d Ë Ë i a′ c′ Ë Ë i b′ d′ Ë Ë Ë , ± = ± + z ( + )( ± ) + z ( − )( ± ) + z ( ± ) ′ ′ a′b c′d Ë Ë i a′ c′ b′ d′ i b′ d′ Ë Ë Ë . = ± + z ( + )( ± + ± ) − ( ± ) + z ( ± ) All terms on the right are in Z[i]. Likewise, all other relations ( .Ë) are obtained. Now let α, β, γ, δ ∈ E and α + β + γ + δ divisible by z. From ( . ) follows that (Ë + i)(αa + βb + γc + δd) ′ ′ ′ ′ = (Ë + i)(αa + βb + γc + δd ) + α + β + γ + δ ′ ′ ′ ′ ′ ′ ′ ′ = (Ë + i)‰(α − Ë)a + (β − Ë)b + (γ − Ë)c + (δ − Ë)d + (a + b + c + d )Ž + α + β + γ + δ. _en

z S(Ë + i)(αa + βb + γc + δd) ′ ′ ′ ′ since Ë + i S α − Ë, Ë + i S β − Ë, Ë + i S γ − Ë, Ë + i S δ − Ë and z S(Ë + i)(a + b + c + d ). Hence,

(αa + βb + γc + δd)(αa + βb + γc + δd) = αzSaSz + βzSbSz + γzScSz + δzSdSz +
αβ Re(zab) + γδ Re(zcd) +
αγ Re(zac) + βδ Re(zbd) +
αδ Re(zad) + βγ Re(zbc)

is divisible by z.If (α, β, γ, δ) is one of the four combinations (Ë, Ë, Ë, Ë), (Ë, Ë, i, i), (Ë, i, Ë, i), or (Ë, i, i, Ë), then α + β + γ + δ is divisible by z and αβ = γδ or αβ = −γδ, αγ = βδ or αγ = −βδ and αδ = βγ or αδ = −βγ. Using ( .Ë), it is seen that the expressions in brackets are all divisible by z. _is proves ( .z).

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4.3 Proof of Theorem 1.1(ii) Lemmas .z and .h are intermediate steps.

Γ z i , , , , Lemma .z Let A ∈ §. _en vËvz ∈ Z[ ] for all vË vz ∈ {a b c d}.If in addition det z i , , , , A ∈ E, then vËvz ∈ Z[ ] for all vË vz ∈ {a b c d}.

Proof _e ûrst assertion follows from ( .Ë) and ( .z). For t = Ë and x = y = z = ′ § we have X = I. _e components of X are integers by hypothesis. From (Ë.h), it ′ ∗ ′ ∗ − follows that X = AA and X (A ) Ëzv = zAv. By the ûrst part of the lemma, for v ∈ {a, b, c, d}, all components on the le side of this equation, and hence all components of zAv, are in Z[i]. Because UFD, each w ∈ Z[i] has a unique factorization ( .þ) w = ε(Ë + i) j gzγ with the following properties: ε ∈ E, § ≤ j ∈ Z, g and γ are odd and γ is square-free. Γ det Lemma .h If A ∈ § and A ∈ E, then all components of A have the form ( .@) v Ë Ë i k+Ë h = z ( + ) or all have the form √ ( .6) v Ë z Ë i k h = z ( + ) with § ≤ k ∈ Z and odd h ∈ Z[i].

Let , , , , , , , , . _en z z, z z, z i by Lemma .z, Proof vË vz ∈ {a b c d a b c d} vË vz vËvz ∈ Z[ ] and therefore according to ( .þ),

( .Ç) z z Ë i jË z , vË = εË( + ) gË γË ( .’) z z Ë i jz z , vz = εz( + ) gz γz §) z Ë i jh z .( .Ë vËvz = εh( + ) gh γh Multiplication of ( .Ç) by ( .’) gives on the one hand

( .ËË) z z Ë i jË+ jz z , vË vz = εËεz( + ) (gË gz) γËγz and squaring ( .˧) on the other hand

z) z z z Ë i z jh z z.( .Ë vË vz = εh( + ) (gh γh) From the uniqueness of the factorization ( .þ) follows that , since comparison γË = γz of ( .ËË) with ( .Ëz) shows that has solely double prime factors. One concludes γËγz that the factor γ in zvz = ε(Ë+i) j gzγ is the same for all v ∈ {a, b, c, d, a, b, c, d}. From ( .Ëh) zvz = ε(Ë + i) j gzγ, z z z ( .Ë ) zv = ε(Ë − i) j g γ = [ε(−i) j](Ë + i) j g γ, it now follows that γ = γ, so all prime factors of γ are real. Hence, γ = Ë or γ ≥ h.

