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```Grassmann and Clifford Algebras
-------------------------------

Grassmann (aka Exterior) Algebra
--------------------------------

The fundamental operation in Exterior Algebra is
the Wedge product.  The wedge product is a completely
antisymmetrized tensor product.  To see this consider
the tensor product of 2 vectors, ui and vj (≡ Tij).

u ⊗ v = (1/2)(uv + vu) + (1/2)(uv - vu)

= symmetric part + antisymmetric part

Lets focus on the antisymmetric part.  We can
now define the wedge product as:

u ∧ v = (1/2)(uv - vu)

= (1/2)[u,v]

At first sight this may look like the Lie algebra.
However, in general, it is not.  Unlike the Lie
algebra, which lives in the tangent space of a
manifold, the exterior algebra lives in a vector
space that is separate from the original space of
vectors, V, (hence exterior).  This space is
referrred to as Λp(V) and is called the pth
EXTERIOR POWER.  In other words the wedge product
is an operation on a vector space where the result
is a member of a new space product and not an element
of the original vector space.  Furthermore, the
Lie algebra satisfies the Jacobi identity whereas
the wedge product generally does not.

Algebraic Properties
--------------------

Associativity:  (u ∧ v) ∧ w = u ∧ (v ∧ w)

u ∧ u = 0

Consider:

(u + v) ∧ (u + v) = u ∧ u + u ∧ v + v ∧ u + v ∧ v

0 = u ∧ v + v ∧ u

Therefore,

u ∧ v = -(v ∧ u)

Thereby, confirming the completely antisymmetric
properties for vectors.

s ∧ t = t ∧ s = st and s ∧ s = s2 for scalars s and t.

s ∧ u = u ∧ s = su for a scalar s and vector, u.

p-vectors
---------

A p-vector is a (p,0) tensor which is completely
antisymmetric.  p-vectors are constructed from
wedge products.  However, the dimension of Λp
(i.e the # of basis elements) for each type of
p-vector has to be taken into account.  The
dimension is given by the formula for combinations:

- -
| D |        D!
| p |  = ----------
- -     p!(D - p)!

The wedge product of a p-vector and q-vector
produces a (p + q) vector.  However, there is a
limitation on p, q and p + q for a given D.
If (p + q) > D then one of the indeces would
have to be duplicated violating the requirement
of complete antisymmetry.  For example, for
D = 3 the wedge product of a 2-basis vector with
a 2-basis vector produces:

Tijkl = ei ∧ ej ∧ ek ∧ el

However, there are only 3 basis vectors in D = 3
therefore el would have to be ei or j or k.

Examples, for D = 3:

0-vector:  f = f(x1,x2,x3)

1-vector:  u = u1e1 + u2e2 + u3e3

2-vector:  v = v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3)

3-vector:  w = w123(e1 ∧ e2 ∧ e3)

What would these look like for D = 4?

0-vector:  f = f(x1,x2,x3,x4)

1-vector:  u = u1e1 + u2e2 + u3e3 + u4e4

2-vector:  v = v12(e1 ∧ e2) + v13(e1 ∧ e3) + v14(e1 ∧ e4)

+ v23(e2 ∧ e3) + v24(e2 ∧ e4) + v34(e3 ∧ e4)

3-vector:  w = w123(e1 ∧ e2 ∧ e3) + w124(e1 ∧ e2 ∧ e4)

+ w134(e1 ∧ e3 ∧ e4) + w234(e2 ∧ e3 ∧ e4)

Note:  The coefficients in each case are f(xμ).

The Wedge Product of p-vectors
------------------------------

We use the example of the wedge product of a
1-vector and a 2-vector in D = 3 to illustrate
the process.

(u1e1 + u2e2 + u3e3) ∧ (v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3))

= u1 ∧ v12 ∧ e1
|
v
= (u1v12e1 ∧ (e1 ∧ e2) + u1v23e1 ∧ (e2 ∧ e3) + u1v13e1 ∧ (e1 ∧ e3)
+
(u2v12e2 ∧ (e1 ∧ e2) + u2v23e2 ∧ (e2 ∧ e3) + u2v13e2 ∧ (e1 ∧ e3)
+
(u3v12e3 ∧ (e1 ∧ e2) + u3v23e3 ∧ (e2 ∧ e3) + u3v13e3 ∧ (e1 ∧ e3)

= u1v23e1 ∧ (e2 ∧ e3) + u2v13e2 ∧ (e1 ∧ e3) + u3v12e3 ∧ (e1 ∧ e2)

= u1v23e1 ∧ (e2 ∧ e3) - u2v13e1 ∧ (e2 ∧ e3) + u3v12e1 ∧ (e2 ∧ e3)

= (u1v23 - u2v13 + u3v12)(e1 ∧ e2 ∧ e3)

Which is a 3-vector.

