Holonomy and combinatorics - Simon Fraser Universitypeople.math.sfu.ca/~mmishna/PUB/holtalk.pdf ·...

Marni Mishna, Freaky Friday the 13th, 2004 Carleton University Holonomy and Combinatorics – p. 1/26

Holonomy and combinatoricsFrom automatic identities to systematic enumeration

Marni MishnaLaBRI, Université Bordeaux I

http://www.labri.fr/~mishna

The Set Up

● A grab bag of problems

● Basic Idea

● Summary of how it works

Creative telescoping

A combinatorial framework

The scalar product

Conclusion


A grab bag of problems

Prove Dixon’s identity:∑

k(−1)k(n+an+k

)(n+bb+k

)(a+ba+k

)= (n+a+b)!

n!a!b!

Calculate the dimensionof one representation ofthe symmetric group inanother

ζ(2, 2) = 1/2(ζ(2)2 − ζ(4)

)

Find a recursive formulafor the number of 4-

regular graphs of size n

Find a formula for standard

Young tableaux of size n

5 7

4 6 8

1 2 3

Enumerate plane parti-tions of shape 2× n

633321

522111

Prove q-Hypergeometric identities

The Set Up


● Basic Idea




The scalar product

Conclusion


Basic Idea

Set of easy identities

A set of recurrences forp(n, k)

A recurrence forφ(p(n, k))

Systems of differentialequations satisfied by f

and g

A differential equationfor ψ(f, g)

The recurrences and DEs are useful! They give sequences,can be solved, allow asymptotic analysis, ...

The Set Up


● Basic Idea




The scalar product

Conclusion


Basic Idea

Set of easy identities Elimination inoperator algebra




and g



The Set Up


● Basic Idea




The scalar product

Conclusion


Basic Idea

Set of easy identities Elimination inoperator algebra Profound identity




and g




Summary of how it works

Dixon’s Identity 4-Regular Graphs

�

k(−1)k

�

n+a

n+k

� �

n+b

b+k

� �

a+b

a+k

�

=(n+a+b)!

n!a!b!

1

2

3 4

5

6

7

Operator Algebra Shift Differential

Closure Operation +, * 〈, 〉

“Easy” Identity recurrence for n! DE for ex

Elimination Answer is in a left ideal. Find a Gröbner basis for this ideal

Output Recurrences for hypergeo-metric sums

Diff eq. satisfied by gen-erating function

The Set Up


● P-Recursive Functions

● Express recurrences as operators

● Elimination

● Zeilberger’s Creative Telescoping

● How does this work in general?


The scalar product

Conclusion



The Set Up




● Elimination




The scalar product

Conclusion


P-Recursive Functions

Def: A function f : N → R is P-recursive if it satisfies a finitelinear recurrence with polynomial coefficients.There exists polynomials γi such that:

γd(n)f(n+ d) + . . .+ γ1(n)f(n+ 1) + γ0(n)f(n) = 0

Ex: f(n) = n! is P-recursive: 1(n+ 1)!− (n+ 1)n! = 0Ex: Rational: p(x)/q(x) =

∑

n f(n)xn =⇒ γi = const.

The Set Up




● Elimination




The scalar product

Conclusion


Express recurrences as operators

Shift operator:

SnP (n, k) = P (n+ 1, k), SkP (n, k) = P (n, k + 1)

Define a (non-commutative) algebra:

Bn,k := C〈n, k, Sn, Sk〉

Snn = (n+ 1)Sn, Skk = (k + 1)Sk, all other variables commute

0 = P (n, k) + P (n, k + 1) + (3n2 + k)P (n+ 2, k + 1)

= P (n, k) + SkP (n, k) + (3n2 + k)S2nSkP (n, k)

= (1 + Sk + (3n2 + k)S2nSk) · P (n, k)

= ψ(n, k, Sn, Sk) · P

■ If ψ, ξ ∈ IP , then A(n, k, Sn, Sk)ψ +B(n, k, Sn, Sk)ξ ∈ IP .■ ψ is in the annihilating (left) ideal IP of P .

