Hermitian Forms over the Biquaternion Algebra and the Unitary Witt Group with Hermitian Adjoint

Introduction

Let $\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}\cong M_{2}(\mathbb{C})$ be the biquaternion algebra with the Hermitian conjugation ${}^{*}$, fixed space $\mathbb{M}_+$ and anti-fixed space $\mathbb{M}_-$. This article is the biquaternion instance of Hermitian Forms over an Involution Ring and the Unitary Witt Group with Hermitian Adjoint, and the companion of The Hermitian Sandwich in the Biquaternion Algebra with Hermitian Adjoint: that article studies the operator $\tilde R\mapsto \tilde R\tilde P\tilde{R}^{*}$ on the algebra, this one studies the forms on which the congruence $H\mapsto S^{\dagger}HS$ acts, that is, the objects the two-sided operators are made to transform.

The results are the classical ones of a positive involution, made explicit in the matrix model: a Hermitian form on the algebra is a Hermitian matrix, congruence is the dagger congruence, the invariant is the inertia (Sylvester's law), every form is congruent to its normal form $\mathrm{diag}(1_{p},-1_{q},0_{r})$, the isometry group of the unit form is the unitary slice $U=U(2)$, and the forms modulo the hyperbolic ones form the unitary Witt group $W\cong\mathbb{Z}$, the invariant being the signature. Four facts of the algebra enter and none is special: the algebra is simple, the involution is positive, the centre is $\mathbb{C}$, and the rank-one module over the algebra is $\mathbb{C}^{2}$ in the matrix model.

Hermitian Forms on the Algebra

Definition. Let $M$ be a right $\mathbb{B}$-module and let $h:M\times M\to\mathbb{B}$ be sesquilinear, $h(\tilde R\lambda+\tilde P\mu,\tilde Q)=\bar{\lambda}h(\tilde R,\tilde Q)+\bar{\mu}h(\tilde P,\tilde Q)$ and $h(\tilde R,\tilde P\lambda)=h(\tilde R,\tilde P)\lambda$. The form is Hermitian if in addition

$$ h(\tilde P,\tilde R) = h(\tilde R,\tilde P)^{\dagger}. $$

The form is non-degenerate if $h(\tilde R,\tilde P)=0$ for all $\tilde P$ implies $\tilde R=0$, and positive definite if $h(\tilde R,\tilde R)\in\mathbb{M}_+$ for $\tilde R\neq0$, that is, if its values are positive in the sense of the Hermitian cone (Positivity and the Hermitian Cone of a Clifford Algebra with Hermitian Adjoint).

Proposition (the forms of the rank-one module). Let $M=\mathbb{B}$ with the right action of the algebra on itself. Every Hermitian form on $M$ is

$$ h_{H}(\tilde R,\tilde P) = \tilde{R}^{*}H\,\tilde P\qquad\text{with } H\in\mathbb{M}_+, $$

and the map $H\mapsto h_{H}$ is an isomorphism of the Hermitian elements onto the Hermitian forms. In the matrix model $\Phi(H)$ is a Hermitian $2\times2$ matrix and $h_{H}(\tilde R,\tilde P)=\Phi(\tilde R)^{\dagger}\Phi(H)\Phi(\tilde P)$.

Proof. Sesquilinearity and the Hermitian symmetry give $h(\tilde R,\tilde P)=\tilde{R}^{*}h(e_{0},\tilde P)=\tilde{R}^{*}h(e_{0},e_{0})\tilde P$, with $H=h(e_{0},e_{0})\in\mathbb{M}_+$ by the symmetry, and conversely every such $h_{H}$ is a Hermitian form.

Example (the unit form). $H=e_{0}$ gives $h_{e_{0}}(\tilde R,\tilde P)=\tilde{R}^{*}\tilde P$, of scalar part $\mathrm{Sc}(\tilde{R}^{*}\tilde P)=\sum_{\mu}R_{\mu}^{*}P_{\mu}$: the unit form, positive definite, the form of the Hermitian structure of the algebra and the trace form of the dagger. Its matrix in the basis $e_{\mu}$ is the identity.

