Positivity and the Hermitian Cone of a Hilbert Algebra with Hermitian Adjoint

Introduction

An involution of an algebra orders it. Once an element $x$ has a conjugate $x^{\dagger}$, the elements fixed by the conjugation play the role of the real numbers of the algebra, the elements of the form $x^{\dagger}x$ play the role of the squares of lengths, and the question whether the latter are "positive" is the question whether the conjugation is a positive involution. On a Clifford algebra over an involutive base all three notions are available, and they are what this article treats: the self-adjoint and skew elements, the decomposition of the algebra into the two, the cone of elements $x^{\dagger}x$, the condition under which the dagger is a positive involution, the polar decomposition and the Cartan involution of the group of units.

The positivity is not automatic, and the article isolates exactly when it holds. The scalar form $\mathrm{Sc}(x^{\dagger}x)$, which is the diagonal of the form of the dagger of The Blade Form and the Hilbert Structure with Hermitian Adjoint, is positive definite exactly when the quadratic form is negative definite; in that case the algebra is a finite-dimensional $C^{*}$-algebra of the canonical anticommutation relations, the cone $\{x^{\dagger}x\}$ is the positive cone, and every element has a polar decomposition $x = u|x|$ with $u$ unitary. For a positive definite quadratic form the good involution is the other one and the dagger is indefinite, so the Euclidean structure and the Hermitian structure are exchanged by a change of signature. This is the same exchange that the previous article recorded at the level of forms and that Hermitian Forms on a Hilbert Algebra with Hermitian Adjoint recorded at the level of signatures; here it becomes the statement that positivity chooses a side.

The algebra-valued forms are Hermitian Forms on a Hilbert Algebra with Hermitian Adjoint; the scalar forms are The Blade Form and the Hilbert Structure with Hermitian Adjoint; the group of units on which the Cartan involution acts is The Unitary Slice and the Compact Real Form with Hermitian Adjoint; the symmetric space it defines belongs to Symmetric Spaces and its Jordan description to Jordan Algebras and the Positive Cone; the infinite-dimensional $C^{*}$-algebra is Infinite-Dimensional Clifford Algebras and CAR with Inner Conjugation; and the spin-factor case is Spin Factors and the Clifford Envelope with Inner Conjugation.

Self-Adjoint and Skew Elements

Definition. Let $\sigma$ be an involution of the base $A$ and let ${}^{\dagger}$ be the induced dagger, $x^{\dagger} = \sigma(\alpha(x^{r}))$. An element $x$ is self-adjoint or Hermitian if $x^{\dagger} = x$, and skew-adjoint if $x^{\dagger} = -x$. Write

$$ \mathrm{Herm}(V,q) = \{\, x : x^{\dagger} = x \,\}, \qquad \mathrm{Skew}(V,q) = \{\, x : x^{\dagger} = -x \,\}. $$

Proposition. The dagger is an anti-involution of order two, so $\mathrm{Herm}$ and $\mathrm{Skew}$ are $A^{\sigma}$-submodules; when $2$ is invertible in $A$,

$$ \mathrm{Cl}(V,q) = \mathrm{Herm}(V,q) \oplus \mathrm{Skew}(V,q), \qquad x = \tfrac12(x + x^{\dagger}) + \tfrac12(x - x^{\dagger}), $$

and the splitting is exactly the eigen-decomposition of the involution.

Proof. $(x^{\dagger})^{\dagger} = x$ gives $\mathrm{Herm} \cap \mathrm{Skew} = 0$, and the displayed formula gives the sum, with the first summand self-adjoint by $\dagger^{2} = \mathrm{id}$ and the second skew.

Proposition. For every $x$, the product $x^{\dagger}x$ is self-adjoint.

Proof. $(x^{\dagger}x)^{\dagger} = x^{\dagger}(x^{\dagger})^{\dagger} = x^{\dagger}x$.

Example (the low-dimensional picture). In the definite negative algebra $\mathrm{Cl}_{0,n}$ the dagger is Clifford conjugation $x^{\dagger} = x^{\natural}$, which fixes the scalars, negates the vectors and acts on a blade by $(-1)^{k(k+1)/2}$; so the scalars and the blades of degree $k \equiv 0,3 \pmod 4$ are self-adjoint and those of degree $k \equiv 1,2 \pmod 4$ are skew. In particular the vectors are skew-adjoint and the bivectors are skew-adjoint in this convention, which is the one in which the spin group is generated by products of pairs of vectors: the Lie algebra of the group lies in the skew part, as it must.

The Cone

The Set of Squares

Definition. The Hermitian cone of the algebra is the image of the squaring map,

$$ P = \{\, x^{\dagger}x : x \in \mathrm{Cl}(V,q) \,\} . $$

Its elements are self-adjoint, it contains $0$ and $1$, and $P \cap (-P) = 0$ when the involution is positive, as the next paragraph shows. It is not automatic that $P$ is closed under addition; that holds when the algebra is a full matrix algebra with the adjoint involution, which is the case for $\mathrm{Cl}_{0,n}$ over $\mathbb{R}$ and for the complexified algebras, and it is the reason the cone is the one of a $C^{*}$-algebra rather than merely a set of squares.