The Set Up




● Elimination




The scalar product

Conclusion


Making new identities using elimination

Recall from your youth:n equations, m unknowns variables + linear algebra =⇒eliminate some unknown variables.

3x+ 4y + z = 2 (1)

5x− 2y + 2z = 3 (2)

5 ∗ (1)− 3 ∗ (2) =⇒

26y − z = 1

We can do this in the setting of non-commutative algebras.(Ore algebra)■ Left ideals will have Gröbner basis (Galligo)■ Modified traditional tools work (Buchbergers + Euclidean

algorithm)

The Set Up




● Elimination




The scalar product

Conclusion


Zeilberger’s Creative Telescoping

F (n, k) =(nk

)

■ Easy identities: F (n+1,k)F (n,k) = n+1

n−k+1F (n,k+1)

F (n,k) = n−kk+1

■ Cross Multiply:

((n− k + 1)Sn − (n+ 1))F = 0 ♥

((k + 1)Sk − (n− k))F = 0 ♣

■ A linear combination: (Sk + 1)♥+ Sn♣ = 0(n+ 1) (SnSk − Sk − 1)F = 0

■ Reorganize this to have only Sn and n on the left, and afactor of (Sk − 1) on the right:

(n+ 1)(Sn − 2)F = (Sk − 1)G(n, k), for some G(n, k)

The Set Up




● Elimination




The scalar product

Conclusion



F (n, k) =(nk

)

(n+ 1)(Sn − 2)F = (Sk − 1)G(n, k) (∗)

■ Take the sum:

(n+ 1)(Sn − 2)

a(n) =∑

k≥0

(n+ 1)(Sn − 2)

F (n, k)

=∑

k≥0

(Sk − 1)(G(n, k))

=∑

k≥0

(G(n, k + 1)−G(n, k)) = 0

■ Gives result about the sum:

a(n+ 1)− 2a(n) = 0 =⇒ a(n) = 2n =⇒∑

k

(nk

)= 2n

The Set Up




● Elimination




The scalar product

Conclusion



F (n, k) =(nk

)

(n+ 1)(Sn − 2)F = (Sk − 1)G(n, k) (∗)

■ Apply the linear combination to this sum:

(n+ 1)(Sn − 2)a(n) =∑

k≥0

(n+ 1)(Sn − 2)F (n, k)

=∑

k≥0

(Sk − 1)(G(n, k))

=∑

k≥0

(G(n, k + 1)−G(n, k)) = 0


a(n+ 1)− 2a(n) = 0 =⇒ a(n) = 2n =⇒∑

k

(nk

)= 2n

The Set Up




● Elimination




The scalar product

Conclusion



F (n, k) =(nk

)

(n+ 1)(Sn − 2)F = (Sk − 1)G(n, k) (∗)

■ Apply (∗):

(n+ 1)(Sn − 2)a(n) =∑

k≥0

(n+ 1)(Sn − 2)F (n, k)

=∑

k≥0

(Sk − 1)(G(n, k))

=∑

k≥0

(G(n, k + 1)−G(n, k)) = 0


a(n+ 1)− 2a(n) = 0 =⇒ a(n) = 2n =⇒∑

k

(nk

)= 2n

The Set Up




● Elimination




The scalar product

Conclusion



F (n, k) =(nk

)

(n+ 1)(Sn − 2)F = (Sk − 1)G(n, k) (∗)

■ Expand:

(n+ 1)(Sn − 2)a(n) =∑

k≥0

(n+ 1)(Sn − 2)F (n, k)

=∑

k≥0

(Sk − 1)(G(n, k))

=∑

k≥0

(G(n, k + 1)−G(n, k)) = 0


a(n+ 1)− 2a(n) = 0 =⇒ a(n) = 2n =⇒∑

k

(nk

)= 2n

The Set Up




● Elimination




The scalar product

Conclusion


How does this work in general?

