A Discrete Dirac–Kähler Equation Using a Geometric Discretisation

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Adv. Appl. Cliﬀord Algebras (2018) 28:72 c The Author(s) 2018 Advances in https://doi.org/10.1007/s00006-018-0889-0 Applied Cliﬀord Algebras

A Discrete Dirac–K¨ahler Equation Using a Geometric Discretisation Scheme

Volodymyr Sushch∗

Communicated by Uwe Kaehler

Abstract. Discrete models of the Dirac–Kähler equation and the Dirac equation in the Hestenes form are discussed. A discrete version of the plane wave solutions to a discrete analogue of the Hestenes equation is established. Mathematics Subject Classification. Primary 39A12; Secondary 81Q05, 39A70. Keywords. Dirac-Kähler equation, Hestenes equation, Discrete models, Clifford multiplication, Plane wave solution.

1. Introduction We study a discrete model of the Dirac–Kähler equation in which key geometric aspects of the continuum counterpart are captured. We pay special attention to the description of some method of construction a discretisation scheme based on the use of the differential forms calculus. The aim of this paper is to establish some discrete version of the plane wave solutions to a discrete Dirac equation in the Hestenes form. To construct these discrete solutions we introduce a Clifford product acting on the space of discrete inhomogeneous forms. This work is a direct continuation of that described in my previous papers [12–15]. In [15], a correspondence between a discrete Dirac–Kähler equation and a discrete analogue of the Hestenes equation was studied. There are several approaches to study of discrete versions of the Dirac–Kähler equation based on the use a discrete Clifford calculus frame- work on lattices. For a review of discrete Clifford analysis, we refer the reader to [4–6,10,16]. It is beyond the scope of this paper to fully discuss differences and intersections between approaches. We first briefly review some definitions and basic facts on the Dirac– Kähler equation [9,11] and the Dirac equation in the spacetime algebra [7,8].

∗Corresponding author.

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Let M = R1,3 be Minkowski space with metric signature (+, −, −, −). Denote by Λr(M) the vector space of smooth differential r-forms, r =0, 1, 2, 3, 4. We consider Λr(M)overC.Letω, ϕ ∈ Λr(M). The inner product is defined by (ω, ϕ)= ω ∧∗ϕ, (1.1) M where ∧ is the exterior product and ∗ is the Hodge star operator ∗ :Λr(M) → Λ4−r(M) with respect to the Lorentz metric. Let d :Λr(M) → Λr+1(M)be the exterior differential and let δ :Λr(M) → Λr−1(M) be the formal adjoint of d with respect to (1.1). We have δ = ∗d ∗ . Denote by Λ(M) the set of all differential forms on M.Wehave Λ(M)=Λev(M) ⊕ Λod(M), where Λev(M)=Λ0(M) ⊕ Λ2(M) ⊕ Λ4(M) and Λod(M)=Λ1(M) ⊕ Λ3(M). Let Ω ∈ Λ(M) be an inhomogeneous differential form, i.e., 4 r Ω= ω, r=0 r where ω ∈ Λr(M). The Dirac–Kähler equation for a free electron is given by i (d + δ)Ω=mΩ, (1.2) where i is the usual complex unit and m is a mass parameter. It is easy to show that Eq. (1.2) is equivalent to the set of equations 1 0 iδω = mω, 0 2 1 i dω + δω = mω, 1 3 2 i dω + δω = mω, 2 4 3 i dω + δω = mω,

3 4 idω = mω. (1.3) Denote by Ωev and by Ωod the even and odd parts of Ω. It is clear that i (d + δ)Ωod = mΩev, i (d + δ)Ωev = mΩod. The operator d + δ is the analogue of the gradient operator in Minkowski spacetime 3 μ ∇ = γμ∂ ,μ=0, 1, 2, 3, μ=0 where γμ is the Dirac gamma matrix. If one think of {γ0,γ1,γ2,γ3} as a vector basis in spacetime, then the gamma matrices γμ can be considered as generators of the Clifford algebra of spacetime C(1, 3) [1,2]. Hestenes [8] calls this algebra the spacetime algebra. Denote by CR(1, 3) (CC(1, 3)) the A Discrete Dirac–Kähler Equation... Page 3 of 17 72 real (complex) Clifford algebra. It is known that an inhomogeneous form Ω can be represented as element of CC(1, 3). Then the Dirac–Kähler equation can be written as the algebraic equation

i∇Ω=mΩ, Ω ∈ CC(1, 3). (1.4) Equation (1.4) is equivalent to the four Dirac equations (traditional column- spinor equations) for a free electron. Let Cev(1, 3) be the even subalgebra of the algebra C(1, 3). The equation ev ev ev ev −∇Ω γ1γ2 = mΩ γ0, Ω ∈ CR (1, 3), (1.5) is called the Hestenes form of the Dirac equation [7,8]. It should be noted that the graded algebra Λ(M) endowed with the Clifford multiplication is an example of a Clifford algebra. In this case the basis covectors eμ = dxμ are considered as generators of the Clifford algebra. Then the Hestenes equation (1.5) can be rewritten in terms of inhomogeneous forms as ev 1 2 ev 0 ev ev − (d + δ)Ω e e = mΩ e , Ω ∈ ΛR (M) . (1.6)

