a scalar function ƒ is, in local coordinates If M is not oriented, then the above calculation carries through exactly as presented, except that the volume form must instead be replaced by a volume element (a density rather than a form). Neither the gradient nor the divergence actually depends on the choice of orientation, and so the Laplace–Beltrami operator itself does not depend on this additional structure. Formal self-adjointness The exterior derivative d and −∇ . are formal adjoints, in the sense that for ƒ a compactly supported function (proof) where the last equality is an application of Stokes' theorem. Dualizing gives for all compactly supported functions ƒ and h. Conversely, () characterizes the Laplace–Beltrami operator completely, in the sense that it is the only operator with this property. As a consequence, the Laplace–Beltrami operator is negative and formally self-adjoint, meaning that for compactly supported functions ƒ and h, Because the Laplace–Beltrami operator, as defined in this manner, is negative rather than positive, often it is defined with the opposite sign. Eigenvalues of the Laplace–Beltrami operator (Lichnerowicz–Obata theorem) Let M denote a compact Riemannian manifold without boundary. We want to consider the eigenvalue equation, where is the eigenfunction associated with the eigenvalue . It can be shown using the self-adjointness proved above that the eigenvalues are real. The compactness of the manifold allows one to show that the eigenvalues are discrete and furthermore, the vector space of eigenfunctions associated with a given eigenvalue , i.e. the eigenspaces are all finite-dimensional. Notice by taking the constant function as an eigenfunction, we get is an eigenvalue. Also since we have considered an integration by parts shows that . More precisely if we multiply the eigenvalue equation through by the eigenfunction and integrate the resulting equation on we get (using the notation ): Performing an integration by parts or what is the same thing as using the divergence theorem on the term on the left, and since has no boundary we get Putting the last two equations together we arrive at We conclude from the last equation that . A fundamental result of André Lichnerowicz states that: Given a compact n-dimensional Riemannian manifold with no boundary with . Assume the Ricci curvature satisfies the lower bound: where is the metric tensor and is any tangent vector on the manifold . Then the first positive eigenvalue of the eigenvalue equation satisfies the lower bound: