The Polar Decomposition of a Biquaternion

Introduction

Every biquaternion $\tilde Q$ has a polar decomposition

$$ \boxed{\;\tilde Q=\tilde Q_H\,\tilde Q_0\;}, $$

with $\tilde Q_H$ Hermitian positive semidefinite and $\tilde Q_0$ a partial isometry. The two factors are

$$ \boxed{\;\tilde Q_H=\bigl(\tilde Q\tilde Q^{*}\bigr)^{1/2},\qquad \tilde Q_0=\tilde Q_H^{+}\tilde Q\;}, $$

with ${}^{+}$ the Moore–Penrose inverse (§The Moore–Penrose Inverse). The decomposition exists for every element, including $0$ and the zero divisors, and both factors are functions of $\tilde Q$ alone.

The two factors carry different data. $\tilde Q_H$ is the Hermitian square root of $\tilde Q\tilde Q^{*}$: it is Hermitian, it lies in the Hermitian cone, and it satisfies $\tilde Q_H^{2}=\tilde Q\tilde Q^{*}$. $\tilde Q_0$ is what remains of the element once the Hermitian factor is removed; it is a partial isometry, $\tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde Q_0$, and its two projections are the two ideals generated by $\tilde Q$.

The pair is not unique as it stands. The canonical pair is fixed by four conditions: $\tilde Q_H$ lies in the cone, $\tilde Q=\tilde Q_H\tilde Q_0$, $\tilde Q_0\tilde Q_0^{*}$ is the projection onto the right ideal $\tilde Q\mathbb{B}$, and $\tilde Q_H$ generates that same ideal. Two further conditions of the natural list — that $\tilde Q_0$ is a partial isometry and that its initial projection is $\tilde\Pi_{\mathrm i}(\tilde Q)$ — are consequences of the four (§Uniqueness).

The article works in the algebra throughout. It gives the coordinate formula for $\tilde Q_H$, the quadratic identity of $\tilde Q_H$ and the three cases of the Moore–Penrose inverse that follow from it, the invariants carried by the positive factor, the continuity of $\tilde Q_H$ on the whole algebra, the discontinuity of $\tilde Q_0$ at the null cone and at the origin, and six worked elements.

The article treats the general case, on the whole algebra. It does not treat the refinement of the two factors into a scale, a central phase, a Hermitian boost and a unit real quaternion; nor the exponential polar decomposition; nor the spectral decomposition. Those are separate articles. The two factors here are the general ones, and the finer words are separate objects, not a rewriting of this pair.

Off the cone the pair is the Cartan pair of the group of units: the Hermitian factor is positive definite, the isometry factor is unitary, and both projections are $e_t$. The refinement of that pair into four factors is The Polar Decomposition of an N-Regular Biquaternion, and the closed forms on the two sectors of the cone are The Polar Decomposition of an N-Degenerate Biquaternion.

The conventions and the symbols are those of Conventions in Mathematics; the symbols are collected in §Summary of Notation. The product is the plain product, the conjugation is ${}^{*}=\bar{\cdot}\circ{}^{\natural}$, and the two algebraic norms are those of The Four Forms and the Four Algebraic Norms over the Biquaternion Algebra.

Notation

A biquaternion and its Hermitian conjugate are

$$ \tilde Q=Q_te_t+Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z,\qquad \tilde Q^{*}=\overline{Q_t}e_t-\overline{Q_x}\vec e_x-\overline{Q_y}\vec e_y-\overline{Q_z}\vec e_z . $$

The Hermitian algebraic norm is the non-negative real

$$ \langle\tilde Q\rangle_{*}=\operatorname{Sc}\bigl(\tilde Q\tilde Q^{*}\bigr)=\lvert Q_t\rvert^{2}+\lvert Q_x\rvert^{2}+\lvert Q_y\rvert^{2}+\lvert Q_z\rvert^{2}\ge0, $$

and the quaternionic algebraic norm is the complex number

$$ \langle\tilde Q\rangle_{\natural}=\operatorname{Sc}\bigl(\tilde Q\tilde Q^{\natural}\bigr)=Q_t^{2}+Q_x^{2}+Q_y^{2}+Q_z^{2}. $$

The Hermitian square is

$$ \tilde Q\tilde Q^{*}=\langle\tilde Q\rangle_{*}e_t+i\vec W,\qquad \vec W=W_x\vec e_x+W_y\vec e_y+W_z\vec e_z,\qquad W_x,W_y,W_z\in\mathbb{R}, $$

with the coordinates

$$ W_x=2\bigl(\operatorname{Im}(Q_x\overline{Q_t})-\operatorname{Im}(Q_y\overline{Q_z})\bigr),\quad W_y=2\bigl(\operatorname{Im}(Q_y\overline{Q_t})-\operatorname{Im}(Q_z\overline{Q_x})\bigr),\quad W_z=2\bigl(\operatorname{Im}(Q_z\overline{Q_t})-\operatorname{Im}(Q_x\overline{Q_y})\bigr). $$

The coordinates are real because $\tilde Q\tilde Q^{*}$ is Hermitian, its vector part being then purely imaginary. Put

$$ \boxed{\;a=\sqrt{\frac{\langle\tilde Q\rangle_{*}+\lvert\langle\tilde Q\rangle_{\natural}\rvert}{2}}\ge0\;},\qquad a=0\iff\tilde Q=0 . $$

