Transcript Aucun titre de diapositive

Graphes Combinatoires et Théorie
Quantique des Champs
Gérard Duchamp, Université de Rouen, France
Collaborateurs :
Karol Penson, Université de Paris VI, France
Allan Solomon, Open University, Angleterre
Pawel Blasiak, Instit. of Nucl. Phys., Cracovie, Pologne
Andrzej Horzela, Instit. of Nucl. Phys., Cracovie, Pologne
Congrès de l’ACFAS, 11 mai 2004
1
Content of talk
 A formula from QFT giving the Hadamard
product of two EGFs
 Development in case F(0)=G(0)=1
 Expression with (Feynman-Bender and al.)
diagrams
 Link with packed matrices
 Back to physics : One parameter groups of
substitutions and normal ordering of boson strings
(continuous case)
Substitutions and the « exponential formula »
(discrete case)
 Lie-Riordan group
 Conclusion
2
The Hadamard product of two sequences
is given by the pointwise product
We can at once transfer this law on EGFs by
but, here, as
we get
3
In case
we can set
if, for example, the Ln are (non-negative)
integers, F(y) is the EGF of set-partitions (see
the talk of M. Rosas yesterday) which
k-blocks can be coloured with Lk different
colours.
As an example, let us take L1, L2 and
Ln=0 for n>2. Then the objects of size n are
the set-partitions of a n-set in singletons and
pairs having respectively L1 and L2 colours
allowed
4
For n=3, we have two types : the type
(three possibilities without the colours, on the
left) and the type
(one possibility without the
colours, on the right).
It turns out that, with the colours, we have
which agrees with the computation.
5
In general, we adopt the denotation
for the type of a (set) partition which means that
there are a1 singletons a2 pairs a3 3-blocks a4 4-blocks
and so on.
The number of set partitions with type  as above is
well known (see Comtet for example)
Thus, using what has been said in the beginning, with
6
one has
Now, one can count in another way the expression
numpart()numpart(), remarking that this is the
number of couples of set partitions (P1,P2) with
type(P1)=, type(P2)=. But every couple of
partitions (P1,P2) has an intersection matrix ...
7
{1,5} {2} {3,4,6}
{1,2}
1
1
0
{3,4}
0
0
2
{5,6}
1
0
1
{1,5}
{1,2}
{2}
{3,4}
{3,4,6}
{5,6}
Packed matrix
see NCSFVI
(GD, Hivert,
and Thibon)
Feynman-Bender (& al.)
diagram
Remark: Juxtaposition of diagrams amounts to do the blockdiagonal
product of the corresponding matrices which are then indexed by 8the
product of set partitions defined by M. Rosas yesterday.
Now the product formula for EGFs reads
The main interest of this new form is that
we can impose rules on the counted graphs !
9
Some Model Graphs*
2. Lines and Vertices
EX: 4 lines
V1*V3
(V2)2
V4
(V1)4
Single
Single
and double
Quadruple
Singles
*C.M. Bender, D.C. Brody, B.K.Meister ,
10
Quantum Field Theory of Partitions, J.Math. Phys. 40, 3239 (1999)
11
12
Back to physics:
One parameter groups of
substitutions and normal ordering of boson strings
(continuous case)
13
Fermion Normal Ordering Problem*
f1 f 2 f 3 f
satisfying the usual

4
f5 f6 f


7

f i f j  f j f i   ij
In this elementary case there are
12 terms f 3
8 terms f+ f 4
1 term
f+2 f 5
The numbers 1,2, 12,.. are combinatorial in origin
(see Navon reference)
14
*Combinatorics and Fermion Algebra, AM Navon, Il Nuovo Cimento 16B,324(1973)
Boson Normal Ordering Problem*
satisfying


bb  b b  1
n
(b b)   S (n, k )b b

n
k
k
k 1
The integers S(n,k) are the Stirling Numbers of the
Second Kind.
15
*Combinatorial Aspects of Boson Algebra, J Katriel, Lett. Nuovo Cimento 10,565(1974)
From now on, we will denote
u=b+ (raising operator) and d=b (lowering op.)
satisfying [d,u]=1.
With w=ud, one has the Stirling matrix
16
In this way, two parameters families of new Bell and
Stirling numbers could be defined by means of the
normal ordering
ns
[(a  ) r a s ]n   (a  ) n ( r s )   S r ,s (n, k )(a  ) k a k
k s
with rs, and
ns
Br ,s (n)   Sr ,s (n, k )
ks
see for example, P Blasiak, KA Penson and AI Solomon,
The Boson Normal Ordering Problem and Generalized Bell Numbers,
Annals of Combinatorics 7, 127 (2003)
17
With w=udu, one has the following matrix
18
With w=udduu, one gets
Each of these matrices are row-finite and
induce a sequence transformation just by
multiplication on the left and they form an algebra.
19
With
the sequence
and the (infinite) matrix
is given by
,
and the transformation induced over the
generating series is f --> g such that
20
We can observe that, if there is only one derivative in
the word w the matrix is a matrix of substitution with
prefactor function i.e. a transformation of the type
this is due to the fact that we can represent u,d by
operators over the functions on the line
(Bargmann-Fock): multiplication by x and
differentiation. The resulting operator being either a
vector field or the conjugate of a vector field by an
automorphism. Let us compute, for example the
substitution corresponding to
21
On gets the first special cases and some others
22
Substitutions and the « exponential
formula » (discrete case)
Well known to enumerative combinatorists:
(For certain classes of graphs)
If C(x) is the EGF of CONNECTED graphs, then
exp(C(x)) is the EGF of all graphs.
This implies that the matrix
M(n,k)=number of graphs with n vertices and
having k connected components is the matrix of a
substitution.
One can prove that if M is such a matrix (with identity
diagonal), then all its powers (positive negative and
fractional) are substitution matrices and form a oneparameter group of substitutions, thus coming from a
vector field on the line which can be computed. But no
nice combinatorics seems to emerge.
23
Conclusion
• We have, following Bender and al., given a « coupled »
decomposition of the product formula. This can be used to
give sections of EGFs (a non-trivial problem, trisection of
Hermite EGF, by Ira Gessel and al. has been obtained very
recently)
• Continuous and Discrete exponentials arising from physics
and combinatorics have been presented. Remains some
problems as, for example, a nice combinatorial description
of the (existing) one-parameter groups associated to a
substitution (say, to begin with, the Stirling substitution,
which seems to induce what I could call a « Lambert
phenomenon »).
24