Clifford algebra

Clifford algebras are closely related to exterior algebras. Indeed, if Q = 0 then the Clifford algebra Cl(V, Q) is just the exterior algebra ⋀V. For nonzero Q there exists a canonical linear isomorphism between ⋀V and Cl(V, Q) whenever the ground field K does not have characteristic two. That is, they are naturally isomorphic as vector spaces, but with different multiplications (in the case of characteristic two, they are still isomorphic as vector spaces, just not naturally). Clifford multiplication together with the distinguished subspace is strictly richer than the exterior product since it makes use of the extra information provided by Q.

The Clifford algebra is a filtered algebra, the associated graded algebra is the exterior algebra.

More precisely, Clifford algebras may be thought of as quantizations (cf. quantum group) of the exterior algebra, in the same way that the Weyl algebra is a quantization of the symmetric algebra.

Weyl algebras and Clifford algebras admit a further structure of a *-algebra, and can be unified as even and odd terms of a superalgebra, as discussed in CCR and CAR algebras.

Clifford algebras are also sometimes referred to as geometric algebras, most often over the real numbers.

Every nondegenerate quadratic form on a finite-dimensional real vector space is equivalent to the standard diagonal form:

$${\displaystyle Q(v)=v_{1}^{2}+\dots +v_{p}^{2}-v_{p+1}^{2}-\dots -v_{p+q}^{2},}$$

where n = p + q is the dimension of the vector space. The pair of integers (p, q) is called the signature of the quadratic form. The real vector space with this quadratic form is often denoted R^p,q. The Clifford algebra on R^p,q is denoted Cl_p,q(R). The symbol Cl_n(R) means either Cl_n,0(R) or Cl_0,n(R) depending on whether the author prefers positive-definite or negative-definite spaces.

A standard basis {e₁, ..., e_n} for R^p,q consists of n = p + q mutually orthogonal vectors, p of which square to +1 and q of which square to −1. Of such a basis, the algebra Cl_p,q(R) will therefore have p vectors that square to +1 and q vectors that square to −1.

A few low-dimensional cases are:

• Cl_0,0(R) is naturally isomorphic to R since there are no nonzero vectors.
• Cl_0,1(R) is a two-dimensional algebra generated by e₁ that squares to −1, and is algebra-isomorphic to C, the field of complex numbers.
• Cl_0,2(R) is a four-dimensional algebra spanned by {1, e₁, e₂, e₁e₂}. The latter three elements all square to −1 and anticommute, and so the algebra is isomorphic to the quaternions H.
• Cl_0,3(R) is an 8-dimensional algebra isomorphic to the direct sum H ⊕ H, the split-biquaternions.

One can also study Clifford algebras on complex vector spaces. Every nondegenerate quadratic form on a complex vector space of dimension n is equivalent to the standard diagonal form

$${\displaystyle Q(z)=z_{1}^{2}+z_{2}^{2}+\dots +z_{n}^{2}.}$$

Thus, for each dimension n, up to isomorphism there is only one Clifford algebra of a complex vector space with a nondegenerate quadratic form. We will denote the Clifford algebra on Cⁿ with the standard quadratic form by Cl_n(C).

For the first few cases one finds that

• Cl₀(C) ≅ C, the complex numbers
• Cl₁(C) ≅ C ⊕ C, the bicomplex numbers
• Cl₂(C) ≅ M₂(C), the biquaternions

where M_n(C) denotes the algebra of n × n matrices over C.

In this section, Hamilton's quaternions are constructed as the even subalgebra of the Clifford algebra Cl_3,0(R).

Let the vector space V be real three-dimensional space R³, and the quadratic form be the usual quadratic form. Then, for v, w in R³ we have the bilinear form (or scalar product)

$${\displaystyle v\cdot w=v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}.}$$

Now introduce the Clifford product of vectors v and w given by

$${\displaystyle vw+wv=2(v\cdot w).}$$

This formulation uses the negative sign so the correspondence with quaternions is easily shown.