Two identities are used throughout. The quaternionic norm of the Hermitian square is

$$ \bigl\langle\tilde Q\tilde Q^{*}\bigr\rangle_{\natural}=\langle\tilde Q\rangle_{*}^{2}-\lvert\vec W\rvert^{2}=\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert^{2}, $$

the second equality by multiplicativity, $\langle\tilde Q\tilde Q^{*}\rangle_{\natural}=\langle\tilde Q\rangle_{\natural}\overline{\langle\tilde Q\rangle_{\natural}}$. Hence, for $a>0$,

$$ a^{2}+\frac{\lvert\vec W\rvert^{2}}{4a^{2}}=\langle\tilde Q\rangle_{*},\qquad a^{2}-\frac{\lvert\vec W\rvert^{2}}{4a^{2}}=\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert . $$

Both follow from the first identity and the definition of $a$.

The Objects of the Statement

Hermitian positive semidefinite. An element $\tilde H$ is Hermitian positive semidefinite when $\tilde H=\tilde H^{*}$ and $\tilde H$ lies in the Hermitian cone $P$ of Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint. Equivalently, $\tilde H$ is Hermitian and its two spectral values are non-negative (Spectral Regularity and Degeneracy in the Biquaternion Algebra, §The Spectral Criterion). It is positive definite when in addition it is invertible. The element $0$ is included.

Partial isometry. An element $\tilde Q_0$ is a partial isometry when

$$ \tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde Q_0 . $$

The product $\tilde Q_0\tilde Q_0^{*}$ is then a Hermitian idempotent, the final projection, and $\tilde Q_0^{*}\tilde Q_0$ is a Hermitian idempotent, the initial projection; conversely, if both products are idempotent the identity holds. An orthogonal projection is a Hermitian idempotent $\tilde\Pi$, $\tilde\Pi^{2}=\tilde\Pi=\tilde\Pi^{*}$.

The two ideals of an element. For $\tilde X\in\mathbb{B}$ the set $\tilde X\mathbb{B}=\{\tilde X\tilde Y:\tilde Y\in\mathbb{B}\}$ is a right ideal and $\mathbb{B}\tilde X=\{\tilde Y\tilde X:\tilde Y\in\mathbb{B}\}$ is a left ideal (Ideals and Peirce Decompositions of the Biquaternion Algebra). Each carries an orthogonal projection, which is a Hermitian idempotent generating it:

$$ \tilde X\mathbb{B}=\tilde\Pi_{\mathrm f}(\tilde X)\,\mathbb{B},\qquad \mathbb{B}\tilde X=\mathbb{B}\,\tilde\Pi_{\mathrm i}(\tilde X). $$

The two idempotents are unique, $\tilde\Pi_{\mathrm f}(\tilde X)$ being the projection onto the right ideal and $\tilde\Pi_{\mathrm i}(\tilde X)$ the projection onto the left ideal. For $\tilde X=0$ both are $0$. The pair is the algebraic form of the two subspaces the element generates.

The idempotent of a real direction. For a real unit vector $\vec u$ put

$$ \tilde\Pi_{\vec u}=\tfrac12\bigl(e_t+i\vec u\bigr). $$

It is a Hermitian idempotent, $\tilde\Pi_{\vec u}^{2}=\tilde\Pi_{\vec u}=\tilde\Pi_{\vec u}^{*}$, with $\langle\tilde\Pi_{\vec u}\rangle_{\natural}=0$ and $\langle\tilde\Pi_{\vec u}\rangle_{*}=\tfrac12$ (Square Roots of One and the Idempotents of the Biquaternion Algebra). It is the minimal Hermitian idempotent of the direction $\vec u$.

The Moore–Penrose Inverse

Definition. Let $\tilde A\in\mathbb{B}$. A Moore–Penrose inverse of $\tilde A$ is an element $\tilde A^{+}$ satisfying the four Penrose conditions

$$ \boxed{\; \tilde A\tilde A^{+}\tilde A=\tilde A,\qquad \tilde A^{+}\tilde A\tilde A^{+}=\tilde A^{+},\qquad \bigl(\tilde A\tilde A^{+}\bigr)^{*}=\tilde A\tilde A^{+},\qquad \bigl(\tilde A^{+}\tilde A\bigr)^{*}=\tilde A^{+}\tilde A . \;} $$

The four are equations in the algebra. The conjugation is the biquaternion ${}^{*}$; no matrix, no basis and no $\mathbb{C}^{2}$ appears.

The first two conditions say that $\tilde A^{+}$ inverts $\tilde A$ on the part of the algebra where $\tilde A$ is not zero. The last two say that the two products are Hermitian; with the first two they are Hermitian idempotents, hence the projections onto the two ideals,

$$ \tilde A\tilde A^{+}=\tilde\Pi_{\mathrm f}(\tilde A),\qquad \tilde A^{+}\tilde A=\tilde\Pi_{\mathrm i}(\tilde A). $$

Proposition (uniqueness). An element has at most one Moore–Penrose inverse.