Now consider the product of two 2-vectors in D = 3:

(u12(e1 ∧ e2) + u23(e2 ∧ e3) + u13(e1 ∧ e3))

∧ (v12(e1 ∧ e2) + v23(e2 ∧ e3) + v13(e1 ∧ e3))

u12v12(e1 ∧ e2) ∧ (e1 ∧ e2) = u12v23(e1 ∧ e2) ∧ (e2 ∧ e3)

= u12v13(e1 ∧ e2) ∧ (e1 ∧ e3)

= 0

u23v12(e2 ∧ e3) ∧ (e1 ∧ e2) = u23v23(e2 ∧ e3) ∧ (e2 ∧ e3)

= u23v13(e2 ∧ e3) ∧ (e1 ∧ e3)

= 0

u13v12(e1 ∧ e3) ∧ (e1 ∧ e2) = u13v23(e1 ∧ e3) ∧ (e2 ∧ e3)

= u13v13(e1 ∧ e3) ∧ (e1 ∧ e3)

= 0

This demonstrates the closure property of the
algebra.

The Cross-Product
-----------------

For D = 3, the cross product is the wedge product
of a 1-vector with a 1-vector.

Proof:

u = aex + bey + cez (1-vector)

v = eex + fey + gez (1-vector)

Here D = 3 and the resulting exterior space is Λ2.

u ∧ v = (bg - cf)(ey ∧ ez)

+ (ag - ce)(ex ∧ ez)

+ (af - be)(ex ∧ ey)

Where the basis is {(ey ∧ ez),(ex ∧ ez),(ex ∧ ey)}

The coefficients are the same as those in the
usual definition of the cross product of vectors
in three dimensions.  However, the wedge product
produces a bivector instead of a vector.

The cross product is a special case where the
wedge product is is the same the Lie bracket.
We can prove this by verifying that the cross
product satisfies the Jacobi identity:

a x (b x c) + b x (c x a) + c x (a x b) = 0

Proof:

Using the definition of the triple vector product,
a x (b x c) = (a.c)b - (a.b)c, this becomes:

(a.c)b - (a.b)c + (b.a)c - (b.c)a + (c.b)a - (c.a)b

= 0 because the dot product is commutative.

Areas in the Plane
------------------

The p-vectors are an algebraic construction used
in geometry to study areas, volumes, and their higher
dimensional analogues.  Consider the following:

For D = 2, the area is the wedge product of a
1-vector with a 1-vector.

Proof:

u = (x3 - x1)ex + (y3 - y1)ey

v = (x2 - x1)ex + (y2 - y1)ey

Here D = 2 and the resulting exterior space is Λ2.

u ∧ v = (x3 - x1)(y2 - y1)(ex ∧ ey)

+ (x2 - x1)(y3 - y1)(ey ∧ ex)

Now (ey ∧ ex) = -(ex ∧ ey).  Therefore,

u ∧ v = (x3 - x1)(y2 - y1)(ex ∧ ey)

- (x2 - x1)(y3 - y1)(ex ∧ ey)

= (x3y2 - x3y1 - x1y2 + x1y1 - x2y3 + x2y1

+ x1y3 - x1y1)(ex ∧ ey)

Let (x1,y1) -> (0,0)

A = (x3y2 - x2y3)(ex ∧ ey)

-     -
= det| x3 y3 |(ex ∧ ey)
| x2 y2 |
-     -

This is the standard determinant form for the
area of a parallelogram.

Differential Forms (aka p-Forms)
--------------------------------

Differential forms are an approach to multivariable
calculus that is independent of coordinates.  They
appear in the mathematics of differential geometry.

A p-form (aka Differential form) is a (0,p) tensor
which is completely antisymmetric.  p-forms are duals
to p-vectors.  The wedge product of p-forms lives in
the space Λp(V*).  Since Λ(V*) is the dual of Λ(V),
dim(Λ(V*)) = dim(Λ(V)).

All of the properties and constructions discussed
previously apply to differential forms:

The Exterior Derivative
-----------------------

p-forms can be both differentiated and integrated.
The EXTERIOR DERIVATIVE, d allows us to differentiate
a p-form to obtain a (p + 1) form.  The exterior
derivative is:

(d)μ ≡ (∂/∂xμ)

Note that:

d(dφ) = 0 since partial derivatives commute.

Forms can be expanded in terms of wedge products
and the exterior derivative.

