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This is from Introduction to Relativity, By Begmann.

enter image description here

After equation 11.9a, Bergmann says:

Along any path $\Delta$ satisfies a linear, homogeneous differential equation of the first order. Therefore, it cannot vanish anywhere on that path if it does not vanish everywhere.

I don't understand why this is so. In particular in the case of a local Riemann normal coordinate system. In such a coordinate system, $\Gamma^i_{jk}=0$, and there is such a coordinate system available at every non-singular point of a Riemannian manifold.

Bergmann's assertion appears to imply that geodesically extending the coordinate curves of a Riemann normal system by a distance at which the effect of curvature cannot be neglected will maintain the vanishing of the affine connection coefficients. In other words, a local Riemann normal system cannot smoothly mesh with coordinate system having $\Gamma^i_{jk}\ne 0$.

To clarify, if $\Gamma^i_{jk}= 0$ then eq. 11.9a gives no information about the non-singularity of$\Delta$. If the curve along which the $b$ basis is parallel transported passes through a neighborhood where the coordinate system is Riemann normal, the connection coefficients vanish.

What am I getting wrong here?

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    $\begingroup$ Please don't post screenshots of text. It breaks search functionality, and it doesn't work for blind users. $\endgroup$ – user4552 Dec 21 '19 at 1:21
  • $\begingroup$ I live in my car, and cannot use a computer when the library is closed. I am posting from an Android device without a keyboard. $\endgroup$ – Steven Thomas Hatton Dec 21 '19 at 1:27
  • $\begingroup$ Android devices do have keyboards. $\endgroup$ – Kyle Kanos Dec 21 '19 at 1:28
  • $\begingroup$ The keyboard stopped working, and Walmart refused to honor their warranty on a Walmart branded tablet. $\endgroup$ – Steven Thomas Hatton Dec 21 '19 at 1:32
  • $\begingroup$ If you have a touch screen, then you have plenty of keyboard apps to choose from. If you don't have a touch screen, how are you typing the responses in? $\endgroup$ – Kyle Kanos Dec 21 '19 at 1:45
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When Bergmann says "Along any path Δ satisfies a linear, homogeneous differential equation of the first order," he does not specify that the equation has constant coefficients. I see nowhere that he states that the connection coefficients are non-vanishing in the neighborhood and on the coordinate system under discussion, but it is clear (to me) that this qualification is necessary in order for $\delta\Delta=\Delta\Gamma^k_{ks}\delta\xi^s\ne0.$ Furthermore, it is possible to parallel transport $\{\overset{i}{b}\}$ to a point where $\Delta\Gamma^k_{ks}=0$ while preserving their linear independence. So, I conclude that Bermann's statement is technically incorrect.

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  • $\begingroup$ There's just one problem with the down-vote. I am right. $\endgroup$ – Steven Thomas Hatton Dec 22 '19 at 9:06
  • $\begingroup$ Note, $\Delta\Gamma^k_{ks}$, are two different objects, $\Delta$ and $\Gamma^k_{ks}$. See equation $(11.9a)$. It should be clear $\Delta$ doesn't vanish. Hence, you still need to prove $\Gamma^k_{ks}$ vanishes. $\endgroup$ – Cinaed Simson Dec 22 '19 at 20:12
  • $\begingroup$ And why is the word "determinant" in the title? The author doesn't calculate a determinant in the page you posted - the author doesn't even mention the word. $\endgroup$ – Cinaed Simson Dec 22 '19 at 20:17
  • $\begingroup$ @CinaedSimson $\Delta$ is a determinant. It is a rank-0 tensor. Summation on the contravariant indices of vector with the indices of the corresponding covariant basis vectors is not the contraction of two rank-1 tensors. There are various interpretations of the result. One interpretation is that it is the contraction of a rank-1 tensor with a rank-2 tensor (formed of the components of the basis). I added a clarifying paragraph to my question. $\endgroup$ – Steven Thomas Hatton Dec 22 '19 at 23:12

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