Clifford algebra

Is the same as Geometric Algebra, but from a different approach.
Let $V$ be a finite-dimensional real (complex) vector space with nondegenerate bilinear form $g$ . The (complex) Clifford algebra $C l (V, g)$ is the algebra generated by $V$ along with the anti-commutation relation

v u + u v = 2 g (u, v),

or equivalently with the relation $u^{2} = g (u, u)$ .
Note: the Clifford algebra can be seen like coming from the tensor algebra $T (V)$ . The Clifford algebra $C ℓ (V, g)$ is the quotient $C ℓ (V, g) = T (V) / I (V)$ of $T (V)$ by the ideal $I (V)$ , generated by the elements of the form $v \otimes u + u \otimes v - 2 g (u, v)$ . This ideal is also generated by $u \otimes u - g (u, u)$ . If you take (u+v)⊗(u+v) = u⊗u + u⊗v + v⊗u + v⊗v we have: g(u+v,u+v) = g(u,u)+ u⊗v + v⊗u + g(v,v), and since g(u+v,u+v) = g(u,u) + g(u,v) + g(v,u) + g(v,v), this gives us: u⊗v + v⊗u = g(u,v) + g(v,u) = 2 g(u,v).

If the signature of $g$ is $(p, q)$ , then we use the notation

C l_{p, q}

C l (p, q) .

Note: If $x, y \in V$ , then $x y$ belong to the subset of $C l (V, g)$ denoted by $V^{2}$ .

Example. ${Cl}_{0, 2} (R)$ is a four-dimensional algebra spanned by ${1, e_{1}, e_{2}, e_{1} e_{2}}$ which behave like quaternions (note that this is not the only way to see quaternions in the framework of Clifford algebras). $◼$

Example: Pauli vectors constitute a representation of the Clifford algebra $C l_{3, 0} = C l (3, 0)$ , since they satisfy

σ_{i} σ_{j} + σ_{j} σ_{i} = 2 I δ_{i j} .

It is called the Algebra of Physical Space (APS). Its matrix representation is given by the Pauli matrices. It is related to Pauli spinors

Pasted image 20231101160624.png|700
$◼$

Example: The Clifford algebra $C l (1, 3)$ is called the Spacetime Algebra (STA) and it is related to the Weyl spinors. See Spacetime Algebra.
$◼$

Rotations

(see this video)
3D Space $C l (3, 0)$ , APS

Rotations in $i j$ -plane:

V \to e^{- σ_{i} σ_{j} \frac{θ}{2}} V e^{σ_{i} σ_{j} \frac{θ}{2}}

4D Spacetime $C l (1, 3)$ , STA

Rotations in $i j$ -plane:

V \to e^{- γ_{i} γ_{j} \frac{θ}{2}} V e^{γ_{i} γ_{j} \frac{θ}{2}}

Boost in $k$ -direction:

V \to e^{- γ_{t} γ_{j} \frac{θ}{2}} V e^{γ_{t} γ_{j} \frac{θ}{2}}

Involutions in $C l (3, 0)$

Lounesto page 56.

Grade involution or inversion. Every vector goes to the opposite, $v \mapsto \hat{v} = - v$ . Then, for a general multivector

\hat{v} = ⟨ u ⟩_{0} - ⟨ u ⟩_{1} + ⟨ u ⟩_{2} - ⟨ u ⟩_{3} . . .

In general, the total space $C (R^{p, q})$ splits into $C^{+} (R^{p, q})$ and $C^{-} (R^{p, q})$ , composed of even and odd multivectors respectively.

Reversion.

\tilde{v} = ⟨ u ⟩_{0} + ⟨ u ⟩_{1} - ⟨ u ⟩_{2} - ⟨ u ⟩_{3}

Clifford conjugation.

\bar{u} = \tilde{\hat{u}} .

The spin group and the spinors

Inside the Clifford algebra $C l (V, g)$ we can find the spin group. These are subgroups of the group of units of the Clifford algebra. In particular, the Spin group, denoted as $Spin (V)$ , consists of products of even numbers of unit vectors.
Observe that the operator $φ (u)$ defined via $φ (u) (x) = u x u^{- 1}$ , $x \in V$ and $u \in S p i n (V, g)$ , is an element of $S O (V, g)$ ; and $φ (u) = φ (- u)$ . So $S p i n (V, g)$ is a double cover of $S O (V, g)$ .

Given a minimal left ideal $S$ of $C l (V, g)$ , we can consider another operator

ρ : S p i n (V, g) \to G L (S)

given by $ρ (u) (s) = u s$ . The elements of $S$ are called spinors.

In the case $n = 3$ (spatial rotations), $S p i n (3)$ consists of [Lounesto page 59]

Spin (3) = {s \in C ℓ_{3} ∣ \tilde{s} s = 1, \bar{s} s = 1}

where the tilde and the bar denotes reversion and Clifford conjugation [Lounesto page 56].
We can consider the action on
In the matrix representation given by the Pauli matrices, the group $S p i n (3)$ corresponds to $S U (2)$ .