2. Discretisation Scheme The starting point for consideration is a combinatorial model of Euclidean space. The proposed approach was originated by Dezin [3]. For the conve- nience of the reader we briefly repeat the relevant material from [13] without proofs, thus making our presentation self-contained. Let the tensor product C(4) = C ⊗ C ⊗ C ⊗ C, of a 1-dimensional complex be a combinatorial model of Euclidean space R4. The 1-dimensional complex C is defined in the following way. Introduce the 0 1 sets {xκ} and {eκ}, κ ∈ Z.LetC and C be the free abelian groups of 0-dimensional and 1-dimensional chains generated by {xκ} and {eκ}.The free abelian group is understood as the direct sum of infinity cyclic groups 0 1 generated by {xκ}, {eκ}. Arbitrary elements (chains) a ∈ C and b ∈ C can be written as the formal sums κ κ κ κ a = a xκ,b= b eκ,a,b ∈ Z. κ κ We define the boundary operator ∂ : C1 → C0,∂: C0 → 0, setting

∂eκ = xκ+1 − xκ,∂xκ =0. (2.1) The definition (2.1) is extended to arbitrary chains by linearity. The direct sum C = C0⊕C1 with the boundary operator ∂ defines the 1- dimensional complex. It is known that a free abelian group is an abelian group with basis. One can regard the sets {xκ}, {eκ} as sets of basis elements of the groups C0 and C1. Geometrically we can interpret the 0-dimensional basis elements xκ as points of the real line and the 1-dimensional basis elements eκ as open intervals between points. We call the complex C a combinatorial real line. 72 Page 4 of 17 V. Sushch Adv. Appl. Clifford Algebras

Multiplying the basis elements xκ, eκ in various way we obtain basis (r) elements of C(4). Let sk be an arbitrary r-dimensional basis element of C(4). Then we have (r) ⊗ ⊗ ⊗ sk = sk0 sk1 sk2 sk3 , ∈ Z where skµ is either xkµ or ekµ and k =(k0,k1,k2,k3), kμ . The dimension (r) r of a basis element sk is given by the number of factors ekµ that appear − in it. The product contains exactly r 1-dimensional elements ekµ and 4 r

0-dimensional elements xkµ , and the superscript (r) indicates also a position (r) of ekµ in sk . For example, the 1- and 2-dimensional basis elements of C(4) can be written as 0 ⊗ ⊗ ⊗ 1 ⊗ ⊗ ⊗ ek = ek0 xk1 xk2 xk3 ,ek = xk0 ek1 xk2 xk3 , 2 ⊗ ⊗ ⊗ 3 ⊗ ⊗ ⊗ ek = xk0 xk1 ek2 xk3 ,ek = xk0 xk1 xk2 ek3 and 01 ⊗ ⊗ ⊗ 12 ⊗ ⊗ ⊗ ek = ek0 ek1 xk2 xk3 ,ek = xk0 ek1 ek2 xk3 , 02 ⊗ ⊗ ⊗ 13 ⊗ ⊗ ⊗ ek = ek0 xk1 ek2 xk3 ,ek = xk0 ek1 xk2 ek3 , 03 ⊗ ⊗ ⊗ 23 ⊗ ⊗ ⊗ ek = ek0 xk1 xk2 ek3 ,ek = xk0 xk1 ek2 ek3 . Let C(4) = C(p) ⊗ C(q), where p + q =4andC(p) is the tensor product of p factors of C = C(1), p =1, 2, 3. The definition (2.1)of∂ is extended to arbitrary chains of C(4) by the rule r ∂ (a ⊗ b)=∂a ⊗ b +(−1) a ⊗ ∂b, (2.2) where a ∈ C(p)andb ∈ C(q)andr is the dimension of the chain a.Itiseasy to check that ∂∂a =0foranya ∈ C(4). Suppose that the combinatorial model of Minkowski space has the same structure as C(4). We will use the index k0 to denote the basis elements of C which correspond to the time coordinate of M. Hence, the indicated basis elements will be written as xk0 , ek0 . Let us introduce the construction of a double complex. This construction generalizes [13]. Together with the complex C(4) we consider its double, namely the complex C˜(4) of exactly the same structure, i.e., the one in the corresponding chains a ∈ C(4) anda ˜ ∈ C˜(4) have the same coefficients. Define the one-to-one correspondence ∗c : C(4) → C˜(4), ∗c : C˜(4) → C(4) by the rule c (r) (4−r) c (r) (4−r) ∗ sk = ε(r)˜sk , ∗ s˜k = ε(r)sk , (2.3) where (4−r) ∗c ⊗∗c ⊗∗c ⊗∗c s˜k = sk0 sk1 sk2 sk3 ∗c ∗c and xkµ =˜ekµ , ekµ =˜xkµ . Here ε(r) is the Levi-Civita symbol, i.e., +1 if ((r) , (4 − r)) is an even permutation of (0, 1, 2, 3) ε(r)= −1if((r) , (4 − r)) is an odd permutation of (0, 1, 2, 3) . A Discrete Dirac–Kähler Equation... Page 5 of 17 72

