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CSM Workshop 1: Zeros of Graph Polynomials
Enumeration of Spanning Subgraphs with Degree
Constraints
Dave WagnerUniversity of Waterloo
I. The Set-Up
graph notation
G=(V,E) a (big) finite graph
graph notation
G=(V,E) a (big) finite graph
a set of edges, i.e. a spanning subgraph
)(GEH
graph notation
G=(V,E) a (big) finite graph
a set of edges, i.e. a spanning subgraph
the degree function of H
)(GEH
NVH :)deg(
graph notation
G=(V,E) a (big) finite graph
a set of edges, i.e. a spanning subgraph
the degree function of H
the set of vertices of degree k in H is
)(GEH
NVH :)deg(
kvHVvHVk ),deg(:)(
energy of a subgraph
J the energy of a single edge
energy of a subgraph
J the energy of a single edge
the “chemical potential” of a vertex of degree kk
energy of a subgraph
J the energy of a single edge
the “chemical potential” of a vertex of degree k
the energy of a (spanning) subgraph H is
k
kk HVHJHU )(##)(
k
partition function
T the absolute temperature
partition function
T the absolute temperature
the inverse temperatureTkB
1
partition function
T the absolute temperature
the inverse temperature
the Boltzmann weight of a subgraph H is)(e HU
TkB
1
partition function
T the absolute temperature
the inverse temperature
the Boltzmann weight of a subgraph H is
the partition function is
)(e HU
TkB
1
E(G)H
HUGZ
)(e
polynomial expression
let andJy e k
kue
polynomial expression
let and
for a subgraph H let
Jy e kku
e
Vv
vHH u ),deg()deg(u
polynomial expression
let and
for a subgraph H let
the partition function is
Jy e kku
e
H
E(G)HHG yyZ #
)deg(),(
uu
Vv
vHH u ),deg()deg(u
multivariate version
let and
Vv
vHv
H x ),deg()deg(x Vvxv :x
multivariate version
let and
the multivariate partition function is
)deg()deg(),(
~ H
E(G)HHGZ xuxu
Vv
vHv
H x ),deg()deg(x Vvxv :x
multivariate version
let and
the multivariate partition function is
then
)deg()deg(),(
~ H
E(G)HHGZ xuxu
Vv
vHv
H x ),deg()deg(x Vvxv :x
),(~
),( 2/1yZyZ GG uu
example
let and for all k>=2
110 uu 0ku
example
let and for all k>=2
110 uu 0ku
otherwise0
matching a is if1),deg()deg(
Hu
VvvHHu
example
let and for all k>=2
and
are, respectively, the multivariate and univariate matching polynomials of G
),(~
xuGZ
110 uu
),( yZG u
0ku
otherwise0
matching a is if1),deg()deg(
Hu
VvvHHu
vertex-dependent activities
the chemical potentials can vary from vertex to vertex:
vertex-dependent activities
the chemical potentials can vary from vertex to vertex:
let where )()(
1)(
0)( ,...,, v
dvvv uuuu ),deg( vGd
vertex-dependent activities
the chemical potentials can vary from vertex to vertex:
let where
and redefine
)()(1
)(0
)( ,...,, vd
vvv uuuu ),deg( vGd
Vv
vvHH u )(
),deg()deg(u
vertex-dependent activities
the chemical potentials can vary from vertex to vertex:
let where
and redefine
the multivariate partition function is still)deg(
)deg(),(~ H
E(G)HHGZ xuxu
)()(1
)(0
)( ,...,, vd
vvv uuuu ),deg( vGd
Vv
vvHH u )(
),deg()deg(u
II. The Results
the key polynomials
for each vertex v of G form the key polynomial
in whichkvk
d
kv zu
k
dzK )(
0
)(
),deg( vGd
the key polynomials
for each vertex v of G form the key polynomial
in which
Since this polynomial depends on T
kvk
d
kv zu
k
dzK )(
0
)(
),deg( vGd
)(
e)( vkv
ku
the key polynomials
for each vertex v of G form the key polynomial
in which
Since this polynomial depends on T
except when all
kvk
d
kv zu
k
dzK )(
0
)(
),deg( vGd
)(
e)( vkv
ku
,0)(vk
the key polynomials
for each vertex v of G form the key polynomial
in which
Since this polynomial depends on T
except when all
that is, when all
kvk
d
kv zu
k
dzK )(
0
)(
),deg( vGd
)(
e)( vkv
ku
1,0)( vku
,0)(vk
first theorem
Assume that all zeros of all the keys are within an angle
of the negative real axis Then…
.2/0
first theorem
Assume that all zeros of all the keys are within an angle
of the negative real axis Then…
1. If for all v then
.2/0
2)arg( vx .0),(
~ xuGZ
first theorem
Assume that all zeros of all the keys are within an angle
of the negative real axis Then…
1. If for all v then
2. If then
.2/0
2)arg( vx .0),(
~ xuGZ
2)arg( y .0),( yZG u
first theorem
Assume that all zeros of all the keys are within an angle
of the negative real axis Then…
1. If for all v then
2. If then
This statement is independent of the size of the graph….
