Derivations of a Sesqualgebra

Introduction

A derivation is an additive map with the Leibniz rule, and in the associative case it is the infinitesimal form of the automorphisms of an algebra. On a sesqualgebra one condition has to be fixed before the rule can be stated, the behaviour under the scalars, because the product is conjugate-linear in the second variable: the rule $x(\lambda y) = \varsigma(\lambda)(xy)$ ties the scalars to the product, and a derivation has to respect the tie. The untwisted choice, $D(\lambda x) = \lambda D(x)$, is the one that needs no hypothesis, and it is the choice made here.

Throughout, $A$ is a sesqualgebra over the datum $(R,\varsigma)$ with a $\varsigma$-semilinear involution $*$, in the sense of Sesqualgebras: the product is additive in each variable, $R$-linear in the first and $\varsigma$-semilinear in the second, and $*$ is additive, of order two, anti-multiplicative and $\varsigma$-semilinear. The product need not be associative, and where an argument uses associativity it is said so. The comparison is with Derivations of a Ring and Star-Derivations and the Skew Derivations in the bilinear layer; Units and the Unitary Elements and The Unitary Lie Algebra carry the group and the bracket that the inner derivations point to.

What is added here: the definition and the scalar discussion, the Lie algebra of the derivations, the star-derivations and the skew derivations with their decomposition and their graded structure, the failure of the Leibniz rule for the derived operation $x\star y = xy^{*}$ and the twisted derivation that restores it, the inner derivations with the centre as their kernel, and the inner derivations generated by the skew-Hermitian elements.

The Definition

The Leibniz Rule

Definition. A derivation of $A$ is an additive map $D : A \to A$ that is $R$-linear, $$ D(\lambda x) = \lambda D(x) \qquad \text{for all } \lambda \in R, \ x \in A, $$ and satisfies the Leibniz rule $$ D(xy) = D(x)y + xD(y) \qquad \text{for all } x, y \in A. $$ The set of the derivations of $A$ is written $\operatorname{Der}(A)$.

Remark (why the scalar condition is untwisted). The two scalar rules have to be tested against the Leibniz rule separately, because the product is not symmetric in the scalars. Applying the rule to $(\lambda x)y = \lambda(xy)$ gives $\lambda\bigl(D(x)y + xD(y)\bigr)$ on both sides, using the first rule for $(\lambda D(x))y$ and $(\lambda x)D(y)$; applying it to $x(\lambda y) = \varsigma(\lambda)(xy)$ gives $\varsigma(\lambda)\bigl(D(x)y + xD(y)\bigr)$ on both sides, using the second rule for $D(x)(\lambda y)$ and $xD(\lambda y)$. Both computations therefore hold for every $R$-linear $D$ and impose no condition. A $\varsigma$-semilinear derivation, $D(\lambda x) = \varsigma(\lambda)D(x)$ together with the Leibniz rule, breaks both: the rule fails at the pair $(\lambda x, y)$ by $(\varsigma(\lambda) - \lambda)xD(y)$ and at the pair $(x, \lambda y)$ by $(\lambda - \varsigma(\lambda))D(x)y$, each failure taken as the left side minus the right. The untwisted condition is the one compatible with both rules without a hypothesis on the product, and it is the one adopted.

Remark. The conjugate-linear maps of Conjugate-Linear Maps and the Conjugate Dual are $R$-linear into the conjugate module and $\varsigma$-semilinear out of $A$; a derivation is $R$-linear as a map of $A$ to itself, and is a different object. In the bilinear case the untwisted condition is the ordinary one and the two remarks above say nothing new.

Sums and Multiples

Proposition. Let $D$ and $E$ be derivations of $A$ and let $\lambda \in R$. Then $D + E$ is a derivation. The multiple $\lambda D$ is a derivation exactly when $$ (\lambda - \varsigma(\lambda))\,A\cdot D(A) = 0, $$ so it is a derivation for every derivation $D$ as soon as $\lambda$ is fixed by $\varsigma$. Hence $\operatorname{Der}(A)$ is a submodule of $\operatorname{End}_{R}(A)$, the $R$-module of the additive $R$-linear maps $A \to A$, over the fixed ring $R^{\varsigma} = \{\lambda \in R : \varsigma(\lambda) = \lambda\}$; it is an $R$-submodule in the bilinear case, and with a nontrivial $\varsigma$ it is one only when $(\lambda - \varsigma(\lambda))A\cdot D(A) = 0$ for every $\lambda$ and every derivation $D$.