For example, for D = 3:

0-form:  γ = f(x1,x2,x3)

1-form:  α = α1dx1 + α2dx2 + α3dx3

2-form:  β = β12(dx1 ∧ dx2) + β23(dx2 ∧ dx3) + β13(dx1 ∧ dx3)

3-form:  δ = δ123(dx1 ∧ dx2 ∧ dx3)

Where the coefficients are f(x1,x2,x3).

In general:

ω = Σfjdxj
j
dω = ΣΣ(∂fj/∂xi)dxi ∧ dxj
ij

Example:  In D = 3, γ = 3x2yz

dγ = (∂(3x2)/∂x)dx + (∂(3x2)/∂y)dy + (∂(3x2)/∂z)dz

+ (∂(y)/∂x)dx + (∂(y)/∂y)dy + (∂(y)/∂z)dz

+ (∂(z)/∂x)dx + (∂(z)/∂y)dy + (∂(z)/∂z)dz

= 6xdx + dy + dz = α

dα = (∂(6x)/∂x)dx + (∂(6x)/∂y)dy + (∂(6x)/∂z)dz) ∧ dy

+ (∂(1)/∂x)dx + (∂(1)/∂y)dy + (∂(1)/∂z)dz) ∧ dx

+ (∂(1)/∂x)dx + (∂(1)/∂y)dy + (∂(1)/∂z)dz) ∧ dz

= 6dx ∧ dy since dx ∧ dx = dy ∧ dy = dz ∧ dz = 0

Leibnitz (Product) Rule
-----------------------

The exterior derivative of the product of a p-form
and a q-form is given by:

d(α ∧ β) = dα ∧ β + (-1)pα ∧ dβ

The result will have dimension, D = p + q + 1.

Proof:

Consider a 1-form and a 2 form (p = 1, q = 2).  The
result will have D = 4.  Therefore, both forms need
to be specified in D = 4.  For brevity we only
consider the first term of each to illustrate the
process.

α = fdx1 + ...

β = g(dx2 ∧ dx3) + ...

d(α ∧ β) = d(fg(dx1 ∧ dx2 ∧ dx3))

= (df.g + f.dg) ∧ dx1 ∧ dx2 ∧ dx3

= (df ∧ dx1) ∧ g(dx2 ∧ dx3) + fdx1 ∧ dg ∧ dx2 ∧ dx3

After reordering the 2nd term to match the order
of the 1st term we get:

d(α ∧ β) = (df ∧ dx1) ∧ g(dx2 ∧ dx3) - dg ∧ fdx1 ∧ dx2 ∧ dx3

Or,

d(α ∧ β) = dα ∧ β + (-1)pα ∧ dβ

Volume (aka Top) Form
---------------------

When p = D the form is called the VOLUME FORM.  It
has a dimension of 1 and establishs an orientation
of the exterior power (manifold).  A coordinate system
(on an oriented manifold) is oriented if:

dx1 ∧ dx2 ∧ ... ∧ dxD

is positive.  A transformation between oriented
coordinate systems has positive Jacobian.

Note that the exterior derivative of the top form
is 0 since it would result in a p + 1 form that is
greater than the dimension of the exterior power.

Example:

For D = 3

ω = dx1 ∧ dx2 ∧ dx3

Change of Coordinates
---------------------

The derivative of a tensor transforms under a
change of coordinates as:

∇μVν = ∂μVμ + ΓνμλVλ

Where Γνμλ are the Christoffel symbols.  Lets look
at how the exterior derivative transforms.  The
definition of the wedge product allows us to
write

dx0 ∧ .... dxn-1 = (1/n!)εμ1 ... μn dxμ1 ∧ .... dxμn

since both the wedge product and the Levi-Civita
symbol are completely antisymmetric.  Under a
coordinate transformation, ε stays the same while
the 1-forms change according to:

Basis:  dxμ' = (∂xμ'/∂xμ)dxμ or dxμ = (∂xμ/∂xμ')dxμ'

Components:  ωμ' = (∂xμ/∂xμ')ωμ'

For convenience we work with the 2-form.  Lets look
at how (dx1 ∧ dx2) part transforms.

dx1 ∧ dx2 = ((∂x1/∂x1')dx1' + (∂x1/∂x2')dx2')
((∂x2/∂x1')dx1' + (∂x2/∂x2')dx2')

= (∂x1/∂x1')(∂x2/∂x1')(dx1' ∧ dx1')

+ (∂x1/∂x2')(∂x2/∂x2')(dx2' ∧ dx2')

+ (∂x1/∂x1')(∂x2/∂x2')(dx1' ∧ dx2')

