Projective Metric Geometry and Clifford Algebras

Each vector space that is endowed with a quadratic form determines its Clifford algebra. This algebra, in turn, contains a distinguished group, known as the Lipschitz group. We show that only a quotient of this group remains meaningful in the context of projective metric geometry. This quotient of the Lipschitz group can be viewed as a point set in the projective space on the Clifford algebra and, under certain restrictions, leads to an algebraic description of so-called kinematic mappings.


Introduction
By a metric vector space we mean a finite dimensional vector space V (over a field F of arbitrary characteristic) that is endowed with a quadratic form Q. The description of orthogonal transformations of a metric vector space (V, Q) in terms of its associated Clifford algebra Cl(V, Q) has a long history.We follow the exposition by E. M. Schröder [50] and provide in Section 2 basic facts about a metric vector space (V, Q) and its weak orthogonal group O ′ (V, Q), which in most cases is generated by reflections.In Section 3, we collect from various sources those results about the Clifford algebra Cl(V, Q) which are needed later on.Section 4 is based on the work of J. Helmstetter as summarised in [25].We recall from there the Lipschitz monoid Lip(V, Q) and the twisted adjoint representation of the Lipschitz group Lip × (V, Q), which provides a surjective homomorphism onto the weak orthogonal group O ′ (V, Q).
The main goal of the present note is the interpretation of the Lipschitz group Lip × (V, Q) in projective terms, that is, we consider the projective metric space P(V, Q) and the projective space on the associated Clifford algebra Cl(V, Q).Thereby one is immediately facing the following problem: if the quadratic form Q is replaced by a non-zero multiple, say cQ with c ∈ F \ {0}, then this does not affect the geometry of P(V, Q), but the Clifford algebras Cl(V, Q) and Cl(V, cQ) need not be isomorphic.Therefore, the usage of Clifford algebras in projective metric geometry at a first sight appears to be problematic.
We start Section 5 by introducing in P Cl(V, Q) point sets M(V, Q) and G(V, Q) that arise from a quotient of the Lipschitz monoid Lip(V, Q) and a quotient of the Lipschitz group Lip × (V, Q).The latter set can be made into a group in a natural way and as such it acts on the initial projective metric space P(V, Q).In Theorems 5.4, 5.5 and 5.6 we carry out a detailed study of this group action and its kernel, thereby extending previous work of C. Gunn [19], [20], R. Jurk [32], M. Hagemann and D. Klawitter [36], [35], E. M. Schröder [49] and others.Since the details are somewhat involved, an alternative point of view is adopted in Tables 1-3.These tables allow us to read off all those instances, where a kind of "kinematic mapping" for the projective weak orthogonal group PO ′ (V, Q) can be obtained.Next, in Section 6, we return to the problem sketched above by comparing the Clifford algebras Cl(V, Q) and Cl(V, cQ).From a result by M.-A.Knus [40,Ch.IV, (7.1.1)],we are in a position to identify the underlying vector spaces of these algebras in such a way that, firstly, their even subalgebras Cl 0 (V, Q) and Cl 0 (V, cQ) coincide (as algebras), secondly, their odd parts Cl 1 (V, Q) and Cl 1 (V, cQ) are the same (as vector subspaces), thirdly, the two multiplications are related in a manageable way.Using this identification, it turns out that all our results from Section 5 remain unaltered when going over from Q to cQ.This is due to the fact that the Clifford algebra Cl(V, Q) just serves as a kind of "container" for its even and odd part, but we never use any element of Cl(V, Q) from outside these two subspaces.Finally, Section 7 provides a list of open questions that may lead to future research.
Let us close by pointing out that our note is not intended to be a critical survey.We therefore mainly quote such work that will clear the way to previous contributions.Also, whenever we just refer to other sources without using them, we usually do not emphasise diverging definitions, differing hypotheses and other deviations from our approach.