Proof. Additivity and $R$-linearity are stable under addition and scalar multiplication. For the sum, $(D+E)(xy) = D(x)y + xD(y) + E(x)y + xE(y) = (D+E)(x)y + x(D+E)(y)$ by the additivity of the product in each variable. For the multiple, $(\lambda D)(xy) = \lambda\bigl(D(x)y\bigr) + \lambda\bigl(xD(y)\bigr) = (\lambda D(x))y + \lambda\bigl(xD(y)\bigr)$, while the right hand side of the rule is $(\lambda D(x))y + x(\lambda D(y)) = (\lambda D(x))y + \varsigma(\lambda)\bigl(xD(y)\bigr)$, the second scalar rule being what moves $\lambda$ across $x$; the two differ by $(\lambda - \varsigma(\lambda))xD(y)$, which vanishes for all $x$ and $y$ exactly under the stated condition, and identically when $\varsigma(\lambda) = \lambda$.

Remark. For a nonzero derivation and a $\lambda$ with $\varsigma(\lambda) \neq \lambda$ the multiple $\lambda D$ is generally not a derivation, because the product is $\varsigma$-semilinear in its second slot; the module structure is over $R^{\varsigma}$, which is all of $R$ only in the bilinear case. The centre is not subject to this: a central element commutes with the element $\lambda x$ itself, and that one extra commutation makes $\lambda z$ central for every scalar, as Subalgebras and the Involution records.

Remark. The composition of two derivations is generally not a derivation: the rule applied twice produces the cross terms $D(x)E(y) + E(x)D(y)$, which are symmetric in the two derivations and do not cancel. The combination that survives their difference is the commutator, treated next.

The Lie Algebra of Derivations

The Commutator

Theorem. Let $D$ and $E$ be derivations of $A$. Then their commutator $$ \boxed{\;[D,E] = D \circ E - E \circ D\;} $$ is a derivation. Consequently $\operatorname{Der}(A)$ is closed under the commutator, so it is a Lie subalgebra, over $R^{\varsigma}$, of the associative algebra $\operatorname{End}_{R}(A)$; no associativity of the product of $A$ is used.

Proof. Additivity and $R$-linearity are clear. For the Leibniz rule, expand $$ \begin{aligned} [D,E](xy) &= D\bigl(E(x)y + xE(y)\bigr) - E\bigl(D(x)y + xD(y)\bigr) \\ &= D(E(x))y + E(x)D(y) + D(x)E(y) + xD(E(y)) \\ &\quad - E(D(x))y - D(x)E(y) - E(x)D(y) - xE(D(y)) \\ &= [D,E](x)\,y + x\,[D,E](y), \end{aligned} $$ the four cross terms cancelling in pairs. Each application of the rule is to a product of two factors, so no associativity is used.

Remark. The bracket exists for every sesqualgebra, associative or not, because $\operatorname{End}_{R}(A)$ is associative under composition and the commutator is its Lie bracket; the derivations are a subspace closed under that bracket, which is all that a Lie subalgebra needs. This is the sesquilinear reading of the theorem of Derivations of a Ring.

Remark (the module structure). Besides the $R^{\varsigma}$-module structure of the previous proposition, $\operatorname{Der}(A)$ is a module over the centre $Z(A)$ when $A$ is associative and the product is bilinear: for central $z$ and a derivation $D$ the map $zD$ is $R$-linear and satisfies $(zD)(xy) = z(D(x)y) + z(xD(y)) = (zD(x))y + x(zD(y))$, the last step moving $z$ across $x$. With a nontrivial $\varsigma$ the second scalar rule gives $(zD)(\lambda x) = \varsigma(\lambda)(zD)(x)$ instead of $\lambda(zD)(x)$, so $zD$ is $\varsigma$-semilinear and is a derivation only when $(\lambda - \varsigma(\lambda))zD(A) = 0$ for every $\lambda$; this is the same defect that governs the inner maps below, and it is why the module structure is stated in the bilinear case. Under the bracket, $\operatorname{Der}(A)$ is also a module over itself, which is the adjoint action $D \mapsto [D,\cdot]$.

