# Scalar vs Vectors

1. ### Steve@lansdownewebdesign.com

0
I don=92t understand something that is very basic. In Geometric Algebra/
Clifford Algebra I don=92t 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 =96 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=92t 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?

2. ### Stephen Blake

0
St...@lansdownewebdesign.com wrote:

> I don=92t understand something that is very basic. In Geometric Algebra/
> Clifford Algebra I don=92t 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 =96 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=92t 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 Blake
http://www.stebla.pwp.blueyonder.co.uk

3. ### Igor

0
On Jun 13, 3:10pm, St...@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.

Please understand that how you use a mathematical set of objects is
not etched in stone. A quaternion, in general, has four independent
degrees of freedom. Depending on how the system is constrained, a
quaternion can be thought of as a 4-vector, a 3-vector and a scalar,
two independent 2-vectors or a 2-vector and two scalars. The
possibilities are almost endless, depending on the context you're
using. Usually in physics, a quaternion is applied as representing a
4-vector (as in relativity) or a 3-vector and a scalar (as in
classical mechanics), But there are other ways of using them.

> 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.

Technically coordinates are not scalars or vector components at all.
They're elements of a language that locally define a system.
Differential coordinates, on the other hand, are usually defined as
making up a vector, since they represent the difference between two
sets of coordinates under the appropriate limit, thus having both
magnitude and direction. And the term scalar does not simply just
relate to a number (if it were, we could say that a vector was made up
of scalars, which is not the case). A scalar also must be invariant
under coordinate transformations and/or rotations. For example, we
could multiply the x components of two independent vectors and derive
a number from it, but it would not be a scalar, since it would change
if we rotated the coordinate system.

> 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?

There's a difference between a scalar and a vector, even in a one
dimensional system. On the number line, we can have a vector directed
forward or backward, but its length is still independent of
direction. In a complex coordinate system, as you describe it above,
the length of the vector would no longer be equal to its component
along the horizontal axis, but would now be given by the Pythagorean
theorem in terms of x and y components, its real and imaginary parts
respectively.

4. ### Rick

0
On Jun 14, 7:42am, Stephen Blake <ste...@blueyonder.co.uk> wrote:
> St...@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 -

When it comes to C, H and O, you would do better to include element
e0, and call "scalar" the product of real component times e0.

Then, it is not "transformation" that defines scalar, but instead
the operation of multiplication as defined by the particular
algebra that indicates what is scalar and what is not in any
possible expression you may wish to form.

Your statement that in C, x of z=x+iy is something different than
w of (possibly) H term w+xe1+ye2+ze1e2 is incorrect, for in the
H case, if y and z are both zero, you have something in the C
subalgebra of H. Surely y and/or z not zero is not required.

Rick Lockyer

www.octospace.com

5. ### ~greg

0
"Stephen Blake" > wrote ....
> 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.

that makes x an eigenvector, with an eigenvalue = +1.

6. ### ~greg

0
> I don't understand why the scalar in an M vector is
> not considered a dimension as the vectors are.

Compare:
( A+Bi ) ( C+Di ) = ( AC-BD ) + ( AD+BC )i

(--the product of two complex numbers )

with

( Ae1 + Be2 ) ( C + D e1^e2 ) = ( AC-BD )e1 + ( AD+BC )e2

(---the geometric product of a vector Ae1+Be2,
with a multi-vector C+De1^e2,
in the geometric-algebra built in an RxR plane
having an orthonormal basis e1,e2.
( This GA has basis {1, e1, e2, e1^e2 } ) )

~~~~~
Ever hear the expression "by an abuse of language"
in math?

7. ### Buddha Buck

0
On Jun 13, 3:10 pm, St...@lansdownewebdesign.com wrote:
> I don=92t understand something that is very basic. In Geometric Algebra/
> Clifford Algebra I don=92t understand why the scalar in an M vector is
> not considered a dimension as the vectors are.

It can be, depending on how you count.

A 3-D GA is an 8-dimensional linear space (the standard term would be
"vector space", but that overloads the use of the term "vector" when
discussing GA.). The basis over this linear space is 1, e1, e2, e3,
i=e1^e2, j=e2^e3, k=e3^e1, and I=e1^e2^e3. In this sense, the scalar
is considered just another dimension.

One reason why users of GA don't refer to the 3D GA as an 8D GA is
because while the full algebra has 8 dimensions that isn't a useful
categorization. A GA, viewed this way, can only have dimensions of 1,
2, 4, 8, 16, ..., 2^N, ... for some N, where N is the number of basis
vectors in the vector subspace of the GA (i.e., e1, e2, ..., eN).
Viewed geometrically, that's the subspace which determines the
geometry of the algebra -- a 3D GA plays in 3D space, a 4D GA plays in
4D space, etc. As such, N is a much more important number determining
the GA than 2^N.

If you take a 3D GA and restrict yourself to looking at M vectors in
it which are multiples of 1, you get a 0D sub-GA. All the rules
follow. Elements of this subalgebra get to be called "scalars" in the
context of the whole GA because:

1. Scalar a1 commutes with all elements M of the whole GA
2. A scalar times a k-vector yields a k-vector, for any scalar and k-
vector in the GA
3. ||au|| = |a|*||u|| for all scalars a and k-vectors u (the norm of
au equals the absolute value of a times the norm of u).
4. In every sense in which a k-vector u can have a direction, au has
the same direction if a>0, the opposite direction if a<0.

They act the same way that "scalars" are expected to act in linear
space -- they change the scale, but not direction (except potentially
by reversal) of elements of the space they act upon.

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