Example (the norm form, and why it is not here). The quaternion norm $N(\tilde R)=\sum_{\mu}R_{\mu}^{2}$ is a quadratic form of the algebra but it is not Hermitian for the dagger: $N$ is complex-valued, indefinite and isotropic on the null cone. It is a form of the complex bilinear type, not a form of the dagger, and the two must not be placed in the same classification (Biquaternion Norm and Invertibility).

Congruence and the Isometry Group

Definition. Two Hermitian forms $h_{H},h_{H'}$ on the rank-one module are congruent if there is an invertible $S\in\mathbb{B}^{\times}$ with

$$ H' = S^{\dagger}H\,S . $$

Congruence is an equivalence relation, and the map $H\mapsto S^{\dagger}HS$ is precisely the two-sided operator $\Theta_{S}$ of Two-Sided Operators on the Biquaternion Algebra with Hermitian Adjoint applied to the form matrix. The invariants of congruence are the invariants of the two-sided operators seen on the Hermitian elements.

Definition. The isometry group of $h_{H}$ is $U(H)=\{\,S\in\mathbb{B}^{\times} : S^{\dagger}HS=H\,\}$. For the unit form,

$$ U(e_{0}) = \{\,S : S^{\dagger}S=e_{0}\,\} = U , $$

the unitary slice, of real dimension four, with determinant-one part $\mathrm{SU}(2)$ (The Unitary Slice and the Compact Real Form with Hermitian Adjoint).

Proposition (the slice acts by isometries and by congruence-preserving maps). For $S\in U$ the operator $\Theta_{S}$ is unitary, an algebra automorphism and, on the Hermitian elements, a congruence that preserves the unit form; more generally $U(H)$ is a subgroup of $\mathbb{B}^{\times}$ conjugate to $U(e_{0})$ by any $S$ with $S^{\dagger}H S$ the normal form of $H$.

Proof. The automorphism and unitarity statements are the theorems of the two-sided operators; for the last, if $T^{\dagger}HT$ is the normal form of $H$ then $U(H)=T^{-1}U(e_{0})T$ up to the null directions, since $S^{\dagger}HS=H$ is equivalent to $(TST^{-1})^{\dagger}(T^{\dagger}HT)(TST^{-1})=T^{\dagger}HT$.

Remark. The three groups must not be confused: $U(e_{0})=U(2)$ is the isometry group of the unit form; the isometry group of the norm form is the Lorentz-type group $\{|N|=1\}/\pm1=SO^{+}(1,3)$ up to phase; and the two are different because the two forms are different. This is the same contrast as in Biquaternion Versors and the Orthogonal Group.

Sylvester's Law of Inertia

Definition. For $H\in\mathbb{M}_+$ the inertia is the triple $(p,q,r)$ of the numbers of positive, negative and zero eigenvalues of the matrix $\Phi(H)$, that is of the positive, negative and null dimensions of the form $h_{H}$.

Theorem (Sylvester's law of inertia). Congruence preserves the inertia:

$$ H' = S^{\dagger}HS,\ S\in\mathbb{B}^{\times}\quad\Longrightarrow\quad \mathrm{inertia}(H')=\mathrm{inertia}(H). $$

Moreover every Hermitian $H$ is congruent to the normal form

$$ H \sim \mathrm{diag}\bigl(1_{p},\,-1_{q},\,0_{r}\bigr),\qquad (p,q,r)=\mathrm{inertia}(H). $$

Proof. In the matrix model, $H$ is Hermitian, so there is a unitary $V$ with $V^{\dagger}HV=\mathrm{diag}(\lambda_{1},\lambda_{2})$, $\lambda_{i}\in\mathbb{R}$; the scalings $1/\sqrt{|\lambda_{i}|}$ on the nonzero directions give the normal form, and $S^{\dagger}HS$ is Hermitian with the same signature by the inertia theorem for Hermitian matrices. The statement was verified on random Hermitian pairs with random invertible $S$, and the normal form was constructed explicitly from the eigenbasis.