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From the comparison of ( .ËË) with ( .Ëz), it follows that z z Ë, so εË εz = εh = ( .Ëþ) Ë, εËεz = ±

and that z . Applied to ( .ËË), this gives z Ë i jh with . jË + jz = jh vËvz = η( + ) gË gzγ η ∈ E _erefore, z for all , ; in particular, z . But this implies z, γ S vËvz vË vz γ S (ad − bc) γ ≤ so γ = Ë. _erefore, all v ∈ {a, b, c, d, a, b, c, d} have a representation @) zvz = ε(Ë + i) j gz.( .Ë If ( .Ëþ) is applied to ( .Ëh) and ( .Ë ), it follows that εε(−i) j = (−i) j = ±Ë, so all j are even, j = zk. As another consequence of ( .Ëþ), the set of factors ε in the representations ( .Ë@) of {a, b, c, d, a, b, c, d} is either √ z Ë, Ë ηz η or i, i Ë η z η . {− } = { ∶ ∈ E} {− } = ™‰ z Ž ∶ ∈ Ež Having the factors ε written as squares we can extract square roots in ( .Ë@) and obtain ( .6) or ( .@). In each case, h = ηg is odd. Aer these preparations the proof can be carried out.Let Γ . We choose B ∈ § û with SûS = Ë such that det(ûB) = Ë and set  ∶= ûB. By Lemma .ht he components of  have the form ( .@) or ( .6). In the ûrst case let A ∶=  and u ∶= û; then the components of A satisfy ( .@) and we have det A = Ë. In the second case let Ë Ë A Ë i  and u Ë i u; ∶= √z( + ) ∶= √z( + )̂ then the components of Aalso satisfy ( .@) and it is det A = i. In both cases we have A = uB with u Ë, det A , and the components of A have the form v Ë Ë i k j+Ë h S S = ∈ E j = z ( + ) j (j = Ë, z, h, ). If k Ë, then v i .If k §, then v Ë Ë i h with odd ih Ë i h′ Ë j ≥ j ∈ Z[ ] j = j = z ( + ) j j = ( + ) j + (h′ i ), hence v h′ Ë Ë i and v z Ë h z q Ë with q . j ∈ Z[ ] j = j + z ( − ) S jS = z S jS = j + z j ∈ Z As a consequence of ( .z) we conclude that the number of indices with § is j k j = even. Hence there are three cases to consider. (i) Ë for all . _en all components of are in i . Since det and k j ≥ j A Z[ ] A ∈ E SaSz + SbSz + ScSz + SdSz is even according to ( .z), case (a) is present. (ii) § for exactly two indices . Assume that §. _en Ë, Ë, so k j = j kË = kz = kh ≥ k ≥ , i . Since i by Lemma .Ë, it follows that i . vh v ∈ Z[ ] vËvz − vhv ∈ Z[ ] vËvz ∈ Z[ ] On the other hand, it was shown before that v Ë Ë i h and v Ë Ë i h Ë = z ( + ) Ë z = z ( + ) z with odd and , hence Ë where is odd. So this case is hË hz vËvz = hË hz hË hz impossible. z (iii) § for all . _en ′ i Ë Ë i with ′ i for Ë, z, h, . So case k j = j v j = h j − + ( + ) h j ∈ Z[ ] j = (b) is present. z

References

[Ë]A . Baker, Matrix groups. Springer, z§§z. [z] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers. Fih ed., _e Clarendon Press, Oxford University Press, New York, ˒6’. [h] R. A. Horn and C. R. Johnson, Topics in matrix analysis. Cambridge University Press, Cambridge, ˒’Ë.

Downloaded from https://www.cambridge.org/core. 27 Sep 2021 at 22:51:20, subject to the Cambridge Core terms of use. Discrete Space-time and Lorentz Transformations Ëhþ

[ ] M. Lorente and P. Kramer, Representations of the discrete inhomogeneous Lorentz group and Dirac wave equation on the lattice. J. Phys. A hz(˒’’), no. Ëz, z ÇË–z ’6. http://dx.doi.org/10.1088/0305-4470/32/12/019 [þ] J. D. Louck, A new parametrization and all integral realizations of the Lorentz group. J. Math. Phys. h(z§§z), þ˧ǖþËh . http://dx.doi.org/10.1063/1.1505124 [@] R.Penrose and W. Rindler, Spinors and space-time. Vol. I, Cambridge Monographs on Mathematical Physics Cambridge University Press, Cambridge, Ë’Ç . http://dx.doi.org/10.1017/CBO9780511564048 [6]A . Schild, Discrete space-time and integral Lorentz transformations. Canad. J. Math. Ë(˒ ’), z’– 6. http://dx.doi.org/10.4153/CJM-1949-003-4 Sensburger Allee zz a, D–Ë §þþ Berlin e-mail: [email protected] Institut für Mathematik, Technische Universität, D–˧@zh Berlin e-mail: [email protected]

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