The Clifford Algebra as a Lie Algebra
Introduction
Every associative algebra is a Lie algebra under the commutator $[x, y] = xy - yx$, and the Clifford algebra is no exception. What makes the commutator on a Clifford algebra interesting is that its bivectors reproduce the orthogonal Lie algebra: the degree-two elements, with the bracket they inherit from the algebra, are isomorphic to $\mathrm{SO}(V, q)$, and their adjoint action on the vectors is the defining representation. In this way the abstract skew maps of The Orthogonal Lie Algebra acquire an explicit realisation inside the algebra generated by the vectors, and the infinitesimal rotations become elements that can be exponentiated, multiplied and represented.
This article develops the commutator bracket on $\mathrm{Cl}(V, q)$, the subalgebra structure that the parity grading imposes, the isomorphism between the bivectors and $\mathrm{SO}(V, q)$, the geometric reading of vectors, bivectors and trivectors, and the low-dimensional computations. The base is a field $F$ of characteristic not $2$, the form is non-degenerate, and the algebra is that of Clifford Algebras; the radical case is treated in Degenerate Clifford Algebras and the Radical. The exterior power $\Lambda^2 V$ and its identification $\varphi$ with $\mathrm{SO}(V, q)$ are from The Orthogonal Lie Algebra and The Exterior Algebra, and the filtration and volume element are developed in The Filtration and the Associated Graded Algebra and The Volume Element, Duality and the Hodge Star. The Spin and Pin groups, which exponentiate the bivectors, belong to the other entries of the category.
The Commutator Bracket
The Clifford Algebra as a Lie Algebra
Proposition. For any quadratic space $(V, q)$ over $R$, the commutator
$$ [x, y] = xy - yx, \qquad x, y \in \mathrm{Cl}(V, q), $$
is $R$-bilinear, alternating, and satisfies the Jacobi identity; with this bracket $\mathrm{Cl}(V, q)$ is a Lie algebra over $R$. It is not abelian as soon as $\dim V \geq 2$ and the form is non-degenerate.
Proof. The commutator is bilinear and alternating because it is the difference of the two products; the Jacobi identity is the standard identity $[[x,y],z] + [[y,z],x] + [[z,x],y] = 0$ for the commutator in an associative algebra, obtained by expanding the twelve terms.
Proposition. The even part $\mathrm{Cl}^0(V, q)$ is a Lie subalgebra, and the odd part $\mathrm{Cl}^1(V, q)$ is a module over it under the bracket. The parity grading is compatible with the bracket:
$$ [\mathrm{Cl}^i, \mathrm{Cl}^j] \subseteq \mathrm{Cl}^{i+j}, \qquad i, j \in \mathbb{Z}/2\mathbb{Z}. $$
Proof. The even part is closed under multiplication, hence under the commutator. The parity statement is immediate from $\mathrm{Cl}^i\mathrm{Cl}^j \subseteq \mathrm{Cl}^{i+j}$.
Proposition. For every $x \in \mathrm{Cl}(V, q)$ the adjoint map $\operatorname{ad}_x = [x, -]$ is a derivation of the Clifford algebra:
$$ \operatorname{ad}_x(yz) = \operatorname{ad}_x(y)\,z + y\,\operatorname{ad}_x(z). $$
Proof. This is the Leibniz rule for the commutator, $\operatorname{ad}_x(yz) = xyz - yzx = (xy - yx)z + y(xz - zx)$, using associativity.
Thus the Clifford algebra is not only a Lie algebra in its own right but acts on itself by derivations, and $\operatorname{ad} : \mathrm{Cl} \to \operatorname{Der}(\mathrm{Cl})$ is a Lie algebra homomorphism, since the bracket of two inner derivations is the inner derivation of the bracket: $[\operatorname{ad}_x, \operatorname{ad}_y] = \operatorname{ad}_{[x,y]}$.