Corollary (additivity and the rank). The signature $\sigma(H)=p-q$ and the rank $p+q$ are additive under the orthogonal sum,

$$ \sigma(H\oplus H')=\sigma(H)+\sigma(H'),\qquad \mathrm{rank}(H\oplus H')=\mathrm{rank}(H)+\mathrm{rank}(H'), $$

and the inertia of the orthogonal sum is the sum of the inertias. This was verified on random pairs in the four-dimensional model.

The Classification and the Unitary Witt Group

Theorem (the classification on the rank-one module). The congruence classes of the Hermitian forms on the rank-one module over $\mathbb{B}$ are exactly the inertias with $p+q+r=2$:

$$ (2,0,0),\quad(1,1,0),\quad(0,2,0),\quad(1,0,1),\quad(0,1,1),\quad(0,0,2). $$

Proof. The normal form is determined by its inertia and the inertia is invariant, so two forms with the same inertia are congruent and two with different inertias are not.

Definition (the hyperbolic plane). The form of inertia $(1,1,0)$,

$$ \mathrm{Hyp} = \mathrm{diag}(1,-1), $$

is the hyperbolic plane; the forms of inertia $(p+k,q+k,r)$ that are orthogonal sums of a form with the hyperbolic plane are hyperbolic. A non-degenerate form with $p=q$ is hyperbolic.

Definition (the unitary Witt group). On the non-degenerate Hermitian forms, call two forms Witt equivalent if they become isometric after adding hyperbolic planes to each. The classes form the unitary Witt group $W(\mathbb{B},{}^{*})$ under the orthogonal sum, the inverse of a class being given by the negation of the form.

Theorem (the Witt group of the biquaternion algebra is $\mathbb{Z}$). The signature is a complete invariant of the Witt class, and

$$ W(\mathbb{B},{}^{*}) \;\cong\; \mathbb{Z},\qquad \text{generated by the class of the unit form } h_{e_{0}} . $$

Proof. The signature is additive, vanishes on the hyperbolic plane and changes sign under the negation, so it descends to a group homomorphism $W\to\mathbb{Z}$. It is injective because a form with signature zero has $p=q$, hence is hyperbolic, hence is the zero class; and it is surjective because the class of the unit form has signature $2$, so its multiples realise every even integer, while the class of the form $\mathrm{diag}(1,-1,-1,-1)$ has signature $-2$ and realises the negative ones. In the rank-two module the same argument applies with the signature of a $4\times4$ Hermitian matrix.

Remark (what the group is not). Over the complex numbers the signature classification collapses, since scaling by $i$ exchanges the signs and there is only one class up to congruence: the interesting group is the one of the real involution, which is the one used here since $\mathbb{M}_+$ is a real form. The Witt group is therefore $\mathbb{Z}$ and not $\mathbb{Z}/2$, and the invariant is the full signature, not only the parity of the rank.

$\varepsilon$-Hermitian Forms and the Signed Involution

Definition. Let $\theta$ be an involution of $\mathbb{B}$ and let $\varepsilon=\pm1$. A form is $\varepsilon$-Hermitian for $\theta$ if $h(\tilde P,\tilde R)=\varepsilon\,\theta(h(\tilde R,\tilde P))$. The case $\varepsilon=+1$ is the Hermitian case of the article; the case $\varepsilon=-1$ is the skew-Hermitian case, and for $\theta=\mathrm{id}$ the two cases are the symmetric and the alternating bilinear forms, whose classification over a field of characteristic not two is again by congruence but with the alternating forms contributing the hyperbolic class.

Remark (the four combinations of the corpus). The corpus's four two-sided operators are the four combinations of the left twist, identity or signed, and the right factor, inverse or dagger: $\Theta_{\tilde R}$ for the twisted cases and $\mathrm{H}_{\tilde R}$ for the Hermitian ones (Mixed Inner Conjugation and Hermitian Adjoint). Each of them acts on the forms of the corresponding type: the congruence $H\mapsto S^{\dagger}HS$ is the Hermitian case, the congruence $H\mapsto S^{\dagger}HS$ with the twist $\alpha$ is the signed case, and the two have the same inertia theory because the twist is an isomorphism of $\mathbb{M}_+$.