But additionally, if we consider a minimal left ideal of $C l (n)$ , denoted by $S$ , then it is a vector space and we have a natural homomorphism

ρ : S p i n (n) \to G L (S)

given by $ρ (g) (s) =$ [TO BE COMPLETED with Lounesto page 60, for n=3] for $g \in S p i n (n)$ and $s \in S$ . We have, then, a group representation of $S p i n (n)$ called the fundamental representation via Clifford algebra, and the elements of $S$ are called spinors.

Change of basis

To understand the change of orthonormal basis in, for example, the Clifford algebra $C l (3)$ , we need to consider the structure of this algebra and how the basis vectors relate to each other. The Clifford algebra $C l (3)$ is the associative algebra generated by three orthogonal unit vectors $e_{1}, e_{2}, e_{3}$ , satisfying the relation $e_{i}^{2} = 1$ for $i = 1, 2, 3$ . The basis of $C l (3)$ consists of 8 elements:

Scalar: 1
Vectors: $e_{1}, e_{2}, e_{3}$
Bivectors: $e_{1} e_{2}, e_{2} e_{3}, e_{3} e_{1}$
Trivector: $e_{1} e_{2} e_{3}$
Now, let's consider another orthonormal basis $f_{1}, f_{2}, f_{3}$ for the same vector space. These vectors can be expressed as linear combinations of the original basis vectors $e_{1}, e_{2}, e_{3}$ :
$f_{1} = a_{11} e_{1} + a_{12} e_{2} + a_{13} e_{3}$
$f_{2} = a_{21} e_{1} + a_{22} e_{2} + a_{23} e_{3}$
$f_{3} = a_{31} e_{1} + a_{32} e_{2} + a_{33} e_{3}$
Here, the coefficients $a_{i j}$ form a 3×3 matrix $A$ , representing the change of basis from $e_{i}$ to $f_{i}$ . To transform an element $x$ in $C l (3)$ from the $e_{i}$ basis to the $f_{i}$ basis, we need to apply the change of basis matrix $A$ to each component of $x$ . Specifically, if we write $x$ as:
$x = α_{0} + α_{1} e_{1} + α_{2} e_{2} + α_{3} e_{3} + α_{12} e_{1} e_{2} + α_{23} e_{2} e_{3} + α_{31} e_{3} e_{1} + α_{123} e_{1} e_{2} e_{3}$
Then, $x$ is given in the $f_{i}$ basis by:
$x = α_{0} + α_{1}^{'} f_{1} + α_{2}^{'} f_{2} + α_{3}^{'} f_{3} + α_{12}^{'} f_{1} f_{2} + α_{23}^{'} f_{2} f_{3} + α_{31}^{'} f_{3} f_{1} + α_{123}^{'} f_{1} f_{2} f_{3}$
where the coefficients $α_{i}^{'}$ and $α_{i j}^{'}$ are obtained by applying the matrix $A$ to the corresponding vector and bivector components of $x$ .

Let's study, for example, how to transform bivectors:

Base Representation:
Recall that each bivector $e_{i} e_{j}$ transforms under the change of basis from ${e_{i}}$ to ${f_{i}}$ via:
$f_{i} = \sum_{k = 1}^{3} a_{k i} e_{k}$
where $a_{k i}$ are the components of the matrix $A$ .
Transformation of Bivectors:
To find how a bivector such as $e_{1} e_{2}$ transforms, consider:
$f_{1} f_{2} = (\sum_{k = 1}^{3} a_{k 1} e_{k}) (\sum_{l = 1}^{3} a_{l 2} e_{l}) = \sum_{k, l}^{3} a_{k 1} a_{l 2} e_{k} e_{l}$
Since the bivectors involving different indices are orthogonal (anticommute) and squares are scalar (since in three dimensions, $e_{i}^{2} = 1$ ), this simplifies to:
$f_{1} f_{2} = (a_{11} a_{22} - a_{12} a_{21}) e_{1} e_{2} + (a_{11} a_{32} - a_{12} a_{31}) e_{1} e_{3} + (a_{21} a_{32} - a_{22} a_{31}) e_{2} e_{3}$
Creating Transformation Matrix $S$ :
By performing a similar expansion for $f_{2} f_{3}$ and $f_{3} f_{1}$ , we construct each column of the transformation matrix $S$ corresponding to each bivector. This involves determining the coefficients for $e_{1} e_{2}, e_{2} e_{3},$ and $e_{1} e_{3}$ as seen in the expansion. This results in a 3x3 matrix $S$ to transform bivectors.

Clifford algebra

Rotations

Involutions in Cl(3,0)

The spin group and the spinors

Change of basis

Involutions in $C l (3, 0)$