/sci/ - Science & Math

>>11472971
[math]\exp[/math] maps algebras to groups (the simples case being the hom [map]\exp(A+B)=\exp(A)\cdot \exp(B)[/math] between addition and multiplication). If E is the two-dimensional identity matrix, then the matrix J=[[0,-1],[1,0]] has J^2=-E so exp maps the matrices generates by J into a group (the generated elements by J are all matrices of the form [math] x*E+y*J [/math], with e.g. real x,y). Since J has no real eigenvalues, the exp(x*J) are just SO(2). Then sin and cos pop up because those are just the two projection of the vectors in the standard representation of SO(2), i.e. if T is a 2-dim vector exp(xJ)*T just circles around the origin and it's components have sin and cos in it.

For the more general low dimensional context in which this split-off of a trigonometric function happens:
if F is characteristic not 2, with particular elements a and b, then if you have any (unital) F-algebra with two anti-commuting elements i, j such that their squares are central (in the sense that i^2=a*e and j^2=b*e, those are constant examples of a Clifford algebra behaviour), then the generated algebra is 4-dimensional.
If you got any 2x2 matrix representation of it, and v=t*e+w is a generic element (w is in the 3 dimensional part, also let r=sqrt(det(w))), then
[math]exp(v) = e^t * (\cosh(r) + \cosh'(r) * w / r)[/math]
Note that cosh' is just sinh.
cos and the Euler formula come into play if a and b is choosen so that the above reduces to the Hamiltonian quaternions.