next up previous
Next: direct sums, tensor products, Up: connection Previous: connection

definition of connections on quasi coherent sheaves

Let $ X$ be a separated scheme over $ S$ . Let $ \mathcal{V}$ be a quasi coherent sheaf on $ X$ . Let $ p_1^{(1)},p_2^{(1)}:\Delta^{(1)}_{X/S} \to X$ be the canonical projections.

There are a several equivalent ways to give a ``connection on $ \mathcal{V}$ ".

One way is to provide an isomorphism

$\displaystyle P: (p_1^{(1)})^*\mathcal{V}\cong (p_2^{(1)}) ^* \mathcal{V}
$

such that $ P\vert _{\Delta_{X/S}}=\operatorname{id}$ . (Since the structure sheaf of $ \Delta^{(1)}_{X/S}$ is an extension of that of $ \Delta_{X/S}$ by nilpotents, we may easily prove that the term ``isomorphism'' here may be safely replaced by a term ``homomorphism''.)

By the adjoint relation, we see that giving $ P$ is equivalent to giving an $ \mathcal{O}_X$ -linear homomorphism

$\displaystyle D: \mathcal{V}\to
(p_1^{(1)})_* (p_2^{(1)}) ^* \mathcal{V}
=\mathcal J_1(\mathcal{V})
$

such that the composition

$\displaystyle \mathcal{V}\overset{D}{\to} \mathcal J_1(\mathcal{V}) \to \mathcal{V}
$

is equal to identity.

Now let us call $ \nabla=jet_1-D$ ``the covariant derivation''. Then $ \nabla$ is $ \mathcal{O}_S$ -linear homomorphism

$\displaystyle \nabla: \mathcal{V}\to \Omega^1_{X/S} \otimes \mathcal{V}.
$

In terms of the covariant derivation $ \nabla$ , the $ \mathcal{O}_X$ -linear is expressed as the following identity.

(Co) $\displaystyle \nabla (f m)=f \nabla(m)+d f\otimes m \quad (f \in \mathcal{O}_X, m\in \mathcal{V})
$

Let us put it in terms of rings and modules. Let $ X=\operatorname{Spec}(B)$ and $ S=\operatorname{Spec}(A)$ . $ I=I_{\Delta}$ be the defining ideal of the diagonal $ X\times_S X$ .

The $ \mathcal{O}_X$ -linear homomorphism $ D$ corresponds to a $ B$ -module homomorphism

$\displaystyle D:M\to ((B\otimes_A B)/I^2) \otimes_B M.
$

such that

$\displaystyle 1\otimes m -D(m) \in (I/I^2)\otimes_B M (\cong \Omega^1_{B/A} \otimes_B M)
$

holds for all $ m\in M$ . The left hand side of the above formula is $ \nabla(m)$ . Namely,

$\displaystyle \nabla(m)=jet_1(m)-D(m)=1\otimes m -D(m).
$

Let us verify the identity (Co).

      $\displaystyle \nabla(fm)=1\otimes fm -D(fm)$
      $\displaystyle =1\otimes fm -f D(m)$
      $\displaystyle =(1\otimes f -f\otimes 1) m+f(1\otimes m -D m)$
      $\displaystyle =d f \otimes m + f \nabla (m)$


next up previous
Next: direct sums, tensor products, Up: connection Previous: connection
2012-02-29