Metric vector spaces
Let V be a vector space with finite dimension n + 1 ≥ 0 over a (commutative) field F. We suppose that V is equipped with a quadratic form Q : V → F; the zero form is not excluded.There is a widespread literature about quadratic forms; see, for example, [8,Ch. 8], [13], [43] or [53].We adopt the terminology from [50] by addressing (V, Q) as a metric vector space.A non-zero vector a ∈ V is called regular if Q(a) 0 and singular otherwise.Observe that none of these attributes applies to the zero vector.A subspace of V is said to be totally singular if all its non-zero vectors are singular. Let Orthogonality is written as ⊥; that is, for all x, y ∈ V, x ⊥ y means B(x, y) = 0.Each subset S ⊆ V determines the subspace Let ( Ṽ, Q) also be a metric vector space over F. A linear bijection ψ : , the scalar c is uniquely determined by ψ and it will be addressed as the ratio of ψ.Whenever Q(V) = {0} we adopt the convention to consider only 1 ∈ F × as being the ratio of ψ.An isometry is understood to be a similarity of ratio c = 1.
We recall that the general orthogonal group GO(V, Q) is that subgroup of the general linear group GL(V) which comprises all similarities of (V, Q) onto itself.All isometries of (V, Q) onto itself constitute the orthogonal group O(V, Q).The weak orthogonal group1 O ′ (V, Q) consists of all isometries of (V, Q) that fix the radical V ⊥ elementwise.Each regular vector r ∈ V determines the mapping We call ξ r the reflection of (V, Q) in the direction of r and note that ξ r ∈ O ′ (V, Q).
Under ξ r all vectors in r ⊥ are fixed and r goes over to −r.Also, ξ r is the identity on V if, and only if, r is a regular vector in the radical V ⊥ ; this can only happen in case of characteristic 2; see [50, 1.6.2].We are now in a position to write up a version of the classical Cartan-Dieudonné Theorem as follows.Each isometry ϕ ∈ O ′ (V, Q) is a product of reflections, except when F and (V, Q) satisfy one of the subsequent conditions (2) or (3) for some basis {e 0 , e 1 , . . ., e n } of V and all x = n j=0 x j e j with x j ∈ F: We refer to [15], [16], [39] for proofs and to [4], [9], [10], [11], [12], [14], [29], [37], [43, p. 18], [46], [50, 1.6.3],[53, pp. 156-159] for further details, generalisations and additional references.Let us just mention that the reflections of (V, Q) generate a proper subgroup of its weak orthogonal group whenever F and (V, Q) meet the requirements of (2) or (3).

Clifford algebras
Let (V, Q) be a metric vector space over F (as in Section 2) and let Cl(V, Q)  [43,Ch. 5].In our study we shall adopt two widely used conventions.Firstly, we identify 1 ∈ F with the unit element of the F-algebra Cl(V, Q) and, secondly, we consider V as being a subspace of Cl(V, Q).So Cl(V, Q) is the universal associative and unital algebra over F that is generated by V and subject to the relations Q(x) = x 2 for all x ∈ V.
Consequently, for all x, y ∈ V, we have B(x, y) = xy + yx.We now write up some well known properties of Cl(V, Q) in order to fix our notation.The Clifford algebra Cl(V, Q) is Z/(2Z)-graded and so it is the direct sum of the even part Cl 0 (V, Q), which is a subalgebra of Cl(V, Q), and the odd part Cl 1 (V, Q).If h ∈ Cl i (V, Q), i ∈ {0, 1}, then we say that h is homogeneous of degree i and write ∂h = i.Given any subset S ⊆ Cl(V, Q) we let S i := S ∩ Cl i (V, Q) for i ∈ {0, 1} and we denote by S × the set of all units (w.r.t.multiplication) in S.
The main involution σ : Cl(V, Q) → Cl(V, Q) is the only algebra endomorphism of Cl(V, Q) such that x → −x for all x ∈ V.Under σ all elements of Cl 0 (V, Q) remain fixed, any h ∈ Cl 1 (V, Q) goes over to −h.The reversal α : Cl(V, Q) → Cl(V, Q) is the only algebra antiendomorphism of Cl(V, Q) such that x → x for all x ∈ V.Each of the mappings σ and α is a bijection leaving invariant Cl 0 (V, Q) and Cl 1 (V, Q).
