beta integrals

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

Beta Integrals

S. Ole Warnaar

Department of Mathematics and Statistics

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Orthogonalpolynomials

Selberg Integral

An SelbergIntegral

Euler Beta Integral

Wallis formula (1656)

1 · 3 · 42

3 · 5 · 62

5 · 7 · · ·

=∞∏

(2n − 1)(2n + 1)

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

Gamma function (Euler 1720s)

Γ(x) = limn→∞

n!nx−1

x(x + 1) · · · (x + n − 1)x 6= 0,−1,−2, . . .

∫ ∞

tx−1e−tdt Re(x) > 0

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

4Wallis’ formula is equivalent to

1 − x2 dx = Γ(1/2)Γ(3/2)

or, by x2 = t, to

t1/2−1(1 − t)3/2−1dt = Γ(1/2)Γ(3/2).

This led Euler to the discovery of a more general integral.

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

Euler beta integral (1730s)

tα−1(1 − t)β−1dt =Γ(α)Γ(β)

Γ(α + β)

for Re(α) > 0, Re(β) > 0.

Replacing (β, t) → (ζ, t/ζ) with ζ ∈ R and letting ζ → ∞ usingStirling formula returns the integral representation of the gammafunction.

Replacing (α, β, t) → (ζ2 + 1, ζ2 + 1, 1/2 − x/(2ζ)) and lettingζ → ∞ yields the Gaussian integral

1√π

∫ ∞

−∞

e−x2

dx = 1.

Much more on this later . . .

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

For those with poor eyesight . . .

Let β = n + 1 with n = 0, 1, 2, . . . .

tα−1(1 − t)ndt =n

(−1)k(

) ∫ 1

tk+α−1dt

(−1)k

k + α

α(α + 1) . . . (α + n)

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

Orthogonal polynomials

Set t = (1 − x)/2 in the Euler beta integral and replace

(α, β) → (α + 1, β + 1).

(1 − x)α(1 + x)βdx = 2α+β+1 Γ(α + 1)Γ(β + 1)

Γ(α + β + 2).

The distribution dw(x) on [−1, 1] given by

dw(x) = (1 − x)α(1 + x)βdx

is that of the Jacobi (orthogonal) polynomials P(α,β)n (x).

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

P(α,β)m (x)P(α,β)

n (x)dw(x)

= δmn

2α+β+1Γ(α + n + 1)Γ(β + n + 1)

n!(2n + α + β + 1)Γ(α + β + n + 1).

Proof of the orthogonality and norm-evaluation follows immediatelyfrom the Rodrigues formula

(1 − x)α(1 + x)βP(α,β)n (x) =

(−1)n

(1 − x)α+n(1 + x)β+n]

(which may be taken as the definition of the Jacobi polynomials) andthe Euler beta integral.

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

The one proof that all (good?) talks are supposed to have . . .

To leading order the Rodrigues formula gives

xα+βP(α,β)n (x) “=”

dxnxα+β+2n

so that

P(α,β)n (x) =

cnk xk

cnn =(α + β + n + 1) · · · (α + β + 2n)

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

Without loss of generality assume that m ≤ n.

P(α,β)m (x)P(α,β)

n (x)dw(x)

(−1)n

(1 − x)α+n(1 + x)β+n]

(Rodrigues)

2nδkn

(1 − x)α+n(1 + x)β+ndx

(k times integration by parts)

= δnm

2α+β+1Γ(α + n + 1)Γ(β + n + 1)

n!(2n + α + β + 1)Γ(α + β + n + 1)

(Euler beta integral & cnn)

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

Of course you should all care about the Jacobi polynomials since theGegenbauer polynomials Cλ

n (x) are nothing but

C (λ)n (x) =

(2λ)(2λ + 1) · · · (2λ + n − 1)

(λ + 1/2)(λ + 3/2) · · · (λ + n − 1/2)P(λ−1/2,λ−1/2)

n (x).

The fabulous Leopold Gegenbauer, Austria’s favourite mathematician.

Beta Integrals

Euler BetaIntegral

Wallis formula

Gamma function

Euler betaintegral

Selberg Integral

An SelbergIntegral

In fact, even non-Austrian’s care (like the French and Russians) . . .

