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According to Lee's book on Riemannian manifolds, a connection $\nabla$ on a smooth manifold $M$ is said to satisfy the flatness criterion if whenever $X,Y,Z$ are smooth vector fields defined on an open subset of $M$, the following identity holds: $$\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z=\nabla_{[X,Y]}Z.\tag{7.3}$$ After introducing this criterion, Lee says:

Proposition 7.2. If $(M,g)$ is a flat Riemannian or pseudo-Riemannian manifold, then its Levi-Civita connection satisfies the flatness criterion.

Its proof is shown below.

Proof. We just showed that the Euclidean connection on $\mathbb{R}^n$ satisfies (7.3). By naturality, the Levi-Civita connection on every manifold that is locally isometric to a Euclidean or pseudo-Euclidean space must also satisfy the same identity.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\square$

I think the naturality here refers to the naturality of the Levi-Civita connection on a manifold:

Proposition 5.13. Suppose $(M,g)$ and $(\widetilde{M},\widetilde{g})$ are Riemannian or pseudo-Riemannian manifolds, and let $\nabla$ denote the Levi-Civita connection of $g$ and $\widetilde{\nabla}$ that of $\widetilde{g}$. If $\phi:M\to\widetilde{M}$ is an isometry, then $\phi^*\widetilde{\nabla}=\nabla$.

But I don't know how to apply this proposition in understanding the proof of Proposition 7.2. Thank you.

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    $\begingroup$ I think what is meant by "by naturality" refers to the naturality of the Lie-bracket: $\phi^*[X,Y] = [\phi^*X,\phi^*Y]$, so that $\phi^*\left(\nabla_{[X,Y]}Z\right) = \left(\phi^*\nabla\right)_{\phi^*[X,Y]}\phi^*Z = \left(\phi^*\nabla\right)_{[\phi^*X,\phi^*Y]}\phi^*Z$, but I might be wrong $\endgroup$
    – Didier
    Sep 6, 2021 at 16:49
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    $\begingroup$ @Masacroso What? Levi-Civita is defined to be torsion-free, not curvature-free! $\endgroup$ Sep 6, 2021 at 17:57
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    $\begingroup$ @Masacroso: In Lee's book, a Riemannian manifold is defined to be flat if it is locally isometric to Euclidean space, i.e. every point has a neighbourhood that is isometric to an open subset of $\mathbb{R}^n$ with its Euclidean metric. Proposition 7.2 shows that this is equivalent to requiring the Levi-Civita connection to have curvature zero (which is what I'm guessing your comment was referring to). $\endgroup$ Sep 6, 2021 at 18:00
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    $\begingroup$ Returning to the question at hand, it follows from Didier's comment that each point $p$ of a flat Riemannian manifold $M$ has an open neighbourhood $U_p$ where $(7.3)$ holds, but this isn't quite sufficient (it has to hold for every open subset $U \subseteq M$, not just the open subsets $U_p$). It should follow from a partition of unity argument that $(7.3)$ holds for all open $U \subseteq M$ (i.e. the Levi-Civita connection satisfies the flatness criterion), but this would effectively boil down to the fact that the curvature is a tensor, which isn't proved until the next section. $\endgroup$ Sep 6, 2021 at 18:11
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    $\begingroup$ @Steve Since we're talking about diffeomorphisms, it's possible to write things equivalently in terms of pushforwards or pullbacks. The pushforward of $\varphi$ is just the pullback of $\varphi^{-1}$. $\endgroup$
    – Kajelad
    Sep 7, 2021 at 2:30

1 Answer 1

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Here's an answer that might be useful to math babies like me. I would only treat the Riemannian case, but the same trick should apply to pseudo-Riemannian manifolds. Let $\overline{g}$ be the Euclidean metric and let $\overline{\nabla}$ be the Euclidean connection. Since $M$ is flat, $\exists F:\mathbb{R}^n\to M$ s.t. $F$ is locally an isometry and each $p\in M$ has a neighborhood $U$ that is isometric to an open subset $V$ of $\mathbb{R}^n$. So there exists a diffeomorphism $\phi:V\to U$ such that $\phi^*g=\overline{g}$. Now pick $X,Y,Z\in\mathfrak{X}(U)$ and keep in mind that the Euclidean connection satisfies the flatness criterion. Then we would agree that $$\overline{\nabla}_{(\phi^{-1}\ )_*X}[\overline{\nabla}_{(\phi^{-1}\ )_*Y}(\phi^{-1})_*Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[\overline{\nabla}_{(\phi^{-1}\ )_*X}(\phi^{-1})_*Z]=\overline{\nabla}_{[(\phi^{-1}\ )_*X,(\phi^{-1}\ )_*Y]}(\phi^{-1})_*Z.$$ By Proposition 5.13, you know $\phi^*\nabla=\overline{\nabla}$. So, in order to show that $\nabla$ satisfies the criterion, you need to strip $\phi$ off the equation. Let's take care of each side of the equation separately. \begin{align} \text{LHS}&=\overline{\nabla}_{(\phi^{-1}\ )_*X}[(\phi^*\nabla)_{(\phi^{-1}\ )_*Y}(\phi^{-1})_*Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[(\phi^*\nabla)_{(\phi^{-1}\ )_*X}(\phi^{-1})_*Z]\\ &=\overline{\nabla}_{(\phi^{-1}\ )_*X}\{(\phi^{-1})_*\nabla_{\phi_*[(\phi^{-1}\ )_*Y]}\phi_*[(\phi^{-1})_*Z]\}\\ &\quad-\overline{\nabla}_{(\phi^{-1}\ )_*Y}\{(\phi^{-1})_*\nabla_{\phi_*[(\phi^{-1}\ )_*X]}\phi_*[(\phi^{-1})_*Z]\}\\ &=\overline{\nabla}_{(\phi^{-1}\ )_*X}[(\phi^{-1})_*\nabla_Y Z]-\overline{\nabla}_{(\phi^{-1}\ )_*Y}[(\phi^{-1})_*\nabla_X Z]\\ &=\text{Again!}\\ &=(\phi^{-1})_*(\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z)\\ \text{RHS}&=\overline{\nabla}_{(\phi^{-1}\ )_*[X,Y]}(\phi^{-1})_*Z\\ &=(\phi^*\nabla)_{(\phi^{-1}\ )_*[X,Y]}(\phi^{-1})_*Z\\ &=(\phi^{-1})_*\nabla_{\phi_*\{(\phi^{-1}\ )_*[X,Y]\}}\phi_*[(\phi^{-1})_*Z]\\ &=(\phi^{-1})_*\nabla_{[X,Y]}Z \end{align} Take a break and then conclude that $$\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z=\nabla_{[X,Y]}Z.$$ Thank all the comments that inspired me. You are the best!

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    $\begingroup$ Sorry, I found a deadly logical flaw in my answer: I didn't show the identity holds for vector fields defined on ANY open subset of $M$. $\endgroup$
    – Boar
    Sep 7, 2021 at 6:32
  • $\begingroup$ Does the answer still contain the logical flaw? $\endgroup$
    – Filippo
    Feb 16 at 8:17

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