Positivity of the Involution

Definition. The dagger is positive if

$$ \mathrm{Sc}\bigl(x^{\dagger}x\bigr) > 0 \qquad \text{for every } x \neq 0 . $$

Theorem. The dagger is positive if and only if the scalar form of the dagger is positive definite, and this holds exactly when the quadratic form is negative definite. In particular

$$ \mathrm{Cl}_{0,n} \text{ over } \mathbb{R}, \quad x^{\dagger} = x^{\natural}, \qquad \mathrm{Sc}(x^{\dagger}x) = \sum_I a_I^{2} \ \ge 0 . $$

Proof. By The Blade Form and the Hilbert Structure with Hermitian Adjoint, the scalar form of the dagger is $\mathrm{Sc}(x^{\dagger}y) = \sum_I(-1)^{|I|}(\prod_{i\in I}e_i^{2})a_Ib_I$ in an orthogonal basis, so $\mathrm{Sc}(x^{\dagger}x) = \sum_I(-1)^{|I|}\prod_{i\in I}e_i^{2}\,a_I^{2}$. For a negative definite form every $e_i^{2} = -1$ and $(-1)^{|I|}\prod_{i\in I}e_i^{2} = (-1)^{|I|}(-1)^{|I|} = 1$, giving the displayed sum. For a positive definite form with $n \ge 1$ the odd part enters with a minus sign, so the form is indefinite; and a mixed signature contains both signs.

Corollary (the $C^{*}$-structure). When the dagger is positive, $\mathrm{Sc}(x^{\dagger}x)$ is a positive definite quadratic form on the algebra, related to the Euclidean form by the diagonal signature signs; the algebra with the involution is a finite-dimensional $C^{*}$-algebra, a full matrix algebra over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ according to the parity of $n$, and $P$ is its cone of positive elements. The completion in infinite dimension carries the canonical anticommutation relations, as in Infinite-Dimensional Clifford Algebras and CAR with Inner Conjugation.

Remark (the exchange of sides). A positive definite quadratic form does not fail to give a cone; it gives the cone of the other involution. For $\mathrm{Cl}_{n,0}$ the form $\mathrm{Sc}(x^{r}x)$ is positive definite, and the involution to use is reversion, not the dagger; the two statements are the same statement after $e_i \mapsto ie_i$, which changes the sign of the quadratic form and the sign of the parity. Over a complex base with a conjugation $\sigma$ the dagger is positive exactly when it induces a positive involution of the complexified algebra, which is the ordinary case of a complex Hilbert space.

The Polar Decomposition

Theorem. Suppose the dagger is positive and $x \in \mathrm{Cl}(V,q)$ lies in a matrix Clifford algebra of the form $\mathrm{Cl}_{0,n}$ over $\mathbb{R}$, or is an invertible element of such an algebra. Then $x$ has a polar decomposition

$$ x = u\,|x|, \qquad |x| = \bigl(x^{\dagger}x\bigr)^{1/2} \in P, \qquad u^{\dagger}u = 1, $$

with $|x|$ the positive square root and $u$ in the unitary slice. The decomposition is unique when $x$ is invertible.

Proof. $x^{\dagger}x$ is self-adjoint and lies in $P$, hence is positive in the matrix algebra, so its square root exists and is positive. Set $u = x|x|^{-1}$ for invertible $x$; then $u^{\dagger}u = |x|^{-1}x^{\dagger}x|x|^{-1} = |x|^{-1}|x|^{2}|x|^{-1} = 1$. Uniqueness is the uniqueness of the positive square root.

Example (the quaternion case). For $\mathrm{Cl}_{0,2} = \mathbb{H}$ the theorem is the polar decomposition of a quaternion: $x^{\dagger}x$ is the scalar $|x|^{2}$, $|x|$ is the real absolute value, and $u = x/|x|$ lies in the unit sphere $\mathrm{Sp}(1) = \mathrm{Spin}(3)$. Computed exactly, $x^{\dagger}x$ is a non-negative real scalar and $u^{\dagger}u = 1$ for every $x$ of this algebra.

The Cartan Involution

Definition. On the group of units of the algebra let

$$ \theta(x) = \bigl(x^{\dagger}\bigr)^{-1}. $$

Proposition. $\theta$ is an automorphism of order two of the group of units, and its fixed set is exactly the unitary slice $U = \{x : x^{\dagger}x = 1\}$. Its differential at the identity is $\mathrm{d}\theta = -{}^{\dagger}$, which is $+1$ on the skew part and $-1$ on the self-adjoint part.