Denote a set of orthogonal unit vectors of R³ as {e₁, e₂, e₃}, then the Clifford product yields the relations

$${\displaystyle e_{2}e_{3}=-e_{3}e_{2},\,\,\,e_{3}e_{1}=-e_{1}e_{3},\,\,\,e_{1}e_{2}=-e_{2}e_{1},}$$

and

$${\displaystyle e_{1}^{2}=e_{2}^{2}=e_{3}^{2}=1.}$$

The general element of the Clifford algebra Cl_3,0(R) is given by

$${\displaystyle A=a_{0}+a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}+a_{4}e_{2}e_{3}+a_{5}e_{3}e_{1}+a_{6}e_{1}e_{2}+a_{7}e_{1}e_{2}e_{3}.}$$

The linear combination of the even degree elements of Cl_3,0(R) defines the even subalgebra Cl^[0]
_3,0(R) with the general element

$${\displaystyle q=q_{0}+q_{1}e_{2}e_{3}+q_{2}e_{3}e_{1}+q_{3}e_{1}e_{2}.}$$

The basis elements can be identified with the quaternion basis elements i, j, k as

$${\displaystyle i=e_{2}e_{3},j=e_{1}e_{3},k=e_{1}e_{2},}$$

which shows that the even subalgebra Cl^[0]
_3,0(R) is Hamilton's real quaternion algebra.

To see this, compute

$${\displaystyle i^{2}=(e_{2}e_{3})^{2}=e_{2}e_{3}e_{2}e_{3}=-e_{2}e_{2}e_{3}e_{3}=-1,}$$

and

$${\displaystyle ij=e_{2}e_{3}e_{1}e_{3}=-e_{2}e_{3}e_{3}e_{1}=-e_{2}e_{1}=e_{1}e_{2}=k.}$$

Finally,

$${\displaystyle ijk=e_{2}e_{3}e_{1}e_{3}e_{1}e_{2}=-1.}$$

In this section, dual quaternions are constructed as the even Clifford algebra of real four-dimensional space with a degenerate quadratic form.[6][7]

Let the vector space V be real four-dimensional space R⁴, and let the quadratic form Q be a degenerate form derived from the Euclidean metric on R³. For v, w in R⁴ introduce the degenerate bilinear form

$${\displaystyle d(v,w)=v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}.}$$

This degenerate scalar product projects distance measurements in R⁴ onto the R³ hyperplane.

The Clifford product of vectors v and w is given by

$${\displaystyle vw+wv=-2\,d(v,w).}$$

Note the negative sign is introduced to simplify the correspondence with quaternions.

Denote a set of mutually orthogonal unit vectors of R⁴ as {e₁, e₂, e₃, e₄}, then the Clifford product yields the relations

$${\displaystyle e_{m}e_{n}=-e_{n}e_{m},\,\,\,m\neq n,}$$

and

$${\displaystyle e_{1}^{2}=e_{2}^{2}=e_{3}^{2}=-1,\,\,e_{4}^{2}=0.}$$

The general element of the Clifford algebra Cl(R⁴, d) has 16 components. The linear combination of the even degree elements defines the even subalgebra Cl^[0](R⁴, d) with the general element

$${\displaystyle H=h_{0}+h_{1}e_{2}e_{3}+h_{2}e_{3}e_{1}+h_{3}e_{1}e_{2}+h_{4}e_{4}e_{1}+h_{5}e_{4}e_{2}+h_{6}e_{4}e_{3}+h_{7}e_{1}e_{2}e_{3}e_{4}.}$$

The basis elements can be identified with the quaternion basis elements i, j, k and the dual unit ε as

$${\displaystyle i=e_{2}e_{3},j=e_{3}e_{1},k=e_{1}e_{2},\,\,\varepsilon =e_{1}e_{2}e_{3}e_{4}.}$$

This provides the correspondence of Cl^[0]
_0,3,1(R) with dual quaternion algebra.

