From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: DeMoivre's Theorem in N-dimensional complex space? Date: 2 Apr 1999 06:51:49 GMT Newsgroups: sci.math Sebastian Mekas wrote: >Hello, I have two questions to ask. First of all,I would like to know if >DeMoivre's theorem can be extended to n-dimensions, for example for an >element of the quaternions, w = a+bi+cj+dk. Well, what exactly does DeMoivre's theorem say? I have to ask because the term is used a little differently by different authors. You can write every nonzero quaternion in a unique way in the form w = r z where r is a positive real number and z is a quaternion of magnitude 1. The pieces of this representation are in fact homomorphisms of groups, that is, if w = w' w" and we write w' = r' z' and w" = r" z", then r = r' r" and z = z' z". All this is comparable to the situation for complex numbers. Now in the case of the complex numbers, we have the additional fact that there is a mapping exp : R -> C which (1) maps onto the complex numbers of magnitude 1, and (2) is again a homomorphism: exp(u+v) = exp(u)*exp(v). Restricting the domain as necessary, we can say loosely that "every complex number has a unique representation w = r exp(u)". You can say it a little differently yet, invoking the trigonometric functions. But really that's (nowadays) nothing more than the _definitions_ of the trig functions, and so not an additional theorem. When combined with (2), however, we get the result cos(nu)+i sin(nu) = (cos(u)+i sin(u))^n, which I guess is also called deM's theorem. Well, does that last paragraph generalize to the quaternions? I'd have to say ... sort of. The set of quaternions of magnitude 1 is the 3-sphere (in R^4), so if you want an essentially one-to-one map "exp" mapping to it, the domain of this map must also be three-dimensional. Indeed, you can find coordinate mappings carrying R^3 to S^3 (e.g. you can generalize the formulas for polar coordinates), but there are fundamental restrictions here: since S^3 is already simply-connected there are no nontrivial local homeomorphisms from a connected space onto S^3. In English: your map will have to misbehave somewhere. You can actually do a bit better than just capturing the topological picture here: since S^3 is a Lie group there is an exponential map associated to this group and its Lie algebra. But that takes us to the other aspect of the complex map "exp": it's a homomorphism of groups. If you want this mythical map "exp": R^3 -> S^3 to be a homomorphism you have to decide what "+" means in a formula "exp"( u "+" v ) = "exp"(u) * "exp"(v). A natural choice is to use the addition of vectors in R^3, but that's doomed from the start, since u+v = v+u would have to imply z z' = z' z, which does not hold for all pairs of unit quaternions z and z'. So this part would have to be abandoned (although the exponential mapping on Lie algebras is "approximately" a homomorphism). Finally there's the trigonometric connections, but here we can punt again: whatever map "exp" we choose to use to represent the unit quaternions, we simply _define_ four "quaternionic-trigonometric" functions to be the coordinates of "exp", that is, exp([a,b,c]) = qt1([a,b,c]) + qt2([a,b,c]) i + qt3([a,b,c]) j + qt4([a,b,c]) k Not much of a theorem there. I should also comment that the original poster's use of the phrase "for example" suggests some kind of structure on R^n like the quaternion (multiplicative) structure on R^4. This doesn't exist except for a few small n, if you want the structure to allow cancellation. It's been a while since that topic came up so I'll give a pointer to more info: index/products.html > My second question is about an interesting property of the equation > z^n=a+bi >The solutions z for this equation have the same modulus and are >separated by an angle of 2pi / n, thus forming a regular n-gon. Can this >result be extended to the n-dimensional case, that is can one construct >4 dimensional regular polygons inside a hypersphere when you work with >the quaternions? What if we are working with octonions? Does being an >alternative algebra, lacking associativity over addition, change >anything? Thank you in advance, When you restrict your attention to the spheres, you're focusing less on the algebra structure of R^k than on the group structure of S^(k-1). Again I should comment that there isn't such a structure unless n=1,2,4,8. But for those values the answer is certainly _yes_, you get regular geometric objects from algebraic ones in those cases. The right way to begin is to note that the complex numbers with z^n=1 actually form a finite subgroup of the set of unit complex numbers. Then you can make a similar statement in R^k: each finite subgroup of S^(k-1) (when the latter is a group) is a set of points whose convex hull is a regular polytope. Or you could be a bit more generous and look at the group O(k) of all rotations, which clearly fixes the spheres at the origin. The finite subgroups of O(k) then become finite automorphism groups of S^(k-1). For most points in the sphere, the translates under the elements of the group are distinct, giving a polytope with a high degree of symmetry. (the group is vertex transitive, not face transitive necessarily). This is the preferred method for making interesting polytopes, I suppose, since there are really very few regular polytopes in dimensions k>4. In the case k=2, O(2) is nearly the same as the set of unit complex numbers anyway, so there is no real difference between the constructions of these two paragraphs. These ideas are all over the mathematical map, but here are some pointers. For polyhedra and so on see Convex geometry: index/52-XX.html especially section 52B. For Lie groups and algebras try index/22-XX.html dave ============================================================================== From: Pertti Lounesto Subject: Re: DeMoivre's Theorem in N-dimensional complex space? Date: Fri, 02 Apr 1999 15:34:12 +0200 Newsgroups: sci.math Sebastian Mekas wrote: > Hello, I have two questions to ask. First of all,I would like to know if > DeMoivre's theorem can be extended to n-dimensions, for example for an > element of the quaternions, w = a+bi+cj+dk. I would try to give the > proof a stab muself, but I don't know the necessary trigonometry > identities, which I think are needed to derive DeMoivre's Theorem. Yes, (cos x+u sin x)^n = cox nx+u sin nx, where u is a unit vector in R^3. > My second question is about an interesting property of the equation > z^n=a+bi > The solutions z for this equation have the same modulus and are > separated by an angle of 2pi / n, thus forming a regular n-gon. Can this > result be extended to the n-dimensional case, that is can one construct > 4 dimensional regular polygons inside a hypersphere when you work with > the quaternions? What if we are working with octonions? Does being an > alternative algebra, lacking associativity over addition, change > anything? Let v be a vector in R^n, a real linear space with a negative definite quadratic form on itself. Consider a+v in R+R^n, a subspace of the Clifford algebra of R^n. The solutions of z^k = a+v form vertices of a regular k-gon in R^n. For more information about Clifford algebras, see my book, with URL http://www.cup.cam.ac.uk/Scripts/webbook.asp?isbn=0521599164