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11). four. three. five Clifford exponentials as opposed to external algebra exponentials We first imagine okay = C. allow A ∈ o(V ), and view the formulation from part four. three. three for C = exp A, D = B ◦ A2 . We receive E1 = 2f (A) ◦ B , E2 = 2B ◦ g(A), T = j R (A) with the subsequent services of z ∈ C, 1 ez − 1 z 1 sinh(z) − z R , j (z) = coth − , g(z) = . 2 2 z z z2 For later use we additionally outline 1 − e−z sinh(z/2) , j L (z) = . j (z) = z/2 z f (z) = 100 four Covariant and contravariant spinors notice that g, j L , j R , and j √ are whole holomorphic capabilities on C, whereas f is meromorphic with poles at 2π −1k with okay ∈ Z − {0}. because f and g are bizarre services, the weather f (A) and g(A) are back in o(V ), whereas j (A) = j (A) and j L (A) = j R (A). moreover, j L (A),√ j R (A), and j (A) are invertible if and provided that A has no eigenvalues of the shape 2π −1k with ok ∈ Z − {0}. The ensuing formulation for the emblem provides: Theorem four. 2 feel ok = R or C. For all A ∈ o(V ) with the valuables has √ no eigenvalue 2π −1k with okay = zero, the logo of exp(γ (A)) ∈ Cl(V ) is given through the formulation σ (exp(γ (A)) = ι(S (A)) exp(λ(A)), the place S : o(V ) → ∧(V ) is the map S (A) = det1/2 (j (A)) exp 4λ(f (A)) . once more, whereas Proposition four. 7 calls for be invertible (which by no means occurs if dim V is odd), the ensuing formulation holds with out this assumption. In case A is invertible we will be able to without delay write sinh(A/2) fourλ 1 coth( A )− 1 2 A . e 2 S (A) = det1/2 A/2 because it stands, S : o(V ) → ∧(V ) is a meromorphic function,√holomorphic at the set whose parts A do not need eigenvalues of the shape 2π −1k with ok ∈ Z − {0}. we'll see less than that it's in truth holomorphic all over. instance four. three We proceed the calculations from instance four. 2, the place exp(γ (A)) = cos(θ/2) − sin(θ/2)e1 e2 . we now have sinh(A/2) A/2 = sin(θ/2) θ/2 I , therefore det1/2 sin(θ/2) sinh(A/2) = . A/2 θ/2 nevertheless, 1 A 1 1 θ 1 f (A) = coth − = − cot − J, 2 2 A 2 2 θ so 4λ(f (A)) = cot 2 θ − e1 ∧ e 2 . 2 θ consequently S (A) = θ 2 sin(θ/2) 1 + cot − e1 ∧ e2 . θ/2 2 θ be aware that S (A) has no poles. Contracting with exp(λ(A)) = 1 − θ2 e1 ∧ e2 , we discover ι(S (A)) exp(λ(A)) = cos as wanted. θ θ − sin e1 ∧ e2 = σ (exp(γ (A))), 2 2 4. three The quantization map revisited a hundred and one four. three. 6 the logo of parts exp(γ (A) − i ei τ i) Theorem four. 2 has an invaluable generalization, taking into account linear phrases. allow P be a vector area of strange parameters. allow ei be a foundation of V , and look at expressions ei ⊗ τ i ∈ V ⊗ P with τ i ∈ P . word that we will be able to view the parameters τ i because the parts τ (ei ) of a linear map τ : V → P . the subsequent theorem indicates that a similar aspect S (A) ∈ ∧(V ) as prior to relates the exponentials of components λ(A) − ei τ i ∈ ∧(V ) ⊗ ∧(P ), i ei τ i ∈ Cl(V ) ⊗ ∧(P ). γ (A) − i particularly remarkably, it's not essential to introduce a τ -dependence into the definition of S (A). Theorem four. three permit okay = R or C. With S (A) as above, eγ (A)− i ei τ i = q ι(S (A))eλ(A)− i ei τ i . facts contemplate Cl(V ) ⊗ ∧(P ) = Cl(V ⊕ P ) because the Clifford algebra for the degenerate bilinear shape B ⊕ zero.