In what way are the Laplacian operator (defined on functions from $\Bbb{R}^n$ to $\Bbb{R}$ which are twice-differentiable) and the Laplacian matrix (defined on simple graphs) similar or related?


1 Answer 1


They are analogues of each other. The continuous Laplacian operator measures how a function changes “on average” as you move away from a given point $\mathbf p$. You can think of it as finding the average value of a function $u$ over all points a small fixed distance away from $\mathbf p$. For a function $u$ defined on the nodes of a graph, the discrete Laplacian operator similarly computes a weighted average of the values of $u$ at a node’s nearest neighbors. Formally, this operator is a linear map from $C^0$, the space of $0$-cochains of the graph, to $C_0$, the space of $0$-chains of the graph, and its matrix relative to a particular pair of bases for these spaces is the Laplacian matrix of the graph.

  • $\begingroup$ But the Laplacian matrix is not defined for a function $u$ on a graph, it's defined for the graph itself (as the rank matrix minus the adjacency matrix) $\endgroup$
    – Lior
    Jan 25, 2017 at 6:33
  • $\begingroup$ @Lior An element $u$ of $C^0$ is a function on the graph: it associates a number with each node. $\endgroup$
    – amd
    Jan 25, 2017 at 7:49
  • $\begingroup$ The answer is not convincing. $\endgroup$
    – keramat
    Jan 24, 2020 at 13:16
  • $\begingroup$ Thanks for the well-considered answer. $\endgroup$
    – khanh
    Apr 25 at 15:32

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