Derivations

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Definition of a Derivation

We now introduce the notion of a derivation. To be more precise, what we will consider is the concept of differential forms, and to treat this we need a graded algebra. Henceforth we denote by \(\Delta\) the abelian group that provides the grading structure of the graded algebra.

Definition 1 For an abelian group \((\Delta, +, 0)\), suppose a function \(\varepsilon : \Delta \times \Delta \to \{ \pm 1 \}\) satisfies the following three conditions.

• \[\varepsilon(\alpha + \alpha', \beta) = \varepsilon(\alpha, \beta)\varepsilon(\alpha', \beta)\]
• \[\varepsilon(\alpha, \beta + \beta') = \varepsilon(\alpha, \beta)\varepsilon(\alpha, \beta')\]
• \[\varepsilon(\beta, \alpha) = \varepsilon(\alpha, \beta)\]

In this case, we call \(\varepsilon\) a commutation factor.

Then in particular \(\varepsilon(2.\alpha, \beta) = \varepsilon(\alpha, 2.\beta) = 1\).

The example of most interest to us is the case \(\Delta=\mathbb{Z}\). In this case, by Definition 1, \(\varepsilon\) is completely determined by its value at \(\varepsilon(1,1)\), and hence the commutation factors defined on \(\Delta=\mathbb{Z}\) are only

\[\varepsilon(p,q)=1,\qquad \varepsilon(p,q)=(-1)^{pq}\]

These two. The commutation factor will appear as the sign that arises when we interchange the order of elements of degree \(p\) and degree \(q\) in a product.

Now let \(A\) be a commutative ring, let \(E\), \(E'\), \(E''\), \(F\), \(F''\) be \(\Delta\)-graded \(A\)-modules, and let

\[\mu: E \times E' \to E'', \qquad \lambda_1: F \times E' \to F', \qquad \lambda_2: E \times F' \to F''\]

be \(A\)-bilinear maps, and consider the \(A\)-linear maps they induce

\[E \otimes_A E' \to E'', \qquad F \otimes_A E' \to F'', \qquad E \otimes_A F' \to F''\]

and assume that these three \(A\)-linear maps are all degree \(0\) graded homomorphisms. These correspond to multiplication operations, and we will simply write, for instance, the image of \(x\otimes x'\) in \(E''\) as \(xx'\). Since the element \(x\otimes x'\) in \(E\otimes_A E'\) lies in degree \(\degree(x)+\degree(x')\), under the above assumption \(xx'\) lies in the degree \(\degree(x)+\degree(x')\) component of \(E''\).

We now define the following.

Definition 2 In addition to the above situation, suppose a commutation factor \(\varepsilon: \Delta \times \Delta \to \{ \pm 1 \}\) is given. Then a triple \(d: E \rightarrow F\), \(d': E' \rightarrow F'\), \(d'': E'' \rightarrow F''\) of degree \(\delta\) graded \(A\)-module homomorphisms satisfying the condition

\[d''(xx') = (dx)x' + \varepsilon(\delta, \deg(x))x(d'x')\]

is called a degree \(\delta\) \((A, \varepsilon)\)-derivation (or simply an \(\varepsilon\)-derivation) from \((E, E', E'')\) to \((F, F', F'')\). If \(\varepsilon\) is always \(1\), so that \(\varepsilon\) can be omitted from the above formula, we simply call \((d,d',d'')\) a derivation.

To avoid confusion in the above definition, it is helpful to check where each term belongs; for example, the term \((dx)x'\) on the right-hand side is the element of \(F''\) obtained by multiplying \(dx\in F\) and \(x'\in E'\) via \(\lambda_1\). However, in practice we are interested in the following two special cases.

1. \(E=F\), \(E'=F'\), \(E''=F''\), and the three bilinear maps \(\mu, \lambda_1, \lambda_2\) are all the same.