The Star-Derivations and the Skew Derivations

The Two Classes

Definition. A derivation $D$ is a $*$-derivation when it commutes with the involution, $$ D(x^{*}) = D(x)^{*} \qquad \text{for all } x, $$ and a skew derivation when it anticommutes with it, $$ D(x^{*}) = -D(x)^{*} \qquad \text{for all } x . $$ The sets are written $\operatorname{Der}_{*}(A)$ and $\operatorname{Der}_{\mathrm{skew}}(A)$.

Proposition (the conjugate derivation). For a derivation $D$ put $$ \Phi(D)(x) = D(x^{*})^{*} . $$ Then $\Phi(D)$ is a derivation, $\Phi$ is an additive map of order two on $\operatorname{Der}(A)$, and $D$ is a $*$-derivation exactly when $\Phi(D) = D$ and a skew derivation exactly when $\Phi(D) = -D$.

Proof. Additivity and $R$-linearity are clear, using $\varsigma^{2} = \mathrm{id}$ for the scalars. For the Leibniz rule, $\Phi(D)(xy) = D(y^{*}x^{*})^{*} = \bigl(D(y^{*})x^{*} + y^{*}D(x^{*})\bigr)^{*} = x\Phi(D)(y) + \Phi(D)(x)y$, the anti-multiplicativity and additivity of $*$ giving the last step. The order two is $\Phi(\Phi(D))(x) = \Phi(D)(x^{*})^{*} = D(x)^{**}$, which is $D(x)$. The two identifications are the definitions read back.

Remark. The map $\Phi$ is the conjugation of a derivation by the involution, and it is an involution of the space $\operatorname{Der}(A)$ rather than of $A$; it is the analogue of the map $D \mapsto \sigma D \sigma$ of Star-Derivations and the Skew Derivations, and it is an automorphism of the Lie algebra, as the next proposition shows.

The Decomposition and the Graded Structure

Proposition. The map $\Phi$ is an automorphism of the Lie algebra $\operatorname{Der}(A)$, that is $\Phi([D,E]) = [\Phi(D),\Phi(E)]$; the $*$-derivations are its fixed part and the skew derivations its anti-fixed part.

Proof. For the automorphism, $[\Phi(D),\Phi(E)](x) = \Phi(D)(\Phi(E)(x)) - \Phi(E)(\Phi(D)(x)) = D(E(x^{*}))^{*} - E(D(x^{*}))^{*} = [D,E](x^{*})^{*}$, which is $\Phi([D,E])(x)$; the last two statements are the previous proposition.

Theorem (the decomposition). Suppose $2$ is invertible in $R$. Then every derivation decomposes uniquely as $$ \boxed{\;D = D_{+} + D_{-}, \qquad D_{+} = \tfrac{1}{2}\bigl(D + \Phi(D)\bigr), \quad D_{-} = \tfrac{1}{2}\bigl(D - \Phi(D)\bigr)\;}, $$ with $D_{+}$ a $*$-derivation and $D_{-}$ a skew derivation.

Proof. The two parts are derivations by the closure under addition and under the scalars of $R^{\varsigma}$, the scalar $\tfrac{1}{2}$ lying in $R^{\varsigma}$ because $\varsigma(1) = 1$; then $\Phi(D_{+}) = \tfrac{1}{2}(\Phi(D) + \Phi^{2}(D)) = D_{+}$ and $\Phi(D_{-}) = \tfrac{1}{2}(\Phi(D) - \Phi^{2}(D)) = -D_{-}$, so they are a $*$-derivation and a skew derivation; they sum to $D$, and the decomposition is unique because a derivation that is both is zero when $2 \neq 0$.