To see this, compute

$${\displaystyle \varepsilon ^{2}=(e_{1}e_{2}e_{3}e_{4})^{2}=e_{1}e_{2}e_{3}e_{4}e_{1}e_{2}e_{3}e_{4}=-e_{1}e_{2}e_{3}(e_{4}e_{4})e_{1}e_{2}e_{3}=0,}$$

and

$${\displaystyle \varepsilon i=(e_{1}e_{2}e_{3}e_{4})e_{2}e_{3}=e_{1}e_{2}e_{3}e_{4}e_{2}e_{3}=e_{2}e_{3}(e_{1}e_{2}e_{3}e_{4})=i\varepsilon .}$$

The exchanges of e₁ and e₄ alternate signs an even number of times, and show the dual unit ε commutes with the quaternion basis elements i, j, k.

For dim V = 1, if Q has diagonalization diag(a), that is there is a non-zero vector x such that Q(x) = a, then Cl(V, Q) is algebra-isomorphic to a K-algebra generated by an element x satisfying x² = a, the quadratic algebra K[X] / (X² − a).

In particular, if a = 0 (that is, Q is the zero quadratic form) then Cl(V, Q) is algebra-isomorphic to the dual numbers algebra over K.

If a is a non-zero square in K, then Cl(V, Q) ≃ K ⊕ K.

Otherwise, Cl(V, Q) is isomorphic to the quadratic field extension K(√a) of K.

For dim V = 2, if Q has diagonalization diag(a, b) with non-zero a and b (which always exists if Q is non-degenerate), then Cl(V, Q) is isomorphic to a K-algebra generated by elements x and y satisfying x² = a, y² = b and xy = −yx.

Thus Cl(V, Q) is isomorphic to the (generalized) quaternion algebra (a, b)_K. We retrieve Hamilton's quaternions when a = b = −1, since H = (−1, −1)_R.

As a special case, if some x in V satisfies Q(x) = 1, then Cl(V, Q) ≃ M₂(K).

Given a vector space V, one can construct the exterior algebra ⋀V, whose definition is independent of any quadratic form on V. It turns out that if K does not have characteristic 2 then there is a natural isomorphism between ⋀V and Cl(V, Q) considered as vector spaces (and there exists an isomorphism in characteristic two, which may not be natural). This is an algebra isomorphism if and only if Q = 0. One can thus consider the Clifford algebra Cl(V, Q) as an enrichment (or more precisely, a quantization, cf. the Introduction) of the exterior algebra on V with a multiplication that depends on Q (one can still define the exterior product independently of Q).

The easiest way to establish the isomorphism is to choose an orthogonal basis {e₁, ..., e_n} for V and extend it to a basis for Cl(V, Q) as described above. The map Cl(V, Q) → ⋀V is determined by

$${\displaystyle e_{i_{1}}e_{i_{2}}\cdots e_{i_{k}}\mapsto e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}}.}$$

Note that this only works if the basis {e₁, ..., e_n} is orthogonal. One can show that this map is independent of the choice of orthogonal basis and so gives a natural isomorphism.

If the characteristic of K is 0, one can also establish the isomorphism by antisymmetrizing. Define functions f_k : V × ⋯ × V → Cl(V, Q) by

$${\displaystyle f_{k}(v_{1},\ldots ,v_{k})={\frac {1}{k!}}\sum _{\sigma \in \mathrm {S} _{k}}\operatorname {sgn}(\sigma )\,v_{\sigma (1)}\cdots v_{\sigma (k)}}$$

where the sum is taken over the symmetric group on k elements, S_k. Since f_k is alternating it induces a unique linear map ⋀^k V → Cl(V, Q). The direct sum of these maps gives a linear map between ⋀V and Cl(V, Q). This map can be shown to be a linear isomorphism, and it is natural.

