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The diameter of a connected finite graph is the maximum distance between two vertices.

A Lévy family of graphs consists of finite connected graphs with their uniform probability measures and graph distances divided by their diameters, whose concentration of measure functions vanish at every fixed positive normalized radius. Equivalently, every family occupying at least half the vertices has a radius-$\lfloor \epsilon D_n\rfloor$ neighbourhood occupying a proportion tending to one, uniformly over such families. Harper theorem and binomial distribution tail bounds prove this for the hypercube graphs.

A graph is bipartite if and only if it contains no odd cycle. One direction follows because every cycle alternates between the two vertex classes. For the converse, in each connected component of a graph choose a root and partition the vertices according to the parity of their graph distance from it. An edge joining equal parities would close an odd cycle, so every edge crosses the partition.

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.

There are only two possibilities for these terminal families for simplicial section compression. If $|X|<|Y|$, consecutiveness makes $X$ the last $k$-set and $Y$ the first $(k+1)$-set. Complementarity forces $n=2k+1$, $X=\{k+2,\dots ,2k+1\}$ and $Y=\{1,\dots ,k+1\}$. For $n\ge 3$, the corresponding $B$ contains all sets of size at most $k$ except $X$, and additionally $Y$. Every $(k+1)$-set has at least two $k$-subsets, so at least one belongs to $B$. The missing $X$ has a $(k-1)$-subset in $B$. Consequently $N[B]$ contains every set of size at most $k+1$, which is exactly $N[I_{|B|}]$.

If $|X|=|Y|$, then $n=2k$. Since $X,Y$ are complementary, the smaller one contains coordinate $1$. Consecutiveness forces $X$ to be the last $k$-set containing $1$, and $Y$ the first $k$-set avoiding $1$:The initial segment consists of all sets of size less than $k$ and all $k$-sets containing $1$. Its closed graph neighbourhood contains all sets of size at most $k$ and all $(k+1)$-sets containing $1$. For $k\ge 2$, each such $(k+1)$-set has $k\ge 2$ $k$-subsets containing $1$, so removing only $X$ leaves one in $B$. All sets of size at most $k$ are still in $N[B]$, because all levels below $k$ belong to $B$. Hence again $N[I_{|B|}]\subseteq N[B]$. For $k=1$, the two families are related by exchanging coordinates and both have the whole $Q_2$ as their closed graph neighbourhood. These comparisons finish the induction and prove Harper theorem.

For the radius-$t$ version, repeatedly use the one-step result. At each step the comparison family remains an initial segment, and its closed graph neighbourhood is monotone in its size. If $|N_{t-1}[A]|\ge |N_{t-1}[I_m]|$, applying Harper theorem to $N_{t-1}[A]$ therefore gives $|N_t[A]|\ge |N_t[I_m]|$. Induction on $t$ proves the asserted extension.

A Lévy family of graphs has vanishing concentration of measure functions after distances have been scaled by the graph diameter. More explicitly, for connected finite graphs $G_n$ of diameter $D_n$, use normalized graph distance $d_n=d_{G_n}/D_n$ and uniform probability measure. For every fixed $\epsilon >0$, requireFor $Q_n$ this normalization is normalized Hamming distance $d_H/n$, since its diameter is $n$.

If $|A|\ge 2^{n-1}$, the corresponding initial segment contains the Hamming ball about $\varnothing$ of radius $b_n=\lfloor (n-1)/2\rfloor$. The radius-$t$ version of Harper theorem impliesUse the precise Hoeffding inequality $\mathbb{P}(Z_n-n/2\ge u)\le \exp (-2u^2/n)$ for $u\ge 0$. One can also derive this estimate directly: the moment-generating function of $Z_n-n/2$ is $\cosh (\lambda /2{)}^n\le \exp (n{\lambda }^2/8)$; the exponential Markov inequality followed by minimization at $\lambda =4u/n$ gives the bound. With $t=\lfloor \epsilon n\rfloor$, we have $b_n+t\ge n/2+\epsilon n-2$. Thus, whenever $\epsilon n>2$,This proves that the hypercube graphs form a Lévy family of graphs. The stronger fixed-positive-mass formulation follows as well. For any $a>0$, choose $c$ with $1/(4c^2)<a$. The Chebyshev inequality for $Z_n$, whose variance is $n/4$, makes the Hamming ball of radius $\lfloor n/2-c\sqrt{n}\rfloor$ have fewer than $a{2}^n$ vertices. An initial segment of size at least $a{2}^n$ contains that ball. Enlarging by $\lfloor \epsilon n\rfloor$ and using the same Hoeffding inequality gives a complement proportion tending to zero uniformly over all such $A$. The normalization matters: with unscaled unit Hamming distance, cubes do not satisfy the metric version at every fixed radius; for a radius below one, a half-cube has no enlargement.

