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curvature

Unlike the exterior derivation $ d$ , the square of $ \nabla$ may not be zero. Nevertheless, for any $ \alpha\in \Omega^k_{X/S}$ and for any $ x \in \Omega^\bullet_{X/S}\otimes_{\mathcal{O}_X}\mathcal{V}$ , we have

      $\displaystyle \nabla\nabla (\alpha x)$
    $\displaystyle =$ $\displaystyle \nabla(d\alpha\wedge x+(-1)^k f \nabla (x))$
    $\displaystyle =$ $\displaystyle (-1)^{k+1} d\alpha \wedge \nabla(x) +(-1)^k d\alpha\wedge \nabla(x) + \alpha \nabla \nabla(x))$
    $\displaystyle =$ $\displaystyle \alpha \nabla\nabla(x).$

So we conclude that there exists

$\displaystyle R \in \Omega^2_{X/S}\otimes_{\mathcal{O}_X}\operatorname{End}_{\mathcal{O}_X}(\mathcal{V},\mathcal{V})
$

such that

$\displaystyle \nabla\nabla(x)=R . x
$

holds.

DEFINITION 4.1   $ R$ is called the curvature tensor of $ \nabla$ .



2012-02-29