Some Representations of Affine Conformal Transformations of Minkowski Space

George Mărgulescu

Abstract We consider the groups G_{1}, G_{2}, G_{3} that are different from affine conformal group just because the space-like or (and) time-like symmetries are accompanied by the inversion I_{0} (charge conjugation operation). For these groups there exist some fundamental spin representations (spin s = \frac{1}{2} ) given by (14); the representations of the subgroup formed from the proper Lorentz group, the homotheties and the considered symmetries, for different couples λ, μ such that μ^{2} − λ^{2} = 1, are equivalent.

Mathematics Subject Classification: 22E43, 53A30, 22E70, 53B50, 53A50. Key words: Conformal group, spin representations, conformal spinors.

1. Let M be Minkowski space, x^{i} (i = 1, 2, 3, 4) being orthogonal coordinates, x^{2} = −(x^{1})^{2} − (x^{2})^{2} − (x^{3})^{2} + (x^{4})^{2}. Let M^{#} be compactified Minkowski space [1], [2], [4] (M with a ”null cone at infinity”) and G the group of conformal transformations of M (more exactly, of M^{#}). We know that the transformations of G may be considered as linear transformations in a six-dimensional pseudo-Euclidean space E_{6}^{2} with the metric G_{ab} = ε_{a}δ_{a}^{b}, ε_{a} = −1, −1, −1, +1, −1, +1 and let O(4, 2) the group of motions of E_{6}^{2}. Then G ≈ O(4, 2)/Z_{2}, where Z_{2} is the center of O(4, 2) which, in the matrix (six-dimensional) representation, consists of I and −I. We can construct an epimorphism η : O(4, 2) → G in the following manner: to every σ ≡ σ(t_{b}^{a}) ∈ O(4, 2), (z^{′a} = t_{b}^{a}z^{b}, G_{ab}t_{c}^{a}t_{d}^{b} = G_{cd}, a, .., d = 1, .., 6) corresponds g ∈ G

(1) g : x^{′i} = \frac{∑ t_{j}^{i}x^{j} + \frac{1}{2}(t_{5}^{i} + t_{6}^{i})x^{2} + \frac{1}{2}(t_{5}^{i} − t_{6}^{i})}{∑(t_{j}^{5} − t_{j}^{6})x^{j} + \frac{1}{2}(t_{6}^{5} − t_{6}^{6} + t_{5}^{5} − t_{5}^{6})x^{2} + \frac{1}{2}(t_{5}^{5} − t_{5}^{6} − t_{6}^{5} + t_{6}^{6})};

(\frac{z^{i}}{z^{5} − z^{6}} = x^{i}, \frac{z^{5}}{z^{5} − z^{6}} = \frac{x^{2} + 1}{2}, \frac{z^{6}}{z^{5} − z^{6}} = \frac{x^{2} − 1}{2}) .

Note that σ ≡ σ(t_{b}^{a}) and σ̃ ≡ σ(−t_{b}^{a}) have the same image, therefore O(4, 2) is a two-fold covering group of G. In particular, if O^{c} and G^{c} denote the identityconnected component of O(4, 2) and G respectively, we see that O^{c} is also a two-fold covering group of G^{c}.

Balkan Journal of Geometry and Its Applications, Vol.2, No.2, 1997, pp. 91-97 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

Let G_{0} ⊂ G be the affine conformal group, for which the ”null cone at infinity” is preserved. G_{0} is formed by the full Lorentz group L, (x^{′i} = t_{j}^{i}x^{j}, 6 parameters), the translation group T (T (a) : x^{′i} = x^{i} − a^{i}, 4 parameters) and the dilatation group H (H(θ) : x^{′i} = e^{−θ}x^{i}, 1 parameter). L is formed from L_{+}^{↑} (identity-connected component of L) by addition of two symmetries, for instance τ_{1} : x^{1} → −x^{1} and τ_{4} : x^{4} → −x^{4.}. The full conformal group G (with 15 parameters) is obtained from G_{0} by addition of the inversion I_{0} : x^{′i} = \frac{x^{i}}{x^{2}} . One obtains a new subgroup with 4 parameters

(2)  G_{s}(a^{i}) : x^{′i} = \frac{x^{i} + a^{i}x^{2}}{1 + 2a^{i}x_{i} + a^{2}x^{2}} .