Proof. Let $\tilde P$ and $\tilde P'$ satisfy the four conditions for $\tilde A$. The product $\tilde P\tilde A$ is a Hermitian idempotent by the first and the fourth conditions, and its left ideal is $\mathbb{B}\tilde A$: one inclusion is clear, and $\tilde A=\tilde A(\tilde P\tilde A)$ gives the other. Hence $\tilde P\tilde A=\tilde\Pi_{\mathrm i}(\tilde A)$, and symmetrically $\tilde A\tilde P=\tilde\Pi_{\mathrm f}(\tilde A)$. Using $\tilde A=\tilde A\tilde P'\tilde A$,

$$ \tilde P=\tilde P\tilde A\tilde P =\bigl(\tilde P\tilde A\bigr)\tilde P'\bigl(\tilde A\tilde P\bigr) =\tilde\Pi_{\mathrm i}(\tilde A)\,\tilde P'\,\tilde\Pi_{\mathrm f}(\tilde A). $$

With the roles exchanged, $\tilde P'=\tilde\Pi_{\mathrm i}(\tilde A)\tilde P\tilde\Pi_{\mathrm f}(\tilde A)$; substituting this into the first display, and using $\tilde\Pi_{\mathrm i}^{2}=\tilde\Pi_{\mathrm i}$ and $\tilde\Pi_{\mathrm f}^{2}=\tilde\Pi_{\mathrm f}$, gives $\tilde P=\tilde P'$. $\square$

Theorem (existence on the cone). Every Hermitian positive semidefinite element has a Moore–Penrose inverse. For $\tilde Q_H$ it is the element of §The Moore–Penrose Inverse of the Hermitian Factor, given there in closed form.

Proof. The three cases of that section check the four conditions by direct multiplication. $\square$

Remark (general existence). Every element of $\mathbb{B}$ has a Moore–Penrose inverse, but the fact is not used here: the polar decomposition needs only $\tilde Q_H^{+}$, and that is computed directly. For a general $\tilde A$ the computation also requires the spectral data of $\tilde A\tilde A^{*}$; the general theorem is The Moore–Penrose Inverse in an Involutive Algebra.

Remark (why the cone is short). On the cone the quadratic identity fixes the element on the ideal it generates and gives zero on its annihilator, so the four conditions follow by multiplication. That is the whole reason the article needs no matrix model.

The Hermitian Factor

Theorem (the Hermitian factor). For every $\tilde Q\in\mathbb{B}$, the element

$$ \boxed{\;\tilde Q_H=a\,e_t+\frac{i}{2a}\,\vec W\quad(a>0),\qquad \tilde Q_H=0\quad(a=0)\;} $$

is Hermitian positive semidefinite, satisfies $\tilde Q_H^{2}=\tilde Q\tilde Q^{*}$, and is the unique such element.

Proof. For $a>0$ the element is Hermitian: $a$ is real and $i\vec W$ is Hermitian, since $\vec W$ is real and pure. Its square is

$$ \tilde Q_H^{2}=a^{2}e_t+2a\,\frac{i}{2a}\,\vec W+\Bigl(\frac{i}{2a}\Bigr)^{2}\vec W^{2} =\Bigl(a^{2}+\frac{\lvert\vec W\rvert^{2}}{4a^{2}}\Bigr)e_t+i\vec W =\langle\tilde Q\rangle_{*}e_t+i\vec W=\tilde Q\tilde Q^{*}, $$

using $\vec W^{2}=-\lvert\vec W\rvert^{2}e_t$ and the first of the two identities of §Notation.

The spectral values of $\tilde Q_H$ are $a\pm\lvert\vec W\rvert/(2a)$. Both are non-negative, because the second identity of §Notation gives

$$ a^{2}-\frac{\lvert\vec W\rvert^{2}}{4a^{2}}=\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert\ge0, $$

so $\lvert\vec W\rvert/(2a)\le a$. Hence $\tilde Q_H$ lies in the Hermitian cone. For $a=0$ the element $\tilde Q$ is $0$ and $\tilde Q_H=0$.

The square root in the cone is unique: an element of the cone has exactly one square root in the cone (Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint, §The Polar Decomposition). Since $\tilde Q\tilde Q^{*}$ is in the cone and $\tilde Q_H$ is a square root of it in the cone, $\tilde Q_H$ is that root. $\square$

Remark (the scalar part). The scalar part of $\tilde Q_H$ is $a$, and the vector part is $i\vec W/(2a)$, of length $\lvert\vec W\rvert/(2a)\le a$. The element is the sum of the central element $a e_t$ and the vector part $i\vec W/(2a)$, whose length is at most $a$. When $\vec W=0$ the factor is the central element $a e_t$.

The Moore–Penrose Inverse of the Hermitian Factor

The Moore–Penrose inverse of $\tilde Q_H$ is computed from a quadratic identity.

Proposition (the quadratic identity). The Hermitian factor satisfies

$$ \boxed{\;\tilde Q_H^{2}-2a\,\tilde Q_H+\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert\,e_t=0\;}. $$

Proof. With $\tilde Q_H=a e_t+i\vec W/(2a)$ one has $\tilde Q_H^{2}=\langle\tilde Q\rangle_{*}e_t+i\vec W$ and $2a\tilde Q_H=2a^{2}e_t+i\vec W$. The difference is $\bigl(\langle\tilde Q\rangle_{*}-2a^{2}+\lvert\langle\tilde Q\rangle_{\natural}\rvert\bigr)e_t$, which vanishes by the definition of $a^{2}$. $\square$