2. \(E=E'=E''\), \(F=F'=F''\), so that \(E\) becomes a graded algebra via \(\mu:E\otimes_A E \rightarrow E\), and

   \[\lambda_1: F \otimes_A E \to F, \qquad \lambda_2: E \otimes_A F \to F\]

   In this case, we call a single \(d:E \rightarrow F\) satisfying the formula

   \[d(xy)=(dx)y+\varepsilon(\delta, \deg(x))x(dy)\]

   an \(\varepsilon\)-derivation from \(E\) to \(F\).

Taking the second case as motivation, we sometimes abuse notation and write all of \(d, d', d''\) with the same symbol \(d\); then the formula in Definition 2 becomes

\[d(xx')=(dx)x'+\varepsilon(\delta,\deg(x))x (dx)\]

and this abuse of notation will not cause confusion in most cases we treat.

The situation that arises most frequently is when both of the above cases hold, i.e., \(E=E'=E''=F=F'=F''\) and \(\lambda_1, \lambda_2\) are multiplication in \(E\), and the derivation is a single graded endomorphism \(d: E \rightarrow E\). Then the \(\varepsilon\)-derivation can be regarded as a map from \(A\) to \(A\), so in this case we simply call it an \(\varepsilon\)-derivation of \(A\).

Meanwhile, we noted earlier that the case \(\Delta=\mathbb{Z}\) is our main interest; considering the non-trivial commutation factor \(\varepsilon(p,q)=(-1)^{pq}\) in this case, we know that any even-degree \(\varepsilon\)-derivation can always ignore the effect of \(\varepsilon\). In the odd-degree case, for any homogeneous element \(x\in E\) the following formula

\[d(xx')=(dx)x'+(-1)^{\deg x}x(dx')\]

holds. In this case we call \(d\) an anti-derivation.

Differential Forms

To see how the preceding discussion can be applied, let us look at a simple example. Here \(\mathbb{K}\) is a field and \(E=\mathbb{K}[\x_1,\ldots, \x_n]\) is a polynomial algebra.

First, since a degree \(0\) derivation can always ignore the commutation factor, if we regard \(E\) as a non-graded \(\mathbb{K}\)-algebra and consider a derivation from \(E\) to \(E\), then \(\varepsilon\) does not appear. Now, for each \(i\), if we define \(\partial_i:E \rightarrow E\) as the partial derivative \(\partial/\partial \x_i\), then the Leibniz rule implies that the equality in Definition 2 is satisfied.

Next, let us see an example of a graded algebra. For the polynomial algebra \(E\) defined as above, let \(M\) be the free \(A\)-module generated by the elements

\[d\x_1,d\x_2,\ldots, d\x_n\]

and consider the exterior algebra \(\bigwedge(M)\); then this exterior algebra is a \(\mathbb{Z}\)-graded \(E\)-algebra given by

\[\bigwedge(M)=\bigoplus_{d=0}^n{\bigwedge}^d(M)\]

where \(\bigwedge^0(M)=A\) and for each \(k\), \(\bigwedge^k(M)\) is the free \(E\)-module generated by elements of the form

\[d\x_J=d\x_{j_1}\wedge d\x_{j_2}\wedge\cdots\wedge d\x_{j_k},\qquad j_1<\cdots< j_k\]

([Multilinear Algebra] §Tensor Algebras, ⁋Proposition 13). Now, for each basis element

\[f\; d\x_{j_1}\wedge d\x_{j_2}\wedge\cdots\wedge d\x_{j_d}\in {\bigwedge}^k(M)\]

the formula

\[d(f\; d\x_{j_1}\wedge d\x_{j_2}\wedge\cdots\wedge d\x_{j_k})=\sum_{i=1}^n\frac{\partial f}{\partial \x_i}d\x_i\wedge d\x_{j_1}\wedge d\x_{j_2}\wedge\cdots\wedge d\x_{j_k}\in{\bigwedge}^{k+1}(M)\]

defines a map which extends to \(d: \bigwedge M \rightarrow \bigwedge M\). Then \(d\) becomes an antiderivation of degree \(1\) on \(\bigwedge(M)\).

