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A connection of a vector bundle $E$ on a manifold $M$ is a map $d_E: \Omega^0(E) \to \Omega^1(E)$ that extends uniquely to a map $d_E: \Omega^p(E) \to \Omega^{p+1}(E)$ while satisfying $$ d_E(\omega \otimes s) = d\omega \otimes s + (-1)^p \omega \wedge d_Es. (*) $$ The curvature tensor is defined to be $d_E \circ d_E.$

Now, in very elementary differential geometry, the curvature (Gaussian or others) can be motivated by second order terms in Taylor series expansion of a parameterisation.

This leads to the question: can we write a Taylor series or something similar with $d_E$? That would really help understanding curvature.

I know that a perfect Taylor series is not going to work, because the definition $(*)$ takes such a "strange" form. But at least, $d_E \circ d_E$ should be able to capture the "curved" part of the second order terms.

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  • $\begingroup$ You can pick a torsion-free connection $\nabla^{TX}$ on $TX$ and consider the iterated connection derivatives $\nabla^{T^*X\otimes\dots\otimes T^*X\otimes E}\circ\dots\circ\nabla^{T^*X\otimes E}\circ \nabla^E:\Gamma(E)\to \Gamma((T^* X)^{\otimes n}\otimes E)$. Note that the antisymmetrization of any two of the $n$ factors of $T^*X$ in the result can be expressed via covariant derivatives of the curvature tensors of $\nabla^{TX}$ and $\nabla^E$. The projections $\Gamma(E)\to \Gamma((\operatorname{Sym}^n T^*X)\otimes E)$ are then the correct generalization of the Taylor coefficients. $\endgroup$
    – Bertram Arnold
    Commented Aug 10, 2021 at 7:37
  • $\begingroup$ @BertramArnold So what's the form of the series? $\endgroup$
    – Ma Joad
    Commented Aug 11, 2021 at 5:07
  • $\begingroup$ It's the usual Taylor series in a local chart, but you re-express the partial derivatives via connection derivatives. The point is that the complicated transformation of iterated derivatives under a change of chart, known as Faà di Bruno's formula, can be absorbed into the transformation of the curvature form. For an intrinsic formulation which does not require a choice of connection, you might want to check out the formalism of jet bundles. $\endgroup$
    – Bertram Arnold
    Commented Aug 11, 2021 at 8:47
  • $\begingroup$ @BertramArnold Are you saying that jet bundles, (which I know a choice of cross section of is essentially a taylor series expansion) already incorporate a covariant derivative into the Taylor expansion? $\endgroup$
    – R. Rankin
    Commented Mar 23 at 4:31

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