The Signed Adjoint of the Left Multiplication on a Clifford Algebra

Introduction

The signed left multiplication of The Signed Left Multiplication on a Clifford Algebra is the operator

$$ \mathrm{L}^{\alpha}_a = L_a\circ\alpha , \qquad \mathrm{L}^{\alpha}_a(y)=a\,\alpha(y) , $$

the twist being on the argument; it agrees with $L_a$ on the even part of the algebra and equals $-L_a$ there where the argument is odd, and it satisfies the composition law $\mathrm{L}^{\alpha}_a\mathrm{L}^{\alpha}_c=L_{a\alpha(c)}$, whose composite is an ordinary left multiplication. The article computes the adjoint of $\mathrm{L}^{\alpha}_a$ for the standard form $\langle x,y\rangle=\operatorname{Sc}(\hat xy)$. The answer is that the adjoint stays inside the signed family,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}=\mathrm{L}^{\alpha}_{\alpha(\hat a)} , $$

the parameter replaced by its grade-twisted Clifford conjugate; the involution of parameters is therefore $a\mapsto\alpha(\hat a)$, the self-adjoint members are those with $\alpha(\hat a)=a$, and the isometric members satisfy the same condition $\hat aa=1$ as for the ordinary left multiplication — the signed and the ordinary left multiplication are isometric simultaneously. For the twisted form the adjoint is $\mathrm{L}^{\alpha}_{\hat a}$, the parameter conjugated by the Clifford conjugation alone.

The boundaries. The signed left multiplication, its composition law and the disambiguation of the two twists are The Signed Left Multiplication on a Clifford Algebra; the ordinary adjoints $L_a^*=L_{\hat a}$ and the Clifford conjugation $\hat x=\alpha(\tilde x)$ are The Adjoint of the Left Multiplication on a Clifford Algebra and The Twisted Adjoint on a Clifford Algebra; the twisted form and the identity $(A^{*\alpha})=\alpha A^*\alpha$ are the latter; the signed sandwich, of which $\mathrm{L}^{\alpha}_a$ is the left factor, is The Signed Sandwich on a Clifford Algebra and The Signed Adjoint Sandwich on a Clifford Algebra. The parameter-twisted family $\Lambda^{\alpha}_x(y)=\alpha(x)y$ of The Graded Multiplication Operators is a different operator and is not treated. The base is a field $F$ of characteristic not $2$, $q$ a non-degenerate quadratic form with $q(u)=B(u,u)$ and $uv+vu=2B(u,v)$.

The Adjoint and Its Involution

Theorem. For the standard form the adjoint of the signed left multiplication is the signed left multiplication by the twisted conjugate,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}=\alpha\,L_{\hat a}=\mathrm{L}^{\alpha}_{\alpha(\hat a)} . $$

The map $a\mapsto\alpha(\hat a)$ is an anti-automorphism and an involution of the algebra, it is the composite $\alpha\circ\hat{}$ of the grade involution with the Clifford conjugation, on a vector it acts by $\alpha(\hat u)=\alpha(-u)=u$, and the adjoint correspondence is an involution of the signed family.

Proof. By definition $\mathrm{L}^{\alpha}_a=L_a\alpha$, so $(\mathrm{L}^{\alpha}_a)^*=\alpha^*L_a^*=\alpha L_{\hat a}$, using the adjoints $L_a^*=L_{\hat a}$ and $\alpha^*=\alpha$ of The Twisted Adjoint on a Clifford Algebra. Now $\alpha L_{\hat a}(y)=\alpha(\hat ay)=\alpha(\hat a)\alpha(y)$, so $\alpha L_{\hat a}=\mathrm{L}^{\alpha}_{\alpha(\hat a)}$. The statements about $a\mapsto\alpha(\hat a)$ are those of the two anti-involutions: $\widehat{\alpha(ab)}=\widehat{\alpha(a)\alpha(b)}=\alpha(b)\alpha(a)=\alpha(\hat b)\alpha(\hat a)$, and $\alpha(\hat{\alpha(\hat a)})=\alpha(\alpha(\hat a))=\hat a$. On a vector, $\hat u=-u$ and $\alpha(-u)=u$.

Corollary. The adjoint of the ordinary left multiplication is ordinary, $L_a^*=L_{\hat a}$, while the adjoint of the signed left multiplication is signed; the two families are each stable under the adjoint correspondence, and the involutions of the parameters are $\hat{}$ and $\alpha(\hat{})$ respectively.