+ (∂x1/∂x2')(∂x2/∂x1')(dx2' ∧ dx1')

= (∂x1/∂x1')(∂x2/∂x2')(dx1' ∧ dx2')

- (∂x1/∂x2')(∂x2/∂x1')(dx1' ∧ dx2')

= (∂x1/∂x1')(∂x2/∂x2') - (∂x1/∂x2')(∂x2/∂x1')(dx1' ∧ dx2')

= (∂x1/∂x1')(∂x2/∂x2') - (∂x1/∂x2')(∂x2/∂x1')(dx1' ∧ dx2')

= det(∂xμ/∂xμ')(dx1' ∧ dx2')

Therefore, the antisymmetric properties of the wedge
product automatically takes care of the need to
include the Christoffel symbols in the calculation.
Since the Christoffel symbols involve the metric
tensor and its derivatives, the implication of this
is that differential forms allow us to work with
manifolds without the help of the metric.  In this
sense they are coordinate independent or 'coordinate
free'.

Relation to Gradient, Curl and Divergence
-----------------------------------------

Recall:

For D = 3:

0-form:  γ = f(xμ)

1-form:  α = α1dx1 + α2dx2 + α3dx3

2-form:  β = β12(dx1 ∧ dx2) + β23(dx2 ∧ dx3) + β13(dx1 ∧ dx3)

3-form:  δ = δ123(dx1 ∧ dx2 ∧ dx3)

Differentiating the above p-forms gives:

-----------------

dγ = (∂f/∂xμ)dxμ  (a 1-form)

Example:

f = x2 + z2

df = 2xdx + 2zdz

The Curl (∇ x α)
---------------

dα = dα1 ∧ dx1 + dα2 ∧ dx2 + dα3 ∧ dx3  (a 2-form)

Example:

α = ydx + z2dy + 0dz

dα = (dy ∧ dx) + 2z(dz ∧ dy)

= -(dx ∧ dy) + 0(dx ∧ dz) - 2z(dy ∧ dz)

= -2z(dy ∧ dz) + 0 -(dx ∧ dy)

The Divergence (∇.β)
--------------------

dβ = dβ12 ∧ dx1 ∧ dx2 + dβ23 ∧ dx2 ∧ dx3

+ dβ13 ∧ dx1 ∧ dx3  (a 3-form)

Example:

β = x2(dy ∧ dz) +  y2(dz ∧ dx) + z2(dx ∧ dy)

dβ = 2x(dx ∧ dy ∧ dz) + 2y(dy ∧ dz ∧ dx)

+ 2z(dz ∧ dx ∧ dy)

= (2x + 2y + 2z)(dx ∧ dy ∧ dz)

Note that:

∇ x ∇f = 0

And, for a 1-form:

∇.(∇ x ) = 0

Integration
-----------

In multivariable calculus we have:

∫∫f(x1,x2)dx1dx2 = ∫∫f(x1',x2')Jx1'dx2'

Where J = det(∂xμ/∂xμ') is the determinant of the
Jacobian of the transformation between the 2
coordinate systems that keeps the area elements
equal.

(dx1 ∧ dx2) = det(∂xμ/∂xμ')(dx1' ∧ dx2')

gμ'ν' = (∂xμ1/∂xμ1')(∂xμ2/∂xμ2')gμν

det(gμ'ν') = (det(∂xμ/∂xμ'))2det(gμν)

In Euclidean space det(gμν) = 1

Therefore,

√(det(gμ'ν')) = det(∂xμ/∂xμ')

Therefore,

(dx1 ∧ dx2) = √(det(gμ'ν'))(dx1' ∧ dx2')

= √(|g'|)(dx1' ∧ dx2') (g' = det(gμ'ν'))

If we make the identication:

dxdy <-> √(|g'|)(dx1 ∧ dx2)

it is easy ro see that integration under a change
of coordinates is properly understood as the
integration of the volume (top) form!

Generalized Stokes' Theorem
---------------------------

M = manifold, ∂M = boundary of the manifold.

∫α = ∫dα
∂M    M

∂M is of dimension p, α is of dimension p.

M is of dimension p + 1, dα is of dimension p.

b
Fundamental Theorem of Calculus: ∫f = ∫df
(p = 1)                          a
∴ f(b) - f(a) = ∫∇f dx
C
Stokes' Theorem:  ∫F.dr = ∫∫∇ x F
(p = 2)           ∂S      S

Gauss' Divergence Theorem: ∫F.dr = ∫∫∫∇.F
(p = 3)                    ∂V V

Role of the Metric
------------------

If V has an inner product defined on it then an
inner product is induced on V* and vice versa.