Tn(x) =22n(n!)2

(2n)!P(−1/2,−1/2)

n (x) Chebyshev I

Un(x) =22n+1((n + 1)!)2

(2n + 2)!P(1/2,1/2)

n (x) Chebyshev II

Pn(x) = P(0,0)n (x) Legendre

L(α)n (x) = lim

β→∞P(α,β)

n (1 − 2x/β) Laguerre

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Macdonald’sconjectures

An−1Bn and DnI2(m)

Exceptionalgroups

An SelbergIntegral

Selberg Integral

Selberg integral (1944)

[0,1]n

tα−1i (1 − ti)

β−1∏

1≤i<j≤n

|ti − tj |2γdt

n−1∏

Γ(α + iγ)Γ(β + iγ)Γ(γ + iγ)

Γ(α + β + (n + i − 1)γ)Γ(γ)

for Re(α) > 0, Re(β) > 0, Re(γ) > · · · .

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

Macdonald’s conjectures (1982)

Let G be a finite reflection group or finite Coxeter group.That is, G is a finite group of isometries of Rn generated byreflections in hyperplanes through the origin.

The reflection group B2 of order 8 (isomorphic to the signedpermutations of (1, 2)), with 4 reflecting hyperplanes.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

Normalise (up to sign) so that each hyperplane is of the form

a1x1 + · · · + anxn = 0

witha21 + · · · + a2

n = 2.

Form the polynomial

P(x) =

a(α)1 x1 + · · · + a(α)

N being the number of hyperplanes.

Geometrically, P(x) gives the product of the distances of the pointx = (x1, . . . , xn) to the hyperplanes (up to a factor 2N/2).

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

By its action on Rn the reflection group G acts on polynomials inx = (x1, . . . , xn).

The G -invariant polynomials form an R-algebra R[f1, . . . , fn]generated by n algebraically independent polynomials f1, . . . , fn.

The f1, . . . , fn are not unique but their degrees d1, . . . , dn are.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

Let ϕ be the Gaussian measure on Rn:

dϕ(x) =e−|x |2/2

(2π)n/2dx .

Macdonald conjectured in 1982 that for every finite reflection group

|P(x)|2γ dϕ(x) =n

Γ(diγ + 1)

Γ(γ + 1).

For the trivial group A0 of order 1 (mapping R to R by the identitymap; i.e., no reflecting hyperplanes), P(x) = 1 and the conjecturecorresponds to the Gaussian integral

dϕ(x) = 1.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

The reflection group An−1

An−1 is the symmetry group of the (n − 1)-simplex.

The 3-simplex or tetrahedron.

It is a group of order n! (isomorphic to the symmetric group Sn)generated by the

hyperplanes

xi − xj = 0 1 ≤ i < j ≤ n.

The polynomial P(x) is given by the Vandermonde product

P(x) =∏

1≤i<j≤n

(xi − xj).

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

The G -invariant polynomials are the symmetric polynomials in x ,generated by the elementary symmetric functions e1, . . . , en:

er (x) =∑

1≤i1<i2<···<ir≤n

xi1xi2 · · · xir

Hence the degrees are given by (d1, d2, . . . , dn) = (1, 2, . . . , n).

Macdonald’s conjecture for An−1 is thus

1≤i<j≤n

|xi − xj |2γ dϕ(x) =

Γ(iγ + 1)

Γ(γ + 1)

better known as Mehta’s integral.

This follows from the Selberg integral by taking

(α, β) = (ζ + 1, ζ + 1) ti =1

1 − xi√2ζ

ζ → ∞.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

The reflection groups Bn and Dn

In these two cases the Macdonald conjecture is

|xi |2γ∏

1≤i<j≤n

|x2i − x2

j |2γ dϕ(x) =

Γ(2iγ + 1)

Γ(γ + 1)

1≤i<j≤n

|x2i − x2

j |2γ dϕ(x) =Γ(nγ + 1)

Γ(γ + 1)

n−1∏

Γ(2iγ + 1)

Γ(γ + 1)

and follows again from the Selberg integral:

Bn : (α, β) = (γ + 1/2, ζ + 1) ti =x2i

2ζζ → ∞

Dn : (α, β) = (1/2, ζ + 1) ti =x2i

2ζζ → ∞

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

The dihedral group I2(m)

I2(m) is the symmetry group of a regular m-gon,

The 3-gon, 4-gon and pentagon.

It is a group of order 2m generated by the m lines of reflection

√2x sin

−√

2y cos( iπ

= 0 0 ≤ i ≤ m − 1.

The polynomial P(x , y) is given by

P(x , y) =

m−1∏

[√2y cos

−√

2x sin( iπ

= −21−m/2(−r)m sin(mφ).

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

For I2(4) (symmetry group of the square) the invariant polynomialsare of the form

cij(xy)2i (x2j + y2j)

generated by x2 + y2 and x2y2 of degree 2 and 4.

