St... at (no spam) lansdownewebdesign.com wrote:
I don't understand something that is very basic. In Geometric Algebra/
Clifford Algebra I don't understand why the scalar in an M vector is
not considered a dimension as the vectors are.
My confusion comes from three different viewpoints, each of which show
the scalar to be a vector:
1. Complex numbers separate the real and imaginary numbers and assign
each a dimension such that the number 3+4i is plotted as 3 units in
the X direction and 4 units in the Y direction. Yet adding two more
complex dimensions j and k, give us three not four dimensions plus a
scalar. An example is a quaternion used for rotations in 3D graphics
which is considered as being 3D plus a scalar which is zero dimension.
2. So a scalar is said not to be a vector because it has magnitude but
no direction and so can't be a reference vector in a coordinate
system. However, the radial direction in circular and spherical
coordinates can be represented by a scalar and is considered a
dimension.
3. In a 1D system, magnitude alone represented by a scalar is
sufficient to give a position. Multiplying by i gives a 90 degree
rotation and a 2D space represented by a complex number (see 1 above).
What am I missing?
A geometry has two ingedients; elements like points, lines, etc and a
group of transformations (e.g. rotations and translations) which act
on the elements to show how an element appears to different observers.
The transformations are such that space looks the same to all the
observers related by the transformations in the group. This is the
content of Klein's Erlangen Programm.
In a geometric algebra the points, lines etc. are represented by
elements of the algebra and they are acted upon by the transformations
in the group.
A scalar is the name we use for a quantity which transforms trivially
under the action of a transformation in the group. In other words, a
scalar is something which is the same to all the observers. Vector and
bivector are names for quantities that transform in different ways
under the group.
Let R be a rotation (say) and let x be some quantity in the algebra.
If x transforms trivially under R so that Rx=x then x is, by
definition, a scalar. So, when confronted by some elements of a
geometric algebra, it is not clear at the outset which elements are
scalars; we have to figure this out by checking which ones transform
trivially.
Suppose we decide to represent the points of the Euclidean plane by
complex numbers z=x+iy. A rotation by 90 degrees is R=i in this
algebra. The element x is not a scalar because Rx=ix and if x was a
scalar we would have Rx=x. In other words, x does not transform
trivially so it is not a scalar.
Now suppose we study the Euclidean plane using the Geometric Algebra/
Clifford Algebra of David Hestenes. The algebra has generators e1,e2
and the basis elements of the algebra are 1, e1,e2 e1e2. A general
element of the algebra is X=w+xe1+ye2+ze1e2 where w,x,y,z are numbers.
A rotation of X is RXR^dagger where R is a "rotor". For a rotation
through 90 degrees, the rotor is R=(1+e2e1)/sqrt(2). If w is to be a
scalar then it must transform trivially which means RwR^dagger=w and
this turns out to be the case upon carrying out the calculation.
The above two examples show that the x in x+iy does not transform
trivially so it is not a scalar, whilst the w in w+xe1+ye2+ze1e2 does
transform trivially so it is a scalar. To summarize, it is not clear
at the outset which elements of a geometric algebra are scalars; we
have to find them by testing which elements transform trivially.
Stephen
Blakehttp://www.stebla.pwp.blueyonder.co.uk- Hide quoted text -
- Show quoted text -