By direct computation one obtains G_{s}(a^{i}) = I_{0}T (−a)I_{0} and also G_{s}(a^{i}) = I(a)τ_{a}, where I(a) is the inversion relative to the hypersphere of center a^{i}/a^{2} and radius 1/\sqrt{a^{2}}, and τ_{a} is the symmetry in the direction of nonisotropic vector

a(τ_{a} : x^{′i} = (δ_{j}^{i} − 2\frac{a^{i}a_{j}}{a^{2}} )x^{j}).

In E_{6}^{2} we have

H(θ) : \{ \begin{matrix}z^{′i} = z^{i} (i = 1, ..., 4) \\ z^{′5} = − sinh θ z^{5} + cosh θ z^{6} \\ z^{′6} = cosh θ z^{5} − sinh θ z^{6}, \end{matrix} L : \{ \begin{matrix}z^{′i} = l_{j}^{i}z^{j} \\ z^{′5} = z^{5} \\ z^{′6} = z^{6}, \end{matrix}

G_{s}(a) : \{ \begin{matrix}z^{′i} = z^{i} + a^{i}(z^{5} + z^{6}) \\ z^{′5} = (1 + \frac{a^{2}}{2} )z^{5} + \frac{a^{2}}{2} z^{6} + a_{i}z^{i} \\ z^{′6} = (1 − \frac{a^{2}}{2} )z^{6} − \frac{a^{2}}{2} z^{5} − a_{i}z^{i}, \end{matrix}

T (a^{i}) : \{ \begin{matrix}z^{′i} = z^{i} + a^{i}(z^{6} − z^{5}) \\ z^{′5} = z^{5} − \frac{1}{2}a^{2}(z^{6} − z^{5}) − a_{i}z^{i} \\ z^{′6} = z^{6} − \frac{1}{2}a^{2}(z^{6} − z^{5}) − a_{i}z^{i}. \end{matrix}

Reflexion τ_{i} : x^{i} → −x^{i} (fixed i) corresponds in E_{6}^{2} to z^{i} → −z^{i}. The reflexion τ_{6} : z^{6} → −z^{6} corresponds, in M^{#},to the inversion I_{0} and τ_{5} is corresponding to I_{0} followed of the ” total reflection” of the axes x^{′i} = −x^{i} (i = 1, ...4).

To obtain the spin representations of G, it suffices to get the spin representations of O(4, 2). For this, we consider the Clifford algebra C_{6} associated to E_{6}^{2}.

2. Let C_{6} be the associative algebra with unity generated (with complex coefficients) by six entities β_{1}, ..., β_{6} satisfying

(3)  β_{a}β_{b} + β_{b}β_{a} = 2G_{ab} · 1

Then β_{1}, ..., β_{6} are linearly independent and let W ⊂ C_{6} be the linear space spanned by the β_{a}. For every two vectors x,y ∈ W we define (x,y) = \frac{1}{2}(xy+yx). Writing down x = x^{i}β_{i}, y = y^{i}β_{i} we will obtain (x,y) = ∑ x^{i}y^{i} such that (x,y) defines a scalar product into W and {β_{a}} is an orthonormal basis. We agree to identify both spaces W and E_{6}^{2} and their orthonormal considered bases. Let Ω be the multiplicative group of the nonisotropic vectors of E_{6}^{2}. Then for the symmetry τ_{a} in the direction of nonisotropic vector a ∈ E_{6}^{2} we get

(4) τ_{a} : z′ = −aza^{−1} or z′ = (aE)z(aE)^{−1} (E = β_{1}...β_{6}), z ∈ E_{6}^{2}).

Because for each σ ∈ O(4, 2) we have σ = τ_{a_{1}}...τ_{a_{k}}, there exists an epimorphism h : Ω → O(4, 2),

(5)  σz = (−1)^{k}ωzω^{−1} or σz = (ωE)z(ωE)^{−1}, (ω = a_{k}...a_{1} ∈ Ω),

ker h = C. Note that χ ≡ η ∘ h is epimorphism of Ω onto G. Let Ω̄ ⊂ Ω be the group [3] of normed elements of Ω such that h̄ ≡ h|Ω̄ has ker h̄ = {±1}, the group Ω̄ being a two-fold covering group of O(4, 2) (cf. [3] there exist four normed groups Ω_{εε′} ⊂ Ω that give us distinct representations of O(4, 2). Let Ω̄ be one of this.) Note that, denoting σ = h(±ω) for ω ∈ Ω̄, we have σ̃ = h(±ωE) whence, for χ̄ ≡ χ|Ω̄ we have ker χ̄ = {±1, ±E} and Ω̄ is a four-covering group of G. Obviously, the matrix representation of C_{6} gives us a four-valued linear representation of G, called fundamental spin representation of G. For the infinitesimal generators of this representation we get