For example, for the 1- and 2-dimensional basis elements we have c 0 123 c 1 023 c 2 013 c 3 012 ∗ ek =˜ek , ∗ ek = −e˜k , ∗ ek =˜ek , ∗ ek = −e˜k and c 01 23 c 02 13 c 03 12 ∗ ek =˜ek , ∗ ek = −e˜k , ∗ ek =˜ek , c 12 03 c 13 02 c 23 01 ∗ ek =˜ek , ∗ ek = −e˜k , ∗ ek =˜ek . The operation (2.3) is linearly extended to chains.

Proposition 2.1. Let ar ∈ C(4) be an r-dimensional chain, then we have c c r ∗ ∗ ar =(−1) ar.

Proof. The proof consists in applying the operation ∗c for basis elements.

Let us now consider a dual complex to C(4). We define it as the complex of cochains K(4) with complex coefficients. The complex K(4) has a similar structure, namely K(4) = K ⊗ K ⊗ K ⊗ K, where K is a dual complex to the 1-dimensional complex C.Letxκ and eκ, κ ∈ Z, be the 0- and 1-dimensional basis elements of K. Then an arbitrary r-dimensional basis element of K(4) k k0 ⊗ k1 ⊗ k2 ⊗ k3 kµ kµ can be written as s(r) = s s s s , where s is either x or kµ e and k =(k0,k1,k2,k3). We will call cochains forms, emphasizing their relationship with differential forms. Denote by Kr(4) the set of all r-forms. Then K(4)canbeexpressedby K(4) = Kev(4) ⊕ Kod(4), where Kev(4) = K0(4) ⊕ K2(4) ⊕ K4(4) and Kod(4) = K1(4) ⊕ K3(4). The r complex K(4) is a discrete analogue of Λ(M). Let ω ∈ Kr(4), then we have 0 0 k 2 μν k 4 4 k ω = ωkx , ω = ωk eμν , ω = ωke , k k μ<ν k 3 1 μ k 3 ιμν k ω = ωk eμ, ω = ωk eιμν , k μ=0 k ι<μ<ν k k k where eμ, eμν and eιμν are the 1-, 2- and 3-dimensional basis elements of K(4), and xk = xk0 ⊗ xk1 ⊗ xk2 ⊗ xk3 , ek = ek0 ⊗ ek1 ⊗ ek2 ⊗ ek3 .The 0 4 μ μν ιμν components ωk, ωk,ωk ,ωk and ωk are complex numbers. We define the pairing (chain-cochain) operation for any basis elements k k ∈ C(4), s ∈ K(4) by the rule k 0, k = sk k,s = (2.4) 1, k = sk. The operation (2.4) is linearly extended to arbitrary chains-cochains. The coboundary operator dc : Kr(4) → Kr+1(4) is defined by

r c r ∂ar+1, ω = ar+1,dω , (2.5) 72 Page 6 of 17 V. Sushch Adv. Appl. Cliﬀord Algebras

c where ar+1 ∈ C(4) is an r+1 dimensional chain. The operator d is an analog of the exterior diﬀerential. From the above it follows that 4 r dcω =0 and dcdcω =0 forany r.

Let the diﬀerence operator Δμ be deﬁned by

(r) (r) (r) μ − Δ ωk = ωτµk ωk , (2.6)

(r) r r where ωk ∈ C is a component of ω ∈ K (4) and τμ is the shift operator which acts as

τμk =(k0, ...kμ +1, ...k3) ,μ=0, 1, 2, 3. Using (2.2) and (2.5) we can calculate 3 c 0 0 k c 1 ν μ k d ω = Δμωk eμ,dω = (Δμωk − Δν ωk ) eμν , (2.7) k μ=0 k μ<ν c 2 12 − 02 01 k d ω = Δ0ωk Δ1ωk +Δ2ωk e012 k + Δ ω13 − Δ ω03 +Δ ω01 ek 0 k 1 k 3 k 013 + Δ ω23 − Δ ω03 +Δ ω02 ek 0 k 2 k 3 k 023 23 − 13 12 k + Δ1ωk Δ2ωk +Δ3ωk e123 , (2.8) c 3 123 023 013 012 k d ω = Δ0ωk − Δ1ωk +Δ2ωk − Δ3ωk e . (2.9) k