More generally, for I2(m) the invariant polynomials are generated by

x2 + y2

so that the degrees are 2 and m.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

Macdonald’s conjecture for I2(m) (in polar coordinates) is thus

22γ−mγ−1

∫ ∞

r2mγ+1e−r2/2dr

∫ 2π

|sin(mφ)|dφ

=Γ(2γ + 1)Γ(mγ + 1)

Γ2(γ + 1)

which is (almost) trivially true.

Beta Integrals

Euler BetaIntegral

Selberg Integral

Selberg integral

Exceptionalgroups

An SelbergIntegral

The exceptional reflection groups

For E6, E7, E8, F4 the proof is hard but follows from a uniform prooffor all crystallographic reflection groups due to Opdam.

For the non-crystallographic groups H3 and H4 the proof is hard(Opdam, Garvan).

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An−1 versusA1The root systemAn

An Selbergintegral

q-BinomialTheorem I

q-BinomialTheorem II

q-BinomialTheorem III

Open Problems

An Selberg Integral

An−1 versus A1

We have seen that the Vandermonde product

∆(t) =∏

1≤i<j≤n

(ti − tj)

and hence also the Selberg integral

[0,1]n

tα−1i (1 − ti )

β−1∏

1≤i<j≤n

|ti − tj |2γdt

are connected to the reflection group An−1.

In the following we are going to depart from this point of view andwill label the Selberg integral by the Lie algebra or root system A1 asexplained below.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The root system An

Recall that the reflection group An is generated by the(

hyperplanesxi − xj = 0 1 ≤ i < j ≤ n + 1.

Let ǫi be the ith standard unit vector in Rn+1.The normals ±(ǫi − ǫj) for 1 ≤ i < j ≤ n + 1 are known as roots andform the root system An.

The roots ai = ǫi − ǫi+1 for 1 ≤ i ≤ n form a basis in the root systemand are known as simple roots.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

ǫ1 − ǫ2

ǫ1 − ǫ3ǫ2 − ǫ3

−(ǫ1 − ǫ2)

−(ǫ2 − ǫ3)−(ǫ1 − ǫ3)

The root system A2 with simple roots in pink.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The Cartan matrix C of An is given by

ai · aj

1≤i ,j≤n=

2 −1

−1 2 −1

−1. . .

. . . −1

−1 2

The An Dynkin diagram encodes the adjaceny matrix 2I − C :

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

To each simple root as attach a set of variables

t(s) = (t(s)1 , . . . , t

such that 0 ≤ k1 ≤ k2 ≤ · · · ≤ kn.

Set k0 = kn+1 = 0 and let α, β1, . . . , βn, γ ∈ C subject to several mildrestrictions, such as

Re(α) > 0, Re(β1) > 0, . . . , Re(βn) > 0.

Set(α1, . . . , αn−1, αn) = (1, . . . , 1, α).

t(1) t(2) t(n)

α1=1 α2=1 αn=α

β1 β2 βn

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Define the generalised Vandermonde product

∆(u, v) =∏

i ,j≥1

(ui − vj)

and letC k1,...,kn [0, 1] ⊆ [0, 1]k1+···+kn

be an integration domain, somewhat too technical for a talk.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

An Selberg integral

C k1,...,kn [0,1]

)αs−1(1 − t

)βs−1

×n−1∏

∣∆(

t(s), t(s+1))∣

−γn

∣∆(

t(s))∣

2γdt(1) · · · dt(n)

1≤s≤r≤n

ks−ks−1∏

Γ(βs + · · · + βr + (i + s − r − 1)γ)

Γ(αr + βs + · · · + βr + (i + s − r + kr − kr+1 − 2)γ)

Γ(αs + (i − ks+1 − 1)γ)Γ(iγ)

Γ(γ).

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The q-binomial theorem I

For k ∈ N and z ∈ C the q-Pochhammer symbols are

(a; q)k = (1 − a)(1 − aq) · · · (1 − aqk−1)

(a; q)∞ = (1 − a)(1 − aq)(1 − aq2) · · ·

(a; q)z =(a; q)∞

(aqz ; q)∞.

Then the q-binomial theorem is given by

∞∑

(b; q)k(q; q)k

zk =(bz; q)∞(z; q)∞

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Let∫ 1

f (x)dqx = (1 − q)

∞∑

f (qi )qi

be the Jackson or q-integral.

f (1)f (q)

f (q2)

x1qq2q3. . .0

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Then the q-binomial theorem with z = qα and b = qβ may bewritten as the q-beta integral

tα−1(tq; q)β−1 dqt =Γq(α)Γq(β)

Γq(α + β),

where Γq is the q-gamma function:

Γq(x) = (1 − q)1−x(q; q)x−1.

