Relations and Comparison Between the Two Products

Introduction

The complex space carries two products, the plain bilinear $AC$ and the plain sesquilinear $A\bar C$, defined in The Two Products of the Complexoid Space, above in this category. This article reads the two on one and the same pair and sets them against each other: the conjugate of a product, the exchange of the two slots, the identities that hold for one product and fail for the other, and the property table of the two with a witness for every failing entry. It closes with the two slots the construction has on $\mathbb{C}$, whose first slot is empty, and with the two pairings and the two adjoints the two products carry.

The article is the complex reading of the relations and the table of the four general products; those have collapsed to the two products of the commutative base, and with them the relations and the table. The definitions are The Two Products of the Complexoid Space above; the forms and the two algebraic norms are The Two Forms and the Two Algebraic Norms of the Complexoid Space, below this article in this category, cited and not repeated. The only computations the article adds are the two adjoints of §The Two Pairings and the Two Adjoints.

The elements are $A = a + ia'$ and $C = c + ic'$; the bar is complex conjugation and the star coincides with it.

The Two Products on One and the Same Pair

On one and the same pair the two products are

$$ AC = (ac - a'c') + i(ac' + a'c), \qquad A\bar C = (ac + a'c') + i(a'c - ac') . $$

Proposition. The two coincide exactly when the second slot is real, and are opposite exactly when it is purely imaginary:

$$ A\bar C = AC \iff C \in \mathbb{R}_{\mathbb{C}}, \qquad A\bar C = -AC \iff C \in i\mathbb{R}_{\mathbb{C}} . $$

Proof. $A\bar C = AC$ is $A(\bar C - C) = 0$, and $\mathbb{C}$ has no zero divisor, so it is $\bar C = C$, that is $c' = 0$. The second is $A(\bar C + C) = 0$, so $\bar C = -C$, that is $c = 0$.

The half sum of the two products splits the second factor into its real and its imaginary part, the complex shadow of the factor-splitting identities:

$$ \tfrac{1}{2}\bigl(AC + A\bar C\bigr) = A\,\operatorname{Re}(C), \qquad \tfrac{1}{2}\bigl(AC - A\bar C\bigr) = i\,A\,\operatorname{Im}(C) . $$

Proof. $C + \bar C = 2\operatorname{Re}(C)$ and $C - \bar C = 2i\operatorname{Im}(C)$, and the multiplication is distributive.

The Conjugate of a Product and the Exchange of the Slots

Proposition. The bar is an automorphism of the bilinear product and converts the sesquilinear product into a bilinear one:

$$ \overline{AC} = \bar A\bar C, \qquad \overline{A\bar C} = \bar A C . $$

Proof. The bar is an involution of the algebra, so it is multiplicative, $\overline{AC} = \bar A\bar C$; composing with the definition gives $\overline{A\bar C} = \overline{A}\,\overline{\bar C} = \bar A C$.

The slot maps are the two ways of conjugating one factor: $\tau_1$ conjugates the first, $\tau_2$ the second. On the same pair,

$$ A\bar C = \tau_2(AC), \qquad \bar A C = \tau_1(AC), \qquad A\bar C = \overline{\tau_1(AC)} = \overline{\bar A C} . $$

The four products have two slot maps acting on them and produce eight conjugate readings; here the four products have collapsed to two, and the eight readings collapse with them to the relations above. The exchange of the two slots gives

$$ AC = CA, \qquad A\bar C = \overline{C\bar A}, $$

the first because the multiplication is commutative and the second because

$$ \overline{C\bar A} = \bar C A = A\bar C . $$

The second identity is the conjugate-commutativity of the sesquilinear product: the two slots may be exchanged at the price of conjugating the value. The bilinear product is not conjugate-commutative, $\overline{CA} = \bar C\bar A \neq AC$ in general, so conjugate-commutativity is the symmetry that the sesquilinear product owns and the bilinear product does not.