■ Begin with a set of “independent” relationsP (n, k, Sn, Sk)F (n, k) = 0 Q(n, k, Sn, Sk)F (n, k) = 0■ Find the good linear combination which eliminates k and has

a factor of Sk − 1

R(n, k, Sn, Sk) = A(n, k, Sn, Sk)P +B(n, k, Sn, Sk)Q

R(n, k, Sn, Sk) = S(Sn, n) + (Sk − 1)R′(n, Sn, Sk)

■ Take the infinite sum over k. We have a telescoping sum:0 =

∑

k R(n, k, Sn, Sk)F (n, k)

0 =∑

k S(Sn, n)F +∑

k(Sk − 1)R′(N,K, n)F︸︷︷︸

G(n,k)

■ ...And we get a recurrence for the sum

The Set Up



● Generating functions

● D-finite functions

● D-finiteness

The scalar product

Conclusion



The Set Up





● D-finiteness

The scalar product

Conclusion


Combinatorial generating functions

Counting information about combinatorial objects can beencoded in formal power series:

Ex. exponential generating function (egf) :

C(z) =∑

n

(number of objects in C of size n)zn

n!

Ex. C = C...

=⇒ C(z) = z + z2/2 + 2z3/3! + 3x4/4! + . . .

Nice: algebraic operations correspond with combinatorialoperations.

■ C = A ∪ B =⇒ C(z) = A(z) +B(z)

■ C = {(a, b), a ∈ A, b ∈ B} =⇒ C(z) = A(z)B(z)

The Set Up





● D-finiteness

The scalar product

Conclusion


Combinatorial generating functions

Counting information about combinatorial objects can beencoded in formal power series:

Ex. exponential generating function (egf) :

C(z) =∑

n

(number of objects in C of size n)zn

n!

Ex. C = C...

=⇒ C(z) = z + z2/2 + 2z3/3! + 3x4/4! + . . .Nice: algebraic operations correspond with combinatorialoperations.

■ C = A ∪ B =⇒ C(z) = A(z) +B(z)

■ C = {(a, b), a ∈ A, b ∈ B} =⇒ C(z) = A(z)B(z)

The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions

Def. A function f(x1, x2, . . . , xn) is Differentiably finite (D-finite)with respect to X = x1, . . . , xn if■ For 1 ≤ j ≤ n, f satisfies n linear differential equations with

polynomial coefficients:

φ0(X)f(X) + φ1(X)∂f(X)∂xj

+ . . .+ φk(X)∂kf(X)

∂xkj

= 0

Examples■ f = sin(2x3 + 3yz) is D-finite with respect to {x, y, z}.xfxx − 2fx + 9x5f = 0, fyy − 9z2f = 0, fzz − 9y2f = 0

■ exp(polynomial) is D-finite■

∑f(n)zn D-finite ⇐⇒ f(n) P-recursive

The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0

Examples■ f = sin(2x3 + 3yz) is D-finite with respect to {x, y, z}.

xfxx − 2fx + 9x5f = 0, fyy − 9z2f = 0, fzz − 9y2f = 0



The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0

Examples■ f = sin(2x3 + 3yz) is D-finite with respect to {x, y, z}.xfxx − 2fx + 9x5f = 0,

fyy − 9z2f = 0, fzz − 9y2f = 0



The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0

Examples■ f = sin(2x3 + 3yz) is D-finite with respect to {x, y, z}.xfxx − 2fx + 9x5f = 0, fyy − 9z2f = 0,

fzz − 9y2f = 0



The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0




The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0


■ exp(polynomial) is D-finite

■∑f(n)zn D-finite ⇐⇒ f(n) P-recursive

The Set Up





● D-finiteness

The scalar product

Conclusion


D-finite functions




+ . . .+ φk(X)∂kf(X)

