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While following a lecture series on General Relativity, an argument was presented that in order the spacetime to be flat, a vector parallel transported along two different paths should yield the same vector. For this the required condition is the integrability of $$\partial _{i}X^{a}+\Gamma ^{a}_{mi}X^{m}=0$$ i.e. the commutation of partial derivatives $\partial _{i}\partial _{j}X^{a}=\partial _{j}\partial _{i}X^{a}$. This condition gives us the fact that for a symmetric connection the Curvature tensor should be zero. My questions are

  1. For a manifold for whom the curvature tensor is non-zero, can we use the commutation of partial derivatives or not? Why or why not so.
  2. Even if the curvature tensor is not zero everywhere on the manifold, it can be zero at a specific point. Does the curvature tensor being zero at a point give any new and further information about the point like about its flatness or something?
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Partial derivatives always commute (for sufficiently smooth functions). The standard calculus proofs apply, and it doesn’t matter that there is this extra Riemannian structure of a metric etc.

The second question is about semantics. If the curvature vanishes at a point, and we drag a vector around a finite size loop, it will generally not return to itself. So there is in general no finite region for which we can say that space is flat. But for very small loops the vector will approximately return to itself, so very close to the point the space will generally be approximately flat.

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  • $\begingroup$ If Partial derivatives always commute then why do we use the commutator condition to prove the flatness of the manifold? $\endgroup$ – Naman Agarwal Jan 9 '18 at 6:03
  • $\begingroup$ I don’t understand the question. If we assume that a vector doesn’t change under parallel transport in a closed loop, then the Riemann tensor vanishes. The calculation relies on partial derivatives commuting, which they always do. If a vector does change under parallel transport in a loop then the same calculation shows that the Riemann tensor does not vanish. $\endgroup$ – Guy Gur-Ari Jan 9 '18 at 6:26
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Partial derivatives commute, but covariant derivative $\nabla_a$ do not in general. (Your integrability equation can be restated as $\nabla_i X^a=0$.) These derivatives are of interest because they send tensors to tensors. If $\phi$ is a scalar and $V_a$ is a vector, $[\nabla_a,\,\nabla_b]\phi=0$ and $[\nabla_a,\,\nabla_b]V_c=R_{abcd}V^d$, where $R_{abcd}$ is the Riemann tensor. When commutators of covariant derivatives act on higher-rank tensors, they generally also include a differentiating term.

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  • $\begingroup$ In Ray d'inverno's book Introducing Einstein's Relativity, he mentions the fact that "A necessary condition for a solution of this first-order partial differential equation is $\partial _{i}\partial _{j}X^{a}=\partial _{j}\partial _{i}X^{a}$" where the equation was $\partial _{i}X^{a}+\Gamma ^{a}_{mi}X^{m}=0$. Now if the partial derivatives commute as such, why is that given as a condition? $\endgroup$ – Naman Agarwal Jan 9 '18 at 8:34
  • $\begingroup$ @NamanAgarwal I'll try to look that up in context if you tell me where he makes that comment. $\endgroup$ – J.G. Jan 9 '18 at 8:36
  • $\begingroup$ On Page no. $79$; under the section name "Affine flatness". Chapter 6 $\endgroup$ – Naman Agarwal Jan 9 '18 at 8:39
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    $\begingroup$ d'Inverno has explained that very badly. Reading his argument, what I think he meant to say is that, since our effort to construct a parallel-transported vector field from $X^a$ at a point requires covariant derivative to commute on the field, the fact that partial derivatives do as well implies $R_{abcd}X^d=0$. $\endgroup$ – J.G. Jan 9 '18 at 8:49
  • $\begingroup$ If this is the case then I think my question has been answered. Thanks $\endgroup$ – Naman Agarwal Jan 9 '18 at 9:04
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  1. If the curvature tensor is not zero, the covariant derivatives do not commute by construction. The partial derivatives do always commute by definition, but by themselves can not say anything on the flatness or curvature of a manifold.
  2. If the curvature tensor is zero at a specific point, we may think that in a small enough region around the point the manifold is about flat, i.e. the connection part is negligible compared to the partial derivative (provided the functions are sufficiently smooth).
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