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

$$\nabla\nabla (\alpha x)$$

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

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

$$= \alpha \nabla\nabla(x).$$

So we conclude that there exists

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

such that

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

holds.

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