Laplace-Beltrami operator

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In differential geometry, the Laplace operator can be generalized to operate on functions defined on surfaces, or more generally on Riemannian and pseudo-Riemannian manifolds. This more general operator goes by the name Laplace-Beltrami operator. As the Laplacian, the Laplace-Beltrami operator is defined as the divergence of the gradient. The operator can be extended to operate on tensors as the divergence of the covariant derivative. Alternatively, the operator can be generalized to operate on differential forms using the divergence and exterior derivative. The resulting operator is called the Laplace-de Rham operator.

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[edit] Laplace-Beltrami operator

One defines the Laplace-Beltrami operator, just as the Laplacian, as the divergence of the gradient. To be able to find a formula for this operator, one will need to first write the divergence and the gradient on a manifold.

If g denotes the (pseudo)-metric tensor on the manifold, one finds that the volume form in local coordinates is given by

\mathrm{vol}_n := \sqrt{|g|} \;dx^1\wedge \ldots \wedge dx^n

where the dxi are the 1-forms forming the dual basis to the basis vectors

\partial_i := \frac {\partial}{\partial x^i}

for the local coordinate system, and \wedge is the wedge product. Here | g | : = | detgij | is the absolute value of the determinant of the metric tensor. The divergence of a vector field X on the manifold can then be defined as

(\mbox{div} X) \; \mathrm{vol}_n := \mathcal{L}_X \mathrm{vol}_n

where LX is the Lie derivative along the vector field X. In local coordinates, one obtains

\mbox{div} X = \frac{1}{\sqrt{|g|}} \partial_i \left(\sqrt {|g|} X^i\right).

Here (and below) we use the Einstein notation, so the above is actually a sum in i. The gradient of a scalar function f may be defined through the inner product \langle\cdot,\cdot\rangle on the manifold, as

\langle \mbox{grad} f(x) , v_x \rangle = df(x)(v_x)

for all vectors vx anchored at point x in the tangent bundle TxM of the manifold at point x. Here, df is the exterior derivative of the function f; it is a 1-form taking argument vx. In local coordinates, one has

\left(\mbox{grad} f\right)^i = \partial^i f = g^{ij} \partial_j f.

Combining these, the formula for the Laplace-Beltrami operator applied to a scalar function f is, in local coordinates

\Delta f = \mbox{div grad} \; f = \frac{1}{\sqrt {|g|}} \partial_i \left(\sqrt{|g|} g^{ij} \partial_j f \right).

Here, gij are the components of the inverse of the metric tensor g, so that g^{ij}g_{jk}=\delta^i_k with \delta^i_k the Kronecker delta.

Note that the above definition is, by construction, valid only for scalar functions f:M\rightarrow \mathbb{R}. One may want to extend the Laplacian even further, to differential forms; for this, one must turn to the Laplace-deRham operator, defined in the next section. One may show that the Laplace-Beltrami operator reduces to the ordinary Laplacian in Euclidean space by noting that it can be re-written using the product and chain rule as

\Delta f = \partial_i \partial^i f + (\partial^i f) \partial_i \ln \sqrt{|g|}.

When | g | = 1, such as in the case of Euclidean space with Cartesian coordinates, one then easily obtains

\Delta f = \partial_i \partial^i f

which is the ordinary Laplacian. Using the Minkowski metric with signature (+++-), one regains the D'Alembertian given previously. Under local parametrization u1,u2, the Laplace-Beltrami operator can be expanded in terms of the metric tensor and Christoffel symbols as follows:

\Delta f = g^{ij}\left(\frac{\partial^2 f}{\partial u^i\, \partial u^j} - \Gamma_{ij}^k \frac{\partial f}{\partial u^k} \right).

Note that by using the metric tensor for spherical and cylindrical coordinates, one can similarly regain the expressions for the Laplacian in spherical and cylindrical coordinates. The Laplace-Beltrami operator is handy not just in curved space, but also in ordinary flat space endowed with a non-linear coordinate system.

Note also that the exterior derivative d and -div are adjoint:

\int_M df(X) \;\mathrm{vol}_n = - \int_M f \mbox{div} X \;\mathrm{vol}_n     (proof)

where the last equality is an application of Stokes' theorem. Note also, the Laplace-Beltrami operator is negative and symmetric:

\int_M f\Delta h \;\mathrm{vol}_n = - \int_M \langle \mbox{grad} f, \mbox{grad} h \rangle \;\mathrm{vol}_n = \int_M h\Delta f \;\mathrm{vol}_n

for functions f and h . For this reason, several authors prefer to define the Laplace-Beltrami operator as the present one with a minus sign in front, so that it is positive .

[edit] Using the covariant derivative

The Laplace-Beltrami operator can be written using the trace of the iterated covariant derivative associated to the Levi-Civita connection. From this perspective, let Xi be a basis of tangent vector fields (not necessarily induced by a coordinate system). Then the Hessian of a function f is the symmetric 2-tensor whose components are given by

H(f)_{ij}=H_f(X_i, X_j) = \nabla_{X_i}\nabla_{X_j} f - \nabla_{\nabla_{X_i}X_j} f

This is easily seen to transform tensorially, since it is linear in each of the arguments Xi, Xj. The Laplace-Beltrami operator is then the trace of the Hessian with respect to the metric:

Δf = ∑ gijH(f)ij.
ij

In abstract indices, the operator is often written

\Delta f = \nabla^a \nabla_a f

provided it is understood implicitly that this trace is in fact the trace of the Hessian tensor.

[edit] Laplace-de Rham operator

More generally, one can define a Laplacian differential operator on the exterior algebra of a differentiable manifold. On a Riemannian manifold it is an elliptic operator, while on a Lorentzian manifold it is hyperbolic. The Laplace-de Rham operator is defined by

\Delta= \mathrm{d}\delta+\delta\mathrm{d} = (\mathrm{d}+\delta)^2,\;

where d is the exterior derivative or differential and δ is the codifferential. When acting on scalar functions, the codifferential may be defined as δ = − * d * , where * ; is the Hodge star; more generally, the codifferential may include a sign that depends on the order of the k-form being acted on.

One may prove that the Laplace-de Rham operator is equivalent to the previous definition of the Laplace-Beltrami operator when acting on a scalar function f; see the proof for details. Notice that the Laplace-de Rham operator is actually minus the Laplace-Beltrami operator; this minus sign follows from the conventional definition of the properties of the codifferential. Unfortunately, Δ is used to denote both; which can sometimes be a source of confusion.

[edit] Properties

Given scalar functions f and h, and a real number a, the Laplace-de Rham operator has the following properties:

  1. \Delta(af + h) = a\,\Delta f + \Delta h\!
  2. \Delta(fh) = f \,\Delta h + 2 (\partial_i f) (\partial^i h) + h\, \Delta f    (proof)

[edit] Laplace operators on tensors

The Laplace-Beltrami operator can be extended to an operator on arbitrary tensors on a pseudo-Riemannian manifold using the covariant derivative associated to the Levi-Civita connection. This extended operator may then act on skew-symmetric tensors. However, the resulting operator is not the same one as that given by the Laplace-de Rham operator: the two are related by the Weitzenböck identity.

[edit] See also

[edit] References

  • Flanders, H (1989). Differential forms with applications to the physical sciences. Dover. ISBN 978-0486661698. 
  • Jürgen Jost, Riemannian Geometry and Geometric Analysis, (2002) Springer-Verlag, Berlin ISBN 3-540-42627-2 . (Provides a general introduction to curved surfaces).