L_{ij} = \frac{1}{2}β_{i}β_{j}, D = \frac{1}{2}β_{6}β_{5}, P_{i} = \frac{1}{2}β_{i}(β_{5} + β_{6}), K_{i} = −\frac{1}{2}β_{i}(β_{5} − β_{6}),

and the operators D(a^{i}) = e^{a^{i}P_{i}}, D(θ) = e^{θ}D, D(c^{i}) = e^{c^{i}K_{i}} of T (a), H(θ), G_{s}(c) are

(6)  D(a^{i}) = 1 + \frac{1}{2}a^{i}β_{i}(β_{5} + β_{6}), D(θ) = cosh \frac{θ}{2}I + sinh \frac{θ}{2}β_{6}β_{5},

D(c^{i}) = 1 − \frac{1}{2}c^{i}β_{i}(β_{5} − β_{6}).

3. We will consider some of the subgroups of G, which contain G_{0}^{c} and which can have applications in the classification of the elementary particles.

Considering the decomposition of the groups G and G_{0} in connected components

G = G^{c} ∪ τ_{1}G^{c} ∪ τ_{4}G^{c} ∪ τ_{1}τ_{4}G^{c}, G_{0} = G_{0}^{c} ∪ τ_{1}G_{0}^{c} ∪ τ_{4}G_{0}^{c} ∪ τ_{1}τ_{4}G_{0}^{c},

we notice that G_{1} = G^{c} ∪ τ_{1}τ_{4}G^{c}, G_{2} = G^{c} ∪ τ_{1}G^{c}, G_{3} = G^{c} ∪ τ_{4}G^{c}, are subgroups of G. For a certain group G_{r} ⊂ G and h_{1}, h_{2} ∈ G, we will denote with {G_{r}; h_{1}, h_{2}} the group formed from the elements of G_{r} and the composition of these, in a certain number of times, with the elements h_{1}, h_{2}. Proposition 1. We have:

1. G_{1} = {G_{0}^{c}; τ_{1}I_{0}, τ_{4}I_{0}};

2. G_{2} = {G_{0}^{c}; τ_{1}, τ_{4}I_{0}};

3. G_{3} = {G_{0}^{c}; τ_{1}I_{0}, τ_{4}}.

We will prove here only the affirmation relative to G_{1}.

Through a direct computation (which can be done in the space M^{#} or E_{6}^{2}) we get

τ_{i}I_{0} = I_{0}τ_{i}, τ_{i}T (a) = T (τ_{i}a)τ_{i}, H(θ)I_{0} = I_{0}H(−θ).

Then, the relation G(a) = I_{0}T (−a)I_{0} can be written

G(a) = (τ_{4}I_{0}) ∘ T (−a) ∘ (τ_{4}I_{0}),

and therefore {G_{0}^{c}; τ_{4}I_{0}}⊇ G^{c}. But τ_{4}I_{0} ∈ G^{c} (because in E_{6}^{2} it represents the transformation τ_{4}τ_{6}; or in M^{#} we can verify that τ_{4}I_{0} = T (b_{0})G(b_{0})T (b_{0}) ∈ G^{c} with b_{0} = (0, 0, 0, 1)). It results

(7)  G^{c} = {G_{0}^{c}; τ_{4}I_{0}}

On the other hand, τ_{1}I_{0} ∉ G^{c} because, in E_{6}^{2} it is the product of the spacelike or time-like symmetries τ_{1} and τ_{6} (in M^{#} we can verify that τ_{1}I_{0} = τ̄ ∘ T (a_{0}) ∘ G(−a_{0}) ∘ T (a_{0}) with a_{0} = (−1, 0, 0, 0) and τ̄ ≡ τ_{1}τ_{2}τ_{3}τ_{4} ∉ G^{c}). For g ∈ G^{c} it results (τ_{1}I_{0})g = τ̄T (a_{0})G(−a_{0})T (a_{0})g = τ_{1}τ_{4}g′, g′ ∈ G^{c}. Using (7) follows G_{1} = {G_{0}^{c}; τ_{1}I_{0}, τ_{4}I_{0}}.