∂xkj

= 0




The Set Up





● D-finiteness

The scalar product

Conclusion


Hierarchy of Power Series

Differentiably Algebraic partitions

D-finite k-regular graphs

Algebraic Dyck Paths

Rationalunrestricted walksin the plane

1−√

1−4x2x

Regular expressionsMotzkin Paths

ex

k × n-latin squares grid walks in the first quadrant

eex−1

Algebraic closure Hadamard product Algebraic substitutionFunctional Inverse Diagonals Hadamard product

Integrals Multiplicative inverse

The Set Up



The scalar product

● Symmetric Series

● D-finite symmetric series

● The Scalar Product


● Encoding combinatorial objects

● Weyl Algebras

● How does it work?

● Holonomic Modules

● Proof

● Other applications

Conclusion


The scalar product

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Symmetric Series

power pk p3 = x31 + x3

2 + x33 + . . .

homogeneous hk h3 = x31 + x3

2 + . . .+ x21x2 + . . .+ x1x2x3 + . . .

elementary en e3 = x1x2x3 + x1x2x4 + x1x3x4 + . . .

monomial mλ m(3,2,1) = x31x

22x3 + x3

2x21x3 + x3

3x22x6 + . . .

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Symmetric Series

power pk p3 = x31 + x3

2 + x33 + . . .

homogeneous hk h3 = x31 + x3

2 + . . .+ x21x2 + . . .+ x1x2x3 + . . .

elementary en e3 = x1x2x3 + x1x2x4 + x1x3x4 + . . .

monomial mλ m(3,2,1) = x31x

22x3 + x3

2x21x3 + x3

3x22x6 + . . .

■ pλ = pλ1pλ2

· · · pλk(e.g. p(3,2,2,1) = p3p

22p1)

■ Our algebra of interest: K[[p1, p2, . . . ]]

■ Ex:H =

∑

n hn = exp(∑

npn

n)

E =∑

n en = exp(∑

n(−1)n+1 pn

n)

■ Scalar product: 〈pλ, pµ〉 = δλµzλ

λ = (1m12m2 · · · kmk ) =⇒ zλ = 1m1m1!2m2m2! · · · kmkmk!

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


D-finite symmetric series

Def. A symmetric series F ∈ K[[p1, p2, . . . ; t]] is D-finite if forany n,■ F (p1, ..., pn, 0, 0, . . . ; t) is D-finite wrt p1, . . . , pn, t.

■ Examples:

H =∑hnt

n = exp(∑

n pntn/n)

E =∑ent

n = exp(∑

n(−1)n+1pntn/n)

S =∑

λ sλt|λ| = exp(

∑

n

p2

nt2n

2n+ p2n−1t2n−1

2n−1 )

∑

λ all parts odd sλt|λ| = SE−1

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


D-finite symmetric series

Def. A symmetric series F ∈ K[[p1, p2, . . . ; t]] is D-finite if forany n,■ F (p1, ..., pn, 0, 0, . . . ; t) is D-finite wrt p1, . . . , pn, t.■ Examples:

H =∑hnt

n = exp(∑

n pntn/n)

E =∑ent

n = exp(∑

n(−1)n+1pntn/n)

S =∑

λ sλt|λ| = exp(

∑

n

p2

nt2n

2n+ p2n−1t2n−1

2n−1 )

∑

λ all parts odd sλt|λ| = SE−1

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


The Scalar Product

■ The scalar product, (symmetric, bilinear form) is defined by

〈pλ, pµ〉 = δλµzλ.

λ = (1m12m2 · · · kmk ) =⇒ zλ = 1m1m1!2m2m2! · · · kmkmk!

■ Equivalent to coefficient extraction:

[xk1

1 xk2

2 . . . xkn

n ]F = 〈F, hk1hk2

· · ·hkn〉

■ Adjunction:〈pnF,G〉 = 〈F, n ∂

∂pnG〉.