Bracket

Meanwhile, assume that the first of the two cases above holds. Then \(d=(d,d',d'')\) can be regarded as a map from \((E,E',E'')\) to itself, so we may also consider the composition of \(\varepsilon\)-derivations. However, looking at the formula in Definition 2 in general, for any two \(\varepsilon\)-derivations \(d_1,d_2\) of degrees \(\delta_1, \delta_2\) and any \(x\in E\), \(x'\in E'\),

\[\begin{aligned}(d_2\circ d_1)(xx')&=d_2((d_1x)x'+\varepsilon(\delta_1, \deg(x))x(d_1'x'))\\&=(d_2d_1x)x'+\varepsilon(\delta_2,\deg(d_1x))(d_1x)(d_2'x')+\varepsilon(\delta_1, \deg(x))(d_2x)(d_1'x')+\varepsilon(\delta_1, \deg(x))\varepsilon(\delta_2, \deg(x))x(d_2' d_1'x')\end{aligned}\]

so in general the composition of \(\varepsilon\)-derivations is not an \(\varepsilon\)-derivation. That is, in general, if we consider the category of triples of \(\Delta\)-graded \(A\)-modules and the endomorphism algebra of a fixed triple \((E,E',E'')\)

\[\End_{\bgr_\Delta \Alg{A}^3}(E, E', E'')\]

then the collection of \(\varepsilon\)-derivations does not define a subalgebra of this endomorphism algebra. However, examining the above computation, it is also clear what kind of product must be defined on this algebra so that the collection of \(\varepsilon\)-derivations becomes closed. In other words, if we eliminate the two middle terms among the four terms on the right-hand side, then \(d_2d_1\) would become an \(\varepsilon\)-derivation of degree \(\delta_1+\delta_2\).

To achieve this, we first define, in the most general setting, for an arbitrary \(\Delta\)-graded algebra \(G\) and a fixed commutation factor \(\varepsilon\), the \(\varepsilon\)-bracket of two homogeneous elements \(x,y\) of \(G\) by the formula

\[[x,y]_\varepsilon=xy-\varepsilon(\deg(x),\deg(y))yx\]

Then through this we can define the \(\varepsilon\)-bracket in \(G=\End_{\bgr_\Delta \Alg{A}^3}(E, E', E'')\).

Proposition 3 Let \(d_1, d_2\) be \(\varepsilon\)-derivations on \((E, E', E'')\). If their degrees are \(\delta_1, \delta_2\), then their \(\varepsilon\)-bracket

\[[d_1, d_2]_\varepsilon = d_1 \circ d_2 - \varepsilon_{\delta_1, \delta_2} \, d_2 \circ d_1\]

is another \(\varepsilon\)-derivation of degree \(\delta_1 + \delta_2\). In particular, if \(d\) is an \(\varepsilon\)-derivation of degree \(\delta\) and \(\varepsilon_{\delta, \delta} = -1\), then \(d^2 = d \circ d\) is a derivation.

The proof of this is obvious from the expansion of \((d_2\circ d_1)(xx')\) computed above.

Then, restricting especially to the case \(\Delta=\mathbb{Z}\), the above proposition yields the following corollary.

Corollary 4 Let \(\Delta = \mathbb{Z}\). Then the following hold.

1. The square of an antiderivation is a derivation.
2. The bracket of two derivations is a derivation.
3. The bracket of an antiderivation and an even-degree derivation is an antiderivation.
4. If \(d_1\) and \(d_2\) are antiderivations, then \(d_1 d_2 + d_2 d_1\) is a derivation.

