general relativity - Is the Invariant interval S between two points independent of the path taken?

Due to a misunderstanding of what I was asking, I'm re-asking this question (Is the Invariant interval S between the singularity and the present, the same for any point in space in an FLRW universe?) in a much more general sense

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather (as per Hamilton's General Relativity, Black holes, and cosmology equation 2.13):

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

Because it is a scalar $ds$ may be written as an exact differential form (As per Hamilton equation 2.11 referenced above):

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}$$

Where summation over \mu is implied. Note that (for geometric consistency) $\frac{\partial s}{\partial x_{\mu}}$ can be identified with the metric:

$$ds^{2}=\left(\frac{\partial s}{\partial x^{\mu}}dx^{\mu}\right)\left(\frac{\partial s}{\partial x^{\nu}}dx^{\nu}\right)$$

$$=\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)dx^{\mu}dx^{\nu}$$

Which implies that:

$$\left(\frac{\partial s}{\partial x^{\mu}}\frac{\partial s}{\partial x^{\nu}}+\frac{\partial s}{\partial x^{\nu}}\frac{\partial s}{\partial x^{\mu}}\right)=\left\{ \frac{\partial s}{\partial x^{\mu}},\frac{\partial s}{\partial x^{\nu}}\right\} =g_{\mu\nu}$$

But this expression is familiar, the generalized gamma matrices $\gamma^{\mu}$ are defined by:

$$\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$$

Which means that $\frac{\partial s}{\partial x_{\mu}}$ can be identified with the generalized gamma matrices:

$$\frac{\partial s}{\partial x^{\mu}}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

The second equation can also be written as:

$$ds=\frac{\partial s}{\partial x_{\mu}}dx_{\mu}=\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Integrating now over some arbitrary interval $\{a,b\}$:

$$S=\intop_{a}^{b}\overrightarrow{\nabla}s\cdot d\overrightarrow{r}$$

Via the fundamental theorem of calculus, it is clear that the interval is independent of the path taken between the points $\{a,b\}$. Did I mess this up somewhere?

EDIT: Here's an approach not assuming s is an exact differential (as per the the objections voiced below)

The invariant interval is defined as:

$$ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$$

or rather:

If we wish, this can be rearranged as:

$$0=dx\cdot g\cdot dx-ds^{2}=dx^{\mu}g_{\mu\nu}dx^{\nu}$$

One can write the metric tensor in terms of local basis:

$$g_{\mu\nu}=e_{\mu}\bullet e_{\nu}$$

Where $\cdot$ denotes the standard dot product and $\bullet$ the tensor product. Used in the preceding equation, the above yields:

$$0=\left(e_{\mu}dx^{\mu}\right)\bullet\left(e_{\nu}dx^{\nu}\right)-ds^{2}$$

(note $dx\cdot e=dx^{\mu}e_{\mu}$) This can be simply factored to obtain:

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\bullet\left(e_{\nu}dx^{\nu}+ds\right)$$ Which, for a given metric, gives two different solutions to $ds$.

$$0=\left(e_{\mu}dx^{\mu}-ds\right)\qquad0=\left(e_{\nu}dx^{\nu}+ds\right)$$

Algebraically, this corresponds to the Clifford algebra as we have the relationship:

$$\left\{ e_{\mu},e_{\nu}\right\} =e_{\mu}e_{\nu}+e_{\nu}e_{\mu}=g_{\mu\nu}$$

Which means that $e_{\mu}$ can be identified with the generalized gamma matrice $\gamma_{\mu}$:

$$e_{\mu}=\frac{1}{\sqrt{2}}\gamma_{\mu}$$

Taking either solution for ds individually, the integral for s between two nearby points appears to be independent of the path taken. Note that either solution to $s$ individually could not be considered as proper time between events, but is simply a geometrical invariant.

The relationship between our starting equation and our two solutions now is entirely analogous to that between the Klein gordon equation and the Dirac equation. Solutions to the former are not necessarily solutions to the latter. Apparently no-one liked my first question using differential forms, so I wrote it up this way.

Also, if it eases concerns of undefined intervals, one can simply consider a flat space, since this argument itself is general.

Answer

I agree with ACM's answer; the conclusion is wrong, and there are many counterexamples.

For example, in the $(+---)$ metric, the invariant interval $\int ds$ is equal to the proper time elapsed, for a purely timelike path. But we know that proper time depends on the path taken, for example in the twin paradox, where the moving twin comes back younger.