# Download Low-dimensional geometry: From Euclidean surfaces to by Francis Bonahon PDF By Francis Bonahon

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26) Case 1: For points on the scalar axis excluding the points (1, 0) and (−1, 0), we have α2 = β2 = 0. In this case x = ±α1 σ1 , z = ±β1 , xz = α1 β1 σ1 . 27 show that whereas the points on the positive scalar axis represent pure dilation by an amount |z| = β1 , those on the negative scalar axis represent rotation through an angle π together with dilation by an amount |z| = β1 . Case 2: For points on the pseudoscalar axis excluding the points (0, i) and (0, −i), we have α1 = β1 = 0. In this case x = ±α2 σ2 , z = ±β2 i, xz = −α2 β2 σ1 .

The unit vector aˆ = a |a |−1 is called the direction of the a-line, whereas aˆ gives the opposite orientation for the line. 2) where β is an arbitrary scalar. 1) by vector a gives x ∧ a = 0. 3) This is a nonparametric equation for the a-line. 3 as x ∧ aˆ = 0. 4) Now we can prove the following theorem. 1 Prove that the equation x∧a =0 41 P1: Binaya Dash October 24, 2006 14:12 C7729 C7729˙C003 42 Geometric Algebra and Applications to Physics has the solution set x = αa . PROOF By definition of the geometric product we have xa = x · a + x ∧ a = x · a .

36) As in the earlier case, vector a is uniquely resolved into a vector a || in the B-space and a vector a ⊥ orthogonal to the B-space as given by a = a || + a ⊥ . 38a) a ⊥ = a ∧ BB−1 . 2. 39a) a ⊥ B = a ∧ B = Ba ⊥ . 39b) The above equations imply that a vector is in the B-space (plane) if and only if it anticommutes with B, and it is orthogonal to the B-space (plane) if and only if it commutes with B. 2 Projection and rejection of vector by a bivector B. Next, we generalize the above case for a multivector M of an arbitrary grade k, which determines the ak-dimensional vector space called M-space.