2
$\begingroup$

The de Rham complex on a smooth manifold $M$ is a complex of sheaves of $\mathbb{R}$-modules, which I will denote by $$\Omega^\bullet \xrightarrow{d} \Omega^{\bullet+1}.$$

A connection (not assumed to be flat!) on a vector bundle $E \to M$ with sheaf of sections $\mathcal{E}$ can be given by a covariant derivative, which is an $\mathbb{R}$-linear sheaf homomorphism:

$$\mathcal{E} \xrightarrow{\nabla} \mathcal{E} \otimes_{\mathcal{C}^\infty} \Omega^1 .$$

Is there an algebraic (in terms of category theory, sheaves and/or homological algebra) way to characterize the construction of the exterior covariant derivative $$\mathcal{E} \otimes_{\mathcal{C}^\infty} \Omega^\bullet \xrightarrow{d^\nabla} \mathcal{E} \otimes_{\mathcal{C}^\infty}\Omega^{\bullet+1}$$ from this data?

Naively, we might just stick stick zero sheaves on either side of the covariant derivative to get a complex, and then tensor with the de Rham complex (over the constant sheaf $\mathbb{R}$). But this is obviously wrong because we get a complex, and the connection is not assumed to be flat. Even in the flat case it is easy to see by looking at the individual sheaves that this approach gives the wrong complex.

$\endgroup$
3
$\begingroup$

For aesthetic reasons, I prefer to use $\Omega^{*} \otimes \mathcal{E}$ (you'll see immediately why) so let me do that. Note that $\Omega^{*} \otimes \mathcal{E}$ has the structure of a left graded $\Omega^{*}$ module and denote the corresponding product between a $k$-form an an $E$-valued $l$-form (resulting in an $E$-valued $(k + l)$-form) by $\wedge$. Then $d^{\nabla}$ is the unique operator which satisfies

$$ d^{\nabla}(\omega \otimes \xi) = d\omega \otimes \xi + (-1)^{|\omega|} \omega \wedge \nabla \xi = d \omega \wedge \xi + (-1)^{|\omega|} \omega \wedge \nabla \xi.$$

for all homogeneous $\omega \in \Omega^{*}(U)$ and $\xi \in \mathcal{E}(U)$. In other words, $d^{\nabla}$ extends $\nabla$ to a "degree $1$ graded derivation with respect to $d$".

$\endgroup$
  • $\begingroup$ Thanks! Does the existence of this unique operator follow from some sort of universal property? $\endgroup$ – ಠ_ಠ Mar 1 '17 at 0:57
  • $\begingroup$ Well, the formula above defines $d^{\nabla}$ explicitly for all $E$-valued $k$-forms. I wrote the "degree 1 graded derivation" in quotes because usually one considers graded derivations on some graded algebra (so that $d(ab) = da \cdot b + (-1)^{|a|} a \cdot db$ makes sense). In our case, this is only a graded left-module. I'm not an algebraist so I don't know if there is a proper terminology for this case. $\endgroup$ – levap Mar 1 '17 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.