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Abstract

The aim of this work is to introduce several different volume computation methods of the graph polytope associated with various type of finite simple graphs. Among them, we obtained the recursive volume formula (RVF) that is fundamental and most useful to compute the volume of the graph polytope for an arbitrary finite simple graph.

graph polytope, volume, generating function
\theoremstyle

definition \theoremstyleremark \numberwithinequationsection

Different Volume Computation Methods of Graph Polytopes]Different Volume Computation Methods of Graph Polytopes

D. Lee]Daeseok Lee H.-K. Ju]Hyeong-Kwan Ju \subjclassPrimary 05C22; Secondary 52B05

1 Introduction

Bóna, Ju and Yoshida  enumerated certain weighted graphs with the following conditions : For a given positive integer , a nonnegative integer and a simple graph with , where and , we consider the set

 W(n;G):={α=(n1,⋯,nk)∈([n]∗)k∣ij∈EG⇒ni+nj≤n}.

We call satisfying the conditions in the set given above (vertex-) weighted graph. In fact, the number of weighted graphs is given by an Ehrhart polynomial of some convex polytope in a unit hypercube. Such a convex polytope is determined uniquely for a given finite simple graph as follows: Let be a simple graph with . Then the graph polytope associated with the graph is defined as

 P(G):={(x1,x2,⋯,xk)∈[0,1]k∣ij∈EG⇒xi+xj≤1}

Our main concerns in this article are the volume computation results of graph polytope associated with many types of graphs using several different methods. In order to obtain the volume of the graph polytope we need a certain kernel function defined by the following:

 K(s,t):={1,s+t≤10,elsewhere.

Then the volume of the polytope is

 vol(G)=∫QnH(x1,x2,⋯,xn)dx1dx2⋯dxn,

where is the -dimensional unit hypercube,

 H(x1,x2,⋯,xn)=∏ij∈EGK(xi,xj).

From now on, all graphs we mentioned will be finite simple graphs and all polytopes are convex. The volume of graph polytope will be denoted by rather than .
Chapter 2 introduces a recursive volume formula for the volume of the graph polytope and volume formulae for the graph polytope associated with various types of graphs. Chapter 3 describes the graph joins and the corresponding volume formula. Volume of the graph polytope associated with the bipartite graph with certain symmetry is dealt in chapter 4. In chapter 5 we use the operator theory to find values for interesting series. In the last chapter we mentioned another way to get the volume of graph polytopes, which uses Ehrhart polynomial and series.

2 Recursive Volume Formula

One of the key and most fundamental techniques is the recursive volume formula, or RVF. It is also useful. Next two lemmas will be used to prove the RVF, and can be shown easily. Applications of RVF are discussed.

Lemma 2.1

(polytope partitioning) Let be a polytope containing a point and
. Then

 vol(G)=∑d(x,F)vol(F)n, (1)

where the sum runs over all facets of .

Lemma 2.2

If a graph has no isolated vertex, then the graph polytope has no facet of the form In other words, P(G) is only composed of facets of form or for .

Theorem 2.3

(RVF) Let be a graph with the vertex set and having no isolated vertex. Then

 vol(G)=n∑i=1vol(G−i)2n, (2)

where is the graph with the vertex set and, accordingly, with the inherited edge set in the original edge set .

Proof: Let Then for the facet in the hyperplane (hence in the hyperplane.) and for all other facets since Note that is the dimensional volume of the graph polytope . By the Lemmas 2.1 and 2.2 we have the desired formula (2).\qed

Corollary 2.4 (path)

Let be the path with Then

 vol(Ln)=Enn!,

where is the th Euler number. Hence its generating function is:

 ∑n≥0vol(Ln)xn=secx+tanx.

Proof: Let . For convenience, we define . It is obvious that .

 Misplaced & (3)

by RVF.

 Misplaced &

This says that satisfies the same recurrence relation as ’s together with . Hence we get the first conclusion Since exponential generating function of the Euler number is , the second conclusion follows. \qed

Corollary 2.5 (cycle)

Let be the path with Then

 vol(Cn)=En−12((n−1)!).

Hence its generating function is:

 ∑n≥1vol(Cn)xn=x(secx+tanx)2.

Proof: Removing an edge in the cycle results in a path . Hence RVF and Corollary 2.4 imply both of conclusions. \qed

Corollary 2.6 (complete graph)

Let be the path with Then

 vol(Kn)=21−n.