Let K˜ (4) be a complex of the cochains over the double complex C˜(4), with the coboundary operator dc defined by (2.5). Hence, K˜ (4) has the same structure as K(4). This means that the corresponding forms ω ∈ K(4) and ω˜ ∈ K˜ (4) have the same components. Let us introduce a discrete version of the Hodge star operator by using the double complex construction. Define the operation ∗ : Kr(4) → K˜ 4−r(4) k ∈ r for the basis element s(r) K (4) by the rule ∗ k k s(r) = Q (k0) ε (r)˜s(4−r), (2.10) where +1 if sk0 = xk0 Q (k )= 0 −1ifsk0 = ek0 . This definition makes sense because the formula (2.10) preserves the Lorentz signature of metric on K(4). More explicitly, we have ∗xk =˜ek, ∗ek = −x˜k, ∗ k − k ∗ k − k ∗ k k ∗ k − k e0 = e˜123, e1 = e˜023, e2 =˜e013, e3 = e˜012, ∗ k − k ∗ k k ∗ k − k e01 = e˜23, e02 =˜e13, e03 = e˜12, ∗ k k ∗ k − k ∗ k k e12 =˜e03, e13 = e˜02, e23 =˜e01, ∗ k − k ∗ k k ∗ k − k ∗ k − k e012 = e˜3 , e013 =˜e2 , e023 = e˜1 , e123 = e˜0 . A Discrete Dirac–Kähler Equation... Page 7 of 17 72

For any r-form the operation (2.10) is extended by linearity. Likewise, the mapping ∗ : K˜ r(4) → K4−r(4) is given by the rule (2.10). r Proposition 2.2. Let ω ∈ Kr(4), then we have r r ∗∗ω =(−1)r+1ω. (2.11) Proof. The operation ∗ is linear. It is easy to check that by deﬁnition, the composition of ∗ with itself gives ∗∗ k − r(4−r)+1 k − r+1 k s(r) =( 1) s(r) =( 1) s(r), k ∈ r foranybasiselements(r) K (4). Hence, the operation ∗2 is either an involution or antiinvolution. It means that the discrete ∗ operation imitates correctly the continuum case.

Proposition 2.3. Let a˜r ∈ C˜(4) be an r-dimensional chain and let ω ∈ K4−r(4). Then we have r c a˜r, ∗ω =(−1) Q (k0) ∗ a˜r,ω . Proof. See Proposition 3 in [13].

Let us consider the 4-dimensional finite chain en ⊂ C(4) of the form: en = ek,k=(k0,k1,k2,k3) ,kμ =1, 2, ..., nμ, (2.12) k where nμ ∈ N is a fixed number for each μ =0, 1, 2, 3. This finite sum of 4-dimensional basis elements of C(4) imitates a domain of M.Weset (r) c (r) Vr = sk ⊗∗ sk ,kμ =1, 2, ..., nμ. k (r) For example, 3 μ c μ 0 123 1 023 2 013 3 012 V1 = ek ⊗∗ ek = ek ⊗ e˜k − ek ⊗ e˜k + ek ⊗ e˜k − ek ⊗ e˜k . k μ=0 k Let 4 V = Vr. r=0 r For any r-forms ϕ, ω ∈ K (4) the inner product over the set en (2.12)is defined by the rule V ⊗∗ (ϕ, ω)en = ,ϕ ω , (2.13) where ω denotes the complex conjugate of the form ω. For forms of different degrees the product (2.13) is set equal to zero. It is clear that (r) c (r) r ⊗∗ ∗ ∗ (ϕ, ω)en = V ,ϕ ω = sk ,ϕ sk , ω . k (r) 1 1 For example, if ϕ, ω ∈ K1(4) then we obtain 1 1 0 0 1 1 2 2 3 3 ϕ, ω = −ϕkωk + ϕkωk + ϕkωk + ϕkωk . e n k 72 Page 8 of 17 V. Sushch Adv. Appl. Clifford Algebras