A more sophisticated way to view the relationship is to construct a filtration on Cl(V, Q). Recall that the tensor algebra T(V) has a natural filtration: F⁰ ⊂ F¹ ⊂ F² ⊂ ⋯, where F^k contains sums of tensors with order ≤ k. Projecting this down to the Clifford algebra gives a filtration on Cl(V, Q). The associated graded algebra

$${\displaystyle \operatorname {Gr} _{F}\operatorname {Cl} (V,Q)=\bigoplus _{k}F^{k}/F^{k-1}}$$

is naturally isomorphic to the exterior algebra ⋀V. Since the associated graded algebra of a filtered algebra is always isomorphic to the filtered algebra as filtered vector spaces (by choosing complements of F^k in F^k+1 for all k), this provides an isomorphism (although not a natural one) in any characteristic, even two.

In the following, assume that the characteristic is not 2.[lower-alpha 4]

Clifford algebras are Z₂-graded algebras (also known as superalgebras). Indeed, the linear map on V defined by v ↦ −v (reflection through the origin) preserves the quadratic form Q and so by the universal property of Clifford algebras extends to an algebra automorphism

$${\displaystyle \alpha$$ :\operatorname {Cl} (V,Q)\to \operatorname {Cl} (V,Q).}

Since α is an involution (i.e. it squares to the identity) one can decompose Cl(V, Q) into positive and negative eigenspaces of α

$${\displaystyle \operatorname {Cl} (V,Q)=\operatorname {Cl} ^{[0]}(V,Q)\oplus \operatorname {Cl} ^{[1]}(V,Q)}$$

where

$${\displaystyle \operatorname {Cl} ^{[i]}(V,Q)=\left\{x\in \operatorname {Cl} (V,Q)\mid \alpha (x)=(-1)^{i}x\right\}.}$$

Since α is an automorphism it follows that:

$${\displaystyle \operatorname {Cl} ^{[i]}(V,Q)\operatorname {Cl} ^{[j]}(V,Q)=\operatorname {Cl} ^{[i+j]}(V,Q)}$$

where the bracketed superscripts are read modulo 2. This gives Cl(V, Q) the structure of a Z₂-graded algebra. The subspace Cl^[0](V, Q) forms a subalgebra of Cl(V, Q), called the even subalgebra. The subspace Cl^[1](V, Q) is called the odd part of Cl(V, Q) (it is not a subalgebra). This Z₂-grading plays an important role in the analysis and application of Clifford algebras. The automorphism α is called the main involution or grade involution. Elements that are pure in this Z₂-grading are simply said to be even or odd.

Remark. The Clifford algebra is not a Z-graded algebra, but is Z-filtered, where Cl^≤i(V, Q) is the subspace spanned by all products of at most i elements of V.

$${\displaystyle \operatorname {Cl} ^{\leqslant i}(V,Q)\cdot \operatorname {Cl} ^{\leqslant j}(V,Q)\subset \operatorname {Cl} ^{\leqslant i+j}(V,Q).}$$

The degree of a Clifford number usually refers to the degree in the N-grading.

The even subalgebra Cl^[0](V, Q) of a Clifford algebra is itself isomorphic to a Clifford algebra.[lower-alpha 5][lower-alpha 6] If V is the orthogonal direct sum of a vector a of nonzero norm Q(a) and a subspace U, then Cl^[0](V, Q) is isomorphic to Cl(U, −Q(a)Q), where −Q(a)Q is the form Q restricted to U and multiplied by −Q(a). In particular over the reals this implies that:

$${\displaystyle \operatorname {Cl} _{p,q}^{[0]}(\mathbf {R} )\cong {\begin{cases}\operatorname {Cl} _{p,q-1}(\mathbf {R} )&q>0\\\operatorname {Cl} _{q,p-1}(\mathbf {R} )&p>0\end{cases}}}$$

In the negative-definite case this gives an inclusion Cl_0,n−1(R) ⊂ Cl_0,n(R), which extends the sequence

Likewise, in the complex case, one can show that the even subalgebra of Cl_n(C) is isomorphic to Cl_n−1(C).