For the group G_{1} we will give the decomposition in connected components:

(8)  G_{1} = G_{0}^{c} ∪ G_{0}^{c}(τ_{4}I_{0})G_{0}^{c}G_{0}^{c}(τ_{1}I_{0})G_{0}^{c} ∪ τ_{1}τ_{4}G_{0}^{c}.

Indeed, (see above) we have G_{s}(a^{i}) = I(a)τ_{a}. But, for a^{2} > 0 we can write τ_{a} = g_{0}τ_{4} (because through a translation and a dilatation, therefore through a proper Lorentz transformation, we can transform the vector a into e_{4}) and for a^{2} < 0 we can write τ_{a} = g_{0}τ_{1}, g_{0} ∈ G_{0}^{c}. Because the inversion

I_{C,R^{2}} : x^{′i} − c^{i} = R^{2} \frac{x^{i} − c^{i}}{x-c^{2}}

can be written, for R^{2} > 0 as I_{C,R^{2}} = T (−c^{i})H(R^{2})I_{0}T (c^{i}) and for R^{2} < 0 as I_{C,R^{2}} = T (−c^{i})τ̄H(R^{2})I_{0}T (c^{i}), it results that for any nonisotropic vector a, we can write G_{s}(a^{i}) = g_{0}(τ_{4}I_{0})g′_{0}, g_{0}, g′_{0} ∈ G_{0}^{c}. From τ_{4}I_{0} ∈ G^{c} it follows

G^{c} = G_{0}^{c} ∪ G_{0}^{c}(τ_{4}I_{0})G_{0}^{c}.

Using the relation which defines G_{1}, we obtain (8).

We notice that the groups G_{1}, G_{2}, G_{3} are different from G_{0} just because of the fact that the symmetries are accompanied by the inversion I_{0}. Thus, it is possible that the elementary particles (for example neutrino) for which the symmetry operations are accompanied in an obligatory way by the operation of charge conjugation, to be the representations of one of these subgroups.

4. Obviously, every representation of G gives us a representation of G_{1} (respectively of G_{2}, G_{3}). We will prove that there exist some representations of G_{1} (also, of G_{2}, G_{3}) that are not obtained by this way. Lemma 2. Let K=λβ_{5} + μβ_{6}, μ^{2} − λ^{2} ≡ ε = ±1. Then the mapping E_{6}^{2} →E_{6}^{2} given by

I(λ, μ) : z′ = −KzK ^{−1}, z = z^{a}β_{a} ∈ E_{6}^{2}, (a = 1, ..., 6; λ, μ ∈ R)

is the inversion with respect to the hypersphere with center O(0,...,0) and the

radius R = \sqrt{\frac{λ+μ}{μ−λ}}.

Proof. Obviously, K is nonisotropic vector and K^{−1} = εK. We have

z′= − KzK^{−1}=z^{i}β_{i} + \frac{1}{ε}[(μ^{2} + λ^{2})z^{5} − 2λμz^{6}]β_{5} + +\frac{1}{ε}[−(μ^{2} + λ^{2})z^{6} + 2λμz^{5}],

and therefore

z^{′i} = z^{i}, z^{′5} = \frac{1}{ε}[(μ^{2} + λ^{2})z^{5} − 2λμz^{6}], z^{′6} = \frac{1}{ε}[−(μ^{2} + λ^{2})z^{6} + 2λμz^{5}].

Since z^{a}z_{a} = z^{′a}z′_{a}, the function I(λ, μ) is an isometry in E_{6}^{2}. Then in M^{#} we have

x^{′i} = \frac{z^{′i}}{z^{′5} − z^{′6}} = \frac{μ + λ}{μ − λ} · \frac{x^{i}}{x^{2}}

and the lemma is proved.

Denoting h_{0} the homothety with the center O and power |\frac{μ + λ}{μ − λ}|, it results

that

(9) I(λ, μ) = h_{0} ∘ I_{0}, if \frac{μ + λ}{μ − λ} > 0; I(λ, μ) = h_{0} ∘ τ̄ ∘ I_{0}, if \frac{μ + λ}{μ − λ} < 0.