It should be noted that the definition of the inner product correctly imitates the case (1.1) and the Lorentz metric structure is still captured here. The inner product (2.13) makes it possible to define the adjoint of dc, denoted δc. Proposition 2.4. For any (r − 1)-form ϕ and r-form ω we have c V ⊗∗ c (d ϕ, ω)en = ∂ ,ϕ ω +(ϕ, δ ω)en , (2.14) where δc =(−1)r ∗−1 dc∗, (2.15) and ∗−1 is the inverse operation of ∗. Proof. The proof is a direct computation. See Proposition 4 in [13]. The relation (2.14) is a discrete analog of the Green formula. From (2.11) we infer ∗−1 =(−1)r(4−r)+1∗ =(−1)r+1 ∗ . Putting this in (2.15) we obtain δc = ∗dc ∗ . (2.16) This makes it clear that the operator δc : Kr+1(4) → Kr(4) is a discrete analog of the codifferential δ.From(2.16) it follows that 0 r δcω =0 and δcδcω =0 forany r. Using the definitions of dc and ∗ we can calculate c 1 0 1 2 3 k δ ω = Δ0ωk − Δ1ωk − Δ2ωk − Δ3ωk x , (2.17) k c 2 01 02 03 k δ ω = Δ1ωk +Δ2ωk +Δ3ωk e0 k + Δ ω01 +Δ ω12 +Δ ω13 ek 0 k 2 k 3 k 1 + Δ ω02 − Δ ω12 +Δ ω23 ek 0 k 1 k 3 k 2 03 − 13 − 23 k + Δ0ωk Δ1ωk Δ2ωk e3 , (2.18) c 3 − 012 − 013 k 012 − 023 k δ ω = Δ2ωk Δ3ωk )e01 +(Δ1ωk Δ3ωk e02 k + Δ ω013 +Δ ω023 ek + Δ ω012 − Δ ω123 ek 1 k 2 k 03 0 k 3 k 12 013 123 k 023 123 k + Δ0ωk +Δ2ωk e + Δ0ωk − Δ1ωk e , (2.19) 13 23 c 4 4 k − 4 k 4 k 4 k δ ω = Δ3ωk e012 Δ2ωk e013 + Δ1ωk e023 + Δ0ωk e123 . k (2.20) The linear map Δc = − (dcδc + δcdc): Kr(4) → Kr(4), is called a discrete analogue of the Laplacian. It is clear that − (dcδc + δcdc)=(dc − δc)2 = − (dc + δc)2 . A Discrete Dirac–Kähler Equation... Page 9 of 17 72

3. Discrete Dirac–Kähler and Hestenes Equations Let us introduce a discrete inhomogeneous form as follows 4 r Ω= ω, r=0 r where ω ∈ Kr(4). A discrete analog of the Dirac–Kähler equation (1.2)is defined to be i (dc + δc)Ω=mΩ, (3.1) where i is the usual complex unit and m is a positive number (mass parameter). It is clear that we can write this equation more explicitly by separating its homogeneous components as the system (1.3). Moreover, by (2.7)–(2.9) and (2.17)–(2.20), this system of equations for each k =(k0,k1,k2,k3)can be expressed in terms of 16 difference equations as follows

0 1 2 3 0 i Δ ω − Δ ω − Δ ω − Δ ω = mωk, 0 k 1 k 2 k 3 k 0 01 02 03 0 i Δ0ωk +Δ1ωk +Δ2ωk +Δ3ωk = mωk, 0 01 12 13 1 i Δ1ωk +Δ0ωk +Δ2ωk +Δ3ωk = mωk, 0 02 12 23 2 i Δ2ωk +Δ0ωk − Δ1ωk +Δ3ωk = mωk, 0 03 13 23 3 i Δ3ωk +Δ0ωk − Δ1ωk − Δ2ωk = mωk,

i Δ ω1 − Δ ω0 − Δ ω012 − Δ ω013 = mω01, 0 k 1 k 2 k 3 k k i Δ ω2 − Δ ω0 +Δ ω012 − Δ ω023 = mω02, 0 k 2 k 1 k 3 k k i Δ ω3 − Δ ω0 +Δ ω013 +Δ ω023 = mω03, 0 k 3 k 1 k 2 k k i Δ ω2 − Δ ω1 +Δ ω012 − Δ ω123 = mω12, 1 k 2 k 0 k 3 k k i Δ ω3 − Δ ω1 +Δ ω013 +Δ ω123 = mω13, 1 k 3 k 0 k 2 k k i Δ ω3 − Δ ω2 +Δ ω023 − Δ ω123 = mω23, 2 k 3 k 0 k 1 k k 12 02 01 4 012 i Δ0ωk − Δ1ωk +Δ2ωk +Δ3ωk = mωk , 13 03 01 4 013 i Δ0ωk − Δ1ωk +Δ3ωk − Δ2ωk = mωk , 23 03 02 4 023 i Δ0ωk − Δ2ωk +Δ3ωk +Δ1ωk = mωk , 23 13 12 4 123 i Δ1ωk − Δ2ωk +Δ3ωk +Δ0ωk = mωk ,