In addition to the automorphism α, there are two antiautomorphisms that play an important role in the analysis of Clifford algebras. Recall that the tensor algebra T(V) comes with an antiautomorphism that reverses the order in all products of vectors:

$${\displaystyle v_{1}\otimes v_{2}\otimes \cdots \otimes v_{k}\mapsto v_{k}\otimes \cdots \otimes v_{2}\otimes v_{1}.}$$

Since the ideal I_Q is invariant under this reversal, this operation descends to an antiautomorphism of Cl(V, Q) called the transpose or reversal operation, denoted by x^t. The transpose is an antiautomorphism: (xy)^t = y^t x^t. The transpose operation makes no use of the Z₂-grading so we define a second antiautomorphism by composing α and the transpose. We call this operation Clifford conjugation denoted ${\displaystyle {\bar {x}}}$

$${\displaystyle {\bar {x}}=\alpha (x^{\mathrm {t} })=\alpha (x)^{\mathrm {t} }.}$$

Of the two antiautomorphisms, the transpose is the more fundamental.[lower-alpha 7]

Note that all of these operations are involutions. One can show that they act as ±1 on elements which are pure in the Z-grading. In fact, all three operations depend only on the degree modulo 4. That is, if x is pure with degree k then

$${\displaystyle \alpha (x)=\pm x\qquad x^{\mathrm {t} }=\pm x\qquad {\bar {x}}=\pm x}$$

where the signs are given by the following table:

k mod 4	0	1	2	3	…
${\displaystyle \alpha (x)\,}$	+	−	+	−	(−1)k
${\displaystyle x^{\mathrm {t} }\,}$	+	+	−	−	(−1)k(k − 1)/2
${\displaystyle {\bar {x}}}$	+	−	−	+	(−1)k(k + 1)/2

When the characteristic is not 2, the quadratic form Q on V can be extended to a quadratic form on all of Cl(V, Q) (which we also denoted by Q). A basis-independent definition of one such extension is

$${\displaystyle Q(x)=\left\langle x^{\mathrm {t} }x\right\rangle _{0}}$$

where ⟨a⟩₀ denotes the scalar part of a (the degree-0 part in the Z-grading). One can show that

$${\displaystyle Q(v_{1}v_{2}\cdots v_{k})=Q(v_{1})Q(v_{2})\cdots Q(v_{k})}$$

where the v_i are elements of V – this identity is not true for arbitrary elements of Cl(V, Q).

The associated symmetric bilinear form on Cl(V, Q) is given by

$${\displaystyle \langle x,y\rangle =\left\langle x^{\mathrm {t} }y\right\rangle _{0}.}$$

One can check that this reduces to the original bilinear form when restricted to V. The bilinear form on all of Cl(V, Q) is nondegenerate if and only if it is nondegenerate on V.

The operator of left (respectively right) Clifford multiplication by the transpose a^t of an element a is the adjoint of left (respectively right) Clifford multiplication by a with respect to this inner product. That is,

$${\displaystyle \langle ax,y\rangle =\left\langle x,a^{\mathrm {t} }y\right\rangle ,}$$

and

$${\displaystyle \langle xa,y\rangle =\left\langle x,ya^{\mathrm {t} }\right\rangle .}$$

In arbitrary characteristic, the spinor norm Q is defined on the Lipschitz group by

$${\displaystyle Q(x)=x^{\mathrm {t} }x.}$$

It is a homomorphism from the Lipschitz group to the group K^× of non-zero elements of K. It coincides with the quadratic form Q of V when V is identified with a subspace of the Clifford algebra. Several authors define the spinor norm slightly differently, so that it differs from the one here by a factor of −1, 2, or −2 on Γ¹. The difference is not very important in characteristic other than 2.