We denote Ω_{0}^{c} the subgroup of Ω̄ for which χ(Ω_{0}^{c}) = G_{0}^{c} ⊂ G^{c}. The group Ω_{0}^{c} does not depend on the choice done for \overline{Ω}. It is the subgroup of the elements like ω = a_{1}a_{2}...a_{2k}, where a_{i} are unit vectors, such that an even number of vectors are space-like among a_{1}, a_{2}, ...a_{2k}. Proposition 3. Let K=λβ_{5} + μβ_{6}, μ, λ ∈ R, μ^{2} − λ^{2} = ±1. Let s=νβ_{1}K, t = ν′β_{4}K, where ν, ν′ are defined by s^{2} = ε, t^{2} = ε′ and ε = ±1, ε′ = ±1(therefore ν, ν′ = ±1, ±i). Let

(10)  Ω_{εε′}^{∗}(λ, μ) = Ω_{0}^{c} ∪ Ω_{0}^{c}sΩ_{0}^{c} ∪ Ω_{0}^{c}tΩ_{0}^{c} ∪ stΩ_{0}^{c}.

Then:

1. Ω_{εε′}^{∗}(λ, μ) is group.

2. χ_{1} = χ|_{Ω_{εε′}^{∗} (λ,μ)} is homomorphism of Ω_{εε′}^{∗}(λ, μ) onto G.

3. There exists U∈ Ω such that UΩ_{εε′}^{∗}(λ, μ)U^{−1} = \{ \begin{matrix}Ω_{εε′}^{∗}(0, 1) for μ^{2} − λ^{2} = 1, \\ Ω_{εε′}^{∗}(1, 0) for μ^{2} − λ^{2} = −1. \end{matrix}

Proof. First, we notice that the four subsets which appear in (10) are disjunctive. So, for example, an equality in the form ω_{0}β_{1}Kω′_{0} = ω_{1}β_{4}Kω′_{1}, ω_{0}, ω′_{0}, ω_{1}, ω′_{1} ∈ Ω_{0} implies ωβ_{1}K = β_{4}Kω′, ω, ω′ ∈ Ω_{0}. Then, through the homomorphism χ, we have gτ_{1}h_{0}I_{0} = τ_{4}h_{0}I_{0}g′, g, g′ ∈ G_{0}^{c}. Because h_{0}τ_{1} = τ_{1}h_{0}, gh_{0} = h_{0}g_{1}, it results g_{1}τ_{1}I_{0} = τ_{4}I_{0}g′, g_{1}, g′ ∈ G_{0}^{c}.

Considering the corresponding transformations in E_{6}^{2}, it result gτ_{1}τ_{6} = τ_{4}τ_{6}g′, where g, g′ belong to the identity-connected component of the group O(4,2); but this relation is impossible because the right member belongs to this connected component and the left member does not belong to it.

To prove that Ω_{εε′}^{∗}(λ, μ) is subgroup of Ω, it is sufficient to show that the product of any two of its elements belongs to Ω_{εε′}^{∗}(λ, μ). Firstly, we mention that if ω ∈ Ω_{εε′}^{∗}(λ, μ), then also ωE ∈ Ω_{εε′}^{∗}(λ, μ). Indeed, ω ∈ Ω_{εε′}^{∗}(λ, μ) implies ω_{0}E ≡ ω_{0}β_{1}...β_{6} ∈ Ω_{0}^{c}, because χ(ω_{0}E) = χ(ω_{0})χ(E) = χ(ω_{0}) ∈ G_{0}^{c}. It follows that each of the four components of Ω_{εε′}^{∗}(λ, μ) (see (10) ) contains, together with the element ω, also the elements −ω, ±ωE. Thus, if ω ∈ Ω_{0}^{c}(β_{4}K)Ω_{0}^{c}, therefore ω = ω_{0}(β_{4}K)ω′_{0}, then ωE = ω_{0}(β_{4}K)ω′_{0}E belongs to Ω_{0}^{c}(β_{4}K)Ω_{0}^{c}, because ω_{0}, ω_{0}E ∈ Ω_{0}^{c}. Whence χ^{−1}(G_{1}) = Ω_{εε′}^{∗}(λ, μ). Then, for ω_{1}, ω_{2} ∈ Ω_{εε′}^{∗}(λ, μ) we have χ(ω_{1}ω_{2}) = χ(ω_{1})χ(ω_{2}); since χ(ω_{1}), χ(ω_{2}) belong to the group G_{1},it follows ω_{1}ω_{2} ∈ χ^{−1}(G_{1}) = Ω_{εε′}^{∗}(λ, μ).

To prove the last point, we denote

K_{1} = cosh φ β_{5} + sinh φ β_{6}, K_{2} = sinh φ β_{5} + cosh φ β_{6}.

We have

K_{1}^{2} = −1, K_{2}^{2} = +1, K_{1}K_{2} + K_{2}K_{1} = 0, K_{α}β_{i} + β_{i}K_{α} = 0, α = 1, 2.

