Social Balance and the Bernoulli Equation

0.13 MB·

by  J. J. P. Veerman

Social Balance and the Bernoulli Equation
J. J. P. Veerman∗
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1 January 25, 2017
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Abstract
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] Since the 1940’s there has been an interest in the question why social networks often give rise to two antagonistic factions.
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p Recentlyadynamicalmodelofhowandwhysuchabalancemightoccurwasdeveloped. Thisnoteprovidesanintroduction to
- the notion of social balance and a new (and simplified) analysis of that model. We show that if we allow more general initial
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conditionsthatpreviouslyconsidered,thetwofactionsgiverisetotwoadditionalfactions. Thenewfactionsareinternallyvery
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s divided but each is attracted to one of theold factions. If theold factions join forces they will havea majority.
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i 1 Introduction
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Sincethe1940’stherehasbeenaninterestinthequestionwhyfinitesocialnetworksoftengiverisetotwoantagonistic
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factions [5, 4, 3]. Motivated by insights from the field of social psychology (most notably by Heider [5]), Harary [4]

1 and Cartwrightand Harary[3] developed a formal graphtheoretical framework for socialbalance that was consistent
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with Heider’s ideas. They were able to provethat a signsymmetric network is balancedif andonly if it has (at most)
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twofactions(seebelowforthe definitions). This becameknowasthe structuretheorem. We willdenotethese factions
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9 by Left and Right. Until recently however a dynamical model of how and why such a balance occurs was lacking.
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0 The field received an impetus in 2011 when in [6] Marvel ea successfully analyzed the dynamics of the matrix
. differential equation
1
0 X˙ =X2 ; X(0)=X0 (1.1)
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Theinterpretationofthismodelisthattheentriesofthen nmatrixX representtheopinionsofindividuals 1, n
1 × { ··· }
towards one another. The value of the entry x of X indicates the strength of the friendliness ([6]) of individual i
: ij
v towards individual j. The modeling becomes clear if we write out the differential equation for one entry:
i
X
n
r x˙ = x x (1.2)
a ij ik kj
k=1
X
Thus individual i communicates somehowwith all individuals k 1, n . If i’s feeling about k is positive, then k’s
∈{ ··· }
feeling about j will pull i’s opinion in the same direction. On the other hand, if i’s opinion about k is negative, then
i’s opinion about j will change in the direction opposite to k’s opinion about j. This implies roughly that ... one’s
friends’ friends will tend to become one’s friends and one’s enemies’ enemies also ones friends, and one’s enemies’
friends and one’s friends’ enemies will tendto become one’s enemies (inthewordsof[8]). Sincefriendlinessisdifficult
to quantize, one is led to study the problem with the initial value X0 being a random matrix.
ToperformtheanalysisofEquation1.1,theauthorsof[6]assumedthe matrixX0 issymmetric. (This waslater
extendedtonormal matrices[12].) Theanalysisitselfnowproceededintwoparts. Thefirstistosolvethedifferential
equation 1.1 by considering it a special case of a matrix Riccati equation (see [9]). This solution is subject to certain
conditions, the most important of which are that X0 has a positive real eigenvalue, and that the largest of these real
∗FariborzMaseehDept. ofMath. andStat.,PortlandStateUniv.;e-mail: [email protected]
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eigenvalues,λ,hasalgebraicandgeometricmultiplicity1. Thesecondpartoftheanalysisistoshowthatforarandom
symmetric matrix X0 these hypotheses holdwith probabilitytending to 1 as the dimensionn growsunbounded. This
involves relying on fairly subtle arguments about eigenvalues of random symmetric matrices (see [1]).
In this note we propose a simpler and more precise treatment of this problem. In Section 2, we give a very
simple proof of a generalization of the structure theorem. In it the hypothesis that the network be sign symmetric
has been dropped. In Section 3, we treat Equation 1.1 not as a Riccati equation but as a special case of a Bernoulli
equation (first mentioned in [2], see also [11] and other contemporary text books). This simplifies our treatment and
has the advantagethat we cancharacterizemuchmoreprecisely thanpreviouslywhathappens fornon-symmetric (or
non-normal)initialconditionsX0. Wemakeuseofarecentstudy([10])ofrandom(non-symmetric)matricestoassert
that also in this case with probability tending to 1 as the dimension n tends to infinity, X0 has a leading real, simple,
positiveeigenvalue. As aresultwe areableto showthatwithoverwhelmingprobabilityarandominitialconditionX0
(not symmetric or normal), will lead to four factions. The old Left and Right factions are now joined by two other
groups that are internally divided and one of which is attracted to the Left (though not vice versa) and similarly the
other to the Right. Finally we show in Section 4 that the if old Left and Right join forces, they can be expected to
have a narrow majority. This last result is new.