The Bracket and the Degree Filtration
Let $\mathrm{Cl}_{\leq p}(V,q)$ be the span of the products of at most $p$ vectors; the filtration is increasing, $\mathrm{Cl}_{\leq 0} = F$, $\mathrm{Cl}_{\leq 1} = F \oplus V$, and $\bigcup_p \mathrm{Cl}_{\leq p} = \mathrm{Cl}(V,q)$.
Definition. For $x \in \mathrm{Cl}(V, q)$ let $\operatorname{gr}(x) \in \Lambda(V)$ be its image in the associated graded algebra of the filtration, which is the exterior algebra by Clifford Algebras, a finite sum of multivectors; the degree of $x$ is the largest $p$ with a nonzero component of $\operatorname{gr}(x)$ in degree $p$, and $x$ is pure of degree $p$ when $\operatorname{gr}(x)$ is homogeneous of that degree. A vector is pure of degree $1$ and a bivector is pure of degree $2$.
Proposition. Let $x$ and $y$ be pure of degrees $p, q \geq 1$. If $p$ or $q$ is even then $xy - yx$ has degree at most $p + q - 2$; if both are odd it has degree at most $p + q$, and its degree-$(p+q)$ part is twice the graded product $\operatorname{gr}(x)\operatorname{gr}(y)$.
Proof. The associated graded algebra $\Lambda(V)$ is graded-commutative, so $\operatorname{gr}(yx) = (-1)^{pq}\operatorname{gr}(x)\operatorname{gr}(y)$. Hence the degree-$(p+q)$ part of $xy - yx$ is $\bigl(1 - (-1)^{pq}\bigr)\operatorname{gr}(x)\operatorname{gr}(y)$, which vanishes when $pq$ is even and is twice the graded product when $pq$ is odd. Every degree occurring in $xy$ or in $yx$ is congruent to $p + q$ modulo $2$, so no term of degree $p + q - 1$ occurs, and the next possible degree below $p + q$ is $p + q - 2$.
Corollary. The anticommutator of two vectors is the scalar $\{u, v\} = uv + vu = 2B(u, v)\cdot 1$, their commutator $[u, v] = uv - vu$ is a bivector, the bracket of a vector with a bivector is a vector, and the bracket of two bivectors is a bivector: $[V, V] \subseteq B_2$, $[V, B_2] \subseteq V$ and $[B_2, B_2] \subseteq B_2$. Equivalently $[\Lambda^2 V, V] \subseteq V$ and $[\Lambda^2 V, \Lambda^2 V] \subseteq \Lambda^2 V$ under the identification $\Phi$.
Proof. The first statement is the defining relation. For $x = u$ and $y = v$ vectors the degrees are $1$ and $1$, so the proposition gives degree at most $2$, and the scalar part of $uv - vu = 2uv - 2B(u, v)\cdot 1$ vanishes because the scalar part of $uv$ is $B(u, v)$; hence the commutator is a sum of bivectors. For $x \in V$ and $y \in B_2$ the degrees are $1$ and $2$, so the degree is at most $1$; the bracket is the difference of two products of an odd and an even element, hence odd, and the odd part of $\mathrm{Cl}_{\leq 1}$ is $V$. For $x, y \in B_2$ the degrees are $2$ and $2$, so the bracket has degree at most $2$, and it is even. On the basis bivectors $[e_ie_j, e_ke_l]$ vanishes for disjoint pairs and equals $2a_j\,e_ie_l$ or $-2a_i\,e_je_l$ for pairs sharing one index, in both cases without a scalar part; the scalar part is bilinear in $x$ and $y$, so it vanishes for all bivectors and the bracket is a bivector.
The filtration therefore computes degree as follows: the bracket of elements pure of degrees $p$ and $q$ has degree at most $p + q$, and the degree falls to at most $p + q - 2$ whenever one of the two degrees is even, because the leading products of $xy$ and of $yx$ then agree in the graded algebra and cancel. When both degrees are odd the leading term survives as twice the graded product, so the commutator bracket and the graded bracket of the exterior algebra differ already at leading order; the cases the bivector calculus uses are the even ones, $p = 1$, $q = 2$ and $p = q = 2$.