Cl(V, Q) is endowed with an (increasing) canonical filtration by subspaces Cl ≤k (V, Q), k ∈ Z, as follows [31, pp. 108-109] generated by all products of at most k vectors from V. Thereby an empty product of vectors is understood to be 1 ∈ F ⊆ Cl(V, Q).If {e 0 , e 1 , . . ., e n } is a basis of V, then we obtain a basis of Cl(V, Q) as Let T be a subspace of V.The restriction Q|T makes T into a metric vector space.The unital subalgebra of Cl(V, Q) generated by T will be considered as the Clifford algebra of (T, Q|T).
Even though the domain of Cl(ψ) is the entire Clifford algebra Cl(V, Q), we shall predominantly apply this mapping to homogeneous elements of Cl(V, Q).In particular, the following formula will turn out crucial, as it describes to which extent Cl(ψ) "deviates" from an isomorphism of algebras.Given homogeneous By the additivity of Cl(ψ) and the law of distributivity, it suffices to verify (9) when m = a 1 a 2 • • • a r and n = a r+1 a r+2 • • • a r+s with a 1 , a 2 , . . ., a r+s ∈ V and r, s ≥ 0.
There are four cases: Case 1: r and s are even.Here (9) holds trivially, since Cl 0 (ψ) is a homomorphism of algebras.
Case 2: r is even and s is odd.By the first equation in (7) and Case 1, Case 3: r is odd and s is even.Writing m = a 1 (a 2 • • • a r ) allows us to proceed in analogy to the previous case, thereby using the second equation in (7).
Next, let m 1 , m 2 , . . ., m k , k ≥ 0, be homogeneous elements of Cl(V, Q) such that precisely p of them are of degree 1.Then there is a unique integer q ≥ 0 with 2q ≤ p ≤ 2q + 1.From ( 9), we therefore obtain There are two immediate consequences of (10).Given a homogeneous ele- where α denotes the reversal on Cl( Ṽ, Q).If, moreover, m is invertible, then which in turn shows that Cl(ψ)(m) is invertible.
Remark 4.1.Any element 1 + st appearing in ( 13) is in the Lipschitz group An easy calculation gives, for all x ∈ V, If s, t are linearly dependent, then st = 0 and so ξ 1+st = id V .Otherwise, ξ 1+st fixes precisely the vectors of the subspace {s, t} ⊥ , which has codimension ≤ 2 in V.
In order to describe the kernel of the twisted adjoint representation (14), we recall the definition of the graded centre of Cl(V, Q).It is defined as where Z g i Cl(V, Q) , i ∈ {0, 1}, comprises precisely those p ∈ Cl i (V, Q) which satisfy pq = (−1) ∂ p∂q q p for all homogeneous q ∈ Cl(V, Q); see [31, (3.5.2)] or [40, p. 152].By (14), for all p ∈ Lip × (V, Q) and all vectors x ∈ V, we have Therefore, using that V generates Cl(V, Q) as an algebra, we readily arrive at the intermediate result From [31, (5.8.7) Lemma], the graded centre of Cl(V, Q) equals the subalgebra generated by V ⊥ , which in turn may be viewed as Cl(V ⊥ , Q|V ⊥ ).We therefore have The above description of the graded centre Z g Cl(V, Q) as the subalgebra of Cl(V, Q) generated by V ⊥ can also be read off from [32, (1.8) a), (1.9) a)].
Below we collect a few more results, which are to be used later on.
Lemma 4.2.Let (V, Q) be a metric vector space.Then the kernel of the twisted adjoint representation ξ of the Lipschitz group Lip × (V, Q) satisfies the following properties.