123 023 013 012 4 i Δ0ωk − Δ1ωk +Δ2ωk − Δ3ωk = mωk. Let us define the Clifford multiplication in K(4) by the following rules: k k k k k k k k (a) x x = x ,xeμ = eμx = eμ, k k k k k (b) eμeν + eν eμ =2gμν x ,gμν = diag (1, −1, −1, −1) , k ··· k k ≤ ··· ≤ (c) eμ1 eμs = eμ1···μs for 0 μ1 < <μs 3, 72 Page 10 of 17 V. Sushch Adv. Appl. Clifford Algebras supposing the product to be zero in all other cases. The operation is linearly extended to arbitrary discrete forms. For ex- 1 1 ample, for ω, ϕ ∈ K1(4) we have 3 3 1 1 µ k µ k 0 0 1 1 2 2 3 3 k ωϕ = ωk eµ ϕk eµ = ωkϕk − ωkϕk − ωkϕk − ωkϕk x k µ=0 k µ=0 k 0 1 1 0 k 0 2 2 0 k 0 3 3 0 k + ωkϕk − ωkϕk e01 + ωkϕk − ωkϕk e02 + ωkϕk − ωkϕk e03 k 1 2 2 1 k 1 3 3 1 k 2 3 3 2 k + ωkϕk − ωkϕk e12 + ωkϕk − ωkϕk e13 + ωkϕk − ωkϕk e23 .

Consider the following unit forms k k k k x = x ,e= e ,eμ = eμ,eμν = eμν . (3.2) k k k k Note that the unit 0-form x plays a role of the unit element in K(4), i.e., for r any r-form ω we have r r r xω = ωx = ω.

Proposition 3.1. The following holds:

eμeν + eν eμ =2gμν x, μ, ν =0, 1, 2, 3.

k k k k k Proof. By the rule eμeν + eν eμ =2gμν x it is obvious.

The following proposition was proven in [15, Proposition 1].

Proposition 3.2. For any inhomogeneous form Ω ∈ K(4) we have 3 c c (d + δ )Ω= eμΔμΩ, (3.3) μ=0 where Δμ is the diﬀerence operator which acts on each component of Ω by the rule (2.6).

Clearly, the discrete Dirac–Kähler equation can be rewritten in the form 3 i eμΔμΩ=mΩ. μ=0 Let Ωev ∈ Kev(4) be a real-valued even inhomogeneous form, i.e., Ωev = 0 2 4 ω + ω + ω. A discrete analogue of the Hestenes equation (1.6) is defined by c c ev ev − (d + δ )Ω e1e2 = mΩ e0, (3.4) where e0,e1,e2 are given by (3.2). From (3.3) it follows that Eq. (3.4)is equivalent to 3 ev ev − eμΔμΩ e1e2 = mΩ e0. μ=0 A Discrete Dirac–Kähler Equation... Page 11 of 17 72

This equation can be expressed in terms of diﬀerence equations as

12 02 01 4 0 Δ0ωk − Δ1ωk +Δ2ωk +Δ3ωk = mωk, 0 02 12 23 01 Δ2ωk +Δ0ωk − Δ1ωk +Δ3ωk = mωk , 0 01 12 13 02 −Δ1ωk − Δ0ωk − Δ2ωk − Δ3ωk = mωk , 23 13 12 4 03 −Δ1ωk +Δ2ωk − Δ3ωk − Δ0ωk = mωk , 0 01 02 03 12 −Δ0ωk − Δ1ωk − Δ2ωk − Δ3ωk = mωk , 23 03 02 4 13 −Δ0ωk +Δ2ωk − Δ3ωk − Δ1ωk = mωk , 13 03 01 4 23 Δ0ωk − Δ1ωk +Δ3ωk − Δ2ωk = mωk , 0 03 13 23 4 Δ3ωk +Δ0ωk − Δ1ωk − Δ2ωk = mωk. In [15, Proposition 5], it is proven that by a solution of the discrete Dirac–K¨ahler equation four independent real-valued solutions of the discrete Hestenes equation (3.4) are constructed. This is a discrete version of the well-known result for corresponding continuum equations [2].

Remark 3.3. It should be noted that the complex K(4) does not give an ex- act geometric counterpart of the Clifford algebra C(1, 3) but a deformation of it. This problem arises because the operators dc and δc contain only finite difference operators of the forward type. The fact that both forward and back- ward differences are needed on the lattice is well-known [11]. For a suitable algebraic structure for lattice problems that reduced to a Clifford algebra in the continuum limit, we refer to [16]. The only way to avoid the deformation gap is to switch to operators which contain both types of differences. This is the subject of current work in progress.