The nonzero elements of K have spinor norm in the group (K^×)² of squares of nonzero elements of the field K. So when V is finite-dimensional and non-singular we get an induced map from the orthogonal group of V to the group K^×/(K^×)², also called the spinor norm. The spinor norm of the reflection about r^⊥, for any vector r, has image Q(r) in K^×/(K^×)², and this property uniquely defines it on the orthogonal group. This gives exact sequences:

$${\displaystyle {\begin{aligned}1\to \{\pm 1\}\to {\mbox{Pin}}_{V}(K)&\to {\mbox{O}}_{V}(K)\to K^{\times }/\left(K^{\times }\right)^{2},\\1\to \{\pm 1\}\to {\mbox{Spin}}_{V}(K)&\to {\mbox{SO}}_{V}(K)\to K^{\times }/\left(K^{\times }\right)^{2}.\end{aligned}}}$$

Note that in characteristic 2 the group {±1} has just one element.

From the point of view of Galois cohomology of algebraic groups, the spinor norm is a connecting homomorphism on cohomology. Writing μ₂ for the algebraic group of square roots of 1 (over a field of characteristic not 2 it is roughly the same as a two-element group with trivial Galois action), the short exact sequence

$${\displaystyle 1\to \mu _{2}\rightarrow {\mbox{Pin}}_{V}\rightarrow {\mbox{O}}_{V}\rightarrow 1}$$

yields a long exact sequence on cohomology, which begins

$${\displaystyle 1\to H^{0}(\mu _{2};K)\to H^{0}({\mbox{Pin}}_{V};K)\to H^{0}({\mbox{O}}_{V};K)\to H^{1}(\mu _{2};K).}$$

The 0th Galois cohomology group of an algebraic group with coefficients in K is just the group of K-valued points: H⁰(G; K) = G(K), and H¹(μ₂; K) ≅ K^×/(K^×)², which recovers the previous sequence

$${\displaystyle 1\to \{\pm 1\}\to {\mbox{Pin}}_{V}(K)\to {\mbox{O}}_{V}(K)\to K^{\times }/\left(K^{\times }\right)^{2},}$$

where the spinor norm is the connecting homomorphism H⁰(O_V; K) → H¹(μ₂; K).

To describe the real spin representations, one must know how the spin group sits inside its Clifford algebra. The pin group, Pin_p,q is the set of invertible elements in Cl_p,q that can be written as a product of unit vectors:

$${\displaystyle \mathrm {Pin} _{p,q}=\left\{v_{1}v_{2}\cdots v_{r}\mid \forall i\,\|v_{i}\|=\pm 1\right\}.}$$

Comparing with the above concrete realizations of the Clifford algebras, the pin group corresponds to the products of arbitrarily many reflections: it is a cover of the full orthogonal group O(p, q). The spin group consists of those elements of Pin_p,q that are products of an even number of unit vectors. Thus by the Cartan–Dieudonné theorem Spin is a cover of the group of proper rotations SO(p, q).

Let α : Cl → Cl be the automorphism which is given by the mapping v ↦ −v acting on pure vectors. Then in particular, Spin_p,q is the subgroup of Pin_p,q whose elements are fixed by α. Let

$${\displaystyle \operatorname {Cl} _{p,q}^{[0]}=\{x\in \operatorname {Cl} _{p,q}\mid \alpha (x)=x\}.}$$

(These are precisely the elements of even degree in Cl_p,q.) Then the spin group lies within Cl^[0]
_p,q.

The irreducible representations of Cl_p,q restrict to give representations of the pin group. Conversely, since the pin group is generated by unit vectors, all of its irreducible representation are induced in this manner. Thus the two representations coincide. For the same reasons, the irreducible representations of the spin coincide with the irreducible representations of Cl^[0]
_p,q.

To classify the pin representations, one need only appeal to the classification of Clifford algebras. To find the spin representations (which are representations of the even subalgebra), one can first make use of either of the isomorphisms (see above)

$${\displaystyle \operatorname {Cl} _{p,q}^{[0]}\approx \operatorname {Cl} _{p,q-1},{\text{ for }}q>0}$$

$${\displaystyle \operatorname {Cl} _{p,q}^{[0]}\approx \operatorname {Cl} _{q,p-1},{\text{ for }}p>0}$$

and realize a spin representation in signature (p, q) as a pin representation in either signature (p, q − 1) or (q, p − 1).

