
Re: Algebra Family multiplication table for 3D vectors
Posted:
Feb 28, 2014 9:24 AM


In article <280220140632289126%edgar@math.ohiostate.edu.invalid>, annu Poropudas <hanporop@luukku.com> wrote:
> On Thursday, February 27, 2014 1:00:50 AM UTC+2, Robin wrote: > > Again you have an algebra in which every element has an inverse, > > except for the (infinitely many) elements that don't :( > > Yes, The smaller space for this special case of my algebra family > seems to be the following: > > R^3\{plane: z=y, x=x,y=y}
It's actually R^3 with both the plane y = z and the line x = y = 0 removed.
> > This plane must be excluded in this special case of my algebra family. > This plane goes through origin. > > One property which you noticed (in your previous post) seems to be that > this set is not an integral domain? (if your calculation: e2*e3=e3 > > (e2e1)*e3=0 is correct.For heaven's sake Mr Poropudas, cannot you even check a simple calculation? Here is the multiplication table, from your previous posting.
> e1, e2, e3 > e2, e1+e3, e3 > e3, e3, 2*e2
From it one has e1*e3 = e3 and e2*e3 = e3, Therefore, (by distributivity) (e2e1)*e3 = e2*e3  e1*e3 = e3  e3 = 0.
Now, have you decided what properties you want of your algebra? You are aware that you cannot have an associative threedimensional algebra over the reals with a multiplicative identity and with the property that all nonzero elements are invertible (there is no threedimensional division algebra over R). If you could tell us what you are aiming to achieve, perhaps we could help you do it.

