In most texts on Riemannian geometry that I've seen, (Riemannian) geodesics are first defined as smooth curves $x: [0, 1] \to M$ on a Riemannian manifold $(M, g)$ satisfying $D_t \dot{x} = 0$, where $D_t$ denotes the Levi-Civita connection corresponding to $g$ taken along $x$. It is then shown via variational analysis that critical points of the length/energy functional (let's say energy functional for convenience) $E(x) = \int_0^1 g(\dot{x}, \dot{x})dt$ are geodesics.

The domain of $E$ is often chosen as piecewise smooth curves satisfying some boundary conditions in position. However, one could reasonably define $E$ over curves with less regularity (for example $H^1$ regularity). To that end, let $\Omega \subset H^1([0, 1], M)$ be the space of Sobolev class $H^1$ curves on $M$ satisfying some fixed boundary conditions in position, and reinterpret $E$ as a functional $E: \Omega \to \mathbb{R}$. $\Omega$ is a smooth Hilbert manifold, and the tangent space $T_x \Omega$ consists of Sobolev class $H^1$ vector fields along $x$ which vanish at the endpoints. Hence, given $x \in \Omega$ and $V \in T_x \Omega$, we can calculate $E'(x)V$ by considering a family of curves $x_s: (-\epsilon, \epsilon) \times [0, 1] \to M$ such that $x_s \in \Omega$ for all $s \in (-\epsilon, \epsilon)$, $x_0 = x$, and $\partial_s x_s \vert_{s=0} = V$, and then calculating $\partial_s E(x_s) \vert_{s=0}$. In particular, we find:

\begin{align*} E'(x)V &= \int_0^1 g(D_s \dot{x}_s, \dot{x}_s)\big\vert_{s=0}dt \\ &= \int_0^1 g(D_t V, \dot{x})dt \end{align*}

Which vanishes for all $V \in T_x \Omega$ if $x$ is a critical point of $E$. An article I read (dealing with a problem that is of higher order, but similar in spirit) seems to suggest that one could prove that any critical point of $E$ is smooth via a "bootstrap method," and from there the proof continues as it normally would (by integrating by parts and then choosing $V = D_t \dot{x}$). I'm not familiar with bootstrap methods in this context, could anyone help show me how such an argument may go in this case?

More concretely, how could I show that the $H^1$ regularity of $x$ together with the condition $E'(x)V = \int_a^b g(\dot{x}, D_t V)dt = 0$ for all $V \in T_x \Omega$ implies that $x$ is $H^2$, and inductively that $x$ is smooth?


1 Answer 1


ADDED: There two key observations.

One is that if $u, u' \in L^1([0,1])$, then $u$ is continuous and satisfies the fundamental theorem of calculus. There are (at least) two ways to show this. One is to apply the fundamental theorem of calculus to a smooth approximation of $f$ and take a limit. The other is to use, for each $\epsilon > 0$, a nonnegative test function $\chi_\epsilon$ that is equal to $1$ on $[a+\epsilon, b-\epsilon]$ and zero at $a$ and $b$. You then show that $$ \lim_{\epsilon\rightarrow 0} \int_0^1 u'\chi_\epsilon\, dt = u(b) - u(a). $$

The second is that, if $V$ is assumed to be smooth and zero at $t= 0, 1$, then in local coordinates, integration by parts implies \begin{align*} 0 &= \int_0^1 g(D_tV,\dot{x})\,dt\\ &= \int_0^1 g_{ij}(\dot{V}^i + \Gamma^i_{pq}V^p\dot{x}^q)\dot{x}^j\,dt\\ &= \int_0^1 -g_{ij}V^j(\ddot{x}^i + \Gamma^i_{pq}\dot{x}^p\dot{x}^q)\,dt \end{align*} Since this holds for any smooth $V$ supported in $(0,1)$, it follows that $x$ is a weak solution to $$ \ddot{x}^i = \Gamma^i_{jk}\dot{x}^j\dot{x}^k. $$ Since the right side is in $L^1$, so is the left. This implies that $$ \dot{x}(t) = \dot{x}(0) + \int_0^t \Gamma^k_{ij}\dot{x}^i\dot{x}^j\,dt. $$ is a continuous function of $t$. This implies that $x$ is a strong solution to the differential equation. Now the argument below applies.

This is covered in most textbooks on ODEs.

The geodesic equation can always be written as a first order system of ODEs, $$ \dot{u} = \Phi(u), $$ by setting $u = (x,\dot{x})$. Integrating this, $u$ satisfies $$ u(t) = u(0) + \int_{s=0}^{s=t} \Phi(u(s))\,ds. $$ Since $u$ is known to be $C^0$, this equation implies that $u$ is $C^1$.

Bootstrapping means applying the above argument to the derivatives of $u$, which themselves are solutions to a system of ODEs obtained by differentiating the original system. For example, $v = (u,\dot{u})$ satisfies $$\dot{v} = \Psi(v),$$ where $\Psi$ is obtained by differentiating the right side of the original system. The argument above shows that $v$ is $C^1$ and therefore $u$ is $C^2$. If $\Phi$ is a smooth function of $u$, then this argument can be continued indefinitely, implying that $u$ is smooth.

  • $\begingroup$ Thank you for the answer, but I don't see the relevance to my particular problem. I'm not trying to show that geodesics are smooth, but rather that the critical points of $E$ over $\Omega$ (curves of class $H^1$) are smooth, from which the geodesic equation will follow. In principle, it is not know that the critical points satisfy the geodesic equation (or any ODE for that matter), only that $\int_0^1 g(\dot{x}, D_t V)dt = 0$ for all vector fields $V \in T_x \Omega$. $\endgroup$ Commented Jan 27, 2023 at 9:28
  • 1
    $\begingroup$ You asked about proving smoothness using bootstrapping (as indicated by the title), so I explained that step. I agree that the first step is a crucial one, but it is not proved using bootstrapping. $\endgroup$
    – Deane
    Commented Jan 27, 2023 at 20:20

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