Self-Adjoint Operators and Isometries

Proposition (self-adjointness). The signed left multiplication $\mathrm{L}^{\alpha}_a$ is self-adjoint exactly when $\alpha(\hat a)=a$. The elements with $\alpha(\hat a)=a$ form a subspace of the algebra, they include the scalars and the unit, and a vector $u$ satisfies the condition because $\alpha(\hat u)=u$: the signed left multiplication by a vector is always self-adjoint, in contrast with the skew-adjointness $L_u^*=-L_u$ of the ordinary left multiplication.

Proof. $\bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}=\mathrm{L}^{\alpha}_{\alpha(\hat a)}$, and the signed left multiplications are faithful, so self-adjointness is $\alpha(\hat a)=a$; the fixed set of an involution is a subspace; the vector case is the computation $\alpha(\hat u)=\alpha(-u)=u$ above.

Proposition (isometry). The signed left multiplication is an isometry of the standard form,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}\mathrm{L}^{\alpha}_a=\operatorname{id} , $$

if and only if $\hat aa=1$; the condition is the same as the orthogonality condition of the ordinary left multiplication $L_a$, so the signed and the ordinary multiplication by $a$ are isometric simultaneously, and the isometric elements form a subgroup of the units containing $\pm1$ and closed under $a\mapsto\hat a$.

Proof. Using the adjoint and the composition law of the signed family,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}\mathrm{L}^{\alpha}_a =\mathrm{L}^{\alpha}_{\alpha(\hat a)}\mathrm{L}^{\alpha}_a =L_{\alpha(\hat a)\alpha(a)}=L_{\alpha(\hat a\,a)} , $$

which equals the identity exactly when $\alpha(\hat aa)=1$, that is $\hat aa=1$; the group statements are those of The Adjoint of the Left Multiplication on a Clifford Algebra.

Remark (the two conditions are different). Self-adjointness is $\alpha(\hat a)=a$ and isometry is $\hat aa=1$; neither implies the other, and they coincide only for the elements with $a^{2}=1$ and $\hat a=a$. The parity of $a$ does not decide either condition; the involution $\alpha(\hat{})$ decides self-adjointness and the conjugation $\hat{}$ decides isometry.

The Twisted Form and the Second Adjoint

Definition. The twisted form is $\langle x,y\rangle_\alpha=\langle\alpha(x),y\rangle$, and the twisted adjoint $A^{*\alpha}$ is the adjoint for it; the identity $A^{*\alpha}=\alpha A^*\alpha$ holds by The Twisted Adjoint on a Clifford Algebra.

Proposition. For the twisted form the adjoint of the signed left multiplication is the signed left multiplication by the Clifford conjugate,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*\alpha}=\alpha\bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}\alpha =\alpha\bigl(\alpha L_{\hat a}\bigr)\alpha=L_{\hat a}\alpha=\mathrm{L}^{\alpha}_{\hat a} , $$

so for the twisted form the parameter involution is the Clifford conjugation $\hat{}$ alone, and the twisted adjoint of the signed family is the signed family with the parameter conjugated. In particular the twisted self-adjointness of $\mathrm{L}^{\alpha}_a$ is $\hat a=a$, and its twisted isometry is again $\hat aa=1$.

Proof. Substitute the standard adjoint and cancel the two grade involutions, $\alpha\gamma=\gamma\alpha$ for $\gamma=\alpha$; the composition $L_{\hat a}\alpha=\mathrm{L}^{\alpha}_{\hat a}$ is the definition of the signed family; the conditions are read off as before.

Remark (the meaning of the two displays). The standard form pairs the signed family with the twisted conjugate and the twisted form with the conjugate; the difference is exactly the twist which defines the form, and it is the operator-level form of the statement that the grading is the twist of the category. In the notation of the graded-algebra article of the category, the same computation reads $(\Sigma^{\alpha}_{a,b})^{*}=\Sigma^{\alpha}_{\alpha(\hat a),\alpha(\hat b)}$ with the right factor absent.

Relation to the Signed Sandwich

Proposition. The signed sandwich factors as $\Sigma^{\alpha}_{a,b}=\mathrm{L}^{\alpha}_a\,R_b$, and the adjoint of the factor is a factor of the adjoint,

$$ \bigl(\Sigma^{\alpha}_{a,b}\bigr)^{*}=\bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}R_b^{*} = \mathrm{L}^{\alpha}_{\alpha(\hat a)}R_{\hat b} , $$

which is the decomposition of the signed-sandwich adjoint $\Sigma^{\alpha}_{\alpha(\hat a),\alpha(\hat b)}$ into its two factors; in particular the self-adjointness of the left factor is the special case $\alpha(\hat a)=a$ of the left parameter of a self-adjoint sandwich.