Proof.(a) This is an immediate consequence of (5), applied to Cl(V ⊥ , Q|V ⊥ ), and (15).(b) By (15), r ∈ V ⊥ implies r ∈ ker ξ, which proves the assertions.(c) We infer from dim V ⊥ ≤ 1 that Cl 0 (V ⊥ , Q|V ⊥ ) = F. Hence (15) gives We distinguish three cases: (i) L is totally singular; (ii) L contains no singular vectors; (iii) L contains a regular and a singular vector.In the first two cases we choose any linearly independent vectors a, b ∈ L. If (iii) applies, we choose a ∈ L regular and b ∈ L singular.Now pick any element x + yab as appearing in (16).In view of (15), it suffices to verify that x + yab ∈ Lip × 0 (V, Q).In case (i), this follows from Otherwise, a is regular and so Char F = 2.By writing x + yab = a(xa −1 + yb), it remains to verify that xa −1 + yb is regular.In case (ii), this turns out trivial.In case (iii), we have

Projective metric geometry
Let V be a vector space over F as described at the beginning of Section 2. By the projective space P(V) we mean here the set of all subspaces of V with incidence being symmetrised inclusion [6, p. 30].The dimension2 of P(V) is one less than the dimension of V; that is, P(V) has projective dimension n.We adopt the usual geometric terms: points, lines and planes are the subspaces of V with (vector) dimension one, two, and three, respectively.Likewise, any subspace T of V gives rise to a projective space P(T), which is a substructure of P(V).The general linear group GL(V) acts in a canonical way on P(V): any κ ∈ GL(V) determines a projective collineation on P(V), which is given by X → κ(X) for all X ∈ P(V).
The kernel of this action of GL(V) equals F × id V .
Next, assume (V, Q) to be a metric vector space.Then Q can been used to furnish the projective space with "additional structure", thus making it into a projective metric space P(V, Q).Thereby, for all c ∈ F × , the spaces P(V, Q) and P(V, cQ) are considered as being equal.We refer to [50] for a detailed description under the assumption Q(V) {0}; otherwise any "additional structure" arising from Q is trivial.Let us recall a few notions derived from (V, Q) that remain unchanged if Q is replaced by cQ.The orthogonality relations of (V, Q) and of (V, cQ) coincide.All points F s with s ∈ V being singular constitute the absolute quadric F(V, Q) of P(V, Q). 3 This quadric does not alter when going over to cQ.Also, we have GO( In contrast, the Clifford algebras Cl(V, Q) and Cl(V, cQ), c ∈ F × , need not be isomorphic; see Example 6.1, where it is also shown that an analogous statement holds for the associated Lipschitz groups.Nevertheless, for the remaining part of this section, we shall make extensive use of the Clifford algebra Cl(V, Q).The problem of how things change when going over to Cl(V, cQ) will be addressed in Section 6.
By the above, any isometry ϕ ∈ O ′ (V, Q) determines a projective collineation of P(V, Q).This action of O ′ (V, Q) on P(V, Q) has the kernel The On the other hand, Now we change over to the projective space on the Clifford algebra Cl(V, Q), where we introduce several point sets.Given such a set, say S, we define S i , i ∈ {0, 1}, to be the subset of S comprising all points that are contained in Cl i (V, Q).We start by defining and proceed by making H(V, Q) into a (multiplicative) group as follows: (F p)(F q) := F(pq) for all F p, F q ∈ H(V, Q).Clearly, there is a canonical isomorphism of groups So, essentially, the two groups from above are the same.The Lipschitz monoid Lip(V, Q) gives rise to the point set By the definition of Lip(V, Q), the point set Remark 5.1.The sets M 0 (V, Q) and M 1 (V, Q) are algebraic varieties of the projective spaces on Cl 0 (V, Q) and Cl 1 (V, Q), respectively.See [25, p. 673] and [24], where a wealth of further properties of these varieties can be found.In particular, all subspaces of P Cl(V, Q) whose point set is contained in M(V, Q) have been determined there.Let us just mention the following particular case.If dim V ≤ 3, then M 0 (V, Q) resp.M 1 (V, Q) comprises all points of P Cl 0 (V, Q) resp.P Cl 1 (V, Q) [23, (31) Lemma].