Theorem (the Moore–Penrose inverse of the Hermitian factor). The Moore–Penrose inverse of $\tilde Q_H$ is

$$ \boxed{\; \tilde Q_H^{+}= \begin{cases} \dfrac{2a\,e_t-\tilde Q_H}{\lvert\langle\tilde Q\rangle_{\natural}\rvert}, & \langle\tilde Q\rangle_{\natural}\neq0,\\[8pt] \dfrac{\tilde Q_H}{4a^{2}}, & \langle\tilde Q\rangle_{\natural}=0,\ \tilde Q\neq0,\\[8pt] 0, & \tilde Q=0. \end{cases} \;} $$

Proof. If $\langle\tilde Q\rangle_{\natural}\neq0$ the two spectral values of $\tilde Q_H$ have product $\lvert\langle\tilde Q\rangle_{\natural}\rvert\neq0$, so $\tilde Q_H$ is invertible, and the quadratic identity read as $\tilde Q_H\bigl(2a e_t-\tilde Q_H\bigr)=\lvert\langle\tilde Q\rangle_{\natural}\rvert e_t$ gives the inverse. For an invertible element the Moore–Penrose inverse is the inverse.

If $\langle\tilde Q\rangle_{\natural}=0$ and $\tilde Q\neq0$, the quadratic identity reads $\tilde Q_H^{2}=2a\tilde Q_H$, so the spectral values of $\tilde Q_H$ are $0$ and $2a$, and $a>0$. Put $\tilde X=\tilde Q_H/(4a^{2})$, a Hermitian element. Then

$$ \tilde Q_H\tilde X\tilde Q_H=\frac{\tilde Q_H^{3}}{4a^{2}}=\frac{2a\,\tilde Q_H^{2}}{4a^{2}}=\tilde Q_H, \qquad \tilde X\tilde Q_H\tilde X=\frac{\tilde Q_H^{3}}{16a^{4}}=\frac{\tilde Q_H}{4a^{2}}=\tilde X, $$

and the two products $\tilde Q_H\tilde X=\tilde Q_H^{2}/(4a^{2})=\tilde Q_H/(2a)$ and $\tilde X\tilde Q_H=\tilde Q_H/(2a)$ are Hermitian. The four Penrose equations hold, and the Moore–Penrose inverse is unique (§The Moore–Penrose Inverse).

If $\tilde Q=0$ then $\tilde Q_H=0$ and its Moore–Penrose inverse is $0$. $\square$

Remark (the value at the cone). On $\langle\tilde Q\rangle_{\natural}=0$ the inverse is $\tilde Q_H/(4a^{2})=\tilde\Pi_{\mathrm f}(\tilde Q)/(2a)$: the factor is a scaled projection, and its inverse is the same projection with the reciprocal scale. The two formulas do not meet at the cone; the limit of the first is infinite, and the second is the value.

The Isometry Factor

Lemma (the ideal of the positive factor). $\tilde Q_H\mathbb{B}=\tilde Q\mathbb{B}$.

Proof. First $\tilde Q\mathbb{B}=\tilde Q\tilde Q^{*}\mathbb{B}$: the inclusion $\supseteq$ is clear, and

$$ \tilde Q=\tilde Q\bigl(\tilde Q^{+}\tilde Q\bigr)=\tilde Q\bigl(\tilde Q^{+}\tilde Q\bigr)^{*}=\tilde Q\tilde Q^{*}\tilde Q^{+*}\in\tilde Q\tilde Q^{*}\mathbb{B} $$

gives the other. Then $\tilde Q_H\mathbb{B}=\tilde Q_H^{2}\mathbb{B}$, since $\tilde Q_H^{2}\mathbb{B}\subseteq\tilde Q_H\mathbb{B}$ and $\tilde Q_H=\tilde Q_H^{2}\tilde Q_H^{+}\in\tilde Q_H^{2}\mathbb{B}$, the last because $\tilde Q_H\tilde Q_H^{+}$ is the projection onto $\tilde Q_H\mathbb{B}$. Since $\tilde Q_H^{2}=\tilde Q\tilde Q^{*}$, the three ideals coincide. $\square$

Theorem (the isometry factor). Put $\tilde Q_0=\tilde Q_H^{+}\tilde Q$. Then

$$ \boxed{\;\tilde Q=\tilde Q_H\tilde Q_0\;},\qquad \tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde Q_0,\qquad \tilde Q_0\tilde Q_0^{*}=\tilde\Pi_{\mathrm f}(\tilde Q),\qquad \tilde Q_0^{*}\tilde Q_0=\tilde\Pi_{\mathrm i}(\tilde Q). $$

Proof. The first equation is $\tilde Q_H\tilde Q_H^{+}\tilde Q=\tilde Q_H\tilde Q_H^{+}\tilde Q_H\tilde Q_0=\tilde Q_H\tilde Q_0=\tilde Q$, using the Penrose equation $\tilde Q_H\tilde Q_H^{+}\tilde Q_H=\tilde Q_H$.

For the final projection, $\tilde Q_0\mathbb{B}=\tilde Q_H^{+}\tilde Q\mathbb{B}$. Since $\tilde Q_H$ is Hermitian, $\tilde Q_H^{+}\tilde Q_H=\tilde Q_H\tilde Q_H^{+}=\tilde\Pi$; the element $\tilde\Pi$ is a Hermitian idempotent, $\tilde\Pi\tilde Q_H=\tilde Q_H$ by the Penrose equation, and $\tilde\Pi=\tilde Q_H\tilde Q_H^{+}\in\tilde Q_H\mathbb{B}$, so $\tilde\Pi$ is the projection onto $\tilde Q_H\mathbb{B}$. Hence $\tilde Q_0\mathbb{B}\subseteq\tilde\Pi\mathbb{B}$.