Meanwhile, considering the partial derivatives defined on the polynomial algebra, they satisfy \(\partial_i\partial_j=\partial_j\partial_i\) for any \(i,j\). Now, writing \(D=\partial_i+\partial_j\) as with general differential operators and considering \(D^2\), we can expand it as

\[D^2=(\partial_i+\partial_j)^2=\partial_i^2+\partial_i\partial_j+\partial_j\partial_i+\partial_j^2\]

and since \(\partial_i\) and \(\partial_j\) commute, this can be written more simply as

\[D^2=\partial_i^2+2\partial_i\partial_j+\partial_j^2\]

The following proposition generalizes this further.

Proposition 5 Under the above assumptions and notation, suppose a polynomial \(F \in A[\x_1, \dots, \x_k]\) in the indeterminates \(T_1, \dots, T_n, T_1', \dots, T_n'\) is given. That is, \(F(T)\) and \(F(T')\) denote

\[F(T) = F(T_1, \dots, T_n), \qquad F(T') = F(T_1', \dots, T_n')\]

respectively. Similarly, define

\[F(T + T') = F(T_1 + T_1', \dots, T_n + T_n')\]

Now, if a polynomial \(P\) satisfies the formula

\[P(T + T') = \sum_i Q_i(T) R_i(T')\]

then for any \(x\in E\), \(x\'\in E'\) the formula

\[P(D)(x x') = \sum_i (Q_i(D) x)(R_i(D) x')\]

holds.

Derivations of \(A\)-Algebras

We now examine the second of the two special cases discussed after Definition 2. That is, let \(E\) be a \(\Delta\)-graded \(A\)-algebra, \(F\) a graded \(A\)-module, and suppose two multiplications \(E\otimes_AF \rightarrow F\) and \(F\otimes_AE \rightarrow F\) are given.

Proposition 6 For an \(\varepsilon\)-derivation \(d:E \to F\) of degree \(\delta\), the kernel \(\ker(d)\) is a graded subalgebra of \(E\), and if \(E\) has a unity then \(1 \in \ker(d)\).

Proof

First, since it is obvious that \(\ker(d)\) is an \(A\)-submodule of \(E\), it suffices to show that \(\ker(d)\) is closed under multiplication. For any homogeneous \(x, y \in \ker(d)\),

\[d(xy) = (dx)y + \varepsilon(\delta, \deg(x))x(dy) = 0\]

we have \(xy \in \ker(d)\). Hence \(\ker(d)\) is a graded subalgebra.

Now, if \(E\) has a unity, then \(1\) has degree \(0\), so

\[d(1) = d(1 \cdot 1) = (d1) \cdot 1 + \varepsilon_{\delta, 0} \cdot 1 \cdot (d1) = d1 + d1 = 2d1\]

and thus we see that \(d(1) = 0\).

Therefore, if \(d_1,d_2\) are \(\varepsilon\)-derivations of degree \(\delta\) from \(E\) to \(F\) and they agree on the generators of \(E\) as an \(A\)-algebra, then \(d_1=d_2\). Meanwhile, the following holds for inverses.

Proposition 7 Let \(E\) be a unital \(\Delta\)-graded \(A\)-algebra and let \(d:E \to F\) be an \(\varepsilon\)-derivation of degree \(\delta\). If \(x\) is an invertible homogeneous element of \(E\), then for its inverse \(x^{-1}\) the following formula

\[d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}(d(x))x^{-1} = -\varepsilon_{\delta, \deg(x)} (d(x)) x^{-2}\]

holds.

Proof

By Proposition 5, \(d(1) = 0\), so

\[0 = d(xx^{-1}) = d(x))x^{-1} + \varepsilon_{\delta, \deg(x)}x(d(x^{-1})\]

Multiplying both sides on the left by \(x^{-1}\),

\[0 = x^{-1}(d(x))x^{-1} + \varepsilon_{\delta, \deg(x)} d(x^{-1})\]

and rearranging,

\[d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}(d(x))x^{-1}.\]

we obtain the first equality. Moreover, using that the degree of \(x^{-1}\) is \(-\deg(x)\) and computing \(d(x^{-1}x)\), we obtain the second equality.