Consider the following:

v = Σvi∂/∂xi   (tangent vector)
i

α = Σαjdxj (cotangent vector)
j

v.α = Σvi∂/∂xj.(Σαjdxj)
i         j

The metric tensor has the effect of converting one
of the input 1-vectors to a 1-form.  Thus, if have
2 vectors, u and v, the 'tensor machine' produces:

g(v,u) -> (vi∂/∂xi)(αjdxj) = viαk

What we want is for the 1-form to 'eat' the 1-vector
and produce the same result as the 1-vector 'eating'
the original function.  This can be expressed as:

df(v) = v(f) ∈ ℝ

f is a scalar function so df is a 1-form. We can
now demonstrate this 'eating' process as follows:

[(∂f/∂xi)dxi)](vj∂j)
----------
^
|
1-form (i.e. f = x2 -> 2xdx)

= vj(∂f/∂xi)dxi.∂j

= vj(∂f/∂xi)δij

= vi(∂f/∂xi)

= vi∂i(f)

= v(f)

The last term is the directional derivative.  It is
a scalar.  It should not be confused with ordinary
derivative which is interpreted as the gradient of
the function at a point. The gradient of a scalar
function is a vector.  However, the two are related:

-----------------------------------------------------

Digression:

The directional derivative represents the instantaneous
rate of change in f along a curve in the direction
of the unit tangent vector.  It is a scalar.  The
directional derivative, D, is related to the gradient
of a scalar function as follows:

Dvf = ∇f.v

Example:

-    -
v = | √2/2 |  (unit tangent vector)
| √2/2 |
-    -

Dvf = ∇f.v

f(x,y) = x2y

-   -   -    -
Dvf.v = | 2xy |.| √2/2 |
| x2  | | √2/2 |
-   -   -    -

At the point (-1,-1) we get:

- -   -    -
| 2 |.| √2/2 | = √2 + √2/2 = scalar
| 1 | | √2/2 |
- -   -    -

The analogy between the DD and the ordinary derivative
is:

f'(a) = lim h->0(f(a + h) - f(a))/h

Dv(a) = lim h->0(f(v + hi) - f(a))/h

Note:  The gradient of a vector is a rank 2 tensor.

v = vii + v2j + v4k

∇v = (∂V/∂x)i + (∂V/∂y)j + (∂V/∂z)k

= i(∂/∂x)(vii + v2j + v4k) + ...

= ii(∂v1/∂x) + ij((∂v2/∂y) + ik((∂v3/∂z)

= eiei(∂vi/∂xj)

-                      -
| ∂v1/∂x1 ∂v2/∂x1 ∂v3/∂x1 |
= | ∂v1/∂x2 ∂v2/∂x2 ∂v3/∂x2 |
| ∂v1/∂x3 ∂v2/∂x3 ∂v3/∂x3 |
-                      -

= Jacobean matrix

-----------------------------------------------------

The Inner Product
-----------------

Consider, the 2D case:

(e1 ∧ e2).(e1 ∧ e2)

(e1e2 - e2e1).(e1e2 - e2e1)

(e1e2 + e1e2).(e1e2 + e1e2)

4(e1e2).(e1e2)

4(e1.e1)(e2.e2)

The is the product of eigenvalues which is equivalent
to the determinant of the matrix, i.e.

-           -
= | e1.e1   0   |
|   0   e2.e2 |
-           -

= det(gij)

= (-1)D-S

Where S is the SIGNATURE of the metric = (# of +
elements, # of - elements).  If D = 2 we get
gij = (1,1) ∴ S = (2,0)

We use these results in the next section.

The Hodge Star (*) Operator
---------------------------

The Hodge Star Operator on an D-dimensional manifold
is a map from p-forms to (D - p) forms.

*:  Λp <-> ΛD-p

The meaning of 'dual' in this context does NOT mean
dual as in vectors -> dual vectors.

If V has an inner product defined on it then an
inner product is induced on Λp(V) and vice versa.

If V* has an inner product defined on it then an
inner product is induced on Λp(V*) and vice versa.

Schematically,

metric
V  <--------------->  V*

metric
Λp(V) <---------------> Λ0(V*)
|                      |
| Hodge *              | Hodge *
|                      |
v                      v
ΛD-p(V) <---------------> ΛD-p(V*)
metric

Consider:

α, β ∈ Λp

Now construct the inner product:

α.β = Y    Y ∈ ℝ

Now take,

α ∈ Λp and γ ∈ ΛD-p

Write,

α ∧ γ = XE

α ∧ γ ∈ ΛpΛD-p = ΛD with dimension = 1 so X is a
single number ∈ ℝ.  E is the unit volume equal to
dx0 ∧ ... dxD and also lives in ΛD so it also 1
dimensional.  Because XE and Y are single numbers
we can equate them.  Therefore,

α ∧ γ = (α.β)E

When this is true, α = *β

Λp     ΛD-p
\     /
α ∧ *β = (α.β)pE
\   \
Λp  ΛD-p

Examples:

Consider *1.  This is the scalar Λ0 with basis = 1.

