even and odd derivations.

Before proceeding further, let us do a little definition and computation on derivations.

DEFINITION 5.1   Let $A$ be a ring. Let $B$ be a $\mathbb{Z}/2\mathbb{Z}$ -graded A-algebra. Let $M$ be a $\mathbb{Z}/2\mathbb{Z}$ -graded $M$ -module. An $A$ -linear map

$$D :B \to M$$

is said to be

1. an even derivation if it satisfies
   $$D(B_i)\subset (M_i) \qquad (i=0,1)$$
   and
   $$D(f g)=(D f)g+f (D g) \qquad (\forall f, g \in B)$$
   holds.

2. an odd derivation if it satisfies
   $$D(B_i)\subset (M_{i+1}) \qquad (i=0,1)$$
   and
   $$D(f g)=(D f)g+(-1)^s f (D g) \qquad (\forall f \in B_s, s=1,2, g \in B)$$
   holds.

Following [3], let us denote by $\hat \bullet$ the ``parity'' of a homogeneous element $\bullet$ . That means,

$$\hat f= s$$    if $$f \in A_s, \hat m= s$$    if $$m \in M_s,$$

$\hat D=0$ for an even derivation $D$ , and $\hat D'=1$ for an odd derivation $D'$ . Then the ``Leibniz rules'' of the definition above may be simply rewritten as

$$D(f g)=(D f) g + (-1)^{\hat D \hat f} f (D g) \qquad (\forall f:$$ homogeneous $$\in B, \forall g\in B.)$$

For convenience, let us call a map $D$ a graded derivation if it is either an even derivation or an odd derivation.

DEFINITION 5.2   Let $A$ be a ring. Let $B$ be a $\mathbb{Z}/2\mathbb{Z}$ -graded $A$ -algebra Let $D_1,D_2:A\to A$ be graded derivations. Then we define their Lie bracket by

$$[D_1,D_2]=D_1 D_2 -(-1)^{\hat {D_1} \hat {D_2}} D_2 D_1.$$

Caution: The bracket defined here differs from the ordinary ``commutator'' if (and only if) both $D_1$ and $D_2$ are odd. In such a case it would be better to write $[D_1,D_2]_+$ instead to avoid confusion.

PROPOSITION 5.3   In the assumption of the definition above, the Lie bracket $[D_1, D_2]$ is an even or odd derivation with the parity $\hat D_1+ \hat D_2$ .

PROOF.. We first compute

$$(D_1 D_2). (f g)$$

$$= D_1( (D_2. f) g +(-1)^{\hat D_2 \hat f} f(D_2.g))$$

$$= (D_1.D_2. f) g +(-1)^{\hat D_1 (\hat D_2+\hat f)} (D_2.f)(D_1.g))$$

$$+ (-1)^{\hat D_2 \hat f}(D_1.f)(D_2. g) + (-1)^{\hat D_2 \hat f}(-1)^{\hat D_1 \hat f} f(D_1.D_2.g).$$

Then by adding this equation with the one with $D_1,D_2$ interchanged, we obtain the required result.

$\qedsymbol$

The following easy lemma is frequently used.

LEMMA 5.4   Let $A$ be a ring. Let $B$ be a $\mathbb{Z}/2\mathbb{Z}$ -graded algebra. Let $M$ be a $\mathbb{Z}/2\mathbb{Z}$ -graded $B$ -module. Then for any graded $A$ -derivation $D: B\to M$ , The kernel of $D$ forms an $A$ -subalgebra of $B$ .

$\qedsymbol$

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