Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $V_{e,1}$): $$ \begin{aligned} \Delta f(V) &= \star d\star df (V) = \frac{1}{|V_{e,3/2} - V_{e,1/2}|}\Big(\star df(V_{e,3/2}) - \star df(V_{e,1/2})\Big) \\ &= \frac{1}{|V_{e,3/2} - V_{e,1/2}|}\left\{\frac{f(V_{e,2}) - f(V_{e,1})}{|e_{1,2}|}- \frac{f(V_{e,1}) - f(V_{e,0})}{|e_{0,1}|}\right\} \end{aligned} $$

share|cite|improve this answer
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
@yasmar you are right, I was wrong, I will edit my answer later. – Shuhao Cao Apr 30 '12 at 18:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.