Hi, I have geometry questions.

I need your help.

Thanks a lot.

Problem 5: The complex numbers C are a field, defined as R2 as a vector space over R but with the additional multiplication (a, b) (c, d) = (ac - bd, ad + bc). An alternative way to say this is that C is a 2-dimensional vector space over R with basis vectors 1 and i, and the product map C x C - C is the bilinear map uniquely determined by 1 . x = x = x . 1 Vx, and i . i = -1. The quaternions H (H is for Hamilton) are defined similarly, they are R4 as a vector space, but if we let { 1, i, j, k} be a basis for R4 we define multiplication to be the bilinear map determined by ' [ - = y . ? ' 2 - = [ . 4' 4 - = 2 . [ '[ = 2 . 4'2 = 4 . 6'4 = 6 .? 'TAI . X = = C . I and i2 = j2 = k2 = -1. So really 1 is shorthand for (1, 0, 0, 0) and i for (0, 1, 0, 0) and so on. So when we write a quaternion as a + bi + cj + dk, we really mean the vector (a, b, c, d) E R4. Given a quaternion q = a + bi + cj + dk, the conjugate is q = a - bi - cj - dk. (a) Verify that conjugation respects multiplication 91 . q2 = 92 . 91. Also check that qq = |q|2, where lal = va2 + 62 + c2 + 12 is the Euclidean norm in R4 (b) A division algebra is an algebraic object that satisfies the axioms of a field, except that multiplication need not be commutative. Verify that the quater- nions H are a division algebra. Check that qx = xq for all real numbers x and quaternions q. (c) Check that the unit-length quaternions, $3 C H with operation $3 x $3 - S' given by quaternion multiplication satisfies the axioms of a Lie group. That is, it is a group, $3 is a manifold, and the group operations of multi- plication and inversion are smooth. (d) Given a unit quaternion q E $3, verify that the left-multiplication map Lq : H - H given by La(p) = qp is an isometry of H = R4 with the Euclidean metric. Similarly for right multiplication R,(p) = pq. Verify that the two maps $3 - 04 given by q - Lq and q - Ro-1 are homomorphisms of groups. (e) By part (d), the map f : $3 - 04 given by f(p) = Lp ORp-1 is a homo- morphism of Lie groups, but by part (b), f(p) fixes the real numbers R and therefore fixes their orthogonal complement {0} x R3 C H and so one can view f as a homomorphism $3 - 03. Since S3 is pathwise connected and f(1) = I, f is a map f : $3 -> SO3. Prove that f is a submersion and therefore by the implicit function theorem, an open map. I suggestcomputing Dfi : T1S3 -+ TISO3, and show that Df, is conjugate to Df1 if q E $3. (f) Argue the map f satisfies f($3) = SO3. Further, show that f is 2 : 1, i.e. that its kernel is {+1} C $3