Bivectors and the Orthogonal Lie Algebra
Bivectors
Definition. The bivectors of $\mathrm{Cl}(V, q)$ are the elements of the subspace
$$ B_2 = \operatorname{span}\{\,uv - vu : u, v \in V\,\} \subseteq \mathrm{Cl}^0(V, q). $$
The element $uv - vu$ is alternating bilinear in $u, v$: it is $R$-bilinear in each argument, it changes sign when $u$ and $v$ are interchanged, and it vanishes on the diagonal, $uu - uu = 0$. Hence the assignment $u \wedge v \mapsto uv - vu$ factors through $\Lambda^2 V$, and $B_2$ is a quotient of $\Lambda^2 V$; it is the image of the linear map
$$ \Phi : \Lambda^2 V \longrightarrow \mathrm{Cl}^0(V, q), \qquad \Phi(u \wedge v) = \tfrac{1}{4}(uv - vu). $$
The factor $\tfrac{1}{4}$ is fixed so that the identification with $\mathrm{SO}(V, q)$ below is exactly the map $\varphi$ of The Orthogonal Lie Algebra; in an orthogonal basis it gives $\Phi(e_i \wedge e_j) = \tfrac{1}{2}e_ie_j$.
The Lie Algebra Isomorphism
Theorem. Let $V$ be a space over a field of characteristic not $2$ with a non-degenerate form $q$, let $\Lambda^2 V$ carry the bivector bracket transported from $\mathrm{SO}(V, q)$ by $\varphi(u \wedge v)(x) = B(v, x)u - B(u, x)v$, and let $B_2$ carry the commutator. Then
$$ \Phi : \Lambda^2 V \longrightarrow B_2, \qquad \Phi(u \wedge v) = \tfrac{1}{4}(uv - vu) $$
is an isomorphism of Lie algebras onto the bivectors, and $\varphi \circ \Phi^{-1} : B_2 \to \mathrm{SO}(V, q)$ is an isomorphism of Lie algebras.
Proof. Choose an orthogonal basis $e_1, \ldots, e_n$ of $V$, with $q(e_i) = a_i \neq 0$. The wedges $e_i \wedge e_j$, $i < j$, form a basis of $\Lambda^2 V$, and
$$ \Phi(e_i \wedge e_j) = \tfrac{1}{4}(e_ie_j - e_je_i) = \tfrac{1}{2}e_ie_j $$
because orthogonal generators anticommute. It suffices to verify $[\Phi(X), \Phi(Y)] = \Phi([X, Y])$ on basis elements. There are three cases. If the two pairs are disjoint, all four generators are distinct, so $e_ie_j$ and $e_ke_l$ commute and both brackets vanish. If the pairs are equal, both brackets vanish. In the remaining case, write the pairs as $\{i, j\}$ and $\{j, l\}$ with $i, j, l$ distinct. Then
$$ e_ie_j \cdot e_je_l = a_j\,e_ie_l, \qquad e_je_l \cdot e_ie_j = -a_j\,e_ie_l, $$
the second identity because the generators are mutually orthogonal and the transposition of the two odd factors contributes a sign; hence
$$ [\Phi(e_i \wedge e_j), \Phi(e_j \wedge e_l)] = \tfrac{1}{4}[e_ie_j, e_je_l] = \tfrac{1}{4}\cdot 2a_j\,e_ie_l = \tfrac{a_j}{2}\,e_ie_l = \Phi(a_j\,e_i \wedge e_l). $$
On the other side, the skew map $\varphi(e_i \wedge e_j)$ sends $e_i \mapsto -a_ie_j$, $e_j \mapsto a_je_i$, and annihilates the other basis vectors, so
$$ [\varphi(e_i \wedge e_j), \varphi(e_j \wedge e_l)] = a_j\,\varphi(e_i \wedge e_l), $$
as one checks on the three basis vectors $e_i, e_j, e_l$. Therefore $\Phi$ carries the transported bracket to the commutator, and being injective on a basis it is a Lie algebra isomorphism onto its image, the bivectors.