Hence its generating function is:

 ∑n≥1vol(Kn)xn=2x2−x.

Proof: Since , the conclusion follows easily.

Corollary 2.7 (complete bipartite graph)

Let be the path with Then

 vol(Ks,t)=1(s+ts). (4)

Proof: Since and

 vol(Ks,t)=svol(Ks−1,t)+tvol(Ks,t−1)2(s+t),

by induction, we have the required formula. \qed

3 Volume of the graph joins

Join of graphs and is the graph where and . We denote the join of graphs and simply by Obviously, the associative law holds. So we can define (add times). The question is that how we can calculate the volume of . Note that where is a null graph with vertices.

Definition 3.1 (sliced volume)

Let be a graph and

 vol(G,a):=∫[0,a]n∏ij∈EGK(xi,xj)dx,

where is defined as:

 K(s,t):={1,s+t≤a0,elsewhere.

It is obvious that Next theorem gives us a volume formula for the graph polytope associated with the joined graph.

Theorem 3.2

The volume of the graph polytope associated with the graph is:

 vol(G+H,c)=∫c0∫c0(ddsvol(G,s))(ddtvol(H,t))K(s,t)dsdt. (5)

Proof:

 Misplaced &
\qed
Theorem 3.3

Let Then the sliced volume of the graph polytope is given by the formula:

 vol(P(Km,n),c)=cn(1−c)m+mi∑i=0(ni)(−1)ncm+i−(1−c)m+Im+i.

Proof:

 Misplaced &
\qed
Corollary 3.4

The volume of the graph polytope associated with the complete bipartite graph is:

 vol(Km,n)=n∑i=0(ni)(−1)imm+i. (6)

Proof: Substitute in the Theorem (3.3). \qed

The following result is immediate by the third term from the bottom in the formula .

Corollary 3.5
 n∑i=0(ni)(−1)imm+i=1(m+nn)=m∑i=0(mi)(−1)inn+i.

According to Rudin(), the beta function is defined as:

 B(r,s)=∫10xr−1(1−x)s−1dx=Γ(r)Γ(s)Γ(r+s),

where is the gamma function. Note that

 vol(Km,n)=mB(m,n+1)=nB(m+1,n)

from the formula 4 and the definition of beta function involving gamma function. Note that we can give another proof of above equality about beta function from the last integration expression in the proof.

Theorem 3.6 (multiple join of a graph)

Let be a graph, and Then, for any positive integer ,

 ddrvol(nG,r)=n(1−r)k(n−1)ddrvol(G,r).

Proof:

 Misplaced &
 Misplaced &

Now, we denote simply by Then

 Misplaced &

Let Then, from the previous recursion formula we get

 F(x,r)=xa1(1−(1−r)kx)2=∑n≥1na1(1−r)(n−1)kxn.

Hence,

 ddrvol(nG,r)=n(1−r)(n−1)kddrvol(G,r).
\qed
Corollary 3.7

For the value we have

 vol(Kn,r)=21−n−(1−r)nandvol(Kn)=21−n. (7)

Proof: Since we have the formula (7) from the Theorem 3.6. \qed

Corollary 3.8

For the value we have

 vol(nDk)=2−kn+n2−kn1(knk)k−1∑i=0(kni).

Proof:

 ddtvol(nDk,t)=n(1−t)k(n−1)ddtvol(Dk,t)=kntk−1(1−t)k(n−1).
 Misplaced &
\qed

4 Volume of bipartite graphs

Definition 4.1

Let be a set of all permutations of and

 [0,1]nσ:={x=(x1,x2,⋯,xn)∈[0,1]n|xσ(1)≤xσ(2)≤⋯≤xσ(n)}.

Note that

 [0,1]n=⋃σ∈Sn[0,1]nσ

and each intersection of two different has measure 0 so that for any measurable function ,

 ∫[0,1]nfdx=∑σ∈Sn∫[0,1]nσfdx

and

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Let be a bipartite graph with .

Theorem 4.2

The volume of the graph polytope associated with the bipartite graph mentioned above is as follows:

 vol(B)=∑σ∈Snn∏i=11i+∑ij=1αj,σ,

where

 αi,σ=numberof({vk∈V2|σ(i)∈Nk}∖∪i−1j=1{vk∈V2|σ(j)∈Nk}).

which means the number of vertices in which the smallest among values of its neighbors is i.

