Differentiation

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition of Derivations

We now introduce the notion of a derivation. More precisely, we shall be concerned with the concept of differential forms, and to handle this we need a graded algebra. Henceforth, we write \(\Delta\) for the abelian group that provides the graded algebra structure.

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

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

Then we call \(\varepsilon\) a commutation factor.

In particular, we then have \(\varepsilon(2.\alpha, \beta) = \varepsilon(\alpha, 2.\beta) = 1\).

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

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

A commutation factor will appear as the sign that arises when we interchange the order of the product of an element of degree \(p\) and an element of degree \(q\).

Now consider a commutative ring \(A\), \(\Delta\)-graded \(A\)-modules \(E\), \(E'\), \(E''\), \(F\), \(F'\), \(F''\), and \(A\)-bilinear maps

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

together with the induced \(A\)-linear maps

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

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

We now make the following definition.

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

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

If \(\varepsilon\) is always \(1\), so that we may omit \(\varepsilon\) from the above formula, we simply call \((d,d',d'')\) a derivation.

To avoid confusion in the above definition, it is worth noting where each term belongs: for example, \((dx)x'\) 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 \(\mu:E\otimes_A E \rightarrow E\) makes \(E\) a graded algebra, and

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

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

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

   for all \(x,y\in E\) is called an \(\varepsilon\)-derivation from \(E\) to \(F\).

Motivated by the second case, we often abuse notation by writing all of \(d, d', d''\) with the same letter \(d\); then the formula of Definition 2 can be written as

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

and in most cases we treat, this abuse of notation will cause no confusion.

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

On the other hand, we noted above that the case \(\Delta=\mathbb{Z}\) is our main interest; in this case, taking the non-trivial commutation factor \(\varepsilon(p,q)=(-1)^{pq}\), we see that any even-degree \(\varepsilon\)-derivation can always ignore the effect of \(\varepsilon\). For odd degree, the formula

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

holds for any homogeneous element \(x\in E\). In this case we call \(d\) an anti-derivation.

Differential Forms

To see how the discussion so far can be applied, let us briefly 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, a degree \(0\) derivation can always ignore the commutation factor, so regarding \(E\) as a non-graded \(\mathbb{K}\)-algebra and considering a derivation from \(E\) to \(E\), \(\varepsilon\) does not appear. Now for each \(i\), define \(\partial_i:E \rightarrow E\) to be the partial derivative \(\partial/\partial \x_i\); then by the Leibniz rule the equality of Definition 2 is satisfied.

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

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

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

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

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

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

(§Tensor Algebras, ⁋Proposition 13) Now for each basis element

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

define

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

and extend this to define \(d: \bigwedge M \rightarrow \bigwedge M\). Then \(d\) is an antiderivation of degree \(1\) of \(\bigwedge(M)\).

Bracket

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

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

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

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

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

To this end, let us first define, for an arbitrary \(\Delta\)-graded algebra \(G\) with 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 \((E, E', E'')\). Denoting their degrees by \(\delta_1\), \(\delta_2\), 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 immediate from the expansion of \((d_2\circ d_1)(xx')\) computed above.

Then restricting to the case \(\Delta=\mathbb{Z}\) in particular, 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 \(d_1\), \(d_2\) are antiderivations, then \(d_1 d_2 + d_2 d_1\) is a derivation.

On the other hand, looking at the partial derivatives defined on the polynomial algebra, they satisfy \(\partial_i\partial_j=\partial_j\partial_i\) for any \(i,j\). Now, as with general differential operators, write \(D=\partial_i+\partial_j\) and consider \(D^2\); this can be expanded 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, we can also write this 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, let a polynomial \(F \in A[\x_1, \dots, \x_k]\) in indeterminates \(T_1, \dots, T_n, T_1', \dots, T_n'\) be given. That is, \(F(T)\), \(F(T')\) mean

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

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 a \(\Delta\)-graded \(A\)-algebra \(E\) and a graded \(A\)-module \(F\) be given, together with two multiplications \(E\otimes_AF \rightarrow F\) and \(F\otimes_AE \rightarrow F\).

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 unit then \(1 \in \ker(d)\).

Proof

That \(\ker(d)\) is an \(A\)-submodule of \(E\) is obvious, so we only need to show that \(\ker(d)\) is closed under multiplication. For arbitrary homogeneous \(x, y \in \ker(d)\),

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

so \(xy \in \ker(d)\). Therefore \(\ker(d)\) is a graded subalgebra.

Now if \(E\) has a unit, then \(1\) is of degree \(0\), so

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

and we obtain \(d(1) = 0\).

Therefore, if \(d_1,d_2\) are degree \(\delta\) \(\varepsilon\)-derivations from \(E\) to \(F\) and they agree on the generators of \(E\) as an algebra over \(A\), then \(d_1=d_2\). On the other hand, 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 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 6 we have \(d(1) = 0\), so

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

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

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

and rearranging yields

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

Using that the degree of \(x^{-1}\) is \(-\deg(x)\) and computing \(d(x^{-1}x)\) gives the second equality.

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

Proof

Let an arbitrary derivation \(d:E \rightarrow f\) be given, and suppose an extension \(\bar{d}\) of \(d\) to \(K\) exists. Applying Proposition 7, we know that the 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 of an element of \(B\) as \(u/v\). That is, even when \(u/v = u'/v'\) we must have

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

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

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

so multiplying both sides by \(vv'\) yields

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

and rearranging gives

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

Thus the definition is well-defined independently of the expression of the element \(u/v\) of \(F\). That \(\bar{d}\) actually satisfies the condition of an \(A\)-derivation from \(K\) to \(F\) is a straightforward computation.

In the next proposition, for notational convenience, for any degree \(\delta\) \(\varepsilon\)-derivation \(d:A \rightarrow E\) define

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

Proposition 9 Let \(A\) be a unital graded associative \(A\)-algebra and let \(E\) be a graded \((A, A)\)-bimodule. Let \(d: A \to E\) be an \(\varepsilon\)-derivation of degree \(\delta\), and let \(a\) be a homogeneous element of \(Z_\varepsilon\) of degree \(\alpha\). 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 arbitrary homogeneous element \(x\) of degree \(\delta'\) and arbitrary \(y\in A\) we have

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

so \(d'\) becomes a degree \(\delta + \alpha\) \(\varepsilon\)-derivation.

On the other hand, if \(E\) is a \(\Delta\)-graded (associative) \(A\)-algebra with an \(\varepsilon\)-bracket given on it, then there is a natural \(\varepsilon\)-derivation on it.

Definition 10 For a homogeneous element \(z\in E\) of a graded \(A\)-algebra \(E\), 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 all homogeneous elements \(z\) of \(E\), we have \({[d, \ad(a)]_\varepsilon = \ad(dz)}\).
2. If \(A\) is associative, then \(\ad(z)\) is an \(\varepsilon\)-derivation of \(A\), and its degree is \(\deg(z)\).

Proof

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

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

   and we obtain the desired result.

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

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

   as desired.

Therefore, if \(E\) is an associative graded \(A\)-algebra, then by 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, replacing \(d\) by an inner \(\varepsilon\)-derivation 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, the equality of the above corollary can be obtained by verifying it for arbitrary homogeneous \(z\in E\), so 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.