Proposition 8 Let \(E\) be an \(A\)-algebra that is an integral domain, and consider its field of fractions \(K=\Frac E\). Regarding any \(K\)-vector space \(F\) as an \(E\)-module and considering an \(A\)-derivation \(d:E \rightarrow F\), then \(d\) extends uniquely to an \(A\)-derivation from \(K\) to \(F\).

Proof

Suppose an arbitrary derivation \(d:E \rightarrow f\) is given, and assume that an extension \(\bar{d}\) of \(d\) to \(K\) exists. Then applying Proposition 7, the following formula

\[\bar{d}(u/v) = v^{-1} d(u) - v^{-2} u\, d(v)\]

must hold, and therefore if \(\bar{d}\) exists its expression is unique.

Let us show that this definition does not depend on the choice of representation \(u/v\). That is, even when \(u/v = u'/v'\),

\[v^{-1} d(u) - v^{-2} u\, d(v) = v'^{-1} d(u') - v'^{-2} u'\, d(v')\]

must hold.

Set \(uv' = u'v\). Applying \(d\) to both sides,

\[v' d(u) + u\, d(v') = v\, d(u') + u'\, d(v)\]

and multiplying both sides by \(vv'\),

\[v v' d(u) - u\, v\, d(v') = v v' d(u') - u'\, v'\, d(v)\]

we obtain, upon rearrangement,

\[v' d(u) - v^{-1} u\, d(v) = v' d(u') - v'^{-1} u'\, d(v')\]

holds. Thus the definition is well defined independent of the expression of the element \(u/v\). Now, that \(\bar{d}\) actually satisfies the conditions of an \(A\)-derivation from \(K\) to \(F\) is a straightforward computation.

In the next proposition, for notational convenience, define

\[Z_\varepsilon=\{a\in A\mid \text{$xa_d=\varepsilon(\deg(a),\deg(x))a_dx$ for all homogeneous component $a_d$ of $a$ and for all homogeneous $x\in E$.}\}\]

for any \(\varepsilon\)-derivation \(d:A \rightarrow E\) of degree \(\delta\).

Proposition 9 Let \(A\) be a unital graded associative \(A\)-algebra and let \(E\) be a graded \((A, A)\)-bimodule. Now let \(d: A \to E\) be an \(\varepsilon\)-derivation of degree \(\delta\), and let \(a\) be a homogeneous element of degree \(\alpha\) of \(Z_\varepsilon\). Then the morphism

\[x \mapsto a (d x)\]

is an \(\varepsilon\)-derivation of degree \(\delta + \alpha\).

Proof

Denoting the given morphism by \(d'\), this morphism is obviously \(A\)-linear. To show that \(d'\) is an \(\varepsilon\)-derivation, for any homogeneous element \(x\) of degree \(\delta'\) and any \(y\in A\),

\[\begin{aligned}d'(xy)&=a(dx)y+\varepsilon(\delta, \delta')a(x(dy))\\&=a(dx)y+\varepsilon(\delta, \delta')\varepsilon(\alpha,\delta')xa(dy)\\&=(d'x)y+\varepsilon(\delta+\alpha,\delta')x(d'y)\end{aligned}\]

and therefore \(d'\) becomes an \(\varepsilon\)-derivation of degree \(\delta + \alpha\).

Meanwhile, if \(E\) is a \(\Delta\)-graded (associative) \(A\)-algebra equipped with an \(\varepsilon\)-bracket, then there exists a natural \(\varepsilon\)-derivation on it.

Definition 10 For a homogeneous element \(z\in E\) of a graded \(A\)-algebra, we write the morphism

\[x\mapsto [z,x]_\varepsilon\]

as \(\ad_\varepsilon(z)\).

Then the following holds.

Proposition 11 Let \(E\) be a graded \(A\)-algebra.