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


The Scalar Product

Theorem (Gessel, 90): If F ∈ K[[p1, p2, . . . , pn; t]] and G ∈ K[[p1, p2, . . . ; t]]are both D-finite symmetric series then the scalar product

〈F,G〉 = φ(t)

is a D-finite function with respect to t.

That is, φ(t) satisfies a linear DE with polynomial coefficients.Infact, this DE can be calculated (Chyzak, M., Salvy 02)

Termination guaranteed by D-finiteness.

Ex:> scalar de(exp(p1+p12/2+p2+p22),exp((p22+p2/2)*t),[t],f);

[(−126 t+ 13− 16 t2

)f (t) +

(−1 + 32 t− 256 t2

)ddtf (t)]


Encoding combinatorial objects

Object Encoding π Sum over objects in the class

MAKE SPACE

t

,,

eee

ll

%%%

,,

t t

tt

t

t2 3

4

6 7

5

1

x31x

42x

23x

54x

55x

26x

27

G =∑

g a graph π(g)

= exp(∑

n

p2

n

2n− p2n

2n)

Set(C) x21x

22x

23x4x5 H[ZC], E[ZC]

C a labelled,finite comb. class

xk

i=⇒ k occurrences

of label i

H =

�

hn, E =

�

en,Z = Polya cycle index,[]= plethystic substitution


Encoding combinatorial objects

Object Encoding π Generating function of regular subclass

MAKE SPACE

t

t

t

t2

4 3

1

x21x

22x

23x

24 Gk(t) = 〈G,

∑hn

2 tn〉

■ Idea: Encode all objects, use scalar product to extract the number with acertain regularity

■ Scalar product with a series gives the egf of a “regular” subclass.

■ Other examples: Young tableaux, k × n-latin squares, hypergraphs

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Weyl Algebras

Ap := K〈p1, . . . , pn, ∂1, . . . , ∂n〉

∂ipi = pi∂i + 1, all other variables commute

f(p1, . . . , pn)

Left annihilating ideal: If

= {φ ∈ Ap|φ · f = 0}

Left module: Apf

= {ψ · f |ψ ∈ Ap} ' Ap/If

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Weyl Algebras

Ap := K〈p1, . . . , pn, ∂1, . . . , ∂n〉

∂ipi = pi∂i + 1, all other variables commute

f(p1, . . . , pn) f = sin(p1 + p2/2)

Left annihilating ideal: If

= {φ ∈ Ap|φ · f = 0} If = Ap〈∂21 + 1, 4∂2

2 + 1, 2∂1 − ∂2〉

Left module: Apf

= {ψ · f |ψ ∈ Ap} ' Ap/If Apf = K[p1, p2]sin(p1 + p2/2)

⊕K[p1, p2]cos(p1 + p2/2)

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?

Like with creative telescoping, we take linear combinations ofequations to eliminate variables and generate new equations.0− 0 = 0〈0, G〉 − 〈F, 0〉 = 0

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?

Like with creative telescoping, we take linear combinations ofequations to eliminate variables and generate new equations.0− 0 = 0〈0, G〉 − 〈F, 0〉 = 0〈φ · F,G〉 − 〈F, ψ ·G〉 = 0 φ ∈ IF , ψ ∈ IG

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?

Like with creative telescoping, we take linear combinations ofequations to eliminate variables and generate new equations.0− 0 = 0〈0, G〉 − 〈F, 0〉 = 0〈φ · F,G〉 − 〈F, ψ ·G〉 = 0 φ ∈ IF , ψ ∈ IG〈F, φ⊥ ·G〉 − 〈F, ψ ·G〉 = 0 p⊥n = n∂n, ∂

⊥n = pn/n

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?


⊥n = pn/n

〈F,(φ⊥ − ψ

)·G〉 = 0 by linearity

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?


⊥n = pn/n

〈F,(φ⊥ − ψ


If,(φ⊥ − ψ

)= γ(t, ∂t) ∈ C[t, ∂t], then...