Corollary (the graded structure). The $*$-derivations form a Lie subalgebra of $\operatorname{Der}(A)$, and the bracket is graded by the two classes: $$ [\operatorname{Der}_{*}, \operatorname{Der}_{*}] \subseteq \operatorname{Der}_{*}, \qquad [\operatorname{Der}_{*}, \operatorname{Der}_{\mathrm{skew}}] \subseteq \operatorname{Der}_{\mathrm{skew}}, \qquad [\operatorname{Der}_{\mathrm{skew}}, \operatorname{Der}_{\mathrm{skew}}] \subseteq \operatorname{Der}_{*} . $$

Proof. For derivations with $\Phi(D) = \varepsilon D$ and $\Phi(E) = \delta E$, with $\varepsilon, \delta \in \{1,-1\}$, the automorphism property gives $\Phi([D,E]) = [\Phi(D),\Phi(E)] = \varepsilon\delta\,[D,E]$. The three inclusions are the three sign combinations, and the first is the closure of the $*$-derivations.

The Scalars on the Star-Derivations

Proposition. The conjugate map satisfies $\Phi(\lambda D) = \varsigma(\lambda)\Phi(D)$; it is $\varsigma$-semilinear and $R^{\varsigma}$-linear. Consequently $\operatorname{Der}_{*}(A)$ and $\operatorname{Der}_{\mathrm{skew}}(A)$ are stable under the scalars of $R^{\varsigma}$, and under all the scalars of $R$ in the bilinear case only.

Proof. $\Phi(\lambda D)(x) = (\lambda D)(x^{*})^{*} = \bigl(\lambda D(x^{*})\bigr)^{*} = \varsigma(\lambda)D(x^{*})^{*} = \varsigma(\lambda)\Phi(D)(x)$, so $\Phi$ is $\varsigma$-semilinear and is linear over the scalars it fixes. The fixed part and the anti-fixed part of an $R^{\varsigma}$-linear involution are $R^{\varsigma}$-stable, and $R^{\varsigma} = R$ exactly when $\varsigma = \mathrm{id}$.

Remark. The fixed ring appears here and in the scalar closure of the derivations above, for the same reason, the $\varsigma$-semilinearity of the product in its second slot; it does not govern the centre, which is closed under every scalar. A scalar with $\varsigma(\lambda) \neq \lambda$ sends a $*$-derivation to $\varsigma(\lambda)D$, which is not a $*$-derivation unless it is not a derivation at all, since $\varsigma(\lambda)D = \lambda D$ would force $(\lambda - \varsigma(\lambda))A\cdot D(A) = 0$ by the previous section.

The Derived Operation

The Failure and the Transfer

Proposition (the transfer of the Leibniz rule). Suppose $A$ unital and associative, let $\star$ be the derived operation $x \star y = xy^{*}$, and let $D$ be a derivation of the algebra $A$ for the ordinary product. Then $D$ satisfies the Leibniz rule for $\star$, $$ D(x \star y) = D(x) \star y + x \star D(y) \qquad \text{for all } x, y, $$ if and only if $D$ is a $*$-derivation.

Proof. By the ordinary rule, $D(x \star y) = D(xy^{*}) = D(x)y^{*} + xD(y^{*})$, while $D(x) \star y + x \star D(y) = D(x)y^{*} + xD(y)^{*}$. The two agree for all $x$ and $y$ exactly when $xD(y^{*}) = xD(y)^{*}$ for all $x$ and $y$, which by the unit is $D(y^{*}) = D(y)^{*}$ for all $y$.

Remark. The Leibniz rule therefore does not transfer from the algebra product to the derived operation: an ordinary derivation is a derivation of $x \star y = xy^{*}$ only when it is also a $*$-derivation, and the general derivation fails the $\star$-rule on the twist produced by the involution.

The Twisted Derivation

Proposition. Let $D$ be an additive $R$-linear map, and use the conjugate $\Phi(D)(x) = D(x^{*})^{*}$ of the previous section. Then $D$ satisfies the twisted rule $$ D(x \star y) = D(x) \star y + x \star \Phi(D)(y) \qquad \text{for all } x, y $$ if and only if $D$ satisfies the ordinary Leibniz rule $D(xy) = D(x)y + xD(y)$.

Proof. Since $x \star \Phi(D)(y) = x\,\Phi(D)(y)^{*} = xD(y^{*})$, the twisted rule reads $D(xy^{*}) = D(x)y^{*} + xD(y^{*})$. As $y^{*}$ runs over all of $A$, this is the ordinary rule read at the pair $(x, y^{*})$.