The Identities That Single Out Each Product

The two products are told apart by the identities each satisfies and the other breaks.

identity	holds for $AC$	holds for $A\bar C$
$\overline{XC} = \bar X\bar C$, the bar multiplicative in both slots	yes	no
$\overline{X\bar C} = \bar X C$	no	yes
$XC = CX$	yes	no
$X\bar C = \overline{C\bar X}$	no	yes
$X\bar C = XC$ when $C$ is real	yes	yes
$X\bar X$ is the definite norm at the two equal arguments	no, the algebraic norm of $AC$ is $X^2$	yes

Proof. The first row is the multiplicativity of the bar; the analogue for the sesquilinear product is $\overline{X\bar C} = \bar X C$, which equals $\bar X\bar C$ only for real $C$, so the sesquilinear product fails the row and satisfies the second instead. The third row is commutativity, owned by the bilinear product; the analogue $X\bar C = C\bar X$ asks for $\bar C = C$, so only the sesquilinear product is saved, by conjugating the value, and it satisfies the fourth. The fifth row is the proposition above. The sixth is the definition of the two algebraic norms, the definite norm $H(X) = X\bar X$ of the sesquilinear product against the bilinear algebraic norm $B(X) = X^2$; the two are the two algebraic norms of The Two Forms and the Two Algebraic Norms of the Complexoid Space, below this article in this category.

The two products have the same vanishing set: $AC = 0$ or $A\bar C = 0$ exactly when $A = 0$ or $C = 0$, since $\mathbb{C}$ is a field. The two differ in which of them carries the algebraic norm that is positive for every nonzero element, and that is the definite norm of the sesquilinear product.

The Comparison

The property table of the two products is the table of the four general products with its four columns collapsed to two. The rows are the properties of an operation, except the last two, which are properties of an algebra.

property	$AC$	$A\bar C$
$\mathbb{C}$-linear in the first argument	yes	yes
$\mathbb{C}$-linear in the second argument	yes	no
associative	yes	no
commutative	yes	no
conjugate-commutative	no	yes
$1$ is a left identity	yes	no
$1$ is a right identity	yes	yes
alternative	yes	no
flexible	yes	no
degree-three identity	yes	no
left multiplications form a monoid	yes	no
multiplication of an algebra over $\mathbb{C}$	yes	no
multiplication of a sesqualgebra over $\mathbb{C}$	no	yes

The entries are established one by one, with the witness for each failure.

Linearity in the Two Slots

Both products are $\mathbb{C}$-linear in the first argument, $(\lambda A)C = \lambda(AC)$ and $(\lambda A)\bar C = \lambda(A\bar C)$; the bilinear product is also $\mathbb{C}$-linear in the second, $A(\lambda C) = \lambda(AC)$, while the sesquilinear product is conjugate-linear there, $A\overline{(\lambda C)} = \bar\lambda(A\bar C)$.

Associativity and the Weaker Laws

The bilinear product is associative, being the multiplication of a field. The sesquilinear product is not, with the witness $A = 1$, $C = 1$, $D = i$:

$$ (A\bar C)\bar D = -i, \qquad A\overline{(C\bar D)} = i . $$

An associative product satisfies the weaker laws of alternation, flexibility and the degree-three identity, so the first column carries them; the second fails each, with these witnesses.

law	identity	witness for $A\bar C$	left side	right side
alternative	$X(XY) = (XX)Y$	$X = 1$, $Y = i$	$i$	$-i$
flexible	$(XY)X = X(YX)$	$X = i$, $Y = 1$	$1$	$-1$
degree-three	$(XX)X = X(XX)$	$X = i$	$-i$	$i$

The degree-three identity is the one that makes the third power well defined; the sesquilinear product breaks it, so its iterated products depend on the parenthesisation. The three failures say the same thing: this product, which conjugates one factor, is not power-associative.

The Unit

The element $1$ is fixed by the bar, so it is a two-sided unit for the bilinear product and a right unit for the sesquilinear one: $A\bar 1 = A$. It is not a left unit for the sesquilinear product, $1\bar C = \bar C \neq C$ for the witness $C = i$. On the right the two agree; on the left they do not, and the difference is exactly the bar.

Commutativity and Conjugate-Commutativity

The bilinear product is commutative and not conjugate-commutative; the sesquilinear product is neither commutative nor free of the symmetry, but conjugate-commutative. The witnesses are $A = 1$, $C = i$ for the failure of commutativity of the sesquilinear product, $A\bar C = -i$ against $C\bar A = i$, and the conjugate-commutativity is the identity proved above.

