Either the differential of the determinant of a basis vanish everywhere or nowhere along a path: what does this imply?

This is from Introduction to Relativity, By Begmann.

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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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.
• 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. – Cinaed Simson Dec 22 '19 at 20:12
• @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. – Steven Thomas Hatton Dec 22 '19 at 23:12