Proof. $\theta(xy) = ((xy)^{\dagger})^{-1} = (y^{\dagger}x^{\dagger})^{-1} = (x^{\dagger})^{-1}(y^{\dagger})^{-1} = \theta(x)\theta(y)$, so $\theta$ is an automorphism. For the order, $((x^{\dagger})^{-1})^{\dagger} = ((x^{\dagger})^{\dagger})^{-1} = x^{-1}$, so $\theta^{2}(x) = (x^{-1})^{-1} = x$. The fixed set is $\{x : (x^{\dagger})^{-1} = x\} = \{x : x^{\dagger}x = 1\} = U$. For the differential, $\theta(1 + tu) \equiv 1 - tu^{\dagger}$ to first order, so $\mathrm{d}\theta(u) = -u^{\dagger}$, which is $u$ on $\mathrm{Skew}$ and $-u$ on $\mathrm{Herm}$.

Theorem (Cartan decomposition). For the Lie bracket $[u,v] = uv - vu$,

$$ [\mathrm{Skew}, \mathrm{Skew}] \subseteq \mathrm{Skew}, \qquad [\mathrm{Skew}, \mathrm{Herm}] \subseteq \mathrm{Herm}, \qquad [\mathrm{Herm}, \mathrm{Herm}] \subseteq \mathrm{Skew} . $$

Proof. The dagger is an anti-automorphism: for $u^{\dagger} = -u$ and $v^{\dagger} = -v$,

$$ [u,v]^{\dagger} = (uv - vu)^{\dagger} = v^{\dagger}u^{\dagger} - u^{\dagger}v^{\dagger} = vu - uv = -[u,v], $$

so $[\mathrm{Skew},\mathrm{Skew}] \subseteq \mathrm{Skew}$; the other two are the same computation with the signs of one or both arguments changed.

Corollary. $\mathrm{Skew}(V,q)$ is closed under the bracket and is the Lie algebra of the unitary slice in The Unitary Slice and the Compact Real Form with Hermitian Adjoint; the complement $\mathrm{Herm}(V,q)$ is the tangent space of the symmetric space of the Cartan involution, in the sense of Symmetric Spaces. When the dagger is positive this is the Cartan decomposition of a compact group and the space is a compact symmetric space; the example worked out in the corpus is $\mathrm{Cl}_{0,3}$, where the skew part is the Lie algebra of $\mathrm{Spin}(3)$ and the Hermitian part carries the spin-factor structure of Spin Factors and the Clifford Envelope with Inner Conjugation.

Remark (the cone and the spin factor). In the three-dimensional definite case the positive cone of the spin factor is the cone over the sphere of radius $1/2$ generated by the idempotents, and the quadratic form of the spin factor is the restriction of the scalar part of the dagger form. This is the point of contact with the Jordan theory of Jordan Algebras and the Positive Cone, and it is the reason the algebra of the three-dimensional Euclidean space carries both a Clifford and a Jordan structure.

Summary

The dagger makes a Clifford algebra an algebra with involution, with self-adjoint and skew parts and the splitting $\mathrm{Cl} = \mathrm{Herm}\oplus\mathrm{Skew}$ when $2$ is invertible, and with $x^{\dagger}x$ self-adjoint for every $x$. The Hermitian cone $P = \{x^{\dagger}x\}$ is the set of squares; the involution is positive exactly when the scalar form of the dagger is positive definite, which happens exactly for a negative definite quadratic form, where $\mathrm{Sc}(x^{\dagger}x) = \sum_I a_I^{2}$; a positive definite form instead makes reversion positive and exchanges the two sides by $e_i \mapsto ie_i$. When the involution is positive the algebra is a finite-dimensional $C^{*}$-algebra, $P$ is its positive cone, and every element has a polar decomposition $x = u|x|$ with $u$ in the unitary slice, which is the quaternion polar decomposition when the algebra is $\mathbb{H}$. The Cartan involution $\theta(x) = (x^{\dagger})^{-1}$ is an automorphism of order two of the group of units whose fixed set is exactly the unitary slice, and the induced Lie-algebra decomposition is that of a symmetric space: $[\mathrm{Skew},\mathrm{Skew}] \subseteq \mathrm{Skew}$, $[\mathrm{Skew},\mathrm{Herm}] \subseteq \mathrm{Herm}$, $[\mathrm{Herm},\mathrm{Herm}] \subseteq \mathrm{Skew}$, so the skew part is the Lie algebra of the unitary group and the self-adjoint part is the tangent space of the symmetric space.

Summary of Notation

Symbol	Meaning
$x^{\dagger}$	The dagger $\sigma(\alpha(x^{r}))$
$\mathrm{Herm}(V,q)$	Self-adjoint elements, $x^{\dagger}=x$
$\mathrm{Skew}(V,q)$	Skew-adjoint elements, $x^{\dagger}=-x$
$P = \{x^{\dagger}x\}$	Hermitian cone of the algebra
$\mathrm{Sc}(x^{\dagger}x)$	Scalar form of the dagger, positive definite iff $q$ is negative definite
$\|x\| = (x^{\dagger}x)^{1/2}$	Absolute value, for a positive involution
$x = u\|x\|$	Polar decomposition, $u^{\dagger}u = 1$
$\theta(x) = (x^{\dagger})^{-1}$	Cartan involution of the group of units
$U$	Its fixed set, the unitary slice