The Two Algebra Rows

The last two rows ask whether the operation is the multiplication of an algebra over $\mathbb{C}$ or the derived operation of a sesqualgebra over $(\mathbb{C}, \bar\cdot)$. The bilinear product is the multiplication of a commutative associative unital $\mathbb{C}$-algebra, indeed a field, so it is the first and not the second; the sesquilinear product is the derived operation $X \star Y = X\bar Y$ of the algebra with the involution bar, so it is the second and not the first, and the two rows are the algebraic summary of the table.

The Two Slots of the Construction

Definition (the general product of two slots). On an algebra with an involution, the general product of two slots is

$$ x\star_{K}y=x\,K(y), $$

where $K$ is the identity or the conjugation and the first slot carries the conjugation of the coefficient ring.

Theorem (the two settings on $\mathbb{C}$). The coefficients of $\mathbb{C}$ are real, so the first slot is the identity and only the second switches, giving the two products of this article:

$$ \star_{\mathrm{id}}=AC, \qquad \star_{\bar{\cdot}}=A\bar C . $$

The construction with both slots alive has four settings; on $\mathbb{C}$ the first slot is empty and the four collapse two-and-two onto the field product and the sesquilinear product.

Theorem (the second slot rules the composition and the form). With $K=\mathrm{id}$ the product is associative, commutative and two-sidedly unital, and the form it carries is symmetric; with $K=\bar{\cdot}$ it is non-associative, conjugate-commutative and right-unital alone, and the form it carries is conjugate-symmetric.

Proof. The two settings are $AC$ and $A\bar C$, and their laws are the two columns of the property table above: the first column is the field, whose laws are The Specificities of the Complex Algebra, and the second is the sesquilinear product of The Sesquilinear Product of the Complexoid Space.

Remark (the empty first slot and the centre). The first slot carries the conjugation of the coefficient ring; the coefficients of $\mathbb{C}$ are real, so that conjugation is the identity and the first slot contributes nothing. What survives of it is the centre: the field is its own centre, the two products are the construction read on the centre, and the collapse of the first slot is the statement that the centre is fixed by the conjugation.

The Two Pairings and the Two Adjoints

Theorem (the two pairings). The two slots carry the two pairings of the space,

$$ B(A,C)=\operatorname{Sc}(AC)=AC, \qquad H(A,C)=\operatorname{Sc}(A\bar C)=A\bar C, $$

the first bilinear and symmetric, the second sesquilinear and conjugate-symmetric, both non-degenerate with Gram matrix $[1]$ in the basis $1$. They are the two forms of The Two Forms and the Two Algebraic Norms of the Complexoid Space, below this article in this category.

Proof. On $\mathbb{C}$ the scalar part of an element is the element itself, so the two pairings are the two products read as forms; the Gram values are $B(1,1)=H(1,1)=1$, so the matrix is $[1]$ and the pairing is non-degenerate.

Definition. For a $\mathbb{C}$-linear map $F$, the transpose $F^{\mathrm T}$ is defined by $B(FX,Y)=B(X,F^{\mathrm T}Y)$ and the adjoint $F^{\dagger}$ by $H(FX,Y)=H(X,F^{\dagger}Y)$.

Theorem (the two adjoints of the ordinary and of the sesquilinear right multiplication). For the ordinary left multiplication $L^{\mathrm{alg}}_{A}(X)=AX$ and the sesquilinear right multiplication $R_{A}(X)=X\bar A$,

$$ \bigl(L^{\mathrm{alg}}_{A}\bigr)^{\mathrm T}=L^{\mathrm{alg}}_{A}, \qquad \bigl(L^{\mathrm{alg}}_{A}\bigr)^{\dagger}=L^{\mathrm{alg}}_{\bar A}, $$

$$ \bigl(R_{A}\bigr)^{\mathrm T}=L^{\mathrm{alg}}_{\bar A}, \qquad \bigl(R_{A}\bigr)^{\dagger}=L^{\mathrm{alg}}_{A}. $$

The transpose keeps the parameter and the adjoint conjugates it, so the transpose and the adjoint of one and the same map agree exactly when $A$ is real.