Remark. The two propositions together say that, among the derivations of the algebra, exactly the $*$-derivations satisfy the plain $\star$-rule, while every derivation satisfies the twisted one, and that the twist $\Phi$ is trivial exactly on the $*$-derivations. The failure of the plain rule and the twisted rule that restores it are the sesquilinear form of the passage, in Star-Derivations and the Skew Derivations, from a derivation to its conjugate.

The Inner Derivations

The Commutator Map

Definition. For $a \in A$ the commutator map or inner map is $$ \mathrm{ad}_a(x) = ax - xa . $$

Proposition. For $a \in A$ the map $\mathrm{ad}_a$ is $R$-linear exactly when $$ (\varsigma(\lambda) - \lambda)\,a\,x = 0 \qquad \text{for all } \lambda \in R, \ x \in A, $$ and it satisfies the Leibniz rule exactly when the associators obey $$ [x,a,y] = [a,x,y] + [x,y,a] \qquad \text{for all } x, y, $$ where $[u,v,w] = (uv)w - u(vw)$. In particular, for an associative $A$, the map $\mathrm{ad}_a$ is a derivation for every $a$ in the bilinear case, and in the general case for the $a$ with $(\varsigma(\lambda) - \lambda)aA = 0$ for all $\lambda$.

Proof. Additivity is clear. For the scalars, $\mathrm{ad}_a(\lambda x) = \varsigma(\lambda)(ax) - \lambda(xa)$ by the two rules, while $\lambda\,\mathrm{ad}_a(x) = \lambda(ax) - \lambda(xa)$; the two differ by $(\varsigma(\lambda) - \lambda)ax$, which vanishes for all $\lambda$ and $x$ under the stated condition. For the rule, expand $\mathrm{ad}_a(xy) - \mathrm{ad}_a(x)y - x\,\mathrm{ad}_a(y)$ as $$ a(xy) - (xy)a - (ax)y + (xa)y - x(ay) + x(ya) = -[a,x,y] + [x,a,y] - [x,y,a], $$ which vanishes for all $x$ and $y$ exactly under the stated identity; every associator vanishes when the product is associative.

Remark. Associativity is sufficient and not necessary: the identity of the proposition can hold with a nonzero associator, because it is a cyclic combination of three associators rather than their separate vanishing. The structural theorems below assume associativity, which is the case that carries the inner structure.

Remark (the twisted scalar condition). In the bilinear case the scalar condition is empty and $\mathrm{ad}_a$ is a derivation for every $a$. With a nontrivial $\varsigma$ the condition bites, and the smallest example shows how hard: in the field $\mathbb{C}$ over $(\mathbb{C},\varsigma)$ with $\varsigma$ the conjugation and the product $x \star y = x\bar y$, the element $\varsigma(\lambda) - \lambda = \bar\lambda - \lambda$ is $2i$ at $\lambda = i$, so the condition forces $a\mathbb{C} = 0$ and hence $a = 0$. The only $R$-linear inner map there is $\mathrm{ad}_0 = 0$, and the inner theory is carried by the bilinear case.

The Centre as the Kernel

Theorem. Suppose $A$ associative, and let $\mathfrak{n} = \{a \in A : (\varsigma(\lambda) - \lambda)aA = 0 \ \text{for all } \lambda \in R\}$, so that $\mathrm{ad}_a$ is a derivation for every $a \in \mathfrak{n}$. Then the inner derivations form a Lie subalgebra of $\operatorname{Der}(A)$ with $$ \boxed{\;[\mathrm{ad}_a, \mathrm{ad}_b] = \mathrm{ad}_{ab - ba}\;}, $$ and the assignment $a \mapsto \mathrm{ad}_a$ is a Lie algebra homomorphism from $\mathfrak{n}$, with the commutator, onto the inner derivations, with kernel $\mathfrak{n} \cap Z(A)$. In the bilinear case $\mathfrak{n} = A$ and the kernel is the centre, so the inner derivations are isomorphic to $A/Z(A)$.