Worked Examples

The unit form and the norm form. The unit form has inertia $(2,0,0)$ and signature $2$; the form $\mathrm{diag}(1,-1)$ has inertia $(1,1,0)$ and signature $0$ and is the hyperbolic plane; the form $\mathrm{diag}(1,0)$ has inertia $(1,0,1)$ and rank one. The norm form $N$ is not in the classification at all.

The congruence by a two-sided operator. For $S=e_{0}+e_{1}$, of determinant two and invertible, and $H=\mathrm{diag}(1,-1)$, the congruent form $S^{\dagger}HS$ has the same inertia $(1,1,0)$. For $S$ a null element, $N(S)=0$, $S$ is not invertible and the congruence is not an equivalence: the form $\mathrm{diag}(1,0)$ is reached, and the rank drops. This is the algebraic content of the collapse worked in Two-Sided Operators on the Biquaternion Algebra with Hermitian Adjoint.

The isometry group of the unit form. $U(e_{0})=U(2)$: the maps $S$ with $S^{\dagger}S=e_{0}$, exactly the unitary slice, of real dimension four. The isometry group of the hyperbolic plane $\mathrm{diag}(1,-1)$ is $U(1,1)$, indefinite, of real dimension four as well but non-compact: the two forms have the same dimension of the isometry group and different topology, which is the group-level shadow of the difference of the signatures.

The Witt class of a sum. $\mathrm{diag}(1,1)\oplus\mathrm{diag}(-1,-1)$ has signature zero, hence is hyperbolic, hence is the zero Witt class, even though each summand is non-degenerate: this is the collapse that the Witt group performs and the inertia does not.

Summary

A Hermitian form on the rank-one module over the biquaternion algebra is a Hermitian matrix $H$, $h_{H}(\tilde R,\tilde P)=\tilde{R}^{*}H\tilde P$; congruence $H\mapsto S^{\dagger}HS$ is the two-sided operator of the corpus applied to the form, and its equivalence classes are the inertias $(p,q,r)$ with $p+q+r=2$, by Sylvester's law, with normal form $\mathrm{diag}(1_{p},-1_{q},0_{r})$. The signature $\sigma=p-q$ and the rank are additive under the orthogonal sum. The isometry group of the unit form $h_{e_{0}}(\tilde R,\tilde P)=\tilde{R}^{*}\tilde P$ is the unitary slice $U=U(2)$, and the isometry group of the quaternion norm is a different, non-compact group, because the two forms are of different types. The hyperbolic plane $\mathrm{diag}(1,-1)$ is the null form of the Witt theory, and the non-degenerate forms modulo the hyperbolic ones form the unitary Witt group $W(\mathbb{B},{}^{*})\cong\mathbb{Z}$, generated by the unit form and computed by the signature. The $\varepsilon$-Hermitian case is the signed analogue, with the same inertia theory because the twist is an isomorphism of the Hermitian cone.

Summary of Notation

Symbol	Meaning
$h_{H}(\tilde R,\tilde P)=\tilde{R}^{*}H\tilde P$	Hermitian form of the rank-one module, $H\in\mathbb{M}_+$
$h_{e_{0}}(\tilde R,\tilde P)=\tilde{R}^{*}\tilde P$	The unit form; positive definite; trace form of the dagger
$N(\tilde R)=\sum_{\mu}R_{\mu}^{2}$	The quaternion norm; not a Hermitian form of the dagger
$H'\sim H \iff H'=S^{\dagger}HS$	Congruence; the two-sided operator on the form matrix
$(p,q,r)$	Inertia: positive, negative and null dimensions
$\mathrm{diag}(1_{p},-1_{q},0_{r})$	Sylvester normal form
$\mathrm{Hyp}=\mathrm{diag}(1,-1)$	The hyperbolic plane; signature $0$, Witt class $0$
$U(H)=\{S:S^{\dagger}HS=H\}$	Isometry group; $U(e_{0})=U(2)$
$\sigma=p-q$	Signature; complete invariant of the Witt class
$W(\mathbb{B},{}^{*})\cong\mathbb{Z}$	The unitary Witt group, generated by the unit form