Application: Morse Index Theorem
Let \((M,g)\) be a smooth Riemannian manifold, with Levi–Civita connection \(\nabla\). See the connection notes for the notation.
The Morse index theorem relates the second variation along a geodesic to its conjugate points. It states that the index of every geodesic is finite, and is equal to the number of its interior conjugate points, counted with multiplicity. The index measures the number of independent directions in which a geodesic \(\gamma\) can be deformed to decrease its length.
Definition 1. For the index form \(I\), its index is the maximal dimension of all subspaces such that \(I(X,X) < 0\) (i.e. negative definite). The nullity of \(I\) is the dimension of the subspace formed by \(V\) such that \(I(V,W)=0\) for all \(W\). Finally, \(I\) is said to be degenerate if its nullity is strictly positive.
Here negative definite means \(I(X,X)<0\) for every nonzero vector of the subspace. We work along a smooth geodesic on a compact interval \([a,b]\), \(a<b\), with fields vanishing at both endpoints. The index is initially the supremum of the possible dimensions; the finiteness proof below makes it an attained finite maximum. The null space means the radical \(\{X:I(X,Y)=0\text{ for all }Y\}\), not the larger cone \(\{X:I(X,X)=0\}\).
Theorem 2 (Morse Index Theorem). The index of the form \(I(X,X)\) is finite. Moreover, this is given by the number of points \(\gamma(t)\) conjugate to \(\gamma(a)\), where \(a < t < b\), each counted with multiplicity.
The form here acts on endpoint-vanishing fields on a compact geodesic interval \([a,b]\), \(a<b\). The full proof, including its Sobolev completion and the endpoint-nullity distinction, is Theorem 10 below. Conjugate parameter times are counted, even if the geodesic revisits the same point of the manifold.
First identify the null space of the index form. Integration by parts shows why the Jacobi equation appears: it is the differential equation left after testing against all fields that vanish at the endpoints.
Proposition 3. A proper normal vector field \(V\) belongs to the null space of \(I\) if and only if \(V\) is a Jacobi field along \(\gamma\).
Proof. Write \(T=\gamma'\), and use the convention \(I(V,W)=\int(\langle V',W'\rangle-\langle R(T,V)T,W\rangle)dt\). For continuous piecewise smooth \(V\), integration on each smooth piece gives
where the endpoint terms vanish because \(W(a)=W(b)=0\). If \(V\) is Jacobi, the integrand and jumps vanish. Conversely choose a nonnegative bump supported in the interior of one piece and set \(W=f(V''+R(T,V)T)\). The expression in parentheses is normal, so this is an allowed test field; the integral of its squared norm is zero and the Jacobi equation holds on that piece. Then test at each junction with a normal field whose value is its derivative jump and whose values at all other junctions vanish. The formula forces every jump to vanish. Matching values and derivatives let ODE uniqueness join the pieces into one smooth Jacobi field. The jump step is essential: being Jacobi separately on pieces alone does not give a global Jacobi field. ◻
We pass to the completion of the space of vector fields in a Sobolev norm. This gives a setting in which bounded sequences have the compactness properties needed to prove finiteness of the index.
Definition 4. Let \(\mathcal E_0(\gamma)\) be the space of smooth vector fields along \(\gamma\) with \(X(a)=X(b)=0\). Define
Its completion is denoted \(\mathscr H_\gamma^1\). We use \(\mathcal E\) here to distinguish arbitrary fields from the Jacobi-field spaces \(\mathscr J^t\) below.
Remark 5. In a parallel orthonormal frame along \(\gamma\), write \(X=x^iE_i\), so \(D_tX=(x^i)'E_i\). This identifies \(\mathscr H_\gamma^1\) with \(H_0^1((a,b);\mathbb R^d)\), whose fields have continuous representatives and zero endpoint traces. Put
The operator \(A(t)\) is smooth and self-adjoint, and \(\|A(t)\|\leq C\) on the compact interval. Hence \(|I(X,Y)|\leq(1+C)\|X\|_{H^1}\|Y\|_{H^1}\), so the form extends continuously to the completion.
For a nonconstant geodesic, split \(X=fT/|T|+X^\perp\). Parallelness of \(T\) and the curvature symmetries give \(I(X,X)=\int|f'|^2+I(X^\perp,X^\perp)\). Thus tangent components contribute neither negative directions nor nullity with zero endpoints. We may use all fields or only normal fields with the same index and nullity. A constant geodesic has \(A=0\) and zero index and nullity.
Passing to the \(H^1_0\) completion does not change the index. Smooth endpoint-zero fields are dense in \(H^1_0\), and the index form is continuous in that norm. A finite-dimensional negative subspace has a uniform strictly negative bound on its unit sphere; approximating a basis by smooth fields preserves linear independence and that bound. The reverse inequality follows by inclusion. The radical is the space of smooth endpoint-zero Jacobi fields, as proved below, so the corresponding extended index agrees as well.
