Diffeomorphism invariance and geodesic action

I'm trying to understand the role of diffeomorphism and isometry invariance in the geodesic action in GR:

$$S = \int_{\tau_1}^{\tau_2} \! d\tau~ g_{ab}(x(\tau)) \frac{dx^a}{d\tau} \frac{dx^a}{d\tau}$$

Now if we transform coordinates with $y = y(x)$ and apply the usual transformation laws of the metric and tangent vectors than it is clear that $$L = g_{ab}(x(\tau)) \frac{dx^a}{d\tau} \frac{dx^a}{d\tau}$$ transforms as a scalar.

I am confused because we can equally consider this coordinate change as an "active" diffeomorphism $\phi: M \to M$ and then the statement that the action transforms as a scalar is that geodesics are mapped to geodesics under an arbitrary diffeomorphism. However I expect that is not true, it should only be true for diffeomorphims such that $\phi^* g = g$ i.e. isometries.

I'd like to understand how we can properly view the transformation of the action and see that geodesics are preserved only under isometries (say one-parameter isometries generated by a vector field $\xi^a$) and not under general diffeomorphisms. I imagine it should be possible to show that this action is preserved under a one parameter group of diffeomorphisms (generated by $\xi^a$) if and only if $\xi^a$ is Killing?

In particular I'm interested to understand how I should properly apply the transformation to $S$ (either active or passive) that corresponds to an isometry? And understanding the distinction between an isometry and a diffeomorphism in both the active and passive picture - i.e. if we can view every diffeomorphism as the identity map in the passive picture then, whilst it's obviously not true, this seems to me at the moment that every diffeomorphism is an isometry - I'd like to see why that is not the case.

• Possible duplicate of The Role of Active and Passive Diffeomorphism Invariance in GR – ACuriousMind Apr 11 '16 at 11:32
• See twistor's answer - in the active p.o.v. on coordinate changes, you have to drag the metric along the diffeomorphism and use the result as the metric after the new transformation. It maps geodesics in the old metric to geodesics in the new metric, but it is not an isometry of $(M,g)$ to itself like the diffeomorphisms generated by Killing fields are. – ACuriousMind Apr 11 '16 at 11:34
• Twistor's answer is great, however I'm really trying to understand the details of this in the particular case of the action and the difference between a diffeomorphism and an isometry (there is no talk of Killing fields in that answer), that's where I'm struggling. What do you mean by drag the metric along in this case? – Wooster Apr 11 '16 at 11:43
• Are you saying that $g_{ab}(x)$ should just become $g_{ab}(y)$? There also seems to be an issue in Twistor's answer in that the new metric under an active diffeomorphism should be written in the same coordinate basis i.e. $dx$ and not $dX$. – Wooster Apr 11 '16 at 11:44

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.