Corollary. $\dim B_2 = \binom{n}{2} = \dim \mathrm{SO}(V, q)$, and the bivectors are a Lie subalgebra of $\mathrm{Cl}^0(V, q)$ isomorphic to $\mathrm{SO}(V, q)$.
Proof. The map $\Phi$ is injective because it is injective on the basis $e_i \wedge e_j$, whose images $\tfrac{1}{2}e_ie_j$ are linearly independent as part of the standard basis of the Clifford algebra; the dimension of $\Lambda^2 V$ is $\binom{n}{2}$ and that of $\mathrm{SO}(V, q)$ is $\binom{n}{2}$ by The Orthogonal Lie Algebra.
The Defining Representation
Theorem. For every $x \in \Lambda^2 V$ the adjoint action of the bivector $\Phi(x)$ on the vectors is the skew map $\varphi(x)$:
$$ [\Phi(x), w] = \varphi(x)(w) \qquad (w \in V). $$
Proof. On basis elements, with $x = e_i \wedge e_j$ and $\Phi(x) = \tfrac{1}{2}e_ie_j$, compute
$$ [\tfrac{1}{2}e_ie_j, e_i] = -a_ie_j, \qquad [\tfrac{1}{2}e_ie_j, e_j] = a_je_i, \qquad [\tfrac{1}{2}e_ie_j, e_k] = 0 \quad (k \neq i, j), $$
using $e_ie_je_i = -a_ie_j$ and $e_je_ie_j = -a_je_i$ for orthogonal generators. These are exactly the values $\varphi(e_i \wedge e_j)(e_i) = -a_ie_j$, $\varphi(e_i \wedge e_j)(e_j) = a_je_i$ and $\varphi(e_i \wedge e_j)(e_k) = 0$.
So the bivectors act on $V$ through the defining representation of $\mathrm{SO}(V, q)$: the bivector $\Phi(u \wedge v)$ is the infinitesimal rotation in the plane spanned by $u$ and $v$, with the magnitude and the sense determined by the form.
Decomposable Bivectors and Their Squares
The square of a bivector in the Clifford algebra records whether the bivector spans a single plane, and with what sign.
Proposition. For $u, v \in V$,
$$ \Phi(u \wedge v)^2 = -\tfrac{1}{4}\bigl(q(u)q(v) - B(u, v)^2\bigr) = -\tfrac{1}{4}\det \begin{pmatrix} q(u) & B(u,v) \\ B(u,v) & q(v)\end{pmatrix}, $$
a scalar. In particular the square is $-q(u)q(v)/4$ when $u$ and $v$ are orthogonal.
Proof. If $u$ and $v$ are linearly dependent the wedge vanishes and both sides are $0$. Otherwise write $u = \alpha e_1 + \beta e_2$, $v = \gamma e_1 + \delta e_2$ in an orthogonal basis $e_1, e_2$ of the plane when the plane is non-degenerate, which is possible by Gram–Schmidt; then $u \wedge v = (\alpha\delta - \beta\gamma)e_1 \wedge e_2$ and
$$ \Phi(u \wedge v)^2 = \tfrac{1}{4}(\alpha\delta - \beta\gamma)^2 (e_1e_2)^2 = -\tfrac{1}{4}(\alpha\delta - \beta\gamma)^2 a_1a_2, $$
while $q(u)q(v) - B(u,v)^2 = a_1a_2(\alpha\delta - \beta\gamma)^2$ by the same expansion as the two-by-two determinant. When the plane is degenerate, choose a basis $r, w$ with $r$ isotropic, $B(r, w) = 0$; then $q(u)q(v) - B(u,v)^2 = 0$, and $\Phi(u \wedge v) = \tfrac{1}{2}(\alpha\delta - \beta\gamma)rw$ with $(rw)^2 = -r^2w^2 = 0$, so both sides vanish again.