Definition 6. The index of \(\gamma\), denoted \(\operatorname{Ind}(\gamma)\), is the dimension of the maximal subspace of \(\mathscr H_\gamma^1\) on which the index form is negative definite. The extended index of \(\gamma\), denoted \(\operatorname{Ind}_0 (\gamma)\), is the dimension of the maximal subspace of \(\mathscr H_\gamma^1\) on which the index form is negative semidefinite. The nullity of \(\gamma\) is defined as \(N(\gamma) \triangleq \operatorname{Ind}_0(\gamma) - \operatorname{Ind}(\gamma)\).
Lemma 7. \(\operatorname{Ind}(\gamma)\) and \(N(\gamma)\) are finite.
Proof. In a nonpositive subspace, every \(L^2\)-unit field satisfies
Thus all such unit fields, from all possible nonpositive subspaces, lie in one bounded \(H_0^1\) set. On an interval this set is precompact in \(L^2\). Indeed Cauchy–Schwarz gives \(|X(t)-X(s)|\leq\|X'\|_{L^2}|t-s|^{1/2}\), and the zero endpoint gives a uniform sup norm bound; Arzelà–Ascoli produces uniform, hence \(L^2\), convergence of a subsequence. This is the interval version of Rellich compactness.
A precompact metric set has a finite cover by balls of radius less than \(\sqrt2/2\). An \(L^2\)-orthonormal family has pairwise distance \(\sqrt2\), so contains at most one point in each such ball. Applying Gram–Schmidt within any finite-dimensional nonpositive subspace gives a uniform bound on its dimension. This proves \(\operatorname{Ind}_0<\infty\), and hence finiteness of the index and its difference \(N\). The uniform bound also rules out arbitrarily large finite-dimensional subspaces, without assuming in advance that they lie in one infinite-dimensional nonpositive subspace. ◻
Definition 8. For each \(t \in (a,b]\), let \(\mathscr J^t (\gamma)\) denote the space of Jacobi fields along \(\gamma : [a,b] \to M\) such that \(X(a) = 0\) and \(X(t) = 0\).
Lemma 9. On \(\mathscr H_\gamma^1\), \(N(\gamma) = \dim \mathscr J^b (\gamma)\).
Proof. First justify that the difference of the two indices equals the dimension of the radical. Choose a maximal negative-definite subspace \(E\), finite-dimensional by Lemma 7. Nondegeneracy of \(I|_E\) gives the algebraic decomposition \(\mathscr H_\gamma^1=E\oplus E^{\perp_I}\). On \(W=E^{\perp_I}\), the form is nonnegative, since a vector with negative square would enlarge \(E\). For a nonnegative bilinear form, \(I(w,w)=0\) implies \(I(w,z)=0\) for all \(z\), by considering \(I(w+tz,w+tz)\) as a quadratic polynomial. Thus the null vectors of \(I|_W\) are precisely the full radical \(K\).
For any nonpositive subspace \(L\), the kernel of its projection to \(E\) is contained in \(K\), because it lies in \(W\). Hence \(\dim L\leq\dim E+\dim K\). Equality is attained by \(E\oplus K\), giving \(N=\dim K\).
In parallel coordinates, an \(H_0^1\) field in \(K\) satisfies \(x''+A(t)x=0\) distributionally, by testing against compactly supported smooth fields. Its continuous representative makes \(Ax\) continuous; integrating twice shows \(x\) is \(C^2\), and the smooth coefficients bootstrap it to smoothness. It is therefore a Jacobi field with the endpoint conditions. Conversely integration by parts shows any such Jacobi field lies in \(K\), first against smooth fields and then against all \(H_0^1\) fields by density. Consequently \(K=\mathscr J^b\), proving the assertion. ◻
Theorem 10 (Morse Index Theorem). Let \(\gamma: [a,b] \to M\) be a geodesic. There are at most finitely many points conjugate to \(\gamma(a)\) along \(\gamma\), and
Proof. Translate the parameter so that \(a=0\). All rescalings below are of coefficient functions in a fixed parallel frame, so vectors in different tangent spaces are compared correctly.
Finitely many conjugate times. For any finite list \(0<t_1<\cdots<t_k\leq b\), choose a basis \(J_{i,r}\) of each \(\mathscr J^{t_i}\) and truncate it to zero after \(t_i\), obtaining continuous \(H_0^1\) fields \(Y_{i,r}\). These fields span a totally isotropic subspace: for \(t_i\leq t_j\), integrate using the Jacobi equation of \(J_{j,s}\) on \([0,t_i]\), obtaining
The fields are independent. On the last interval \((t_{k-1},t_k)\), any linear relation involves only the last group of Jacobi fields; ODE uniqueness forces that group's coefficients to vanish. Continue backwards. Thus \(\sum_i\dim\mathscr J^{t_i}\leq\operatorname{Ind}_0(\gamma)\). Lemma 7 bounds this sum for every finite list, proving finiteness of the conjugate times, with their multiplicities. Merely having individual zero squares would not suffice; the cross terms above matter.
