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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

$\displaystyle D :B \to M
$

is said to be
  1. an even derivation if it satisfies

    $\displaystyle D(B_i)\subset (M_i) \qquad (i=0,1)
$

    and

    $\displaystyle D(f g)=(D f)g+f (D g) \qquad (\forall f, g \in B)
$

    holds.

  2. an odd derivation if it satisfies

    $\displaystyle D(B_i)\subset (M_{i+1}) \qquad (i=0,1)
$

    and

    $\displaystyle 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,

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

$\displaystyle D(f g)=(D f) g + (-1)^{\hat D \hat f} f (D g)
\qquad (\forall f:$ homogeneous $\displaystyle \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

$\displaystyle [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

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

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$

ARRAY(0x9620c48)


next up previous
Next: pairing of exterior algebras Up: some linear algebra Previous: some linear algebra
2012-02-29