Remark 5.2.In general, the point set M(V, Q) cannot be made into a monoid by following the path taken above.This is because the product of two non-zero elements of Lip(V, Q) may be the zero vector, which fails to span a point.Our third point set is which is a subgroup of H(V, Q).The canonical isomorphism from (20) determines (by restriction) the isomorphism of groups The Lipschitz group Lip × (V, Q) contains F × as a normal subgroup.The representation (14) factors through the canonical homomorphism Lip × (V, Q) → Lip × (V, Q)/F × .We therefore have a surjective homomorphism of groups the twisted adjoint representation of the quotient group Lip × (V, Q)/F × .By virtue of the inverse of ( 23), the twisted adjoint representation (24) and the canonical action of O ′ (V, Q) on P(V, Q), the group G(V, Q) as in (22) acts on the projective space P(V, Q).Explicitly, for all F p ∈ G(V, Q) and all X ∈ P(V, Q), this action of G(V, Q) takes the form Furthermore, the action of G(V, Q) on P(V, Q) yields a surjective homomorphism of groups see (17).By our construction, ker θ is just the kernel of the group action described in (25).This means Remark 5.3.Let any F m ∈ H(V, Q) be given.The (group theoretic) left translation by F m, that is the mapping F q → (F m)(F q) for all F q ∈ H(V, Q), extends to a projective collineation of the ambient projective space.Obviously, the left translation λ m ∈ GL Cl(V, Q) , which is given by z → mz for all z ∈ Cl(V, Q), provides a solution.The same properties hold, mutatis mutandis, for the right translation by F m and its counterpart ρ m : z → zm on the Clifford algebra.
Given any F m ∈ G(V, Q) the above observations clearly remain true when replacing H(V, Q) with G(V, Q).However, G(V, Q) satisfies yet another property, which appears to be more substantial.The mapping that sends any F q ∈ G(V, Q) to its inverse (F q) −1 also extends to a projective collineation of the ambient projective space.Such a collineation is determined by the reversal α, since qα(q) ∈ F × for all q ∈ Lip × (V, Q); see the noteworthy properties (i)-(iii) of the Lipschitz monoid mentioned at the beginning of Section 4.
We proceed by examining in detail the kernel of the surjective homomorphism θ appearing in (26).We also investigate whether or not the subgroup θ G 0 (V, Q) coincides with its ambient group PO ′ (V, Q).Clearly, whenever G 0 (V, Q) = G(V, Q), the answer to the latter question is affirmative.The large number of cases makes us split our findings into three theorems, according to the dimension of the radical V ⊥ .Theorem 5.4.Let (V, Q) be a metric vector space with dim V ⊥ = 0. Then the surjective homomorphism θ : (c) If dim V is odd and Char F 2, then ker θ comprises precisely two points, namely F1 ∈ G 0 (V, Q) and one more point in Proof.To begin with, we note that as follows readily from Lemma 4.2 (a) and (c).
(a) The assertions hold, since (b) Due to Char F = 2 and ( 18), we have I ′ (V, Q) = {id V }.Therefore ( 27) and (28) give ker θ = {F1}.As V ⊥ = {0} and dim V > 0, there exists a regular vector r ∈ V and so There exists an orthogonal basis {e 0 , e 1 , . . ., e n } of (V, Q) and we put From V ⊥ = {0} we obtain e ∈ Lip × (V, Q).Together with dim V ≥ 1 and Char F 2 this shows ξ e = − id V id V .So, from ( 27) and ( 28), ker θ = {F1, Fe} is a group of order two.Clearly, F1 ∈ ker 0 θ.As dim V is odd, Fe ∈ ker 1 θ.We have We may repeat the reasoning from (c) up to the end of the first paragraph.By contrast, now dim V > 0 is even, whence Fe ∈ ker 0 θ.In analogy with (b), there is a regular vector r ∈ V and so Note that under the hypotheses of Theorem 5.4 (b) the bilinear form B is nondegenerate and alternating.Therefore, dim V is necessarily even.Theorem 5.5.Let (V, Q) be a metric vector space with dim V ⊥ = 1.Then the surjective homomorphism θ : Proof.First of all, from (18), we have that (a) From Lemma 4.2 (a) and (c), we have ker ξ = ker 0 ξ = F × .Thus, using ( 27) and (31), we arrive at ker θ = {F1}.