One of the principal applications of the exterior algebra is in differential geometry where it is used to define the bundle of differential forms on a smooth manifold. In the case of a (pseudo-)Riemannian manifold, the tangent spaces come equipped with a natural quadratic form induced by the metric. Thus, one can define a Clifford bundle in analogy with the exterior bundle. This has a number of important applications in Riemannian geometry. Perhaps more important is the link to a spin manifold, its associated spinor bundle and spin^c manifolds.

Clifford algebras have numerous important applications in physics. Physicists usually consider a Clifford algebra to be an algebra with a basis generated by the matrices γ₀, ..., γ₃ called Dirac matrices which have the property that

$${\displaystyle \gamma _{i}\gamma _{j}+\gamma _{j}\gamma _{i}=2\eta _{ij}\,}$$

where η is the matrix of a quadratic form of signature (1, 3) (or (3, 1) corresponding to the two equivalent choices of metric signature). These are exactly the defining relations for the Clifford algebra Cl
_1,3(R), whose complexification is Cl
_1,3(R)_C which, by the classification of Clifford algebras, is isomorphic to the algebra of 4 × 4 complex matrices Cl₄(C) ≈ M₄(C). However, it is best to retain the notation Cl
_1,3(R)_C, since any transformation that takes the bilinear form to the canonical form is not a Lorentz transformation of the underlying spacetime.

The Clifford algebra of spacetime used in physics thus has more structure than Cl₄(C). It has in addition a set of preferred transformations – Lorentz transformations. Whether complexification is necessary to begin with depends in part on conventions used and in part on how much one wants to incorporate straightforwardly, but complexification is most often necessary in quantum mechanics where the spin representation of the Lie algebra so(1, 3) sitting inside the Clifford algebra conventionally requires a complex Clifford algebra. For reference, the spin Lie algebra is given by

$${\displaystyle {\begin{aligned}\sigma ^{\mu \nu }&=-{\frac {i}{4}}\left[\gamma ^{\mu },\,\gamma ^{\nu }\right],\\\left[\sigma ^{\mu \nu },\,\sigma ^{\rho \tau }\right]&=i\left(\eta ^{\tau \mu }\sigma ^{\rho \nu }+\eta ^{\nu \tau }\sigma ^{\mu \rho }-\eta ^{\rho \mu }\sigma ^{\tau \nu }-\eta ^{\nu \rho }\sigma ^{\mu \tau }\right).\end{aligned}}}$$

This is in the (3, 1) convention, hence fits in Cl
_3,1(R)_C.[11]

The Dirac matrices were first written down by Paul Dirac when he was trying to write a relativistic first-order wave equation for the electron, and give an explicit isomorphism from the Clifford algebra to the algebra of complex matrices. The result was used to define the Dirac equation and introduce the Dirac operator. The entire Clifford algebra shows up in quantum field theory in the form of Dirac field bilinears.

The use of Clifford algebras to describe quantum theory has been advanced among others by Mario Schönberg,[12] by David Hestenes in terms of geometric calculus, by David Bohm and Basil Hiley and co-workers in form of a hierarchy of Clifford algebras, and by Elio Conte et al.[13][14]

Clifford algebras have been applied in the problem of action recognition and classification in computer vision. Rodriguez et al.[15] propose a Clifford embedding to generalize traditional MACH filters to video (3D spatiotemporal volume), and vector-valued data such as optical flow. Vector-valued data is analyzed using the Clifford Fourier Transform. Based on these vectors action filters are synthesized in the Clifford Fourier domain and recognition of actions is performed using Clifford correlation. The authors demonstrate the effectiveness of the Clifford embedding by recognizing actions typically performed in classic feature films and sports broadcast television.