Number of non-isomorphic graphs

This expository essay is to test my understanding of the techniques used in More Bricks – More Walls?, Thirty-three Miniatures by Jiří Matoušek’s.

We shall prove the sequence g_n(0),\ldots, g_n({n\choose 2}) is unimodal, i.e., it is first nondecreasing and then, from some point on, non-increasing, where g_n(k) is the number of non-isomorphic graphs with n vertices and k edges. In particular, we shall prove \begin{gathered}g_n(0)\leq g_n(1)\leq\ldots\leq g_n(\lfloor m/2\rfloor) \\ =g_n(\lceil m/2\rceil)\geq\ldots\geq g_n(m-1)\geq g_n(m),\end{gathered} where m={n\choose 2}.

Throughout the article, the number of vertices, which is denoted as n, is always fixed. And m always stands for the maximal number of edges between n vertices.

Notice that taking the complement of graphs establishes a bijection between graphs with k edges and graphs with m-k edges. Moreover, two graphs are isomorphic if and only if their complement graphs are isomorphic. Thus we have g_n(k)=g_n(m-k). Hence it is enough to show that g_n(k)\leq g_n(l) for all k\leq l\leq m/2.

Denote U, V be the sets of all graphs with k, l edges on the fixed vertex set [n] respectively. Let r, s denote the number of non-isomorphic graphs in U, V. By our notation above, r=g_n(k), s=g_n(l). We shall show r\leq s. The graph G is the bipartite graph between U and V with u\sim v if and only if u is a subgraph of v.

Let B=(b_{uv})_{u\in U, v\in V} be the bipartite adjacent matrix of G, where b_{uv}=1 if u and v are adjacent in G, otherwise 0.

We claim the matrix B is of full rank. As it is easy to see the size of U is no greater than the size of V, we shall prove an equivalent statement that the matrix B is of full row rank, that is the rows of B are linearly independent.

Suppose not. There is a non-zero row vector y in \mathbb{R}^{U} such that yB=0. Notice the coordinates of y are indexed by the elements in U. Let K^*\in U such that y_{K^*}\neq 0.

Now we partition U, V into k+1 according to the number of edges in the intersection of the graph with K^*: K_i is the set of graphs in U who share i common edges with K^* and L_j is the set of graphs in V who share j common edges with K^* for all i, j=0,\ldots, k. Remember the number of edges in K^* is k and K_k contains only one element K^*. Also all K_i and L_j defined above are non-empty because of the assumption k<l\leq m/2.

Note that for any i, j\in \{0,\ldots, k\}, all vertices in K_i have the same degree to L_j in the bipartite graph G. This is because the ways to extend a graph with k edges and i edges in common with K^* to graphs with l edges and j edges in common with doesn’t specifically depend the graph we start with. Denote this number as d_{ij} and let D=(d_{ij}) be the matrix. Apparently, D is an upper triangular matrix with d_{ii}\neq 0. Thus D is non-singular. On the other hand, as k_i\cdot d_{ij}=\sum_{K\in K_i, L\in L_j}B_{KL} where k_i is the size of K_i, D=\mathrm{diag}(k_0^{-1},\ldots, k_k^{-1})FBG where F=(f_{i, u})_{i\in\{0,\ldots, k\}, u\in U} and G=(g_{v, j})_{v\in V, j\in\{0,\ldots, k\}} are matrices with f_{i, u}=1 if and only if u\in K_i and g_{v, j}=1 if and only if v\in L_j otherwise 0.

Let x be a k+1 dimensional row vector such that x=yE, where E=(e_{ui})_{u\in U, i\in\{0,\ldots, k\}} is matrix with e_{ui}=1 if u\in K_i otherwise 0. In fact, x_i=\sum_{u\in U}y_ue_{ui}=\sum_{K\in K_i}y_K. Hence x_k=y_{K^*}\neq 0 as K_k contains only one element K^*.

Now we have xD=yE\mathrm{diag}(k_0^{-1},\ldots, k_k^{-1})FBG. However it is easy to check EF=\mathrm{diag}(k_0,\ldots, k_k) and yB=0. So xD=yBG=0 contradicting to the non-singularity of D as x\neq 0. According to the graph isomorphism the graphs in U, V are classified into equivalent classes, denoted as U_1,\ldots, U_r, V_1, \ldots, V_s.

Similarly, we observe that all vertices in V_j have the same degree to U_i since the number of ways to restrict two isomorphic graphs to a certain class of isomorphic graphs is the same. Denote this degree as d_{ij} as well. And again we claim the matrix D(different from the one we define before) is of full row rank where D=(d_{ij})_{r\times s}. This will finish the proof since an r\times s matrix of full row rank implies r\leq s.

Suppose not. We have a non-zero r dimensional row vector y such that yD=0. Let x\in\mathbb{R}^U be the row vector indexed by U such that x_u=y_i for all u\in U_i. Thus for all u\in V_j we have \begin{aligned}(xB)_v &=\sum_{u\in U}x_uB_{uv}=\sum_{i\in[r]}\sum_{u\in U_i}x_uB_{uv} \\ &=\sum_{i\in[r]}y_i\sum_{u\in U_i}B_{uv}=\sum_{i\in[r]}y_id_{ij}=(yD)_j=0.\end{aligned} Notice x\neq 0 but xB=0 which contradicts with the non-sigularity of B.

Remark. This proof appears to me as an intricate idea similar to finding an injection. To prove one set is smaller than the other, one would just hope to find an injection from one set to the other. But this is hard in this problem because those two sets we are considering are sets of equivalent classes. On the other hand, in this problem, the inclusion mapping works just fine on the level of graphs though it breaks down on the level of equivalent classes. Thus we should come up with a delicate analysis on the bipartite graph induced by the inclusion mapping. And I think this is the motivation behind the proof.