1. For any \(\varepsilon\)-derivation \(d : E \rightarrow E\) and every homogeneous element \(z\) of \(E\), \({[d, \ad(a)]_\varepsilon = \ad(dz)}\) holds.
2. If \(A\) is associative, then \(\ad(z)\) is an \(\varepsilon\)-derivation of \(A\), and its degree is \(\deg(z)\).

Proof

1. Suppose \(d\) is an \(\varepsilon\)-derivation of degree \(\delta\) and let \(\zeta = \deg(z)\). Now set \(f = [d, \ad(z)]_\varepsilon\); then for every homogeneous element \(x \in A\) of degree \(\xi\),

   \[\begin{aligned}f(x)&=d(z x - \varepsilon(\zeta, \xi) x z) - \varepsilon(\delta, \zeta) (z (dx) - \varepsilon(\zeta, \delta+\xi) (dx) z) \\&= d(z x) - \varepsilon(\zeta, \xi) d(x z) - \varepsilon(\delta, \zeta) z (dx) + \varepsilon(\zeta,2.\delta+\xi) d(x) z \\&=(dz)x+\varepsilon(\delta, \zeta)z(dx)-\varepsilon(\zeta,\xi)((dx)z+\varepsilon(\delta, \xi)x(dz))- \varepsilon(\delta, \zeta) z (dx) + \varepsilon(\zeta,2.\delta+\xi) (dx) z\\&=(dz)x+\varepsilon(\delta,\zeta)z(dx)-\varepsilon(\zeta,\xi)(dx)z-\varepsilon(\delta+\zeta,\xi)x(dz)-\varepsilon(\delta,\zeta)z (dx)+\varepsilon(\zeta,\xi)(dx)z\\&=(dz)x-\varepsilon(\delta+\zeta,\xi)x(dz)=[dz,x]_\varepsilon=\ad_\varepsilon(dz)(x)\end{aligned}\]

   thus we obtain the desired result.

2. For every homogeneous element \(x \in A\) of degree \(\xi\) and homogeneous element \(y \in A\) of degree \(\eta\),

   \[\begin{aligned}\ad(z)(x y) &= z(x y) - \varepsilon(\zeta, \xi + \eta)(x y) z \\&= (z x) y - \varepsilon(\zeta, \xi) x z y + \varepsilon(\zeta, \xi) x z y - \varepsilon(\zeta, \xi + \eta) x y z \\&= (ax-\varepsilon(\zeta,\xi xz)y+\varepsilon(zeta,\xi)x(ay-\varepsilon(\zeta,\eta)ya)\\&=\ad(z)(x) \cdot y + \varepsilon(\zeta, \xi) x \cdot \ad(z)(y)\end{aligned}\]

   holds.

Therefore, if \(E\) is an associative graded \(A\)-algebra, then through Definition 10 any homogeneous element of \(E\) defines an \(\varepsilon\)-derivation from \(E\) to itself, and we call this an inner \(\varepsilon\)-derivation.

When this holds, substituting an inner \(\varepsilon\)-derivation for \(d\) in the above formula yields the following corollary.

Corollary 12 For two homogeneous elements \(x,y\) of an associative graded algebra \(E\), the formula

\[{[\ad_\varepsilon(x), \ad_\varepsilon(y)]_\varepsilon = \ad_\varepsilon([x,y]_\varepsilon)}\]

always holds.

Moreover, since the equality in the above corollary can be obtained by verifying it for any homogeneous \(z\in E\), if \(x,y,z\) are homogeneous elements of degrees \(\xi,\eta,\zeta\) respectively, then the formula

\[{\varepsilon}_{\xi, \zeta} [[x, [y,z]_{\varepsilon}]_{\varepsilon} + \varepsilon_{\eta,\xi} [y, [z,x]_{\varepsilon}]_{\varepsilon} + \varepsilon_{\zeta,\eta} [z, [x,y]_{\varepsilon}]_{\varepsilon} = 0\]

holds, and we call this the Jacobi identity.