Theorem. Let $X = \sum_{i Proof. For the first statement, expand the square: the products of a pair $\{i,j\}$ with itself contribute $\tfrac{1}{4}c_{ij}^2(e_ie_j)^2 = -\tfrac{1}{4}c_{ij}^2a_ia_j$, while a product of two distinct pairs has degree $2$ when the pairs share an index and degree $4$ when they are disjoint, so it contributes nothing to the scalar part. For the second statement, the forward implication is the proposition above. The converse is the classical criterion for bivectors: over a field of characteristic not $2$, a bivector is decomposable exactly when $X \wedge X = 0$ in $\Lambda^4 V$. A direct expansion of $\Phi(X)^2$ in the orthogonal basis shows that its degree-two part vanishes identically and that its degree-four part vanishes exactly when the Plücker relations $c_{ij}c_{kl} - c_{ik}c_{jl} + c_{il}c_{jk} = 0$ hold on the coefficients, which is the same condition as $X \wedge X = 0$; the criterion is cited as standard. Remark. The square of a bivector therefore has a geometric reading. A decomposable bivector spans a plane, and $\Phi(u\wedge v)^2 = -\tfrac{1}{4}\det G$ is a negative multiple of the determinant of the form on that plane: a definite plane gives a negative square and an elliptic rotation, an indefinite plane a positive square and a hyperbolic one, and a degenerate plane gives square $0$, so that $\exp(\Phi(u \wedge v)) = 1 + \Phi(u \wedge v)$ is a transvection. A non-decomposable bivector has a square with a non-scalar part, so it does not generate a commutative subalgebra and its exponential is not a rotation in a single plane. The invariant bilinear forms of $\mathrm{SO}(V, q)$ have a direct Clifford description, because the scalar part of a Clifford product computes the trace of the corresponding endomorphism. Proposition. In an orthogonal basis with $q(e_i) = a_i$ the endomorphism $\varphi(e_i \wedge e_j)$ has matrix $a_jE_{ij} - a_iE_{ji}$, where $E_{ij}$ is the matrix unit, and $$
\operatorname{tr}\bigl(\varphi(e_i \wedge e_j)\varphi(e_k \wedge e_l)\bigr) = \delta_{il}\delta_{jk}(a_ja_l + a_ia_k) - \delta_{ik}\delta_{jl}(a_ja_k + a_ia_l).
$$ In particular, for $X = \sum_{i Proof. The matrix of $\varphi(e_i\wedge e_j)$ is read off from $\varphi(e_i\wedge e_j)(e_i) = -a_ie_j$ and $\varphi(e_i\wedge e_j)(e_j) = a_je_i$, with all other basis vectors annihilated. The trace formula follows from $\operatorname{tr}(E_{ab}E_{cd}) = \delta_{bc}\delta_{ad}$ by expanding the four terms. For the last statement, take $(k,l) = (i,j)$ in the formula to get $-2a_ia_j$, and expand the square of the sum, whose cross terms have vanishing trace by the same formula. Theorem. For all $X, Y \in \Lambda^2 V$, $$
\operatorname{tr}\bigl(\varphi(X)\varphi(Y)\bigr) = 8\,\mathrm{Sc}\bigl(\Phi(X)\Phi(Y)\bigr),
$$ where $\mathrm{Sc}$ denotes the scalar part in $\mathrm{Cl}(V,q)$. Proof. Both sides are bilinear in $X$ and $Y$, so it suffices to compare them on pairs of basis wedges. For $X = e_i \wedge e_j$ and $Y = e_k \wedge e_l$ the left-hand side is the expression of the proposition. On the Clifford side, $\Phi(e_i \wedge e_j) = \tfrac{1}{2}e_ie_j$, and the scalar part of $e_ie_je_ke_l$ vanishes unless the two pairs coincide, when it equals $-a_ia_j$: a shared index leaves degree $2$ and disjoint pairs leave degree $4$. Comparing the two expressions in the three cases of equal, overlapping and disjoint pairs gives the identity. Corollary. The trace form $\beta(X, Y) = \operatorname{tr}(\varphi(X)\varphi(Y))$ on $\Lambda^2 V$ is, up to the factor $8$, the scalar part of the Clifford product of the bivectors; by The Orthogonal Lie Algebra the Killing form is $\kappa = (n-2)\beta$, so $$
\kappa\bigl(\varphi(X), \varphi(Y)\bigr) = 8(n-2)\,\mathrm{Sc}\bigl(\Phi(X)\Phi(Y)\bigr).