Proof. The sandwich is the composite of the signed left multiplication with the ordinary right multiplication, and the adjoint of a composite is the composite of the adjoints in the reverse order; the right multiplications commute with the left ones, so the product of the two adjoints is the signed sandwich with the two parameters as displayed, matching The Signed Adjoint Sandwich on a Clifford Algebra.

Worked Cases

A Vector

For a vector $u$ the twisted conjugate is $\alpha(\hat u)=u$, so $(\mathrm{L}^{\alpha}_u)^{*}=\mathrm{L}^{\alpha}_u$: the signed left multiplication by a vector is self-adjoint. It is an isometry exactly when $\hat uu=-u^2=-q(u)=1$, that is $q(u)=-1$; in the negative-definite normalisation every unit vector gives a self-adjoint isometry.

The Quaternions

For $\mathrm{Cl}\cong\mathbb{H}$ with $e_1^2=e_2^2=-1$ and the quaternionic conjugation $\hat x=\bar x$, the twisted conjugate of a pure quaternion $a$ is $\alpha(\hat a)=\alpha(\bar a)=-\bar a=-a^{-1}$, so $(\mathrm{L}^{\alpha}_a)^{*}=\mathrm{L}^{\alpha}_{-a^{-1}}$; a pure unit quaternion has $a^{-1}=-a$, hence $\alpha(\hat a)=a$ and $\mathrm{L}^{\alpha}_a$ is self-adjoint.

An Even Element

For an even element $a$ one has $\alpha(a)=a$ and $\alpha(\hat a)=\hat a$, so the standard adjoint of $\mathrm{L}^{\alpha}_a$ is $\mathrm{L}^{\alpha}_{\hat a}$ and self-adjointness is $\hat a=a$; for $a=e_1e_2$ in the negative-definite plane, $\hat a=-e_1e_2=-a$ and the signed left multiplication is not self-adjoint, while it is an isometry exactly when $\hat aa=-a^2=1$.

Summary

For the standard form the adjoint of the signed left multiplication is the signed left multiplication by the twisted conjugate,

$$ \bigl(\mathrm{L}^{\alpha}_a\bigr)^{*}=\mathrm{L}^{\alpha}_{\alpha(\hat a)} , \qquad \hat a=\alpha(\tilde a) , $$

so the adjoint stays inside the signed family and its parameter involution is $a\mapsto\alpha(\hat a)$. The operator is self-adjoint exactly when $\alpha(\hat a)=a$ — a condition met by every vector — and it is an isometry exactly when $\hat aa=1$, the same condition as for the ordinary left multiplication $L_a$, so signedness does not change the orthogonal elements. For the twisted form the adjoint is $\mathrm{L}^{\alpha}_{\hat a}$, the parameter conjugated by $\hat{}$, the two twists of the form and of the operator cancelling; and the factorisation $\Sigma^{\alpha}_{a,b}=\mathrm{L}^{\alpha}_aR_b$ makes the adjoint of the signed sandwich the product of the adjoints of its factors, matching The Signed Adjoint Sandwich on a Clifford Algebra. The ordinary adjoint is The Adjoint of the Left Multiplication on a Clifford Algebra, the disambiguation of the twists is The Signed Left Multiplication on a Clifford Algebra, and the module-level story is The Graded Adjoint Action on a Module over a Clifford Algebra, the last entry of the group.

Summary of Notation

Further Reading

• Pertti Lounesto, Clifford Algebras and Spinors, 2nd ed. (Cambridge University Press, 2001), for the signed multiplications, the conjugations and their adjoints in the low-dimensional cases.
• Ian R. Porteous, Clifford Algebras and the Classical Groups, Cambridge Studies in Advanced Mathematics 50 (Cambridge University Press, 1995), for the twisted conjugation and the groups defined by the isometry conditions.
• H. Blaine Lawson and Marie-Louise Michelsohn, Spin Geometry (Princeton University Press, 1989), for the signed Clifford multiplication and its self-adjointness up to sign.
• Claude Chevalley, The Algebraic Theory of Spinors and Clifford Algebras, Collected Works vol. 2 (Springer, 1997), for the graded structure of the Clifford algebra and the two conjugations.