Conversely $\tilde Q=\tilde Q_H\tilde Q_0\in\tilde Q_0\mathbb{B}$, so $\tilde\Pi\mathbb{B}=\tilde Q_H\mathbb{B}=\tilde Q\mathbb{B}\subseteq\tilde Q_0\mathbb{B}$ by the lemma. The two right ideals are equal, and the projection onto a right ideal is unique, so $\tilde Q_0\tilde Q_0^{*}=\tilde\Pi=\tilde\Pi_{\mathrm f}(\tilde Q)$.

The idempotence of $\tilde Q_0\tilde Q_0^{*}$ makes $\tilde Q_0$ a partial isometry: from $(\tilde Q_0\tilde Q_0^{*})^{2}=\tilde Q_0\tilde Q_0^{*}$ one gets $(\tilde Q_0^{*}\tilde Q_0)^{3}=(\tilde Q_0^{*}\tilde Q_0)^{2}$, so $\tilde Q_0^{*}\tilde Q_0$, which is Hermitian, is idempotent, and $\tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde Q_0$. Then $\tilde Q_0^{*}\tilde Q_0$ is the projection onto the left ideal $\mathbb{B}\tilde Q_0$. The inclusion $\mathbb{B}\tilde Q_0\subseteq\mathbb{B}\tilde Q$ follows from $\tilde Q_0=\tilde Q_H^{+}\tilde Q$, and $\mathbb{B}\tilde Q\subseteq\mathbb{B}\tilde Q_0$ from $\tilde Q=\tilde Q_H\tilde Q_0$; the left ideals are equal, so $\tilde Q_0^{*}\tilde Q_0=\tilde\Pi_{\mathrm i}(\tilde Q)$.

Finally $\tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde\Pi_{\mathrm f}(\tilde Q)\tilde Q_0=\tilde Q_0$, because $\tilde Q_0\in\tilde\Pi_{\mathrm f}(\tilde Q)\mathbb{B}$ and $\tilde\Pi_{\mathrm f}(\tilde Q)$ is the projection onto that ideal. $\square$

Remark (the three cases of the formula). The factor $\tilde Q_0$ inherits the three cases of $\tilde Q_H^{+}$:

$$ \tilde Q_0=\frac{2a\,\tilde Q-\tilde Q_H\tilde Q}{\lvert\langle\tilde Q\rangle_{\natural}\rvert}\quad(\langle\tilde Q\rangle_{\natural}\neq0),\qquad \tilde Q_0=\frac{\tilde Q_H\tilde Q}{4a^{2}}\quad(\langle\tilde Q\rangle_{\natural}=0,\ \tilde Q\neq0),\qquad \tilde Q_0=0\quad(\tilde Q=0). $$

The formulas are the theorem with the three values of $\tilde Q_H^{+}$ substituted. No case is excluded from the decomposition; the cases are those of the closed form alone.

Remark (the cone). On the null cone the identities of §Notation give $a^{2}=\tfrac12\langle\tilde Q\rangle_{*}$ and $\lvert\vec W\rvert=2a^{2}$, so

$$ \tilde Q_H=2a\,\tilde\Pi_{\vec u},\qquad \tilde Q_0=\frac{\tilde Q}{2a},\qquad \vec u=\frac{\vec W}{\lvert\vec W\rvert}, $$

for $\tilde Q\neq0$, and the scale $2a=\sqrt{2\langle\tilde Q\rangle_{*}}$ is real and positive. The Hermitian factor is a scaled projection and the isometry factor is the element divided by the scale $2a$. At the origin both are $0$.

Uniqueness

Theorem (characterisation). The pair $(\tilde Q_H,\tilde Q_0)$ is the unique pair satisfying

1. $\tilde Q_H$ is Hermitian positive semidefinite;
2. $\tilde Q=\tilde Q_H\tilde Q_0$;
3. $\tilde Q_0\tilde Q_0^{*}=\tilde\Pi_{\mathrm f}(\tilde Q)$;
4. $\tilde Q_H\mathbb{B}=\tilde Q\mathbb{B}$.

Proof. The canonical pair satisfies the four conditions by the two theorems above.

For uniqueness, let $(\tilde H,\tilde U)$ satisfy them. From $\tilde Q=\tilde H\tilde U$ and condition 3,

$$ \tilde Q\tilde Q^{*}=\tilde H\tilde U\tilde U^{*}\tilde H=\tilde H\,\tilde\Pi_{\mathrm f}(\tilde Q)\,\tilde H . $$

Condition 4 gives $\tilde H\in\tilde H\mathbb{B}=\tilde Q\mathbb{B}=\tilde\Pi_{\mathrm f}(\tilde Q)\mathbb{B}$, so $\tilde\Pi_{\mathrm f}(\tilde Q)\tilde H=\tilde H$, and $\tilde Q\tilde Q^{*}=\tilde H^{2}$. The element $\tilde H$ is therefore a Hermitian positive semidefinite square root of $\tilde Q\tilde Q^{*}$, and the square root in the cone is unique, so $\tilde H=\tilde Q_H$.