Finallyweremarkonalaterdevelopment. In[12]aslightlydifferentmodelforsocialbalanceisproposed,namely
the differential equation in 1.1 is replaced by X˙ =XXT. This model gives rise to two factions with high probability
under generalinitial conditions. The analysis is substantially more complicated. It appearsthat the simplificationwe
propose here does not help in the study of this model.
Acknowledgements: I am grateful to Patrick de Leenheer for his insightful comments.
2 What is Social Balance?
Thesigneddirected graph GonnverticesisacollectionV ofnverticestogetherwithasetE ofdirectededgesbetween
certain ordered pairs of vertices whose weights are positive or negative. The graph G is sign symmetric if for every
pair u and v in V, all weights of any edges between u and v are either all positive or all negative. An undirected
path or cycle in a directed graph G is a non self-intersecting path or cycle following edges without any regard for
their direction. A cycle or path is called positive if the product of the weights encountered along the cycle or path is
positive, and negative if that product is negative. An undirected component of G is a maximal subgraph all of whose
vertices can be connected by undirected paths.
Definition 2.1 A directed graph G is called balanced if every undirected cycle is positive.
In the context of social dynamics, this definition is essentially due to Harary [4] and Cartwrightand Harary [3]
expanding on earlier concepts by Heider [5]. The idea is that presumably in a state of imbalance pressures will arise
to change it toward a state of balance in the words of [3].
Definition 2.2 A directed graph G is said to have two factions if the vertices can be partitioned into (at most) two
sets U0 and U1, such that all edges entirely within U0 or within U1 have positive weights, while those that connect the
two have negative weights.
The main result here is the simple but surprising conclusion that these two definitions are equivalent for undi-
rected graphs. This result, the structure theorem, was proved in [3, 4] for undirected graphs. We give a slight
generalization here.
Theorem 2.3 A directed graph with signed weights G is balanced if and only if it has two factions U0 and U1.
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Proof: Without lossofgeneralitywe may assume that Ghas one undirected component. By definition Gis balanced
iff every undirected cycle is positive. But that is true iff for all u and v in V any two undirected paths connecting u
and v have the same sign. In turn this is equivalent to partitioning the vertices in V as follows. Start with a vertex u
and call its faction U0. For any v V, v is in Uπ where π is the (unique) parity of the paths from u to v.
∈
Historical remark: From a graph G with two factions we obtain a bipartite graph H by deleting all its positive
edges. OnethusseeswithoutmuchtroublethattheaboveresultisinfactequivalenttoaresultpublishedbyD.K¨onig
in 1916 [13], namely that a graph is bipartite if and only if it has no odd cycles.
3 When does Social Balance Evolve?
Definition 3.1 An n n matrix X is called typical if it is a real matrix that satisfies all of the following conditions:
×
1) detX0 =0.
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2) X0 has a positive real eigenvalue.
3) The largest positive real eigenvalue λ of X0 is simple.
Let M be an ensemble of real n n matrices so whose entries have independent Gaussian distribution with
n
×
mean zeroand variance one. Tao and Vu provedthat if n is even then matrices in M will havesome realeigenvalues
n
with probability tending to 1 as n tends to infinity ([10], Corollary 17). In addition, most of these eigenvalues will be
simple(Corollary18). They addacommentinwhichtheyconjecturethatinfactwithoverwhelmingprobabilitynone
of the eigenvalues should be repeated. For this reason we use the characterizationtypical in the above definition.
To facilitate the proof of the next result, we introduce some notation. Let X0 be a typical n n matrix with
×
leading (real, simple, positive) eigenvalue λ. The unit eigenvector associated with λ will be called v. Denote by W
the span of all (generalized) eigenspaces other than span v . Let w be the vector determined by:
{ }
w W⊥ and (w,v)=1
∈
where (.,.) is the standard inner-product in Rn (see Figure 3.1). Finally we choose an orthonormal basis w2, wn
{ ··· }
for W. Note that w is not a unit vector unless v happens to be in W⊥. In that case w =v.
w
v
W
Figure 3.1: Schematic representation of W, w, and v in Rn.