For D = 3 we have:

Λ0     Λ3-0
\     /
α ∧ *1 = (α.1)pE
\   \
Λ0  Λ3-0

α is a scalar so α ∧ *1 = α*1.  Therefore,

α*1 = αE

So,

*1 = e1 ∧ e2 ∧ e3)

Now consider *E

E ∧ *E = (E.E)NE

For D = 3 we have:

Λ3     Λ3-3
\     /
E ∧ *E = (E.E)pE
\   \
Λ3  Λ3-3

E is a scalar since it has dimension = 1.  So,

*(e1 ∧ e2 ∧ e3) = 1

Now consider *e2:

e2 ∧ *e2 = (e2.e2)pE

e2 ∧ *e2 = -(e2 ∧ e1 ∧ e3)

For D = 3 we have:

Λ1      Λ3-1
\      /
e2 ∧ *e2 = -(e2 ∧ e1 ∧ e3)
\       \
Λ1      Λ3-1

Compare the Λ3-1 spaces to get *e2 = -(e1 ∧ e3)

As a final example consider *(e1 ∧ e3):

(e1 ∧ e3) ∧ *(e1 ∧ e3) = (e1.e1)(e3.e3)(e1 ∧ e2 ∧ e3)

(e1 ∧ e3) ∧ *(e1 ∧ e3) = -(e1 ∧ e3 ∧ e2)

For D = 3 we have:

Λ2              Λ3-2
\              /
(e1 ∧ e3) ∧ *(e1 ∧ e3) = -(e1 ∧ e3 ∧ e2)
\       \
Λ2      Λ3-2

Compare the Λ3-1 spaces to get *(e1 ∧ e3) = -e2

In D dimensions we can generalize these operations
as follows:

Let I = i1 ... ip and H = j1 ... jD-p. I spans the
space Λp and H spans the space ΛD-p.  Therefore,
together I and H span the entire space ΛD.  We
can then write:

*[ei1 ∧ ... ∧ eip]

= εI,H[ei1.ei1 ... eip.eip]ej1 ∧ ... ∧ ejD-p

Example:

For D = 3

*e1 = ε1,23(e1.e1)e2 ∧ e3

= e2 ∧ e3

*e2 = ε2,13(e2.e1)e2 ∧ e3

= -(e1 ∧ e3)

Geometrically, the Hodge dual captures the idea
that planes can be identified with their normals
and so forth.

Grassmann Numbers
-----------------

Grassmann numbers, θ, are anti-commuting numbers that
are individual elements of the exterior algebra.
Crudely speaking, they can be viewed as differential
forms without the notion of a manifold or exterior
deivative.

Grassmann numbers are used in Physics to describe
Fermionic fields where the particles have antisymmetric
wavefunctions.

Geometric (aka Clifford) Algebra
--------------------------------

The fundamental operation in Geometric Algebra is
the Geometric Product.  It is defined as:

uv = u.v + u ∧ v

u.v = (1/2)(uv + vu)

u ∧ v = (1/2)(uv - vu)

Geometric algebra can be regarded as an extension
of exterior algebra to include scalars and vectors.
The scalars and vectors have their usual interpretation
and live subspaces of the geometric algebra whereas
the wedge products are multivectors that live in the
exterior space, Λ(V) of a vector space, V.

Axioms:

(uv)w = u(vw)

u(v + w) = uv + uw

2 Dimensions
------------

Consider the basis vectors e1 and e2:

e1.e1 = (1/2)(e1e1 + e1e1) = 1

e1 ∧ e1 = (1/2)(e1e1 - e1e1) = 0

e1e1 = e1.e1 + e1 ∧ e1  = 1 + 0 = 1

Likewise,

e2e2 = 1

e1 ∧ e2 = (1/2)(e1e2 - e2e1)

= -(1/2)(e2e1 - e1e2)

= -(e2 ∧ e1)

e1e2 = e1.e2 + e1 ∧ e2

= e1 ∧ e2

e2e1 = e2 ∧ e1

= -(e1 ∧ e2)

= -e1e2

Summary:

eμeμ = 1

eμeν = -eνeμ

eμ ∧ eμ = 0

eμ ∧ eν = -(eν ∧ eμ)

uv = u.v + u ∧ v

uv = (u1e1 + u2e2)(v1e1 + v2e2)