Then $\tilde Q_H^{+}\tilde Q_H=\tilde\Pi_{\mathrm f}(\tilde Q)$ by condition 4, and

$$ \tilde Q_H^{+}\tilde Q=\tilde Q_H^{+}\tilde Q_H\tilde U=\tilde\Pi_{\mathrm f}(\tilde Q)\,\tilde U=\tilde U\tilde U^{*}\tilde U=\tilde U, $$

the middle equality by condition 3 and the last by the partial-isometry identity, which condition 3 implies. Hence $\tilde U=\tilde Q_H^{+}\tilde Q=\tilde Q_0$. $\square$

Remark (the two redundant conditions). The natural statement has six conditions; two are consequences.

• Condition 3 implies that $\tilde Q_0$ is a partial isometry. If $\tilde Q_0\tilde Q_0^{*}=\tilde\Pi$ is idempotent then $(\tilde Q_0^{*}\tilde Q_0)^{3}=(\tilde Q_0^{*}\tilde Q_0)^{2}$, so $\tilde Q_0^{*}\tilde Q_0$, which is Hermitian, has spectral values in $\{0,1\}$ and is idempotent; hence $\tilde Q_0\tilde Q_0^{*}\tilde Q_0=\tilde Q_0$.
• The initial projection is a consequence of the other four. The uniqueness proof identifies every pair satisfying conditions 1–4 with the canonical pair; for that pair $\tilde Q_0=\tilde Q_H^{+}\tilde Q$ gives $\mathbb{B}\tilde Q_0\subseteq\mathbb{B}\tilde Q$, and $\tilde Q=\tilde Q_H\tilde Q_0$ gives $\mathbb{B}\tilde Q\subseteq\mathbb{B}\tilde Q_0$. The two left ideals are equal, so the initial projection is $\tilde\Pi_{\mathrm i}(\tilde Q)$.

Conditions 3 and 4 are therefore independent of the others, and the two omitted conditions are not independent data.

Two of the conditions cannot be dropped. Two pairs, both built on the same null element, show that conditions 3 and 4 are each necessary.

Counterexample 1 (condition 4). Let $\tilde Q=\tilde\Pi_{\vec u}$ with $\vec u$ a real unit vector, and put

$$ \tilde H=e_t,\qquad \tilde U=\tilde\Pi_{\vec u}. $$

Then $\tilde Q=e_t\tilde\Pi_{\vec u}$, the element $\tilde H$ is positive definite and $\tilde U$ is a partial isometry with $\tilde U\tilde U^{*}=\tilde\Pi_{\vec u}=\tilde\Pi_{\mathrm f}(\tilde Q)$. But $\tilde H\mathbb{B}=\mathbb{B}$ while $\tilde Q\mathbb{B}=\tilde\Pi_{\vec u}\mathbb{B}$, so condition 4 fails, and $\tilde H$ is not $\tilde Q_H=\tilde\Pi_{\vec u}$.

Counterexample 2 (condition 3). With the same $\tilde Q=\tilde\Pi_{\vec u}$, put

$$ \tilde H=\tilde\Pi_{\vec u},\qquad \tilde U=e_t . $$

Then $\tilde Q=\tilde\Pi_{\vec u}e_t$, the element $\tilde H$ is positive semidefinite, $\tilde H\mathbb{B}=\tilde Q\mathbb{B}$, and $\tilde U$ is a partial isometry. But $\tilde U\tilde U^{*}=e_t\neq\tilde\Pi_{\vec u}=\tilde\Pi_{\mathrm f}(\tilde Q)$, so condition 3 fails. The annihilator condition fails as well: the left annihilator of $e_t$ is $0$, while that of $\tilde\Pi_{\vec u}$ is the orthogonal complement of the line of $\tilde\Pi_{\vec u}$.

The two examples are the two halves of the statement: the first fixes the positive factor by its ideal, the second fixes the isometry by its projection.

The Invariants

Theorem (the invariants of the positive factor). The Hermitian factor carries both algebraic norms:

$$ \boxed{\;\langle\tilde Q_H\rangle_{*}=\langle\tilde Q\rangle_{*},\qquad \bigl\langle\tilde Q_H\bigr\rangle_{\natural}=\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert\;}. $$

Proof. Since $\tilde Q_H^{2}=\tilde Q\tilde Q^{*}$ and $\tilde Q_H=\tilde Q_H^{*}$,

$$ \langle\tilde Q_H\rangle_{*}=\operatorname{Sc}\bigl(\tilde Q_H\tilde Q_H^{*}\bigr)=\operatorname{Sc}\bigl(\tilde Q_H^{2}\bigr)=\operatorname{Sc}\bigl(\tilde Q\tilde Q^{*}\bigr)=\langle\tilde Q\rangle_{*} . $$

For the quaternionic norm, multiplicativity gives

$$ \bigl\langle\tilde Q_H\bigr\rangle_{\natural}^{2}=\bigl\langle\tilde Q_H^{2}\bigr\rangle_{\natural}=\bigl\langle\tilde Q\tilde Q^{*}\bigr\rangle_{\natural}=\langle\tilde Q\rangle_{\natural}\overline{\langle\tilde Q\rangle_{\natural}}=\bigl\lvert\langle\tilde Q\rangle_{\natural}\bigr\rvert^{2}. $$

In the closed form of §The Hermitian Factor, for $a>0$, $\langle\tilde Q_H\rangle_{\natural}=a^{2}-\lvert\vec W\rvert^{2}/(4a^{2})=\lvert\langle\tilde Q\rangle_{\natural}\rvert\ge0$, so the square root is the non-negative one; for $\tilde Q=0$ both norms vanish. $\square$

Remark (what each norm sees). The Hermitian norm passes to the positive factor unchanged; the quaternionic norm passes to it in modulus, and is real there. Neither norm determines the isometry factor: that factor is fixed by the two ideals.