〈F, γ(t, ∂t) ·G〉 = 0

=⇒ γ(t, ∂t) · 〈F,G〉 = 0

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?


⊥n = pn/n

〈F,(φ⊥ − ψ


If,(φ⊥ − ψ

)= γ(t, ∂t) ∈ C[t, ∂t], then...

〈F, γ(t, ∂t) ·G〉 = 0 =⇒ γ(t, ∂t) · 〈F,G〉 = 0

This gives a differential equation satisfied by 〈F,G〉.

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


How does it work?


⊥n = pn/n

〈F,(φ⊥ − ψ


If,(φ⊥ − ψ

)= γ(t, ∂t) ∈ C[t, ∂t], then...

〈F, γ(t, ∂t) ·G〉 = 0 =⇒ γ(t, ∂t) · 〈F,G〉 = 0

This gives a differential equation satisfied by 〈F,G〉.

Any non-trivial element γ of(

I⊥f + Ig

)⋂C[t, ∂t] provides a

differential equation satisfied by 〈f, g〉.

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Holonomic Modules

Ax := K〈x1, . . . , xn, ∂1, . . . , ∂n〉∂ixi = xi∂i + 1, all other variables commute

Filtration: A(d)x = {ψ ∈ Ax| degψ ≤ d}

Def. A module Axf is holonomic if

dimA(d)x · f = O(dn).

✔ K[x1, ..., xn] = Ax • 1: dimA(d)x · 1 = O(dn)

✘ Ax: dimA(d)x = O(d2n)

✔ Ax · sin(x) ' K[x] sin(x)⊕K[x] cos(x)

dimA(d)x · sin(x) = 2 dimK[x](d) = O(d1)

Theorem: (Kashiwara) f(x) D-finite ⇐⇒ Axf holonomic

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


(I⊥f + Ig

)∩ At is not empty!

Idea: model adjunction p⊥n = n∂pnby a tensor product

■ Define surjection:Ap,tf ⊗

⊥C[t] Ap,tg −→ Ap,t〈f, g〉

a⊗⊥C[t] b 7→ 〈a, b〉

Prop: I⊥f + Ig contains AnnAt(f ⊗⊥C[t] g)

■ Thm: Ap,tf ⊗⊥C[t] Ap,tg is a holonomic Ap,t-module

■

(

Ap,tf ⊗⊥C[t] Ap,tg

)

∩ “At” is a holonomic At-module

isomorphic to At/AnnAt(f ⊗⊥

C[t] g) as At-modules

Cor: AnnAt(f ⊗⊥

C[t] g) is non-trivial.

The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


(I⊥f + Ig





a⊗⊥C[t] b 7→ 〈a, b〉



key fact: ⊥ is degree preserving

■

(


)




Cor: AnnAt(f ⊗⊥


The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


(I⊥f + Ig





a⊗⊥C[t] b 7→ 〈a, b〉



■

(


)




Cor: AnnAt(f ⊗⊥


The Set Up



The scalar product






● Weyl Algebras



● Proof


Conclusion


Other applications

Algorithm computes scalar products whose adjoint operation isa degree preserving operation:➢Related to Macdonald Polynomials

〈pλ, pµ〉q,t = zλδµλ

∏

λi∈λ

1− qλi

1− tλi

➢MacMahon symmetric functions

〈p(a,b)2(c,d), p(a,b)2(c,d)〉 = 2

(a!b!

(a+ b− 1)!

)2 (c!d!

(c+ d− 1)!

)

The Set Up



The scalar product

Conclusion

● Lots to do...


Lots to do...

■ Other D-finite closure properties?■ Effective closure properties??■ Classifying families of combinatorial objects e.g. walks in the

plane, families of generalized partitions■ Improving the Gröbner basis calculations

Holonomy and combinatorics - Simon Fraser Universitypeople.math.sfu.ca/~mmishna/PUB/holtalk.pdf ·...

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