beta integrals
TRANSCRIPT
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)
π
2=
22
1 · 3 · 42
3 · 5 · 62
5 · 7 · · ·
=∞∏
n=1
(2n)2
(2n − 1)(2n + 1)
Beta Integrals
Euler BetaIntegral
Wallis formula
Gamma function
Euler betaintegral
Orthogonalpolynomials
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, . . .
=
∫ ∞
0
tx−1e−tdt Re(x) > 0
Beta Integrals
Euler BetaIntegral
Wallis formula
Gamma function
Euler betaintegral
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
Since
1
=π
4Wallis’ formula is equivalent to
2
∫ 1
0
√
1 − x2 dx = Γ(1/2)Γ(3/2)
or, by x2 = t, to
∫ 1
0
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
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
Euler beta integral (1730s)
∫ 1
0
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
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
For those with poor eyesight . . .
Let β = n + 1 with n = 0, 1, 2, . . . .
∫ 1
0
tα−1(1 − t)ndt =n
∑
k=0
(−1)k(
n
k
) ∫ 1
0
tk+α−1dt
=n
∑
k=0
(−1)k
k + α
(
n
k
)
=n!
α(α + 1) . . . (α + n)
Beta Integrals
Euler BetaIntegral
Wallis formula
Gamma function
Euler betaintegral
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
Orthogonal polynomials
Set t = (1 − x)/2 in the Euler beta integral and replace
(α, β) → (α + 1, β + 1).
Then
∫ 1
−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
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
∫ 1
−1
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
2nn!
dn
dxn
[
(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
Orthogonalpolynomials
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) “=”
1
2nn!
dn
dxnxα+β+2n
so that
P(α,β)n (x) =
n∑
k=0
cnk xk
with
cnn =(α + β + n + 1) · · · (α + β + 2n)
2nn!.
Beta Integrals
Euler BetaIntegral
Wallis formula
Gamma function
Euler betaintegral
Orthogonalpolynomials
Selberg Integral
An SelbergIntegral
Without loss of generality assume that m ≤ n.
Then
∫ 1
−1
P(α,β)m (x)P(α,β)
n (x)dw(x)
=
m∑
k=0
cmk
(−1)n
2nn!
∫ 1
−1
xk dn
dxn
[
(1 − x)α+n(1 + x)β+n]
dx
(Rodrigues)
=
m∑
k=0
cmk
2nδkn
∫ 1
−1
(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
Orthogonalpolynomials
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
Orthogonalpolynomials
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
n∏
i=1
tα−1i (1 − ti)
β−1∏
1≤i<j≤n
|ti − tj |2γdt
= n!
n−1∏
i=0
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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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.
x1
x2
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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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) =
N∏
α=1
(
a(α)1 x1 + · · · + a(α)
n xn
)
,
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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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
∫
Rn
|P(x)|2γ dϕ(x) =n
∏
i=1
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
∫
R
dϕ(x) = 1.
Beta Integrals
Euler BetaIntegral
Selberg Integral
Selberg integral
Macdonald’sconjectures
An−1Bn and DnI2(m)
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
(
n2
)
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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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
∫
Rn
∏
1≤i<j≤n
|xi − xj |2γ dϕ(x) =
n∏
i=1
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
2
(
1 − xi√2ζ
)
ζ → ∞.
Beta Integrals
Euler BetaIntegral
Selberg Integral
Selberg integral
Macdonald’sconjectures
An−1Bn and DnI2(m)
Exceptionalgroups
An SelbergIntegral
The reflection groups Bn and Dn
In these two cases the Macdonald conjecture is
∫
Rn
n∏
i=1
|xi |2γ∏
1≤i<j≤n
|x2i − x2
j |2γ dϕ(x) =
n∏
i=1
Γ(2iγ + 1)
Γ(γ + 1)
and
∫
Rn
∏
1≤i<j≤n
|x2i − x2
j |2γ dϕ(x) =Γ(nγ + 1)
Γ(γ + 1)
n−1∏
i=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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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
( iπ
m
)
−√
2y cos( iπ
m
)
= 0 0 ≤ i ≤ m − 1.
The polynomial P(x , y) is given by
P(x , y) =
m−1∏
i=0
[√2y cos
( iπ
m
)
−√
2x sin( iπ
m
)]
= −21−m/2(−r)m sin(mφ).