Proof. For the bracket, $\mathrm{ad}_a \mathrm{ad}_b(x) - \mathrm{ad}_b \mathrm{ad}_a(x) = a(bx - xb) - (bx - xb)a - b(ax - xa) + (ax - xa)b = (ab - ba)x - x(ab - ba) = \mathrm{ad}_{ab-ba}(x)$, associativity being what expands the four general products; and $ab - ba$ lies in $\mathfrak{n}$ because $(\varsigma(\lambda) - \lambda)(ab - ba)A$ is contained in $(\varsigma(\lambda) - \lambda)aA + (\varsigma(\lambda) - \lambda)bA = 0$. The kernel is the set of the $a$ that commute with every element, which is $\mathfrak{n} \cap Z(A)$, and the centre is the kernel in the bilinear case because then $\mathfrak{n} = A$.

Remark. The theorem is the sesquilinear form of the corresponding statement of Derivations of a Ring: the inner derivations are the image of the commutator bracket of the algebra, the centre is exactly what the image forgets, and the quotient $A/Z(A)$ is a Lie algebra under the commutator.

The Skew-Hermitian Generators

Proposition. For $a \in A$ the commutator map satisfies $\mathrm{ad}_a(x^{*}) = \mathrm{ad}_a(x)^{*}$ for all $x$ exactly when $a + a^{*}$ is central, and it satisfies $\mathrm{ad}_a(x^{*}) = -\mathrm{ad}_a(x)^{*}$ for all $x$ exactly when $a - a^{*}$ is central.

Proof. $\mathrm{ad}_a(x^{*}) = ax^{*} - x^{*}a$, while $\mathrm{ad}_a(x)^{*} = (ax - xa)^{*} = x^{*}a^{*} - a^{*}x^{*}$. The first equality holds for all $x$ exactly when $ax^{*} + a^{*}x^{*} = x^{*}a + x^{*}a^{*}$, that is $(a + a^{*})x^{*} = x^{*}(a + a^{*})$; as $x^{*}$ runs over $A$ this is the centrality of $a + a^{*}$. The second is the same computation with the signs of the $a^{*}$ terms reversed, giving the centrality of $a - a^{*}$.

Corollary (the skew-Hermitian generators). Suppose $A$ associative, the bilinear case, and $2$ invertible in $R$, so that every $\mathrm{ad}_a$ is a derivation. A Hermitian $a$ gives a skew derivation and a skew-Hermitian $a$ gives a $*$-derivation; the skew-Hermitian elements $a^{*} = -a$ are closed under the commutator, and the map $a \mapsto \mathrm{ad}_a$ carries them onto the inner $*$-derivations with kernel the central skew-Hermitian elements. By the transfer proposition these maps are derivations of the derived operation $x \star y = xy^{*}$ as well.

Proof. A Hermitian $a$ has $a - a^{*} = 0$ central and a skew-Hermitian $a$ has $a + a^{*} = 0$ central, so the proposition gives the two classes. For $a$, $b$ skew-Hermitian, $(ab - ba)^{*} = b^{*}a^{*} - a^{*}b^{*} = ba - ab = -(ab - ba)$, so the commutator is skew-Hermitian again and the set is a Lie algebra; its image is a Lie subalgebra of $\operatorname{Der}_{*}(A)$ by the bracket theorem, and the kernel is the intersection with the centre. It is onto because an $a$ with $a + a^{*} = z$ central is carried to $\mathrm{ad}_a = \mathrm{ad}_{a - z/2}$ and $a - z/2$ is skew-Hermitian when $2$ is invertible. The last statement is the transfer proposition read on the $*$-derivations of this form.

Remark. The corollary is the infinitesimal form of the unitary group of Units and the Unitary Elements: the skew-Hermitian elements are the elements whose exponential is unitary, the map $a \mapsto \mathrm{ad}_a$ is their action on the algebra, and the bracket they carry is the subject of The Unitary Lie Algebra. In the matrix case the skew-Hermitian elements are the skew-adjoint matrices and their bracket is the Lie algebra of the unitary group.

Examples

The Matrix Algebra

Proposition. Let $A = M_{n}(\mathbb{R})$ over the datum $(\mathbb{R},\mathrm{id})$ with the transpose as the involution, and $n \geq 2$. Then every derivation of $A$ is inner (standard), the derivations are $\operatorname{ad}_a$ with $a$ an arbitrary matrix and $\operatorname{Der}(A)$ is isomorphic to the quotient of $A$ by the scalar matrices; a derivation is a $*$-derivation exactly when it is $\operatorname{ad}_a$ with $a + a^{\mathsf{T}}$ a scalar matrix, and a skew derivation exactly when $a - a^{\mathsf{T}}$ is a scalar matrix. So the symmetric matrices generate the skew derivations and the skew-symmetric matrices generate the $*$-derivations.