Theorem 3.2 Let n be even and let X0 Mn be a typical n n matrix (see Definition 3.1). Then with probability
∈ ×
tending to 1 as n tends to , the (matrix valued) initial value problem
∞
X˙ =X2 ; X(0)=X0 (3.1)
has a unique solution for t [0,λ−1) and near t=λ−1 the solution diverges as follows:
∈
lim λ−1 t X =vwT (3.2)
tրλ−1 −
(cid:0) (cid:1)
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Proof: Upon substitution of Z =X−1, the differential equation given in 3.1 transforms into
Z˙ = I ; Z(0)=X−1
0
−
(where we used condition 1). The solution is:
Z(t)=X−1 tI or X(t)= X−1 tI −1 (3.3)
0 0
− −
It is unique and exists until it diverges. (cid:0) (cid:1)
Nowrecallthe notationin the paragraphpriorto the theorem. LetP be the matrix whosefirstcolumnis v and
whose i-th column is w for i>1:
i
P =[v,w2, wn]
···
By its definition P is invertible and using conditions 2 and 3 we have
λ 0T
X0 =P 0 X˜0 P−1 (3.4)
(cid:18) (cid:19)
Thisuncouplesthefirstcomponentfromtherestofthesystem. ByusingEquation3.3),weobtainthatfort [0,λ−1)
∈
λ−1 t −1 0T
X(t) = P − −1 P−1
 (cid:0) 0 (cid:1) X˜0−1 tI 
−
= P (cid:18) (cid:0)λ−1−0 t(cid:1)−1 0(cid:16)0T (cid:19) P−1+(cid:17) P  00 X˜0−10−TtI −1 ! P−1
(cid:16) (cid:17)
ByhypothesisthelargestrealeigenvalueofX˜0islessthanλ,andsointheinterval[0,λ−1]thesecondtermisuniformly
bounded by a constant.
Finally P−1P =1 implies that the first row r of P−1 satisfies:
(r,v)=1 and i>1: (r,w )=0
i
∀
Therefore the first row of P−1 equals w. The above relation for X(t) now gives:
X(t)= λ−1 t −1vwT + (1)
− O
Multiplying both sides by λ−1 t implies the th(cid:0)eorem. (cid:1)
−
(cid:0) (cid:1)
Remark: By transposing Equation3.4, we see that in fact w is a eigenvectorof XT associated with λ. Equivalently,
0
it is a left eigenvector of X0.
Corollary 3.3 If X0 satisfies the hypotheses of Theorem 3.2, then
lim λ−1 t X =vvT
tրλ−1 −
(cid:0) (cid:1)
if and only if the left and right eigenvectors associated to λ of X0 are the same.
This easily implies that for t close to λ−1 and up to permutation of the coordinates, X(t) has the block-form
+
sgnX(t)= −
+
(cid:18) − (cid:19)
This means that the individuals have split into two factions (one possibly being empty). Using the terminology of
the previous section, we say that in this case X has two factions. Individuals from each faction like each other
among themselves, but they dislike individuals from the other faction. The cases discussed in the literature are X0 is
symmetric ([6]) or normal([12]). In these cases alleigenspacesareorthogonal,andso we easily recoverthat w W⊥.
∈
One does need to prove that in the more restricted setting of symmetric or normal matrices, the typical case still has
overwhelming probability. This is done in the literature cited.
We can improve Corollary 3.3 somewhat noticing the following.
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Corollary 3.4 If X0 satisfies the hypotheses of Theorem 3.2, then for t close to λ−1, X(t) has 2 factions if and only
if the left and right eigenvectors associated with λ fall in the same orthant, ie: their components have the same signs.
We note finally that the vectors v is as likely a choice for the eigenvector associated to λ as it its negative v.
It follows that they are equally probable and so the Left and Right factions both have expected size 1(n+1). −
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4 What if Social Balance Does Not Evolve?
Letv andw be definedasinthe previoussection. Theorem3.2implies thatfortcloseto λ−1 the matrixX(t)hasthe
sign pattern of vwT. Permute the components so that the first K n components of v are positive and the others
≈ 2
negative. Afterthat,permutethefirstK componentssothatwithinit,thepositivecomponentsofw andthenegative
ones are grouped together, and so forth. It follows that for t close to λ−1, X has the following sign pattern:
+ +
− −
+ +
sgnX(t)= − − 
+ +
− −
 + + 
 − − 
 
Notice that this matrix is not sign symmetric and thus the system is not balanced.