= u1v1e1e1 + u2v2e2e2 + u1v2e1e2 + u2v1e2e1

= u1v1 + u2v2 + e1e2(u1v2 - u2v1)

= u.v + e1e2(u1v2 - u2v1)

Therefore,

e1e2(u1v2 - u2v1) ≡ u ∧ v

Proof:

u ∧ v = u1v1(e1 ∧ e1) + u2v2(e2 ∧ e2)

+ u1v2(e1 ∧ e2) + u2v1(e2 ∧ e1)

= u1v2(e1 ∧ e2) + u2v1(e2 ∧ e1)

= (u1v2 - u2v1)(e1 ∧ e2)

This is a bivector.  Bivectors are often referred
to as axial or pseudovectors.  Geometrically e1e2
represent oriented areas.

The algebra should be closed meaning that the
product of 2 elements is another element.  If we
consider all possible combinations of e1 and
e2 we obtain a 4-dimensional space spanned by
{1,e1,e2,e1 ∧ e2}.

Bivectors (grade 2):  e1 ∧ e2

In Clifford Algebra dimensions are referred to

Relation to Complex Numbers
---------------------------

(e1e2)2 = e1e2e1e2

= -e1e1e2e2

We can therefore identify e1e2 with i since i2 = -1

We say that Cl2 is isomorphic to the complex
numbers.

3 Dimensions
------------

We can extend the basis vector constructions to
any number of dimensions.  However, we need to be
careful about how these rules are applied.  For
example,

e1e2e1e3 = e1 ∧ e2 ∧ e1 ∧ e3 = -(e1 ∧ e1 ∧ e2 ∧ e3)

= -(0 ∧ e2 ∧ e3)

= 0

Is not a valid construction.  To do this correctly
we need to go back to the formula uv = u.v + u ∧ v
and write:

e1e2e1e3 = (-e1e1)(e2e3) = uv

Then:

uv = (-e1e1).(e2e3) + (-e1e1) ∧ (e2.e3)

= -1.(e2e3) + -1 ∧ e2 ∧ e3

= 0 - (e2 ∧ e3)

= -(e2 ∧ e3)

Of course, the shorter calculation is just to
simplify the original expression before constructing
the wedge product.  Thus,

e1e2e1e3 = -e1e1e2e3 = -e2e3 = -(e2 ∧ e3)

Again, we look for the closure requirement by
considering at all possible combinations of e1, e2
and e3:

(e1e2)e1 = -e2
(e1e2)e2 = e1
(e1e2)e3 = e1e2e3

(e2e3)e1 = e2e3e1 = e1e2e3)
(e2e3)e2 = -e3
(e2e3)e3 = e2

(e1e3)e1 = -e3
(e1e3)e2 = e1e3e2 = -(e1e2e3)
(e1e3)e3 = e1

Therefore we obtain an 8-dimensional space spanned
by {1,e1,e2,e3,e1 ∧ e2,e2 ∧ e3,e1 ∧ e3,e1 ∧ e2 ∧ e3}

Vectors (grade 1):   e1, e2, e3

Bivectors (grade 2):  e1 ∧ e2, e2 ∧ e3, e1 ∧ e3

Trivectors (grade 3):  e1 ∧ e2 ∧ e3

Trivectors are often referred to as pseudoscalars.
Geometrically, pseudoscalars e1e2e3 represent oriented
volumes.

It is worth noting that the bases of the algebra
follows Pascal's triangle.

0:           1
1:         1   1
2:       1   2   1
3:     1   3   3   1
1   4   6   4   1
1   5  10  10   5   1

Which is tantamount to saying that the dimension of
the Clifford algebra generated by a vector space of
dimension, D, is given by 2D.

Relation to Quaternions
-----------------------

i2 = j2 = k2 = ijk = -1

i2 = e1e2e1e2 = -e1e1e2e2 = -1

j2 = e2e3e2e3 = -e2e2e3e3 = -1

k2 = e1e3e1e3 = -e1e1e3e3 = -1

ijk = e1e2e2e3e1e3

= e1e3e1e3

= -e1e1e3e3

= -1

We say that Cl3 is isomorphic to the quaternions.