.2/0
2)arg( vx .0),(
~ xuGZ
2)arg( y .0),( yZG u
first theorem
Assume that all zeros of all the keys are within an angle of the negative real axis
Then…
1. If for all v then
2. If then
This statement is independent of the size of the graph….so it can be used for thermodynamic limits.
.2/0
2)arg( vx .0),(
~ xuGZ
2)arg( y .0),( yZG u
first theorem
first theorem
first theorem
Consider the case :0
first theorem
Consider the case
Assume that all zeros of all the keys are nonpositive real numbers. Then…
:0
first theorem
Consider the case
Assume that all zeros of all the keys are nonpositive real numbers. Then…
1. If for all v then
:0
2)arg(
vx .0),(
~ xuGZ
first theorem
Consider the case
Assume that all zeros of all the keys are nonpositive real numbers. Then…
1. If for all v then
(the half-plane property)
:0
2)arg(
vx .0),(
~ xuGZ
first theorem
Consider the case
Assume that all zeros of all the keys are nonpositive real numbers. Then…
1. If for all v then
(the half-plane property)
2. All zeros of are nonpositive real numbers.
:0
2)arg(
vx .0),(
~ xuGZ
),( yZG u
the Heilmann-Lieb (1972) theorem
let and for all k>=2
110 uu 0ku
the Heilmann-Lieb (1972) theorem
let and for all k>=2
for each vertex v,
has only real nonpositive zeros….
110 uu 0ku
),deg( vGd dzzKv 1)(
the Heilmann-Lieb (1972) theorem
let and for all k>=2
for each vertex v,
has only real nonpositive zeros.…
1. The multivariate matching polynomial has the half-plane property.
),(~
xuGZ
110 uu 0ku
),deg( vGd dzzKv 1)(
the Heilmann-Lieb (1972) theorem
let and for all k>=2
for each vertex v,
has only real nonpositive zeros….
1. The multivariate matching polynomial has the half-plane property.
2. The univariate matching polynomial has only real nonpositive zeros.
),(~
xuGZ
110 uu
),( yZG u
0ku
),deg( vGd dzzKv 1)(
a generalization
fix functions such that(at every vertex)
N)(:, GVgf 1 fgf
a generalization
fix functions such that(at every vertex)
choose vertex chemical potentials so that
otherwise0
)()( if1)( vgkvfu vk
N)(:, GVgf 1 fgf
a generalization
fix functions such that(at every vertex)
choose vertex chemical potentials so that
Then every key has only real nonpositive zeros, so that
1. has the half-plane property (new)
2. has only real nonpositive zeros (W. 1996)
),(~
xuGZ
),( yZG u
otherwise0
)()( if1)( vgkvfu vk
N)(:, GVgf 1 fgf
a theorem of Ruelle (1999)
let and for all k>=3
1210 uuu 0ku
a theorem of Ruelle (1999)
let and for all k>=3
for each vertex v,
has all its zeros within of the negative real axis
1210 uuu
2
21)( z
ddzzKv
0ku
4/
a theorem of Ruelle (1999)
let and for all k>=3
for each vertex v,
has all its zeros within of the negative real axis
1. If for all v then (new)
1210 uuu
2
21)( z
ddzzKv
0ku
4/
4)arg(
vx .0),(
~ xuGZ
a theorem of Ruelle (1999)
let and for all k>=3
for each vertex v,
has all its zeros within of the negative real axis
1. If for all v then (new)
2. If then
1210 uuu
2
21)( z
ddzzKv
0ku
4/
4)arg(
vx .0),(
~ xuGZ
2)arg(
y .0),( yZG u
a theorem of Ruelle (1999)
let and for all k>=3
for each vertex v,
has all its zeros within of the negative real axis
1. If for all v then (new)
2. If then
(Ruelle proves that for 2. it suffices thatfor a graph with maximum degree .)
