Postgresql 11 logical replication - stuck in `catchup` state

The question is subject to interpretation, because the Langlands program was never a clearly delimited set of conjectures to begin with, and moreover many things were added during the following decades, and the answer depends if you want to consider these innovations as part of the Langlands program or not.

I believe nevertheless that essentially everything in the original Langlands program (of the 70's) is known for $GL_1$, and actually was known long before Langlands began working on his program. Every instance of Langlands functoriality involving only $GL_1$ results easily from Class field theory. As for reciprocity, if the existence of the Langlands group is still unknown, only its abelianization matters for $GL_1$, and we know, again by CFT, what this abelianization is.

If we add to the Langlands program the correspondance of algebraic automorphic forms with Galois representations, as for example formulated in Clozel's 1988 paper "Motifs et Formes Automorphes" in Clozel-Milne, then again the case of $GL_1$ was well-known long before, and due to André Weil (that's his theory of algebraic Grössencharacters).

But if you are willing to consider the theory of $p$-adic automorphic forms and their families as part of the Langlands program, then there is one thing that is still unknown: the dimension of the eigenvariety of automorphic forms for $GL_1$ over a number field. Determining this dimension is equivalent to proving Leopoldt's conjecture.

A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples

• You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point $x$ and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing through point $x$ will certainly be mapped to some weird curve, no longer a geodesic on the plane.
• Consider the upper half plane $\mathbb E^+ \equiv \mathbb R^+ \times \mathbb R$, with the Euclidean metric. Consider the poincare half plane $\mathbb H^2$ with the hyperbolic metric. There is an obvious diffeomorphism between the two - the identity map. Under the identity map, straight lines gets mapped to ... well, themselves. So geodesics on the plane do not get mapped to geodesics in hyperbolic plane under this diffeomorphism.

In particular, the diffeomorphism invariance of the geodesic functional, which you (pretty much) correctly showed certainly doesn't imply geodesics get mapped to geodesics. So let's see what this actually implies.

Diffeomorphisms do not map geodesics to geodesics

Let $M$ be our manifold. Let $g$ be a Riemannian metric on $M$. Let $S_g$ denote the energy functional (that you wrote down) using the metric $g$. That is, let $\gamma: [0, 1] \to M$ be a smooth curve,

$$ S_g[\gamma] \equiv \int_{[0, 1]} g_{\gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau $$

You found (by computing in local coordinates) that this is invariant under a diffeomorphism $\phi: M \to M$. This statement reads

$$ \int_{[0, 1]} g_{\gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau=\int_{[0, 1]} g_{\phi \circ\gamma(t)}(\phi_*\gamma'(\tau), \phi_*\gamma'(\tau))d\tau $$

RHS can be rewritten in terms of the pullback metric

$$ \int_{[0, 1]} g_{\phi \circ\gamma(t)}(\phi_*\gamma'(\tau), \phi_*\gamma'(\tau))d\tau = \int_{[0, 1]} \phi^*g_{\phi \circ \gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau $$

Comparing with our definition of $S_g$, what you have shown is $$ S_g [\gamma]= S_{\phi^*g}[\phi \circ \gamma] $$

In particular this implies, for the special case that you have a curve $\gamma$ that minimizes $S_g$ within a variational family of curves:

$$ \text{ $\gamma$ minimizes $S_g$ } \implies \text{ $\phi\circ\gamma$ minimizes $S_{\phi^*g}$} $$

Observe this does not mean $\phi \circ \gamma$ is a geodesic for the metric $g$. This says $\phi \circ \gamma$ is a geodesic for the (in general, different) metric $\phi^*g$. Therefore a general diffeomorphism does not map geodesics to geodesics.

Isometries do!

Now observe there is a special case when $\gamma$ actually does get mapped to a geodesic of $g$. Namely, when we have $$\phi^*g = g \implies S_g = S_{\phi^*g}$$

And the above implication becomes

$$ \text{ $\gamma$ minimizes $S_g$ } \implies \text{ $\phi\circ\gamma$ minimizes $S_{g}$} $$

Observe the condition $\phi^*g = g$ is exactly the statement that $\phi$ is an isometry.