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Let $X$ be a smooth algebraic variety with tangent sheaf $\Theta$ and $M$ a sheaf of modules on $X$. Let $D$ be the sheaf of differential operators on $X$.

Then giving a left $D-$ module structure on $M$ is equivalent to giving a homomorphism

$\nabla: \Theta \rightarrow End(M)$ with

1: $\nabla_{f \theta}(s)=f\nabla_\theta(s)$

2: $\nabla_\theta (fs)=\theta(f)s+f\nabla_\theta(s)$

3: $\nabla_{[\theta_1,\theta_2]}(s)=[\nabla_{\theta_1},\nabla_{\theta_2}](s)$,

where $\theta \in \Theta, f \in O_X, s \in M$.

The module structure is given by $\theta s=\nabla_\theta(s)$ I checked all conditions except associativity, i.e. why can one deduce from 3: that $(\theta_1 \theta_2)s = \theta_1(\theta_2s)$ holds?

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    Haven't we assumed that $\nabla$ is a homomorphism? Doesn't that imply the condition you want?2011-10-06

1 Answers 1

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I don't think you should "deduce" the formula from 3. Instead consider the formula as a definition of how generators of $\mathcal{D}_X$ act on $M$. What you should do then is check that this definition is compatible with relations in $\mathcal{D}_X$ using your condition 3 (the fact that $\nabla$ is a morphism of Lie algebras). But this is almost tautological since these relations are $\theta_1\theta_2 - \theta_2\theta_1 = [\theta_1,\theta_2]$ and $\theta f - f\theta = \theta(f)$.

It helps to think of $\mathcal{D}_X$ as the universal envelopping algebra of the Lie algebra $\Theta_X$.