# Is the laplacian operator for meshes just the sum of the differences of neighbouring vertices?

I am starting to learn about the laplacian operator $$\Delta = \nabla \cdot \nabla\phi(p)$$

Which can be described as the divergence of the gradient of a scalar function $$\phi$$. This is equivalent to the sum of the second order partial derivatives of the function.

However, it seems its graph equivalent is just the sum of the differences of adjacent vertices. Am I reading that right? In my head that should be equivalent to the first order partial derivatives, not the second order ones.

(I didn't know whether to ask here or in math, I decided here is more relevant because this is in the context of geometry processing)

You can see why the sum of differences is giving a second derivative if you look at the derivative of a one-dimensional function $$f$$. The derivative of $$f$$ at $$x_0$$ is $$\frac{df}{dx}(x_0) \approx \frac{f(x_0 + dx) - f(x_0)}{dx}$$ The second derivative is then $$\begin{eqnarray*} \frac{d^2f}{dx^2}(x_0) &\approx& \frac{df/dx(x_0) - df/dx(x_0-dx)}{dx} \\ &\approx& \frac{\left(f(x_0+dx) - f(x_0)\right) - \left(f(x_0) - f(x_0-dx)\right)}{dx^2} \\ &=& \frac{\left(f(x_0+dx) - f(x_0)\right) + \left(f(x_0-dx) - f(x_0)\right)}{dx^2}, \end{eqnarray*}$$ i.e. the sum of the local differences, scaled by a distance factor $$dx^2$$. In a graph Laplacian, you don't have any distances, just graph connections, so it's just the sum of the local differences.