# Laplace-Beltrami operator for curves

I'm CS major and have used discrete Laplace-Beltrami operator for 2D-manifold (surface meshes). I'm wondering if it is possible to define Laplace-Beltrami operator for 1D-manifold. If this is possible, I will try to discretize it for polylines.

Any help is appreciated.

-

The differential operation on discrete mesh is close relate to discrete exterior calculus. For a collection of directional line segments presumably on a surface mesh, the Laplacian operator is defined as $\Delta = \delta d + d\delta$, where $d$ is the exterior derivative and $\delta$ is the co-derivative in the adjoint sense $\langle d \alpha,\beta\rangle = \langle \alpha,\delta\beta\rangle$, on a 1-chain, $\alpha$ is a 0-form, $\beta$ is a dual 1-form.
Consider a 0-form $f$ defined on a collection of edges $E$, then $d \delta f$ vanishes since $d$ acts on $\delta f$ which is a 1-form, you get a 2-form, 2-form on a 1-chain is zero. Now we have
$$\Delta f = \delta d f = \star d \star d f$$
where $\star$ is the Hodge dual operator that maps 1-forms on the primal edges $E$ to the 0-forms on the dual edges $\star E$, in which the dual of the edges are the collection of midpoints of the edges.
Let us ignore the boundary for a moment, in the discrete sense on some interior edge $e_{0,1}$ with unit directional vector $l_{0,1}$, then the discrete version of $d$ reads: $$df = \frac{1}{|e_{0,1}|}\Big(f(V_{e,1}) - f(V_{e,0})\Big)\,dl_{0,1}$$ apply the Hodge dual operator on this 1-form we get back to a 0-form which is defined on midpoint $V_{e,1/2}$ of the this $e_{0,1} = \overrightarrow{V_{e,0} V_{e,1}}$: $$\star df(V_{e,1/2}) = \frac{1}{|e_{0,1}|}\Big(f(V_{e,1}) - f(V_{e,0})\Big)$$ Do the same for the neighboring edge $e_{1,2} = \overrightarrow{V_{e,1} V_{e,2}}$, we have $$\star df(V_{e,3/2}) = \frac{1}{|e_{1,2}|}\Big(f(V_{e,2}) - f(V_{e,1})\Big)$$ Now apply $d$ again: $$d\star df = \frac{1}{|V_{e,3/2} - V_{e,1/2}|}\Big(\star df(V_{e,3/2}) - \star df(V_{e,1/2})\Big)\,dl_{1/2,3/2}$$ Lastly apply the Hodge dual on above 1-form, we get the Laplacian of $f$ defined on the midpoint $V$ of $V_{e,1/2}$ and $V_{e,3/2}$(not necessarily $V_{e,1}$): \begin{aligned} \Delta f(V) &= \star d\star df (V) = \frac{1}{|V_{e,3/2} - V_{e,1/2}|}\Big(\star df(V_{e,3/2}) - \star df(V_{e,1/2})\Big) \\ &= \frac{1}{|V_{e,3/2} - V_{e,1/2}|}\left\{\frac{f(V_{e,2}) - f(V_{e,1})}{|e_{1,2}|}- \frac{f(V_{e,1}) - f(V_{e,0})}{|e_{0,1}|}\right\} \end{aligned}
The Laplacian of a $0$-form is not a $2$-form. There is no problem describing a discrete Laplacian operator for a $1$-manifold. Even if you use the discrete exterior calculus formalism, you will arrive at the formula that you would get with a finite difference approximation to the second derivative (with respect to arc length). – yasmar Apr 30 '12 at 17:33