Thus the old Left and Right cohesive factions now give rise to two additional factions that are internally very
divided but of which one is attracted to the old Left and the other to the old Right. An interesting prediction is that
if in this situation the old Left and Right join forces they will have a majority.
Theorem 4.1 Let X0 satisfy the hypotheses of Theorem 3.2. Assume furthermore that v is uniformly distributed in
N . Then the expected number k of sign disagreements between u and v satisfies:
u
n+1 n 11 n+1
1+ <k <
2 − 2π 12n 2
r (cid:18) (cid:19)
TheTheoremisa directconsequenceofthe Propositionbelow. Letv andw asinSection3 anddefine u= w .
||w||
Consider the “Northern” hemisphere
N = v Sn (u,v) 0
u
{ ∈ | ≥ }
of the standard sphere Sn embedded in Rn. Furthermore define f :N Z by
u u
→
n+1
f (v)= sgnv sgnu =n+1 2k
u i i
−
i=1
X
where k is the number of sign disagreements between the components of the vectors v and u. Then if v is uniformly
distributed in N , the expected value of n+1 2k equals
u
−
1
I+ = f (v)dSn
u volN u
u ZNu
where dSn is the density of the Lebesgue measure on Sn.
Proposition 4.2 Let n even. Given u Sn, then
∈
2n 11
0<I+ < 1+
u π 12n
r (cid:18) (cid:19)
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Proof: LetI− betheaverageoff overthe“Southern”hemisphere v Sn (u,v) 0 . WefirstshowthatI+ >I−.
u u { ∈ | ≤ } u u
Supposetheangleφparametrizessomegeodesicthatconnectsthe“Southpole”, uwhereφ=0,tothe“Northpole”,
−
uwhereφ=π. Itiseasytoseethatforanyi, sgnu sgnv (φ)isanincreasingfunction,andinfactthefunctionhasa
i i
strictlypositiveincreaseoveritsdomain. Thesamethereforeistrueforf (v). ThusI+ >I−. Sincef ( v)= f (v)
we have that the integral of f over Sn yields zero, or I++I− = 0. Tougether withuthe inuequality, tuhi−s impli−es tuhat
u u u
0<I+.
u
To get the other inequality, we first partition Sn into orthants O . For every σ 1,+1 n+1:
σ
∈{− }
Oσ = v Sn (sgnu1 sgnv1, sgnu2 sgnv2, , sgnun+1 sgnvn+1)=σ
{ ∈ | ··· }
NowletQbe the quotientof 1,+1 n+1 andmultiplicationby-1. Itcanbe parametrizedbythosebinarysequences
{− }
s that have positive average. Define
Z =(O O ) N
s s −s u
∪ ∩
The sets Z partition the N into 2n sets of equal measure (up to measure zero). For all v in the set Z , f (v) equals
s u s u
either n+1 s or n+1( s ). Thus
i=1 i i=1 − i
n+1
P P I+ 2−n s
u ≤ i!
s∈Q i=1
X X
The sum n+1 s is equal to n+1 2k where k is the number of signdifferences. There are n+1 distinct sequences
i=1 i − k
s that have k sign differences. We get by telescoping
P (cid:0) (cid:1)
n+1 n2 n+1 n
2−n s =2−n [(n+1 k) k]=2−n(n+1)
s∈Q i=1 i! k=0 (cid:18) k (cid:19) − − (cid:18)n2(cid:19)
X X X
Now use Stirling’s Formula in the following form ([7], section 3.6):
n n 1 1
n!=√2πn 1+ + +
e 12n 288n2 ···
(cid:16) (cid:17) (cid:18) (cid:19)
The Proposition follows after some straightforwardalgebra.
The prediction of Theorem 4.1 depends on the the assumption that v is uniformly distributed in N . It would
u
seem reasonable that for any distribution dρ of v in N compatible with the problem, we would have that the
u
expectation of the number of sign disagreements between u and v is strictly less than n. If the distribution dρ of v is
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morebiasedtowardsthe vicinity ofthe pole u,then the majorityofa united LeftandRightwillbe morepronounced.
On the contrary, if the distribution is more biased towards the equator, then that majority will be narrower.
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