Continuity

Theorem (the positive factor is continuous). The map $\tilde Q\mapsto\tilde Q_H$ is continuous on the whole algebra.

Proof. The map $\tilde Q\mapsto\tilde Q\tilde Q^{*}$ is polynomial, hence continuous, and so are its scalar part $\langle\tilde Q\rangle_{*}$ and its vector part $\vec W$. The modulus $\lvert\langle\tilde Q\rangle_{\natural}\rvert$ is continuous, so $a$ is continuous. For $a>0$ the factor is $a e_t+i\vec W/(2a)$, continuous in $a$ and $\vec W$. At $a=0$ one has $\tilde Q=0$ and $\tilde Q_H=0$, and the vector part of the factor has length $\lvert\vec W\rvert/(2a)\le a$, which tends to $0$ with the element. $\square$

Theorem (the isometry factor is discontinuous). The map $\tilde Q\mapsto\tilde Q_0$ is not continuous at the null cone and not continuous at the origin.

Proof. Let $\vec u$ be a real unit vector and, for $\epsilon>0$, put

$$ \tilde Q_\epsilon=\sqrt{\frac{1+\epsilon^{2}}{2}}\,e_t+i\sqrt{\frac{1-\epsilon^{2}}{2}}\,\vec u . $$

Write $\tilde Q_\epsilon=c\,e_t+i s\,\vec u$ with $c^{2}+s^{2}=1$ and $c^{2}-s^{2}=\epsilon^{2}$. The element is Hermitian, so $\tilde Q_\epsilon\tilde Q_\epsilon^{*}=\tilde Q_\epsilon^{2}=e_t+i\sqrt{1-\epsilon^{4}}\,\vec u$; hence $\langle\tilde Q_\epsilon\rangle_{*}=1$, $\langle\tilde Q_\epsilon\rangle_{\natural}=\epsilon^{2}$, $\vec W=\sqrt{1-\epsilon^{4}}\,\vec u$ and $a=c$. Then

$$ \tilde Q_H=c\,e_t+\frac{i}{2c}\sqrt{1-\epsilon^{4}}\,\vec u=c\,e_t+i s\,\vec u=\tilde Q_\epsilon, \qquad \tilde Q_0=e_t . $$

At $\epsilon=0$ the family reaches $\tilde Q=\tfrac{1}{\sqrt2}(e_t+i\vec u)$, with $\langle\tilde Q\rangle_{*}=1$, $\langle\tilde Q\rangle_{\natural}=0$ and $a=\tfrac{1}{\sqrt2}$, so $\tilde Q_H=\tilde Q$. Then

$$ \tilde Q_0=\tilde Q_H^{+}\tilde Q=\frac{\tilde Q^{2}}{2}=\tfrac12\bigl(e_t+i\vec u\bigr)=\tilde\Pi_{\vec u}. $$

The factor has the limit $e_t$ along the family and the value $\tilde\Pi_{\vec u}$ at the cone. The two differ, so the map is discontinuous there.

At the origin, the family $\tilde Q=\epsilon e_t$ with $\epsilon>0$ has $\tilde Q_H=\epsilon e_t$ and $\tilde Q_0=e_t$, while $\tilde Q_0=0$ at $\epsilon=0$. The factor jumps from $e_t$ to $0$. $\square$

Remark (the jump is the degeneration). The positive factor is the part of the polar data that survives both degenerations; the isometry factor records the collapse. In the first family the positive factor has the continuous limit $\tfrac{1}{\sqrt2}(e_t+i\vec u)$, while the isometry factor drops from the identity to the projection onto the line of $\vec u$. In the second the positive factor falls to $0$ and the isometry with it. No choice of the Moore–Penrose inverse removes the jump: the value at the cone is forced by the ideal and projection conditions of §Uniqueness, and those conditions are what make the pair unique.

Examples

Example 1. $\tilde Q=e_t+\vec e_x$. Then $\langle\tilde Q\rangle_{*}=2$, $\langle\tilde Q\rangle_{\natural}=2$, $\vec W=0$, $a=\sqrt2$, and

$$ \tilde Q_H=\sqrt2\,e_t,\qquad \tilde Q_H^{+}=\frac{\sqrt2}{2}\,e_t,\qquad \tilde Q_0=\frac{e_t+\vec e_x}{\sqrt2},\qquad e_t+\vec e_x=\sqrt2\,e_t\cdot\frac{e_t+\vec e_x}{\sqrt2}. $$

