CS 408 Spectral Graph Theory Abhiram

We often think of graphs geometrically, i.e. we see vertices as points on a plane or in space, with edges as lines connecting them. When speci ed algebraically (e.g. by its adjacency matrix), the geometrical imagery is absent. Can we somehow create it from the algebraic description? The geometric representation might be useful in itself, say because we want to draw the graph on paper. But once a graph is embedded in space, in a nice manner, say whereby neighbouring vertices are placed close, perhaps problems such as partitioning a graph might be solved by partitioning the associated volume. We can start with any algebraic representation of the graph, but a convenient one is the node-edge incidence matrix B of a graph G with n vertices and m edges. B has n rows and m columns, and bie = 1 if vertex i is an endpoint of edge e, and 0 otherwise. Clearly, every column of B has exactly two 1s and rest 0s. Note that BB = A + D where A is the adjacency matrix, in which aij = 1 if (i; j) is an edge and 0 otherwise, and D a diagonal matrix in which the iith entry equals the degree di of vertex i. For d-regular graphs, the matrix B readily yields a nice geometric representation: we consider the elements of the ith row Bi of the incidence matrix B as the coordinates of vertex i in m dimensional space. If (i; j) is an edge, then the distance between i; j is p 2d 1, whereas otherwise it is p 2d. That neighbours are located nearer than non neighbours seems like a good property. Suppose now that we want to partition this graph into equal sized subgraphs (as possible) by removing minimum number of edges. We instead look to partition the point cloud formed by the vertices. It turns out (exercises) that simply by slicing the cloud by a hyperplane, we can generate essentially any partition. The key question then is how do we nd the hyperplane which gives us the partition we want. If our point cloud is in some sense homogeneous, it may seem natural to nd the direction in which the point cloud is long, and cut perpendicular to that direction, expecting we will cut a small cross section of the cloud and hence presumably a small number of edges. Likewise, suppose we want to draw a picture of the graph, Let Bi denote the ith row of B. Then (BB T )ij = BiB T j . This evaluates to the degree di of vertex i if i = j, to 1 if (i; j) is an edge, and 0 otherwise. It is not clear, a priori, that it is even possible to get the best cut using just a