Equality in Harper theorem does not determine a down-set up to isomorphism. In $Q_4$, takeThis is a down-set of size nine. Its closed graph neighbourhood contains every set of size at most two, because all singletons belong to $A$. Every triple contains an edge of the four-cycle among the listed pairs, so every triple also belongs to $N[A]$. The four-element set does not. Hence $|N[A]|=1+4+6+4=15$.

The size-nine simplicial order on the discrete cube initial segment is all sets of size at most one together with the pairs $12,13,14,23$. Every triple contains one of these pairs, and the four-element set is again absent from its closed graph neighbourhood. Its neighbourhood therefore also has size $15$, proving extremality of $A$. Nevertheless the induced graphs are not isomorphic: in $A$, the empty set has degree four, the four singletons have degree three, and the four pairs have degree two. The initial segment has two vertices of degree four, namely $\varnothing$ and $\{1\}$. Since a graph automorphism preserves induced degree of a vertex, no automorphism of the cube can carry one family to the other. This gives nonunique down-set extremizers for Harper theorem even under the stronger test of induced graph isomorphism.

Let $d=d_G(a,b)>0$, and for $i=1,\dots ,d$ putEach edge cutset ${\Pi }_i$ separates $a$ from $b$. Adjacent vertices have graph distances from $a$ differing by at most one. Hence an edge can cross at most one of these level boundaries, so the edge cutsets are pairwise edge-disjoint. Write $k_i=|{\Pi }_i|\ge 1$; then ${\sum }_i{k}_i\le |E|$.

The Nash-Williams inequality for unit conductances giveswhere the middle inequality is the Cauchy-Schwarz inequality. One can also derive the first inequality directly: every unit flow has signed net flux one through each ${\Pi }_i$, so its energy of a flow on that cut is at least $1/k_i$; sum over the disjoint cuts and use the Thomson principle.

Combining with the commute time identity yieldsThe case $a=b$ is trivial. A path graph, with $a,b$ its endpoints, attains equality.

Suppose first that $G$ is bipartite. As one traverses any cycle in a graph, its vertices alternate between the two vertex classes. Returning to the initial class therefore requires an even number of edges, so $G$ contains no odd cycle.

Conversely, suppose that $G$ has no odd cycle. Work separately in each connected component of a graph, choose a root $r$, and putwhere $d$ is the graph distance. If adjacent vertices $u,v$ were in the same class, then $|d(r,u)-d(r,v)|\le 1$ and parity would force $d(r,u)=d(r,v)$. Choose a breadth-first spanning tree. The two tree paths from $r$ to $u$ and $v$ diverge at some last common vertex; their remaining segments have equal length, and adjoining the edge $uv$ gives an odd cycle. This contradiction shows that every edge joins $A$ to $B$, proving the odd-cycle characterization of bipartite graphs.

The family $(Y_x)$ is a Gaussian random field. For distinct $x,x^{\prime }$ its covariance isTwo distinct vertices of ${\mathbb{Z}}^d$ have a common neighbour exactly when their graph distance, equivalently their ${\ell }^1$ distance, is two. Since jointly Gaussian variables are independent exactly when they are uncorrelated,Thus the field has finite-range dependence, even though nearest-neighbour values are independent.

Write $p(a)=P(Y_0\ge a)$; because $Y_0\sim N(0,1+1/(2d))$, $p(a)\to 0$ as $a\to \infty$. Every path in a graph of length $n$ contains, by a greedy selection, at least $n/M_d$ vertices at mutual graph distance greater than two, where $M_d=|B_2(0)|$. The corresponding field values are jointly independent by part (a). Hence the probability that a fixed path lies in the superlevel set is at mostThere are at most $(2d{)}^n$ length-$n$ paths from the origin. The union bound therefore givesChoose a finite $a$ for which $(2d)p(a{)}^{1/M_d}<1$ and let $n\to \infty$. There is then no unbounded component through the origin, and translation invariance rules out an unbounded component anywhere almost surely. Thus the critical threshold for level-set percolation satisfies $a_c(d)<\infty$.

A path graph $P_n$ has $n$ vertices in a single chain and $n-1$ edges. Its endpoint graph distance is $n-1$ and, with unit resistances, its endpoint effective resistance is also $n-1$.

A rooted tree is a tree with a distinguished vertex called the root. Its levels consist of the vertices at a fixed graph distance from the root.