1210 uuu
2
21)( z
ddzzKv
0ku
4/
4)arg(
vx .0),(
~ xuGZ
2)arg(
y .0),( yZG u
2)1(
2)Re(
y
second theorem
Assume that all zeros of all the keys havemodulus at least . Then…
1. If for all v then
2. If then
vx .0),(~ xuGZ
2y .0),( yZG u
third theorem
Assume that all zeros of all the keys havemodulus at most , and that the degree of each
key equals the degree of the corresponding vertex.
Then…
1. If for all v then
2. If then
vx .0),(~ xuGZ
2y .0),( yZG u
corollary
If all zeros of all keys are on the unit circle, and all keys
have the same degree as the corresponding vertex, then
every zero of is on the unit circle.
),( yZG u
corollary
If all zeros of all keys are on the unit circle, and all keyshave the same degree as the corresponding vertex, thenevery zero of is on the unit circle.
For any graph G, every zero of
is on the unit circle.
),( yZG u
1
),deg(
),deg(
)(
#
VvGEH
H
vH
vGy
application
consider a sequence of graphs G whose union isan infinite graph
application
consider a sequence of graphs G whose union isan infinite graph
assume that each graph G is d-regular
application
consider a sequence of graphs G whose union isan infinite graph
assume that each graph G is d-regularthat all keys are the same
),( zK
application
consider a sequence of graphs G whose union isan infinite graph
assume that each graph G is d-regularthat all keys are the sameand that the thermodynamic limit free energy
exists:
),(log)(#
1lim),,( yZ
GVJf G
Guμ
),( zK
application
consider a sequence of graphs G whose union isan infinite graph
assume that each graph G is d-regularthat all keys are the sameand that the thermodynamic limit free energy exists:
If the free energy is non-analytic at a nonnegative realthen has a zero not at the origin with
nonnegativereal part.
),(log)(#
1lim),,( yZ
GVJf G
Guμ
*
),( zK
),( * zK
example 1.
let and for all k>=310 u 0ku
example 1.
let and for all k>=3
the key is
10 u 0ku
221 2
1)( zud
zduzK
example 1.
let and for all k>=3
the key is
if then the zeros of K(z) have negative real part…. No phase transitions for any physical (J,T)
10 u 0ku
221 2
1)( zud
zduzK
0e 11 u
example 1.
let and for all k>=3
the key is
if then the zeros of K(z) have negative real part…. No phase transitions for any physical (J,T)
from the second theorem it follows that whenthere is no phase transition for
10 u 0ku
221 2
1)( zud
zduzK
0e 11 u
01 u
2
2
log dB
JTk
example 2.
fix functions such that(at every vertex)
N)(:, GVgf 3 fgf
example 2.
fix functions such that(at every vertex)
choose vertex chemical potentials so that
otherwise0
)()( if1)( vgkvfu vk
N)(:, GVgf 3 fgf
example 2.
fix functions such that(at every vertex)
choose vertex chemical potentials so that
When the thermodynamic limit exists
it is analytic for all physical values of (J,T).(no phase transitions)
otherwise0
)()( if1)( vgkvfu vk
N)(:, GVgf 3 fgf
),,( μJf
example 3.
in a 2d-regular graph, consider the key
dd zuzd
dzK 22
1)(
example 3.
in a 2d-regular graph, consider the key
for a thermodynamic limit of these
a phase transition with
can only happen at
dd zuzd
dzK 22
1)(
),,( μJf
e
22
2
1log)2(
d
ddJ
4log2
log2
d
dTkB
III. Summary
summary
* very general set-up, but it records no global structure
summary
* very general set-up, but it records no global structure
* unifies a number of previously considered things
summary
* very general set-up, but it records no global structure
* unifies a number of previously considered things
* very mild hypotheses, but similarly weak conclusions about absence of phase transitions:
summary
* very general set-up, but it records no global structure
* unifies a number of previously considered things
* very mild hypotheses, but similarly weak conclusions about absence of phase transitions:
* many general “soft” results
summary
* very general set-up, but it records no global structure
* unifies a number of previously considered things
* very mild hypotheses, but similarly weak conclusions about absence of phase transitions:
* many general “soft” results* some quantitative “hard” versions of qualitatively
intuitive results
summary
* very general set-up, but it records no global structure
* unifies a number of previously considered things
* very mild hypotheses, but similarly weak conclusions about absence of phase transitions:
* many general “soft” results* some quantitative “hard” versions of qualitatively
intuitive results
* proofs are short and easy: (half-plane property/polarize & Grace-Walsh-Szego/ “monkey business”/diagonalize)