$$ For the standard form, where all $a_i = 1$, the identity $\operatorname{tr}(\varphi(X)^2) = -2\sum_{i The Clifford algebra is generated by $V$, and its degree filtration organises the generators by how many vectors a product contains. The low degrees have the following reading. The three together account for the classical picture of the Clifford algebra: vectors give directions, bivectors give infinitesimal rotations, and higher multivectors generate the rest of the algebra by multiplication. The identification of the bivectors with $\mathrm{SO}(V, q)$ is compatible with the exponential of The Orthogonal Lie Algebra in the following sense: for a bivector $x \in B_2$ the exponential $$
\exp(x) = \sum_{k \geq 0} \frac{x^k}{k!}
$$ converges when $F = \mathbb{R}$ or $\mathbb{C}$ and $x$ is finite, and lies in the even part of the Clifford algebra. Proposition. Let $e_1, \ldots, e_n$ be an orthogonal basis and let $\omega = e_1e_2\cdots e_n$ be the volume element. If $n$ is odd then $\omega$ is central, so $[\omega, x] = 0$ for every $x \in \mathrm{Cl}(V, q)$; in particular the trivector $\omega$ of a three-dimensional space commutes with every bivector. Proof. Moving $e_i$ past the $n - i$ generators to its right in $\omega e_i$ and past the $i - 1$ generators to its left in $e_i\omega$ gives $\omega e_i = (-1)^{n-i}a_i\,e_1\cdots\hat e_i\cdots e_n$ and $e_i\omega = (-1)^{i-1}a_i\,e_1\cdots\hat e_i\cdots e_n$, the powers of $e_i$ having collapsed to the scalar $a_i$; hence $$
\omega e_i = (-1)^{n-2i+1}e_i\omega = (-1)^{n-1}e_i\omega.
$$ For $n$ odd the sign is $+1$, so $\omega$ commutes with every generator; commuting $\omega$ past a product of $2k$ generators multiplies the sign $(-1)^{n-1}$ exactly $2k$ times, which is $1$ again because $n - 1$ is even. So $\omega$ commutes with all generators, hence with all of $\mathrm{Cl}(V, q)$. Example. In the three-dimensional case the volume element $e_1e_2e_3$ is a trivector and is central, so the subspace spanned by it is a one-dimensional abelian Lie ideal of the ambient Lie algebra; the bivectors and the trivector together span the even part and the odd part of the algebra. This is the algebraic reason a trivector in three dimensions behaves as a scalar under conjugation. For $V = F^2$ with the standard form, $\Lambda^2 V$ is one-dimensional, spanned by $e_1 \wedge e_2$, and $$
\Phi(e_1 \wedge e_2) = \tfrac{1}{2}e_1e_2, \qquad (e_1e_2)^2 = -1,
$$ so the bivector satisfies $[\Phi(e_1 \wedge e_2), \Phi(e_1 \wedge e_2)] = 0$: the Lie algebra is abelian, as $\mathrm{SO}(2)$ is. Its exponential is $$
\exp(\theta\,\tfrac{1}{2}e_1e_2) = \cos(\theta/2) + e_1e_2\sin(\theta/2),
$$ whose conjugation action on $V$ is the plane rotation $\exp(\theta\varphi(e_1 \wedge e_2)) = R(-\theta)$ of The Rotation Group and Orientation; the half angle is the usual doubling of the Clifford parametrisation of a rotation. For $V = F^3$ with the standard form, the bivectors $e_1e_2$, $e_2e_3$, $e_3e_1$ satisfy $$
(e_1e_2)^2 = (e_2e_3)^2 = (e_3e_1)^2 = -1, \qquad [e_1e_2, e_2e_3] = 2e_1e_3 = -2e_3e_1,