Proof. That every derivation of a full matrix algebra over a field is inner is standard, and it identifies $\operatorname{Der}(A)$ with the image of $a \mapsto \mathrm{ad}_a$, whose kernel is the centre, the scalar matrices, by the kernel theorem. The two class statements are the proposition on the commutator map, and the symmetric matrices have $a - a^{\mathsf{T}} = 0$ while the skew-symmetric ones have $a + a^{\mathsf{T}} = 0$.

Remark. The skew-symmetric matrices are closed under the commutator, and the map to the $*$-derivations has kernel the central skew-symmetric matrices, which are zero for $n \geq 2$; so the $*$-derivations of $M_{n}(\mathbb{R})$ with the transpose are the skew-symmetric matrices themselves, the orthogonal Lie algebra of the transpose.

The Quaternions

Proposition. Let $A = \mathbb{H}$ over the datum $(\mathbb{R},\mathrm{id})$ with the quaternion conjugation. Then every derivation of $A$ is inner (standard), $\operatorname{Der}(A)$ is isomorphic to the quotient of $\mathbb{H}$ by the real centre, and the skew-Hermitian elements are the pure imaginary quaternions, which are closed under the commutator and map onto the $*$-derivations with kernel zero.

Proof. Every derivation of $\mathbb{H}$ is inner because $\mathbb{H}$ is a central division algebra over $\mathbb{R}$ (standard); the kernel of $a \mapsto \mathrm{ad}_a$ is the centre $\mathbb{R}$, the real multiples of the unit. The conjugate of a pure imaginary quaternion is its negative, so the skew-Hermitian elements are exactly the pure imaginary ones, they are closed under the commutator because a commutator of two of them is again pure imaginary, and no nonzero pure imaginary quaternion is central.

Remark. The pure imaginary quaternions with the commutator are the Lie algebra of the unit quaternions, and the $*$-derivations of $\mathbb{H}$ are a copy of it; this is the compact case of the unitary picture of the previous section.

The Sesquilinear Field

Proposition. Let $A = \mathbb{C}$ over the datum $(\mathbb{C},\varsigma)$ with $\varsigma$ the conjugation and the product $x \star y = x\bar y$. Then $A$ has exactly one derivation, the zero map.

Proof. The unit $1$ is a right unit of the product, since $x \star 1 = x\bar 1 = x$. A derivation $D$ satisfies $D(x) = D(x \star 1) = D(x) \star 1 + x \star D(1) = D(x) + x \star D(1)$, so $x \star D(1) = 0$ for every $x$, which forces $D(1) = 0$. An $R$-linear $D$ is determined by $D(1)$, so $D = 0$.

Remark. The triviality is a feature of the field and not of the twisted datum: the proof uses that an $R$-linear $D$ is determined by the value $D(1)$, and that fails as soon as the algebra is larger. The next subsection exhibits the nonzero derivations of the twisted matrix algebra.

The Twisted Matrix Algebra

Proposition. Let $A = M_{n}(\mathbb{C})$, $n \geq 2$, with the conjugate transpose $\dagger$ for the involution, the product $x \star y = xy^{\dagger}$, and the datum $(\mathbb{C},\varsigma)$ with $\varsigma$ the conjugation. For a skew-Hermitian $a$, that is $a^{\dagger} = -a$, and outside the scalar matrices, the map $$ D(x) = ax - xa, $$ with the product on the right the ordinary one, is a nonzero derivation of the $\star$-product and a $*$-derivation. For a scalar $\lambda$ with $\varsigma(\lambda) \neq \lambda$ the multiple $\lambda D$ is not a derivation.