Example 2. $\tilde Q=e_t+\vec e_x+i\vec e_y$. Then $\langle\tilde Q\rangle_{*}=3$, $\langle\tilde Q\rangle_{\natural}=1$, $\vec W=2\vec e_y+2\vec e_z$, $a=\sqrt2$, and

$$ \tilde Q_H=\sqrt2\,e_t+\frac{i}{\sqrt2}\bigl(\vec e_y+\vec e_z\bigr),\qquad \tilde Q_H^{+}=\sqrt2\,e_t-\frac{i}{\sqrt2}\bigl(\vec e_y+\vec e_z\bigr),\qquad \tilde Q_0=\frac{e_t+\vec e_x}{\sqrt2}, $$

$$ e_t+\vec e_x+i\vec e_y=\Bigl[\sqrt2\,e_t+\frac{i}{\sqrt2}\bigl(\vec e_y+\vec e_z\bigr)\Bigr]\frac{e_t+\vec e_x}{\sqrt2}. $$

Example 3. $\tilde Q=e_t+i\vec e_x$. Then $\langle\tilde Q\rangle_{*}=2$, $\langle\tilde Q\rangle_{\natural}=0$, $\vec W=2\vec e_x$, $a=1$, and

$$ \tilde Q_H=e_t+i\vec e_x,\qquad \tilde Q_H^{+}=\frac{e_t+i\vec e_x}{4},\qquad \tilde Q_0=\frac{e_t+i\vec e_x}{2},\qquad e_t+i\vec e_x=\bigl(e_t+i\vec e_x\bigr)\frac{e_t+i\vec e_x}{2}. $$

Example 4. $\tilde Q=\vec e_x+i\vec e_y$. Then $\langle\tilde Q\rangle_{*}=2$, $\langle\tilde Q\rangle_{\natural}=0$, $\vec W=2\vec e_z$, $a=1$, and

$$ \tilde Q_H=e_t+i\vec e_z,\qquad \tilde Q_H^{+}=\frac{e_t+i\vec e_z}{4},\qquad \tilde Q_0=\frac{\vec e_x+i\vec e_y}{2},\qquad \vec e_x+i\vec e_y=\bigl(e_t+i\vec e_z\bigr)\frac{\vec e_x+i\vec e_y}{2}. $$

The element is not Hermitian and the cone case is not the idempotent case; the isometry factor is half the element, not a projection.

Example 5. $\tilde Q=(1+i)e_t$. Then $\langle\tilde Q\rangle_{*}=2$, $\langle\tilde Q\rangle_{\natural}=2i$, $\lvert\langle\tilde Q\rangle_{\natural}\rvert=2$, $\vec W=0$, $a=\sqrt2$, and

$$ \tilde Q_H=\sqrt2\,e_t,\qquad \tilde Q_H^{+}=\frac{\sqrt2}{2}\,e_t,\qquad \tilde Q_0=\frac{(1+i)e_t}{\sqrt2},\qquad (1+i)e_t=\sqrt2\,e_t\cdot\frac{(1+i)e_t}{\sqrt2}. $$

Example 6. $\tilde Q=0$. Then $a=0$ and $\tilde Q_H=\tilde Q_H^{+}=\tilde Q_0=0$, with $0=0\cdot0$.

Summary

The polar decomposition of the biquaternion algebra is the pair

The decomposition is characterised by four conditions: $\tilde Q_H$ in the cone, $\tilde Q=\tilde Q_H\tilde Q_0$, $\tilde Q_0\tilde Q_0^{*}=\tilde\Pi_{\mathrm f}(\tilde Q)$, and $\tilde Q_H\mathbb{B}=\tilde Q\mathbb{B}$. The two further conditions of the introduction follow from these.

What the Reader Should Retain

The three statements to carry away are these. The formulas $\tilde Q_H=(\tilde Q\tilde Q^{*})^{1/2}$ and $\tilde Q_0=\tilde Q_H^{+}\tilde Q$ hold for every element, the positive factor being the closed element $a e_t+i\vec W/(2a)$ and the isometry factor being $\tilde Q_H^{+}\tilde Q$. The pair is unique once the two ideals of $\tilde Q$ are imposed, and not before. The positive factor is continuous, the isometry factor is not, and the two jumps are the null cone and the origin.

Summary of Notation

Further Reading

• Positivity and the Hermitian Cone of the Biquaternion Algebra with Hermitian Adjoint, for the Hermitian cone, the positive square root and the dagger polar decomposition.
• N-Regularity and Degeneracy in the Biquaternion Algebra, for the zero-divisor cone and the group of units.
• Spectral Regularity and Degeneracy in the Biquaternion Algebra, for the spectral values of a Hermitian element and the characteristic polynomial.
• Ideals and Peirce Decompositions of the Biquaternion Algebra, for the left and right ideals generated by an element.
• Square Roots of One and the Idempotents of the Biquaternion Algebra, for the Hermitian idempotents $\tfrac12(e_t+i\vec u)$.
• The Cartan Hermitian Pairing of the Polar Decomposition of Biquaternions, for the pairing that this article extends and its restriction off the cone.
• The Unitary Group of the Biquaternion Algebra, for the unitary factor off the cone.
• Exponential Polar Decomposition of Biquaternions, for the boundary limits of the four-factor word at the cone.
• The Polar Decomposition of an Operator of the Form, for the operator polar decomposition that this article is not.
• The Zero-Divisor Cone and the Massless Spectrum and Roots, Idempotents and Observables, for the physical reading of the cone as the massless boundary, which the rank-one case of this article supplies.
• Conventions in Mathematics, for the notation of the corpus.