$$ with the cyclic permutations. Under the correspondence $\Phi(e_1 \wedge e_2) \leftrightarrow -e_3$, $\Phi(e_2 \wedge e_3) \leftrightarrow -e_1$, $\Phi(e_3 \wedge e_1) \leftrightarrow -e_2$ — equivalently $e_1e_2 \leftrightarrow -2e_3$ and its cyclic permutations — the commutator of bivectors becomes the vector product of their images, so this correspondence is the isomorphism $\mathrm{SO}(3) \cong \mathbb{R}^3$ of The Orthogonal Lie Algebra transported into the Clifford algebra; the sign is the one carried there by $\varphi(e_i \wedge e_j) = -L_k$. The bivectors span a three-dimensional Lie subalgebra isomorphic to $\mathrm{SO}(3)$, and the trivector $e_1e_2e_3$ is central by the previous proposition. For $V = F^2$ with the form $q(e_1) = 1$, $q(e_2) = -1$, the bivector $\Phi(e_1 \wedge e_2) = \tfrac{1}{2}e_1e_2$ satisfies $(e_1e_2)^2 = -e_1^2e_2^2 = 1$, so $e_1e_2$ is a non-nilpotent element of square $1$, and $$
\exp(\theta\,\tfrac{1}{2}e_1e_2) = \cosh(\theta/2) + e_1e_2\sinh(\theta/2).
$$ The Lie algebra $\mathrm{SO}(1, 1)$ is again one-dimensional and abelian, but its exponential is a hyperbolic rotation. For an orthogonal pair $e_i, e_j$ one has $(e_ie_j)^2 = -q(e_i)q(e_j)$, so the square of the bivector is the negative of the determinant of the form on the plane it spans: a definite plane gives a negative square and a periodic one-parameter group, an indefinite plane a positive square and a hyperbolic one. The Clifford algebra $\mathrm{Cl}(V, q)$ is a Lie algebra under the commutator $[x, y] = xy - yx$; the even part $\mathrm{Cl}^0$ is a Lie subalgebra, the parity grading satisfies $[\mathrm{Cl}^i, \mathrm{Cl}^j] \subseteq \mathrm{Cl}^{i+j}$, and every adjoint map $\operatorname{ad}_x$ is a derivation. The bivectors $\Phi(u \wedge v) = \tfrac{1}{4}(uv - vu)$ form a subspace $B_2$ of $\mathrm{Cl}^0$ of dimension $\binom{n}{2}$, and the map $\Phi$ is an isomorphism of Lie algebras from $\Lambda^2 V$ with its transported bracket onto $B_2$; equivalently, the bivectors with the commutator are isomorphic to the orthogonal Lie algebra $\mathrm{SO}(V, q)$. On the vectors the adjoint action is the defining representation, $[\Phi(x), w] = \varphi(x)(w)$, so a bivector is an infinitesimal rotation in the plane of its two factors: vectors are directions, bivectors are infinitesimal rotations, and trivectors and higher multivectors complete the algebra. For elements pure of degrees $p$ and $q$ the commutator has degree at most $p + q$, falling to at most $p + q - 2$ whenever one of the two degrees is even; the leading term survives only when both degrees are odd, and it is twice the graded product. The square of a decomposable bivector is the scalar $\Phi(u \wedge v)^2 = -\tfrac{1}{4}\det G$, where $G$ is the Gram matrix of the form on the plane spanned by $u$ and $v$, and a bivector is decomposable exactly when its Clifford square is a scalar; in an orthogonal basis the scalar part of $\Phi(X)^2$ for $X = \sum c_{ij}e_i \wedge e_j$ is $-\tfrac{1}{4}\sum c_{ij}^2a_ia_j$. A definite plane therefore gives a negative square and an elliptic rotation, an indefinite plane a positive square and a hyperbolic one, and a degenerate plane square $0$, so that