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curvature revisited

In this subsection we prove an important proposition on curvature. But before we do that, we do two things. Firstly, we prove the following lemma.

LEMMA 5.7   Let $ A$ be a commutative ring. Let $ B$ be a commutative $ A$ -algebra. Then for any $ X,Y\in \operatorname{Der}_A(B)$ and for any $ \omega\in \Omega^1_{X/S}$ , we have

$\displaystyle i_Y i_X (d \omega)
= X. \langle \omega,Y\rangle
- Y.\langle \omega,X \rangle
-i_{[X,Y]}\omega .
$

PROOF..

      $\displaystyle i_Y i_X d \omega$
    $\displaystyle =$ $\displaystyle i_Y (L_X-d i_X)\omega$
    $\displaystyle =$ $\displaystyle i_Y (L_X \omega)-i_Y d \langle \omega,X \rangle$
    $\displaystyle =$ $\displaystyle L_X (i_Y \omega)-i_{[X,Y]}\omega - Y.\langle \omega,X \rangle$
    $\displaystyle =$ $\displaystyle X. \langle \omega,Y\rangle -i_{[X,Y]}\omega - Y.\langle \omega,X \rangle$

$ \qedsymbol$

Secondly, we add a notation.

DEFINITION 5.8   Let $ A$ be a commutative ring. Let $ B$ be a commutative $ A$ -algebra. Let $ M$ be a $ B$ -module with a connection

$\displaystyle \nabla: M\to \Omega^1_{B/A}\otimes_B M.
$

Then for any $ X\in \operatorname{Der}_A(B)$ ,

$\displaystyle (i_X\otimes 1)\nabla : M \to M
$

is an $ A$ -linear map. We shall denote it by $ \nabla_X$ .

PROPOSITION 5.9   Let $ A$ be a commutative ring. Let $ B$ be a commutative $ A$ -algebra. Let $ M$ be a $ B$ -module with a connection

$\displaystyle \nabla: M\to \Omega^1_{B/A}\otimes_B M.
$

Let $ R$ be the connection $ 2$ -form of $ \nabla$ . Then for any $ X,Y\in \operatorname{Der}_A(B)$ , we have

$\displaystyle \langle R, X \wedge Y\rangle_\wedge =[\nabla_X,\nabla_Y]-\nabla_{[X,Y]}
$

PROOF.. Let $ v$ be an element of $ M$ . Since $ \nabla v$ is an element of $ \Omega^1_{B/A}\otimes_B M$ , we may write it as :

$\displaystyle \nabla v=\sum_j \omega_j \otimes v_j
$

for some $ \omega_j \in \Omega^1_{B/A}, v_j \in M$ . Then we compute

$\displaystyle \nabla \nabla v
=\sum_j d \omega_j \otimes v_j - \sum_j \omega_j \wedge \nabla v_j.
$

      $\displaystyle i_Y i_X \nabla \nabla v$
    $\displaystyle =$ $\displaystyle ((i_Y i_X )(d \omega_j)) \otimes v_j -i_Y \sum_j \langle \omega_j,X\rangle \nabla v_j +\sum_j \omega_j \nabla_X v_j$
    $\displaystyle =$ $\displaystyle (i_Y i_X d \omega_j)\otimes v_j -\sum_j \langle \omega_j , X) \nabla_Y v_j +\sum_j \langle \omega_j , Y) \nabla_X v_j$

On the other hand, we have

$\displaystyle \nabla_X(v)
=(i_X\otimes 1)\nabla(v))=\sum_j \langle \omega_j,X\rangle v_j
$

So we may proceed

$\displaystyle \nabla \nabla_X(v)
=\sum_j d \langle \omega_j,X\rangle v_j
+\sum_j \langle \omega_j,X\rangle \nabla v_j
$

$\displaystyle \nabla_Y \nabla_X(v)
=\sum_j Y. \langle \omega_j,X\rangle v_j
+\sum_j \langle \omega_j,X\rangle \nabla_Y v_j
$

Thus, together with the Lemma above, we see

$\displaystyle i_Y i_X \nabla \nabla (v)+ \nabla_Y \nabla_X(v)-\nabla_X\nabla_Y(v)
=-\sum_j i_{[X,Y]}\omega_j \otimes v_j=-\nabla_{[X,Y]} v
$

$ \qedsymbol$

COROLLARY 5.10   If the curvature $ R$ of $ \nabla$ is equal to zero, then

$\displaystyle \nabla_{\bullet}: \operatorname{Der}_A \to \operatorname{End}_A(M)
$

is a Lie algebra homomorphism.


next up previous
Next: a formula for -powers Up: some linear algebra Previous: Relations of derivations.
2012-02-29