4. Plane Wave Solutions Let us consider the real-valued forms ± ± k Ψ = Ψk x , (4.1) k where

± k0 k1 k2 k3 Ψk =(x ± p0e12) (x ± p1e12) (x ± p2e12) (x ± p3e12) , (4.2) and pμ ∈ R. Recall that e12 is the unit 2-form given by (3.2). It is easy to check that ± ± ΔμΨk = ±pμΨk e12,μ=0, 1, 2, 3. (4.3) Therefore 3 3 c c ± ± ± (d + δ )Ψ = eμΔμΨ = ± eμpμΨ e12. (4.4) μ=0 μ=0 72 Page 12 of 17 V. Sushch Adv. Appl. Cliﬀord Algebras

It should be noted that the components Ψ± can be represented as k ± ± ± Ψk = (ψk x + φk e12), k where ± ± ± ± ψk = ψk (p0,p1,p2,p3)andφk = φk (p0,p1,p2,p3) . Hence Ψ± are inhomogeneous real-valued even forms of the form Ψ± = ψ± + φ±, where ± ± k ± ± k ψ = ψk x ,φ= φk e12. k k We wish to ﬁnd a solution of the discrete Hestenes equation (3.4)oftheform Ω± = AΨ±, (4.5) where A ∈ Kev(4) is a constant real-valued form. Hence A can be expanded as 0 μν 4 A = α x + α eμν + α e, (4.6) μ<ν 0 μν 4 where α ,α ,α ∈ R and x, eμν , e are the unit forms given by (3.2). We have 3 c c ± c c ± ± (d + δ )Ω =(d + δ ) AΨ = eμΔμ AΨ μ=0 3 3 ± ± k = eμA ΔμΨ = eμA ΔμΨk x μ=0 μ=0 k 3 3 ± k ± = ± eμpμA Ψk x e12 = ± eμpμAΨ e12. μ=0 k μ=0 Substituting this into the discrete Hestenes equation (3.4) we obtain 3 ± ± ∓ eμpμAΨ e12 e12 = mAΨ e0. μ=0

By definition, e12e12 = −x. This yields 3 ± ± ± eμpμAΨ = mAΨ e0. μ=0 ± ± Since Ψ e0 = e0Ψ , the equation above reduces to 3 eμpμA = mAe0, (4.7) μ=0 in the case of the form Ψ+ and to 3 − eμpμA = mAe0, (4.8) μ=0 A Discrete Dirac–Kähler Equation... Page 13 of 17 72 in the case of the form Ψ−. Firstly, we take the form Ψ−. Equation (4.8) implies − (p0x + p1e0e1 + p2e0e2 + p3e0e3) A = me0Ae0, (4.9) since e e = x. By trivial computation one finds that 00 3 3 3 − 2 − 2 p0x pμe0eμ p0x + pμe0eμ = p0 pμ x. μ=1 μ=1 μ=1

Then multiplying both sides of Eq. (4.9)by−(p0x−p1e0e1 −p2e0e2 −p3e0e3) we obtain 3 3 2 − 2 − − p0 pμ xA = m p0x pμe0eμ e0Ae0, μ=1 μ=1 or equivalently, 3 3 2 − 2 − p0 pμ A = m eμpμA e0. μ=1 μ=0 Applying (4.8) gives 3 2 − 2 2 p0 pμ A = m Ae0e0, μ=1 or equivalently, 3 2 − 2 2 p0 pμ A = m A. μ=1 Thus Eq. (4.8) has a non-trivial solution if and only if 2 − 2 − 2 − 2 2 p0 p1 p2 p3 = m , (4.10) or ± 2 2 2 2 p0 = m + p1 + p2 + p3.

Proposition 4.1. The form AΨ− is a non-trivial solution of the discrete Hestenes equation (3.4) if and only if the condition (4.10) holds. It is clear that the same is true for AΨ+ in place AΨ−, i.e., if we take Eq. (4.7) we obtain the condition (4.10) again. Let p = {p0,p1,p2,p3} be the energy-momentum vector of a particle with (proper) mass m. Then the relation (4.10) is the energy-momentum relation. It is known that in the continuum theory the Hestenes equation (1.5) admits the plane wave solutions of the form ± Φ = A exp(±γ2γ1p · x). Thus the forms Ω± = AΨ± are discrete versions of the plane wave solutions Φ±. Let us represent the even real-valued form (4.6)as

A = A+ + A−, 72 Page 14 of 17 V. Sushch Adv. Appl. Cliﬀord Algebras where 0 12 13 23 A+ = α x + α e12 + α e13 + α e23, (4.11) 01 02 03 4 A− = α e01 + α e02 + α e03 + α e. (4.12)

It is easy to check that A+ commutes with e0 and A− anticommutes with it, i.e., e0A± = ±A±e0. (4.13)