In the q → 1− limit the q-binomial theorem thus yields the Eulerbeta integral.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Macdonald polynomials

Let x be an n-letter alphabet with letters x1, x2, . . . , xn and letλ = (λ1, . . . , λn) be a partition

The q-shift operator Tq,xiis defined as

f (x))

= f (x1, . . . , xi−1, qxi , xi+1, . . . , xn).

and Macdonald’s commuting family {Dr}nr=0 of q-difference

operators is given by

Dr = t(r2)

I⊆[n]|I |=r

i∈Ij 6∈I

txi − xj

xi − xj

Tq,xi.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Defining the generating series of the Dr as

D(u; q, t) =

the Macdonald polynomials Pλ(x ; q, t) are the eigenfunctions ofD(u; q, t) with eigenvalue

(1 + utn−iqλi ).

For q = t the Macdonald polynomials simplify to the well-knownSchur functions

Pλ(x ; t, t) = sλ(x) =det1≤i ,j≤n(x

λj+n−j

det1≤i ,j≤n(xn−ji )

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Cauchy identity

Given a partition λ, each of its squares s is assigned four integers,known as the arm-length a(s), leg-length l(s), arm-colength a′(s)and leg-colength l ′(s).

The arm-length of is 4. The leg-length of is 3.The arm- and leg-colengths of are both 2.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The Cauchy identity for Macdonald polynomials is

Pλ(x ; q, t)Pλ(y ; q, t)∏

s∈λ

1 − qa(s)t l(s)+1

1 − qa(s)+1t l(s)

i ,j≥1

(txiyj ; q)∞(xiyj ; q)∞

When q = t this reduces to the well-known Cauchy determinant

det1≤i≤j≤n

1 − xiyj

=∆(x)∆(y)n

i ,j=1

(1 − xiyj)

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The q-binomial theorem II

The power sums pr are given by p0 = 1 and

pr (x) =∑

x ri .

The map ǫb,t — acting on symmetric functions of y — is defined byits action on the pr :

pr (y))

=1 − br

1 − tr.

A theorem of Macdonald states that

Pλ(y ; q, t))

s∈λ

t l′(s) − b qa′(s)

1 − qa(s)t l(s)+1.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

It may also be shown that

i ,j≥1

(txiyj ; q)∞(xiyj ; q)∞

(b xi ; q)∞(xi ; q)∞

Applying the map ǫb,t to the Cauchy identity we thus obtain ann-dimensional analogue of the q-binomial theorem:

Pλ(x ; q, t)∏

s∈λ

t l′(s) − b qa′(s)

1 − qa(s)+1t l(s)=

(b xi ; q)∞(xi ; q)∞

If n = 1 then x = (x1), λ = (k) and

P(k)(x ; q, t) = xk1 ,

s∈λ

t l′(s) − b qa′(s)

1 − qa(s)+1t l(s)=

(b; q)k(q; q)k

so that we recover the classical q-binomial theorem (with z → x1).

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Taking

xi = qα+γ(n−i) for 1 ≤ i ≤ n

t = qγ

b = qβ

in the n-dimensional q-binomial theorem yields an n-dimensionalq-integral, generalising the q-beta integral.

In the q → 1− limit this gives the Selberg integral.

To prove the An Selberg integral we need a further generalisation ofthe q-binomial theorem!

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The q-binomial theorem III

One may prove a q-binomial theorem of the form

λ(1),...,λ(n)

Pλ(1)(x (1); q, t) · · ·Pλ(n)(x (n); q, t)

stuff with arms and legs)

= infinite product

with x (s) = (x(s)1 , . . . , x

) and k1 ≤ k2 ≤ · · · ≤ kn.

⇒ A (k1 + · · · + kn)-dimensional q-integral

⇒ The An Selberg integral.

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

Open problems

Can we evaluate the integral

Ck1,...,kn [0,1]

´αs−1`

1 − t(s)i

´βs−1

n−1Y

˛∆`

, t(s+1)

−γn

˛∆`

(1)· · · dt

when(α1, . . . , αn−1, αn) 6= (1, . . . , 1, α)?

Can we remove the ordering

0 ≤ k1 ≤ k2 ≤ · · · ≤ kn ?

Can we generalise to other root systems and/or reflectiongroups?

Beta Integrals

Euler BetaIntegral

Selberg Integral

An SelbergIntegral

An Selbergintegral

q-BinomialTheorem I

Open Problems

The End

beta integrals

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