Therefore the systems of vectors {β_{i}, β_{5}, β_{6}} and {β_{i}, K_{1}, K_{2}} satisfy the same commutation rules, and hence there exists U∈ Ω such that

Uβ_{i}U^{−1} = β_{i}, UK_{1}U^{−1} = β_{5}, UK_{2}U^{−1} = β_{6}.

Let us consider two choices of the vectors K, namely K = K_{2}, K′ = β_{6}. Then s= νβ_{1}K_{2}, t= ν′β_{4}K_{2} and s′ = νβ_{1}β_{6}, t′ = ν′β_{4}β_{6}. Because UsU^{−1} = s′, UtU^{−1} = t′, it results

(11)  UΩ_{εε′}^{∗}(sinh φ, cosh φ)U^{−1} = Ω_{εε′}^{∗}(0, 1).

Analogously,

(12)  UΩ_{εε′}^{∗}(cosh φ, sinh φ)U^{−1} = Ω_{εε′}^{∗}(1, 0).

Indeed, the operators D(l_{j}^{i}), D(a^{i}) and D(θ) for the proper Lorentz transformation, translation T (a^{i}) respectively for the homothety H(θ) in the matrix representation of C_{6} are given by (6) and using the relations

K_{1} + K_{2} = e^{φ}(β_{5} + β_{6}), K_{1}K_{2} = β_{5}β_{6}

we obtain

(13)  UD(l_{j}^{i})U^{−1} = D(l_{j}^{i}), UD(θ)U^{−1} = D(θ), UD(a^{i})U^{−1} = D(e^{−φ}a^{i}),

and therefore UΩ_{0}^{c}U^{−1} = Ω_{0}^{c}.

Analogously, since {β_{i}, β_{5}, β_{6}} and {β_{i}, iβ_{6}, iβ_{5}} satisfy (3) there exists V ∈ Ω such that

V β_{i}V^{−1} = β_{i}, V β_{5}V^{−1} = iβ_{6}, V β_{6}V^{−1} = iβ_{5}.

Then

V β_{i}β_{5}V^{−1} = iβ_{i}β_{6}, V β_{5}β_{6}V^{−1} = β_{5}β_{6}, V (β_{5} + β_{6})V^{−1} = i(β_{5} + β_{6}).

Therefore

V Ω_{εε′}^{∗}(1, 0)V^{−1} = Ω_{−ε,−ε′}^{∗}(0, 1).

The relations (13) suggest to consider the subgroup G′_{1} ≡ {L_{+}^{↑}, H, τ_{1}I_{0}, τ_{4}I_{0}}. It is the Weyl group {L, H}, where the symmetries are accompanied by the inversion I_{0}. Conclusion. Considering the matrix (spin) representation of C_{6}, one obtains:

For given ε, ε′ (ε, ε′ = 1 or −1) there exist four classes of fundamental spin representations (spin s = \frac{1}{2}) of the group G_{1} such that to the symmetries in the direction of vectors a,b∈ M^{#} (a^{2} = 1, b^{2} = −1, a= a^{i}β_{i}, b= b^{i}β_{i}) correspond the operators

(14)  s=νa(λβ_{5} + μβ_{6}), t = ν′b(λβ_{5} + μβ_{6}), μ^{2} − λ^{2} = 1.

For different couples λ, μ such that μ^{2} − λ^{2} = 1 one obtain equivalent representations of the subgroup G′_{1} ⊂ G_{1} (for given ε, ε′). For this subgroup, the substitution μ^{2} − λ^{2} = 1 → μ^{2} − λ^{2} = −1 is equivalent to ε, ε′ → −ε, −ε′.

It is possible that an elementary particle with isospin corresponds to such a representation.

Acknowledgements. A version of this paper was presented at the First Conference of Balkan Society of Geometers, Politehnica University of Bucharest, September 23-27, 1996.

References

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[2] R.Penrose,W.Rindler, Spinors and space-time, Cambridge Univ.Press, 1986.

[3] G.Mărgulescu, Representations spinorielles du groupe conforme de l’espace de Minkowski, Lucrările Colocviului Naţional de Geometrie şi Topologie, 1981, 226-233.

[4] b4 H.S.M.Coxeter, Representation of conformal space on a quadratic, Ann. Math, 1936, 37,416-26.

Department of Mathematics, Tehnical University of Civil Engineering of Bucharest Bul. Lacul Tei, nr. 124, Bucharest, 72302, Romania