Proof. The map $D$ is a derivation of the ordinary product, by the associativity of the matrix product, and it is linear as the ordinary product is linear, hence $R$-linear. It is a $*$-derivation, since $D(x^{\dagger}) = ax^{\dagger} - x^{\dagger}a$ while $D(x)^{\dagger} = (ax - xa)^{\dagger} = x^{\dagger}a^{\dagger} - a^{\dagger}x^{\dagger} = ax^{\dagger} - x^{\dagger}a$ by $a^{\dagger} = -a$. The transfer proposition then gives the $\star$-rule for $D$, and $D \neq 0$ because $a$ is outside the scalar matrices. For the scalar, the proposition on the sums and the multiples gives $$ (\lambda D)(x \star y) - \bigl((\lambda D)(x) \star y + x \star (\lambda D)(y)\bigr) = (\lambda - \varsigma(\lambda))\,x \star D(y), $$ which is nonzero for a $\lambda$ with $\varsigma(\lambda) \neq \lambda$, because $D \neq 0$ and $z \star z = zz^{\dagger}$ vanishes only at $z = 0$.

Remark. This is the nonzero twisted example that the field does not give: the derivations do not collapse, they form a module over the reals only, and the scalar defect is read on an explicit element. The maps $D$ are inner for the ordinary product but not of the inner form $x \mapsto b \star x - x \star b$ of the $\star$-product, which is $R$-linear only for $b = 0$. The centre of the twisted algebra and the annihilators that its zero divisors carry are the subject of The Centre and the Zero Divisors of a Sesqualgebra.

Summary

A derivation of a sesqualgebra is an additive $R$-linear map with $D(xy) = D(x)y + xD(y)$; the untwisted scalar condition is the one compatible with both scalar rules of the product, and the derivations form the $R^{\varsigma}$-module $\operatorname{Der}(A)$ (an $R$-module in the bilinear case), closed under the commutator, which makes it a Lie algebra with no associativity assumed, and a module over the centre when $A$ is associative and the product is bilinear. The conjugate $\Phi(D)(x) = D(x^{*})^{*}$ is an involutive automorphism of that Lie algebra; its fixed part is the $*$-derivations and its anti-fixed part the skew derivations, which decompose every derivation when $2$ is invertible and carry a graded bracket. The Leibniz rule does not transfer from an associative algebra to its derived operation $x \star y = xy^{*}$: among the derivations of the algebra it holds exactly for the $*$-derivations, and the twisted rule, in which the conjugate of the derivation acts on the second factor, holds exactly for the ordinary derivations. The inner maps $\mathrm{ad}_a(x) = ax - xa$ are derivations when $A$ is associative, when they are $R$-linear and when the associators obey the cyclic identity; for an associative $A$ in the bilinear case with $2$ invertible they satisfy $[\mathrm{ad}_a,\mathrm{ad}_b] = \mathrm{ad}_{ab-ba}$ and their kernel is the centre, so the inner derivations are the algebra modulo its centre, and the skew-Hermitian elements map onto the inner $*$-derivations, which by the transfer proposition are derivations of the derived operation as well.

Summary of Notation

symbol	meaning
$D(\lambda x) = \lambda D(x)$	the untwisted scalar condition on a derivation
$D(xy) = D(x)y + xD(y)$	the Leibniz rule
$\operatorname{Der}(A)$	the $R^{\varsigma}$-module and the Lie algebra of the derivations
$[D,E] = DE - ED$	the commutator, a derivation for derivations $D, E$
$Z(A)$	the centre, the kernel of the inner derivations
$\Phi(D)(x) = D(x^{})^{}$	the conjugate derivation, an involution of $\operatorname{Der}(A)$
$\operatorname{Der}_{*}(A)$	the $$-derivations, $D(x^{}) = D(x)^{*}$
$\operatorname{Der}_{\mathrm{skew}}(A)$	the skew derivations, $D(x^{}) = -D(x)^{}$
$R^{\varsigma} = \{\lambda : \varsigma(\lambda) = \lambda\}$	the fixed ring of $\varsigma$, the scalars acting on $\operatorname{Der}(A)$
$\lambda D \in \operatorname{Der}(A) \iff (\lambda - \varsigma(\lambda))A\cdot D(A) = 0$	the scalar defect, empty in the bilinear case
$\mathrm{ad}_a(x) = ax - xa$	the inner map of $a$
$[u,v,w] = (uv)w - u(vw)$	the associator
$\mathfrak{n} = \{a : (\varsigma(\lambda) - \lambda)aA = 0\}$	the elements whose inner map is $R$-linear
$\{\mathrm{ad}_a : a^{*} = -a\}$	the inner $*$-derivations of the bilinear associative case