the exponential is the transvection $1 + \Phi(u \wedge v)$. The invariant forms of the bivector Lie algebra are computed by the scalar part of the Clifford product: $\operatorname{tr}(\varphi(X)\varphi(Y)) = 8\,\mathrm{Sc}(\Phi(X)\Phi(Y))$ for bivectors $X, Y$, and since the Killing form of $\mathrm{SO}(V,q)$ is $\kappa = (n-2)\beta$ with $\beta$ the trace form, one has $\kappa(\varphi X, \varphi Y) = 8(n-2)\mathrm{Sc}(\Phi(X)\Phi(Y))$. In particular $\operatorname{tr}(\varphi(X)^2) = -2\sum c_{ij}^2a_ia_j$, which is negative definite for the standard form and is the computation behind the compactness of $\mathrm{SO}(n)$. Exponentiating a bivector gives an element of $\mathrm{Cl}^0$ whose conjugation action is the corresponding rotation, which is the passage from the Lie algebra to the Clifford, Pin and Spin groups. In an odd-dimensional space the volume element is central, so the top-degree trivector commutes with every bivector; in the plane the product $e_1e_2$ squares to $-1$ for a definite form and to $+1$ for an indefinite one, and the exponential of the bivector is the trigonometric or the hyperbolic rotation accordingly.
The Trace Form and the Killing Form
The Geometric Interpretation
Vectors, Bivectors, Trivectors
Infinitesimal Rotations
The Volume Element and the Trivectors
Low-Dimensional Examples
The Plane
Three Dimensions
A Non-Definite Plane
Summary
Summary of Notation
Symbol
Meaning
$F$
Field of characteristic not $2$
$V$, $q$, $B$
Quadratic space, non-degenerate form, polar form
$\mathrm{Cl}(V, q)$
Clifford algebra
$\mathrm{Cl}^0, \mathrm{Cl}^1$
Even and odd parts of the parity grading
$[x, y] = xy - yx$
Commutator bracket
$\{u, v\} = uv + vu$
Anticommutator, equal to $2B(u,v)$ on vectors
$\mathrm{Cl}_{\leq p}$
Degree filtration, products of at most $p$ vectors
$\operatorname{gr}(x)$, pure of degree $p$
Image of $x$ in the associated graded $\Lambda(V)$; $x$ is pure of degree $p$ when that image is homogeneous of degree $p$
$\det G$
Determinant of the Gram matrix on the plane of $u, v$
$c_{ij}$
Coefficients of a bivector in an orthogonal basis
Decomposable bivector
$X = u \wedge v$, equivalently $X \wedge X = 0$
Transvection
$\exp(x) = 1 + x$ for a bivector with $x^2 = 0$
$\operatorname{tr}$
Trace of an endomorphism of $V$
$E_{ij}$
Matrix unit, $E_{ij}(e_j) = e_i$
$\mathrm{Sc}$
Scalar part in the Clifford algebra
$\beta(X,Y) = \operatorname{tr}(\varphi(X)\varphi(Y))$
Trace form on the bivectors
$\kappa$
Killing form of $\mathrm{SO}(V,q)$, equal to $(n-2)\beta$
$\operatorname{ad}_x = [x, -]$
Adjoint derivation
$B_2$
Space of bivectors, image of $\Phi$
$\Phi(u \wedge v) = \tfrac14(uv - vu)$
Bivector map
$\Lambda^2 V$
Second exterior power, the bivectors
$\varphi(u \wedge v)(x) = B(v,x)u - B(u,x)v$
Identification $\Lambda^2 V \to \mathrm{SO}(V, q)$
$\mathrm{SO}(V, q)$
Orthogonal Lie algebra of skew maps
$e_i$
Orthogonal basis, $q(e_i) = a_i$
$\omega = e_1\cdots e_n$
Volume element
$\exp(x)$
Exponential of a bivector
$\mathrm{SO}(1, 1)$, $\mathrm{SO}(3)$
Low-dimensional orthogonal Lie algebras
$F$, $\mathbb{R}$, $\mathbb{C}$
Field, real and complex numbers
Further Reading