Examples in 3D
--------------

Vector:  u = (3e2 + 4e3)

Bivector:  v = (2e1e2 + 3e2e3 + 4e1e3)

uv = (3e2 + 4e3)(2e1e2 + 3e2e3 + 4e1e3)

= 6e2e1e2 + 9e2e2e3 + 12e2e1e3 + 8e3e1e2 + 12e3e2e3 + 16e3e1e3

= -6e1 + 9e3 - 12e1e2e3 + 8e1e2e3 - 12e2 - 16e1

= -22e1 - 12e2 + 9e3 - 4e1e2e3

Compare this result to the Grassman algebra where
cannot simplify the basis vectors, e1 ... :

u ∧ v = (3e2 + 4e3) ∧ (2e1e2 + 3e2e3 + 4e1e3)

= (3e2 ∧ 2e1e2) + (3e2 ∧ 3e2e3) + (3e2 ∧ 4e1e3)

+ (4e3 ∧ 2e1e2) + (4e3 ∧ 3e2e3) + (4e3 ∧ 4e1e3)

= (3 ∧ e2 ∧ 2 ∧ e1 ∧ e2) + (3 ∧ e2 ∧ 3 ∧ e2 ∧ e3)

+ (3 ∧ e2 ∧ 4 ∧ e1 ∧ e3) + (4 ∧ e3 ∧ 2 ∧ e1 ∧ e2)

+ (4 ∧ e3 ∧ 3 ∧ e2 ∧ e3) + (4 ∧ e3 ∧ 4 ∧ e1 ∧ e3)

= 12(e2 ∧ e1 ∧ e3) + 4(e3 ∧ 2e1 ∧ e2)

= -12(e1 ∧ e2 ∧ e3) + 8(e1 ∧ e2 ∧ e3)

= -4(e1 ∧ e2 ∧ e3)

Bivector:  u = (5 + 2e1e2)

Bivector:  v = (2 + 2e1 + e1e3)

uv = (5 + 2e1e2)(2 + 2e1 + e1e3)

= 10 + 10e1 + 5e1e3 + 4e1e2 + 4e1e2e1 + 2e1e2e1e3

= 10 + 10e1 - 4e2 + 5e1e3 + 4e1e2 - 2e2e3
--   ----------   --------------------
^         ^                ^
|         |                |
|      vector          bivectors
scalar

We see that the geometric product of a bivector
with a bivector does not expand to a quadrivector,
but rather it reduces to a bivector.

This 'wrapping' is a form of closure in that runaway
expansion into ever higher dimensions is prevented.
We can see this in 3D as follows:

(vector)(vector) = (e1)(e2) = e1e2 (bivector)

(vector)(bivector) = (e1)(e2e3) = e1e2e3 (trivector)

(vector)(trivector) = (e1)(e1e2e3) = e2e3 (bivector)

(bivector)(bivector) = (e1e2)(e1e3) = -e2e3 (bivector)

(bivector)(trivector) = (e1e2)(e1e2e3) = -e3 (vector)

Bivector:  u = (5 + 2e1e2)

Trivector:  v = (2 + e1e2e3)

uv = (5 + 2e1e2)(2 + e1e2e3)

= 10 + 5e1e2e3 + 4e1e2 + 2e1e2e1e2e3

= 10 + 5e1e2e3 + 4e1e2 - 2e3

Now, for an orthogonal basis e1e2 = e1 ∧ e2 etc.

uv = 10 + 5(e1 ∧ e2 ∧ e3) + 4(e1 ∧ e2) - 2e3

= 10 - 2e3 + 4(e1 ∧ e2) + 5(e1 ∧ e2 ∧ e3)
--   ---   ----------   ---------------
^    ^         ^              ^
|    |         |              |
|  vector   bivector       trivector
scalar

Matrix Representation
---------------------

Consider the Pauli matrices.

σ1σ2 - σ2σ1 = 2iσ3

-   -  -    -     -    -  -   -     -      -
| 0 1 || 0 -i | - | 0 -i || 0 1 | = | 2i  0  |
| 1 0 || i  0 |   | i  0 || 1 0 |   | 0  -2i |
-   -  -    -     -    -  -   -     -      -

If we let σ1 ≡ e1, σ2 ≡ e2 and σ3 ≡ e3

(1/2)(e1e2 - e2e1) = e1 ∧ e2

≡ e1e2e3e3

≡ Ie3

Where I = e1e2e3

Or,

(e1e2 - e2e1) = 2Ie3

The Clifford Algebra for the Pauli matrices is
often written as:

{σμ,σν} = 2ημνI

Which yields:

σμσν + σνσμ = 2ημνI

μ = ν => σ12 = σ22 = σ32 = 1

μ ≠ ν => σ1σ2 + σ2σ1 = 2η12I = 0

∴ σ1σ2 = -σ2σ1

It should not be surprising that the Dirac matrices
also also obey the Clifford algebra.

{γμ,γν} = 2ημνI

```