(b) By our hypotheses, we have (c) Due to V ⊥ V, there exists a regular vector in r ∈ V and so (d) There is a regular vector r ∈ V ⊥ .Using Lemma 4.2 (b) and (c), we obtain4 ker ξ = F × ∪ F × r.Together with (27) and (31) (by replacing e with r) in analogy to (30) .Regarding Theorem 5.5 (d), it seems worth pointing out that Q(V) {0} implies Char F = 2. Together with dim V ⊥ = 1 this forces dim V to be odd, since B induces a non-degenerate alternating bilinear form on the quotient space V/V ⊥ .Theorem 5.6.Let (V, Q) be a metric vector space with dim V ⊥ ≥ 2. Then the surjective homomorphism θ : Proof.(a) Lemma 4.2 (a) gives ker ξ = ker 0 ξ and, from (18), we have I ′ (V, Q) = {id V }.Thus (27) shows ker θ = ker 0 θ ⊆ G 0 (V, Q).Let L be any two-dimensional subspace of V ⊥ .By adopting the terminology from Lemma 4.2 (d) and by substituting x := 1 in ( 16), we arrive at (b) The assertion follows from ( 5).
(c) From dim V > dim V ⊥ , there exists a regular vector r ∈ V and so (d) There exists a two-dimensional subspace L of V ⊥ that contains a regular vector a, say.We pick any vector b ∈ L such that a, b are linearly independent.According to the proof of Lemma 4.2 (d) we now use these vectors to obtain (16) and, as in (a), we substitute there x := 1.In this way we get a point set as in (32).This implies that ker 1 θ contains at least |F| points, namely all points of the form F a + yQ(a)b with y varying in F.
Our description of ker θ in Theorems 5.4, 5.5 and 5.6 improves [32, (2.3) Satz] in two ways: The result b) from there describes an analogue of our surjective homomorphism θ onto the group PO ′ (V, Q); however, it is based upon a subgroup of H(V, Q) that in general is larger than our G(V, Q).The result c) from there coincides with our findings whenever Lip × (V, Q) is generated by all regular vectors of (V, Q), but provides no information about the exceptional cases (2) and (3).
Clearly, the surjective homomorphism θ as in (26) turns out to be an isomorphism of G(V, Q) onto PO ′ (V, Q) if, and only if, ker θ contains no point other than F1.There are few possibilities for this to happen.All of them can be read off from Table 1.The first entry in each row (other than the header) provides a reference to the corresponding theorem, the remaining entries summarise the conditions that have to be met.Entries in braces are redundant and could be omitted.
Likewise, there is a rather small number of instances such that θ|G 0 (V, Q) establishes an isomorphism of G 0 (V, Q) onto PO ′ (V, Q).An exhaustive summary is given in Table 2.Note that there is a single overlap between Table 1 and Table 2.It pertains the trivial case dim V = 0, where There is one more noteworthy situation, where θ fails to be injective, but ker θ is a group of order two; the details are displayed in Table 3.Here the group G 0 (V, Q) is equipped with the distinguished point Fe, which does not depend on the choice of the orthogonal basis {e 0 , e 1 , . . ., e n } of V that has been used in (29) when defining e.The left translation λ e (right translation ρ e ) acts on P Cl(V, Q) as a projective collineation; see Remark 5.3.It is easy to verify that, for all j ∈ {0, 1, . . ., n}, we have ee j = −e j e.Using the basis (4) of Cl(V, Q), we therefore obtain Thus, even though λ e and ρ e act differently on P Cl(V, Q) , their actions on P Cl 0 (V, Q) and P Cl 1 (V, Q) coincide.
= 0 (= {0}) > 0 and even 2 Table 3: To conclude this section, let us point out the following.If one of the situations from Table 1 occurs, then we may consider θ −1 as being a bijective "kinematic mapping" for the group PO ′ (V, Q).Note that this just a name for a series of examples rather than a general definition of such a mapping.Also, if one of the situations from Table 2 occurs, we have a bijective "kinematic mapping" for the group PO ′ (V, Q) given by θ|G 0 (V, Q) −1 .Under the restrictions of Table 3 we still have a kind of "kinematic mapping", but here one element of PO ′ (V, Q) is represented by an unordered pair of points from G(V, Q).Some of the examples in [35, 3.4] and [36,Sect. 6] fit into the above concepts.However, the quoted works should be read with caution due to several misprints.