Proof:

 Misplaced &

where represents the opposite order of in the fifth term. \qed

An automorphism of a simple graph is a permutation of which has the property that is an edge of if and only if is an edge of .

Theorem 4.3

Assume that the bipartite graph is symmetric, by the sense that for any permutation on , there exists a permutation such that the combination of and induces an automorphism on . Then,

 vol(P(B))=n!n∏i=11i+∑ij=1αj

Proof:The symmetry of the graph implies that all s are same for different The conclusion follows from the Theorem 4.2. \qed

Corollary 4.4

Let be the graph that is obtained from the complete bipartite graph by deleting disjoint edges. Then,

 vol(Bn)=(1+1n)1(2nn).

In particular, Note that the bipartite graph is the graph obtained from the 1-skeleton of the cube.

5 An Application related to the Operator Theory

We introduce here another interesting fact that uses the linear operator theory to obtain the value of a series described in the theorem below. One of the results related with the operator theory is the computation of the , which is referred from Elkies . We will restate the lemma regarding this.

We define by the following:

 K(s,t):={1,s+t≤10,elsewhere.

Then the volume of the polytope is

 vol(G)=∫QnH(x1,x2,⋯,xn)dx1dx2⋯dxn,

where is the -dimensional unit hypercube,

 H(x1,x2,⋯,xn)=∏ij∈EK(xi,xj).

We are interested in the computation of the volume of the polytope for a given simple graph .

We define inductively as in the following:

 K1(t,s):=K(t,s),

and

 Kn(t,s):=∫10K1(t,x1)Kn−1(x1,s)dx1(n≥2).

Let be a linear operator with kernel on defined by:

 (Tg)(t)=∫10K1(t,s)g(s)ds=∫1−t0g(s)ds. (8)

From the definition of we see that is the kernel function of the linear operator as follows:

 (Tng)(t)=∫10Kn(t,s)g(s)ds. (9)

The next lemma gives the spectral decomposition of the linear operator , and also of . Its proof is immediate from the standard linear operator theory.(See also Elkies  or Hutson. et. al. .)

Lemma 5.1

The linear operator is compact and self-adjoint on Its eigen values are ; the corresponding eigenfunctions are Moreover, The linear operator is compact and self-adjoint on Its eigen values are with same corresponding eigenfunctions . Each of its eigenvalues for and is simple.

Our main goal here is to find the value of certain formula using the operator theory. In fact, it is the which is obtained from the RVF. By the simple calculations we can get the following formula from the definition of (see ):

 vol(Cn)=∫10Kn(t,t)dt. (10)

It turned out that the right hand side of the formula (10) is the trace of a trace-class operator over the diagonal, and is equal to

 ∞∑k=−∞2n(π(4k+1))n.

Note that this series is absolutely convergent for . As a summary we have a theorem:

Theorem 5.2

For any integer the following holds:

 ∞∑k=−∞1(4k+1)n=πnvol(Cn)2n.
\qed

For the case

 Misplaced &

meanwhile, for the case

 Misplaced &

6 Concluding Remarks and Further Problems

In fact, we have another volume computation method which comes from the Ehrhart polynomial of . Let be an integral convex polytope in . Then we call the Ehrhart polynomials of It is known that, for a given -polytope ,

 vol(P)=limt→∞LP(t)td,whered=dim(P),

or

 f(1)d!,where∞∑t=0LP(t)xt=f(x)(1−x)d+1.

(Refer  or  about this.) If is a bipartite graph with vertices, then its graph polytope is a -polytope of dimension . Hence, we can get the volume from the Ehrhart polynomial , which we can get by using divided difference technique. (See Bóna et. al.  for details.)

References

1. M. Beck and R. Sinai, Computing the Continuous Discretely, Springer, 2007.
2. M. Bóna, H.-K. Ju and R. Yoshida, On the enumeration of a certain weighted graphs, Discrete Applied Math., 155(2007), 1481-1496.
3. N. Elkies, On the sums Amer. Math. Monthly, 110, no. 7(Aug.-Sep. 2003), 671-573.
4. V. Hutson, J. Pym and M. Cloud, Applications of Functional Analysis and Operator Theory (2nd ed.), Elsevier Science, 2005.
5. W. Rudin, Principles of Mathematical Analysis (3rd ed.), McGraw-Hill, 1976.
6. R. Stanley, Enumerative Combinatorics, vol.1(2nd ed.), Cambridge Univ. Press, 2012.
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