Lemma 4.2. The form e0μA− commutes with e0 and has the view (4.11), while e0μA+ anticommutes with e0 and has the view (4.12) for any μ =1, 2, 3. Proof. For μ =1wehave

01 02 03 4 e01A− = e01 α e01 + α e02 + α e03 + α e 01 02 03 4 = α x − α e12 − α e13 + α e23 and

0 12 13 23 e01A+ = e01 α x + α e12 + α e13 + α e23 0 12 13 23 = α e01 − α e02 − α e03 + α e. The same proof remains valid for all other cases. Theorem 4.3. The forms Ω± = AΨ± are non-trivial solutions of the discrete Hestenes equation if and only if the conditions p1e01 + p2e02 + p3e03 A± = A∓, (4.14) m − p0 hold, or equivalently, p1e01 + p2e02 + p3e03 A∓ = − A±. (4.15) m + p0 − Proof. Let AΨ satisfy Eq. (3.4). Then A = A+ + A− satisﬁes Eq. (4.8): 3 − eμpμ (A+ + A−)=m(A+ + A−)e0. μ=0 From this we obtain 3 − e0eμpμ (A+ + A−)=p0(A+ + A−)+me0(A+ + A−)e0. μ=1 Applying (4.13) we can rewrite the above relationship as 3 − e0eμpμ (A+ + A−)=(p0 + m)A+ +(p0 − m)A−. μ=1 By Lemma 4.2, collecting like terms gives

− (e0e1p1 + e0e2p2 + e0e3p3) A+ =(p0 − m) A−, (4.16)

− (e0e1p1 + e0e2p2 + e0e3p3) A− =(p0 + m) A+. (4.17) Conversely, substituting (4.14)into(4.15) yields the condition (4.10). It follows that AΨ− is a non-trivial solution of Eq. (3.4). The same is true for AΨ+ in place AΨ−. A Discrete Dirac–K¨ahler Equation... Page 15 of 17 72

It is clear that the conditions (4.14) and (4.15) can be rewritten as systems of four linear algebraic equations. For example, from (4.16) we obtain

01 0 12 13 (m − p0) α − p1α − p2α − p3α =0, 02 0 12 23 (m − p0) α − p2α + p1α − p3α =0, 03 0 13 23 (m − p0) α − p3α + p1α + p2α =0, 4 12 13 23 (m − p0) α − p3α + p2α − p1α =0. From this the form (4.12) can be written as

0 12 13 0 12 23 p1α + p2α + p3α p2α − p1α + p3α A− = e01 + e02 m − p0 m − p0 0 13 23 12 13 23 p3α − p1α − p2α p3α − p2α + p1α + e03 + e. m − p0 m − p0 Hence

A = a1 ((m − p0) x + p1e01 + p2e02 + p3e03)

+a2 ((m − p0) e12 + p2e01 − p1e02 + p3e)

+a3 ((m − p0) e13 + p3e01 − p1e03 − p2e)

+a4 ((m − p0) e23 + p3e02 − p2e03 + p1e) , (4.18) where α0 α12 α13 α23 a1 = ,a2 = ,a3 = ,a4 = . m − p0 m − p0 m − p0 m − p0 Thus the general plane wave solution of Eq. (3.4)isΩ− = AΨ−, where A is given by (4.18). Note that for given pμ, μ =1, 2, 3, there are four linearly independent solutions of the form (4.5) for each positive and negative p0. Similarly, from (4.17) we obtain

0 01 02 03 (m + p0) α + p1α + p2α + p3α =0, 12 02 01 4 (m + p0) α − p1α + p2α + p3α =0, 13 01 03 4 (m + p0) α + p3α − p1α − p2α =0, 23 02 03 4 (m + p0) α + p3α − p2α + p1α =0. This leads to

A = b1 ((m + p0) e01 − p1x − p2e12 − p3e13)

+b2 ((m + p0) e02 − p2x + p1e12 − p3e23)

+b3 ((m + p0) e03 − p3x + p1e13 + p2e23)

+b4 ((m + p0) e − p3e12 + p2e13 − p1e23) , (4.19) where α01 α02 α03 α4 b1 = ,b2 = ,b3 = ,b4 = . m + p0 m + p0 m + p0 m + p0 72 Page 16 of 17 V. Sushch Adv. Appl. Cliﬀord Algebras

By Theorem 4.3, (4.19) is equivalent to (4.18). In the same way, the explicit formula for A can be written in the case of the plane wave solution Ω+ = AΨ+. Acknowledgements The author would like to thank N. Faustino for valuable discussions. The author also thanks the referees for careful reviews and for fruitful suggestions.

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Volodymyr Sushch Koszalin University of Technology Sniadeckich 2 75-453 Koszalin Poland e-mail: [email protected]

Received: January 29, 2018. Accepted: June 25, 2018.