A comparison of Clifford algebras
We now switch back to a problem that we encountered in Section 5. Given a metric vector space (V, Q) and a constant c ∈ F × what is the relationship between the Clifford algebras Cl(V, Q) and Cl(V, cQ)?For a closer look, we take into account that the identity id V is a similarity of ratio c from (V, Q) onto (V, cQ).(Recall our convention that c = 1 whenever Q(V) = {0}.)Therefore, according to (8), we obtain a linear bijection This linear bijection allows us pulling back the algebra structure from Cl(V, cQ) to Cl(V, Q), which amounts to introducing a "new" multiplication ⊙ c on the vector space Cl(V, Q).The algebra obtained in this way is isomorphic to Cl(V, cQ) and will be abbreviated as Cl(V, Q, ⊙ c ).A bridge between the initial and the new multiplication is provided by ( 6) and (7).We read off from there, for all x, y ∈ V and all p ∈ Cl 0 (V, Q): Similarly, one may write up analogues of ( 9), ( 10), ( 11) and (12).In what follows right now, we shall adopt a slightly different point of view.We investigate the Clifford algebras of metric vector spaces (V, Q) and ( Ṽ, Q) admitting a similarity ψ of ratio c ∈ F × with Cl(ψ) playing the role of the linear bijection (33).We shall return to Cl(V, Q, ⊙ c ) only at the end of this section.
Example 6.1.Let V be a one-dimensional vector space over the field R of real numbers and let i ∈ V be non-zero.We define a quadratic form Q : V → R by Q(i) = −1.Then Cl(V, Q) and the field C of complex numbers are isomorphic as R-algebras, as follows from i 2 = −1.Furthermore, let ( Ṽ, Q) be isometric to (V, −Q), whence there is a similarity ψ : So the Lipschitz groups of (V, Q) and ( Ṽ, Q) cannot be isomorphic either.In contrast, the quotient groups Lip × (V, Q)/R × and Lip × ( Ṽ, Q)/R × both have order two and so they are isomorphic; see Theorem 6.3.Theorem 6.2.Let ψ : V → Ṽ be a similarity of ratio c ∈ F × of metric vector spaces (V, Q) and ( Ṽ, Q).Then the Clifford extension Cl(ψ) has the following properties.(b) To begin with, choose any m ∈ Cl × 0 (V, Q) ∪ Cl × 1 (V, Q).From ( 12), the element Cl(ψ)(m) is in Cl × 0 ( Ṽ, Q) ∪ Cl × 1 ( Ṽ, Q).We now show that Cl(ψ) sends any generator of Lip(V, Q), that is to mean any element g from F, V or the set (13), to a generator of Lip( Ṽ, Q) of the same kind.If g is in F ∪ V, then this is obvious.If g belongs to the set (13) or, more explicitly, if g = 1 + st with s, t ∈ V subject to Q(s) = Q(t) = B(s, t) = 0, then (9) implies Cl(ψ)(1 + st) = 1 + c −1 ψ(s)ψ(t).As ψ is a similarity, we obtain Q c −1 ψ(s) = Q ψ(t) = B c −1 ψ(s), ψ(t) = 0, whence Cl(ψ)(g) has the required property.
Finally, (b) follows by repeating the above considerations with the similarity ψ −1 instead of ψ.
Another step, still to be taken in a general context, is the inclusion of affine metric geometry.Over the real numbers this task has been accomplished quite a while ago and leads to what is called a homogeneous model.Related work can be read off from [35, 3.4.2],[36, 6.2], [20] and [21].However, the approach used there relies on the signature of a real quadratic form, a notion which is no longer available over an arbitrary field.
Last, but not least, also the general theory should allow for amplification.The results in [48], where points and planes of a three-dimensional projective space are used to represent motions of metric planes, suggest to investigate under which conditions the subspaces Cl 0 (V, Q) and Cl 1 (V, Q) of the Clifford algebra Cl(V, Q) can be made into a dual pair of vector spaces in some meaningful way.