The Index Form

This post has been migrated from my old blog, the math-physics learning seminar.

Let \(f: [0,T] \times (-\epsilon, \epsilon) \to M\) be a family of parametrized curves in a Riemannian manifold \((M, g)\). To simplify this calculation, we assume that \(f(0,s) = p, f(T, s) = q\) for some \(p,q \in M\) and all \(s \in (-\epsilon, \epsilon)\). (This assumption is not necessary, but without it our variational formulae will have additional boundary terms.)

For convenience, set \(\dot f = \partial f / \partial t\) and \(f' = \partial f / \partial s\). For each \(s \in (-\epsilon, \epsilon)\) we define the energy functional \(E = E(s)\) to be

\[ E(s) = \frac{1}{2} \int_0^T |\dot f|^2 dt. \]

The first variation is

\[ \begin{aligned} \frac{dE}{ds} &= \int_0^T \langle \nabla_{f'} \dot f, \dot f \rangle dt \cr &= \int_0^T \langle \nabla_{\dot f} f', \dot f \rangle dt \cr &= -\int_0^T \langle f', \nabla_{\dot f}\dot f \rangle dt \end{aligned} \]

Set \(\gamma(t) := f(t,0)\) and \(X(t) = f'(t)\) (thought of as a vector field supported on \(\gamma\)). Evaluating the above at \(s=0\) we obtain

\[ \left.\frac{dE}{ds}\right|_{s=0} = -\int_0^T \langle X, \nabla_{\dot \gamma} \dot \gamma \rangle dt, \]

which shows immediately that

Theorem. \(\gamma\) is a critical point of the energy functional if and only if \(\nabla_{\dot \gamma} \dot \gamma = 0\).

The second variation is

\[ \begin{aligned} \frac{d^2 E}{ds^2} &= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle + \langle f', \nabla_{f'}\nabla_{\dot f}\dot f \rangle dt \cr &= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle + \langle f', \nabla_{\dot f}\nabla_{f'}\dot f \rangle dt + \langle f', R(f', \dot f)\dot f \rangle dt \cr &= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle - \langle \nabla_{\dot f}f', \nabla_{f'}\dot f \rangle dt + \langle f', R(f', \dot f)\dot f \rangle dt \cr &= -\int_0^T \langle \nabla_{f'}f', \nabla_{\dot f}\dot f \rangle - \langle \nabla_{\dot f} f', \nabla_{\dot f} f'\rangle dt + \langle f', R(f', \dot f)\dot f \rangle dt \end{aligned} \]

Assume now that \(\gamma\) is a geodesic, i.e. \(\nabla_{\dot \gamma} \dot \gamma = 0\). Then evaluating the above at \(s=0\), we obtain

\[ \frac{d^2 E}{ds^2} = \int_0^T |\nabla_{\dot \gamma} X|^2 - \langle X, R(X, \dot \gamma) \dot \gamma \rangle dt. \]

Definition. Let \(\gamma\) be a geodesic. The index form associated to variations \(X,Y\) of \(\gamma\) is

\[ \begin{aligned} I(X,Y) &= \int_0^T \langle \nabla_{\dot \gamma} X, \nabla_{\dot \gamma} Y \rangle dt - \langle Y, R(X, \dot \gamma) \dot \gamma \rangle \cr &= -\int_0^T \langle Y, \nabla_{\dot \gamma}^2 X + R(X, \dot\gamma)\dot \gamma \rangle \end{aligned} \]

It follows from symmetries of the Riemann tensor that \(I(X,Y) = I(Y, X)\) and also \(I(X,X) = E''\) as above.

Theorem. Suppose that \(X\) is the infinitesimal variation of a family of affine geodesics about a fixed geodesic \(\gamma\). Then

\[ \nabla_{\dot \gamma}^2 X + R(X, \dot\gamma)\dot\gamma = 0. \]

In particular, \(I(X, -) = 0\).

Proof. Let \(f(t,s)\) denote the family as above. By hypothesis, we have that \(\nabla_{\dot f} \dot f = 0\) for all \(s\), so that

\[ \nabla_{f'} \nabla_{\dot f} \dot f = 0. \]

Commuting the derivatives using the curvature tensor, we have

\[ 0 = \nabla_{\dot f} \nabla_{f'} \dot f + R(f', \dot f) \dot f. \]

Now use \(\nabla_{\dot f} f' = \nabla_{f'} \dot f\) and evaluate at \(s=0\) to obtain

\[ 0 = \nabla_{\dot \gamma}^2 X + R(X, \dot \gamma)\dot\gamma. \]