Proof. For the first pair, $B(AX,Y)=AXY$ and $B(X,AY)=XAY$ are equal because the multiplication is commutative, so $L^{\mathrm{alg}}_{A}$ is its own transpose; and $H(AX,Y)=AX\bar Y$ while $H(X,\bar AY)=X\overline{\bar AY}=XA\bar Y$, so the adjoint of $L^{\mathrm{alg}}_{A}$ is $L^{\mathrm{alg}}_{\bar A}$. For the second pair, $B(X\bar A,Y)=X\bar AY$ and $B(X,\bar AY)=X\bar AY$, so the transpose of $R_{A}$ is $L^{\mathrm{alg}}_{\bar A}$; and $H(X\bar A,Y)=X\bar A\bar Y$ while $H(X,AY)=X\bar A\bar Y$, so the adjoint of $R_{A}$ is $L^{\mathrm{alg}}_{A}$. The two agree exactly at $A=\bar A$, that is at the real $A$.

Remark (the conjugate-linear family). The sesqualgebra left multiplication $L_{A}(X)=A\bar X$ is conjugate-linear and self-adjoint for $H$ in the convention of The Sesquilinear Product of the Complexoid Space, which owns that computation together with the adjoints of the right multiplication and of the sandwich.

Summary

The two products of the complex space are read on one and the same pair. They coincide exactly when the second slot is real and are opposite exactly when it is purely imaginary, and their half sum and half difference split the second factor into its real and imaginary parts. The bar is an automorphism of the bilinear product and converts the sesquilinear product into a bilinear one of the conjugate pair, $\overline{A\bar C} = \bar A C$; the exchange of the two slots gives $AC = CA$ for the bilinear product and $A\bar C = \overline{C\bar A}$ for the sesquilinear one, the conjugate-commutativity that the bilinear product does not have.

The property table of the two is the table of the four general products collapsed to two columns. Both products are $\mathbb{C}$-linear in the first argument; the bilinear product is also $\mathbb{C}$-linear in the second, associative, commutative, unital on both sides, alternative, flexible and degree-three, and its left multiplications form a monoid; the sesquilinear product is conjugate-linear in the second slot, non-associative, non-commutative but conjugate-commutative, right-unital and not left-unital, and fails alternation, flexibility and the degree-three identity. The first is the multiplication of a field over $\mathbb{C}$; the second is the derived operation of a sesqualgebra over $(\mathbb{C}, \bar\cdot)$. The four columns collapse to two, and the complex space is the degenerate base of the construction. The two products are the two settings of the two-slot construction on a real base: the first slot is empty, the second decides the composition and the symmetry of the form, and what remains of the first slot is the centre. The two pairings $B$ and $H$ carry two adjoints of a linear map, computed here: the transpose keeps the parameter and the adjoint conjugates it, so the ordinary multiplication by $A$ has transpose itself and adjoint the multiplication by $\bar A$, and the two agree exactly for real $A$.

Summary of Notation

symbol	meaning
$AC$, $A\bar C$	the plain bilinear and the plain sesquilinear product
$A = a + ia'$, $C = c + ic'$	the elements, with real coordinates
$\tau_1, \tau_2$	the slot maps, conjugating the first and the second factor
$\overline{AC} = \bar A\bar C$	the bar is multiplicative on the bilinear product
$\overline{A\bar C} = \bar A C$	the bar converts the sesquilinear product into a bilinear one
$A\bar C = \overline{C\bar A}$	conjugate-commutativity of the sesquilinear product
$A\bar C = AC$, $A\bar C = -AC$	the coincidence on a real second slot and the opposition on a purely imaginary one
$B(X) = X^2$, $H(X) = X\bar X$	the bilinear algebraic norm and the definite norm, the two algebraic norms
left multiplications form a monoid; multiplication of an algebra; multiplication of a sesqualgebra	the operation-level and algebra-level rows of the table
$x\star_{K}y=xK(y)$	the general product of two slots
$K\in\{\mathrm{id},\bar\cdot\}$	the switch of the second slot, the first slot being the identity
$B(A,C)=AC$, $H(A,C)=A\bar C$	the two pairings, of Gram matrix $[1]$
${}^{\mathrm T}$, ${}^{\dagger}$	the transpose for $B$ and the adjoint for $H$