Moving the endpoint. Set \(m(\tau)=\operatorname{Ind}(\gamma|_{[0,\tau]})\) and \(m_0(\tau)=\operatorname{Ind}_0(\gamma|_{[0,\tau]})\). Extension by zero embeds the form on a shorter interval into that on a longer interval, so both functions are nondecreasing. After rescaling to \([0,1]\), the form is
It depends continuously on \(\tau>0\) in the norm of bilinear forms on \(H_0^1\). On the unit sphere of a fixed finite-dimensional negative subspace, negativity is uniform, so it persists for \(\sigma\) sufficiently close to \(\tau\). Combined with monotonicity this proves \(m(\tau^-)=m(\tau)\).
To prove \(m_0(\tau^+)=m_0(\tau)\), suppose instead that a sequence \(\tau_n\downarrow\tau\) has nonpositive subspaces of dimension \(r=m_0(\tau)+1\). Choose \(L^2\)-orthonormal bases \(X_{n,1},\ldots,X_{n,r}\) and extend them by zero to \([0,b]\). Nonpositivity bounds each derivative norm by \(C\). Pass to a common subsequence converging weakly in \(H_0^1\) and strongly in \(L^2\) to \(X_1,\ldots,X_r\). The limits remain \(L^2\)-orthonormal and vanish on \((\tau,b]\); continuity gives zero trace at \(\tau\). For every coefficient vector \(c\), weak lower semicontinuity of the derivative norm and strong convergence of the curvature term give
This constructs an \(r\)-dimensional nonpositive subspace on \([0,\tau]\), a contradiction. Passing a whole orthonormal basis to the limit, rather than one vector, is what proves the dimension assertion.
Counting jumps. On a sufficiently short interval, \(\int_0^\tau|X|^2\leq\tau^2\int_0^\tau|X'|^2/2\), so \(I(X,X)\geq(1-C\tau^2/2)\int|X'|^2>0\) for nonzero \(X\). Thus \(m=m_0=0\) initially. Lemma 9 gives \(m_0(\tau)-m(\tau)=\dim\mathscr J^\tau\). Away from the finitely many conjugate times this difference is zero, and the two one-sided continuity statements and monotonicity force the common integer value to be locally constant. At a conjugate time \(t_i\), the value before it is \(m(t_i)\), and the value after it is \(m_0(t_i)\); their difference is exactly its multiplicity. Summing these jumps proves
This includes the cases of no conjugate times and of an endpoint conjugate time. ◻
Finally, comparison of the curvature terms compares index forms after identifying parallel frames. The next lemma records the resulting inequality for indices.
Lemma 11. Let \(M\) and \(\widetilde M\) be differentiable manifolds of the same dimension, such that their section curvatures \(K\) and \(\widetilde K\) satisfy \(\sup \widetilde K \leq \inf K\), where the supremum and infimum are taken over all 2-planes of \(TM\) and \(T \widetilde M\). Let \(\sigma : [0,b] \to M\) be a geodesic such that \(\sigma(0) = p\), and fix \(\widetilde p \in \widetilde M\). Let \(i : T_p M \to T_{\widetilde p} \widetilde M\) be a linear isometry, and \(\widetilde \sigma : [0,b] \to \widetilde M\) be given by \(\widetilde \sigma (t) = \exp_{\widetilde p} (t i(\sigma'(0)))\). Then, \(\operatorname{Ind}(\sigma) \geq \operatorname{Ind}(\widetilde \sigma)\).
It is assumed that the target geodesic exists for all \(0\leq t\leq b\), for example if \(\widetilde M\) is complete. The two velocities have equal lengths. The proof applies to the endpoint-vanishing Sobolev spaces as well as smooth fields by density.
Proof. Define \(B_t=P_t i^{-1}\widetilde P_t^{-1}\), which sends the target velocity to the source velocity and commutes with covariant differentiation. Thus \(W=B_t\widetilde W\) has the same norm, derivative norm, endpoints and tangential component. The squared area \(|\sigma'|^2|W|^2-\langle\sigma',W\rangle^2\) is preserved. Multiplying the sectional-curvature bound by this nonnegative quantity gives
For a zero area both sides vanish by curvature antisymmetry. Therefore \(I(W,W)\leq\widetilde I(\widetilde W,\widetilde W)\). The map \(\widetilde W\mapsto W\) is a linear bijection of field spaces, so it takes every negative-definite subspace for \(\widetilde I\) to a negative-definite subspace of the same dimension for \(I\). Taking maximum dimensions proves the comparison. ◻
Reading: Carmo (1992), Riemannian geometry.