Beta Integrals
Euler BetaIntegral
Selberg Integral
Selberg integral
Macdonald’sconjectures
An−1Bn and DnI2(m)
Exceptionalgroups
An SelbergIntegral
For I2(4) (symmetry group of the square) the invariant polynomialsare of the form
∑
i ,j
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
and
xm∑
i≥0
(
−y2
x2
)i(
m
2i
)
so that the degrees are 2 and m.
Beta Integrals
Euler BetaIntegral
Selberg Integral
Selberg integral
Macdonald’sconjectures
An−1Bn and DnI2(m)
Exceptionalgroups
An SelbergIntegral
Macdonald’s conjecture for I2(m) (in polar coordinates) is thus
22γ−mγ−1
π
∫ ∞
0
r2mγ+1e−r2/2dr
∫ 2π
0
|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
Macdonald’sconjectures
An−1Bn and DnI2(m)
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 I
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
n∏
i=1
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
The root system An
Recall that the reflection group An is generated by the(
n+12
)
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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 :
1 2 n
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
To each simple root as attach a set of variables
t(s) = (t(s)1 , . . . , t
(s)ks
)
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
An Selberg integral
∫
C k1,...,kn [0,1]
n∏
s=1
ks∏
i=1
(
t(s)i
)αs−1(1 − t
(s)i
)βs−1
×n−1∏
s=1
∣
∣∆(
t(s), t(s+1))∣
∣
−γn
∏
s=1
∣
∣∆(
t(s))∣
∣
2γdt(1) · · · dt(n)
=∏
1≤s≤r≤n
ks−ks−1∏
i=1
Γ(βs + · · · + βr + (i + s − r − 1)γ)
Γ(αr + βs + · · · + βr + (i + s − r + kr − kr+1 − 2)γ)
×n
∏
s=1
ks∏
i=1
Γ(αs + (i − ks+1 − 1)γ)Γ(iγ)
Γ(γ).
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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) · · ·
and
(a; q)z =(a; q)∞
(aqz ; q)∞.
Then the q-binomial theorem is given by
∞∑
k=0
(b; q)k(q; q)k
zk =(bz; q)∞(z; q)∞
.
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
Let∫ 1
0
f (x)dqx = (1 − q)
∞∑
i=0
f (qi )qi
be the Jackson or q-integral.
f (1)f (q)
f (q2)
. . .
f (x)
x1qq2q3. . .0
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
Then the q-binomial theorem with z = qα and b = qβ may bewritten as the q-beta integral
∫ 1
0
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
Macdonald polynomials
Let x be an n-letter alphabet with letters x1, x2, . . . , xn and letλ = (λ1, . . . , λn) be a partition
λ =
The q-shift operator Tq,xiis defined as
Tq,xi
(
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
∏
i∈I
Tq,xi.
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
Defining the generating series of the Dr as
D(u; q, t) =
n∑
r=0
Drur
the Macdonald polynomials Pλ(x ; q, t) are the eigenfunctions ofD(u; q, t) with eigenvalue
n∏
i=1
(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
i )
det1≤i ,j≤n(xn−ji )
.
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
The Cauchy identity for Macdonald polynomials is
∑
λ
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
1 − xiyj
)
=∆(x)∆(y)n
∏
i ,j=1
(1 − xiyj)
.
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
The q-binomial theorem II
The power sums pr are given by p0 = 1 and
pr (x) =∑
i≥1
x ri .
The map ǫb,t — acting on symmetric functions of y — is defined byits action on the pr :
ǫb,t
(
pr (y))
=1 − br
1 − tr.
A theorem of Macdonald states that
ǫb,t
(
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
It may also be shown that
ǫb,t
(
∏
i ,j≥1
(txiyj ; q)∞(xiyj ; q)∞
)
=∏
i≥1
(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:
∑
λ
Pλ(x ; q, t)∏
s∈λ
t l′(s) − b qa′(s)
1 − qa(s)+1t l(s)=
n∏
i=1
(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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
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
(s)ks
) and k1 ≤ k2 ≤ · · · ≤ kn.
⇒ A (k1 + · · · + kn)-dimensional q-integral
⇒ The An Selberg integral.
Beta Integrals
Euler BetaIntegral
Selberg Integral
An SelbergIntegral
An−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
Open problems
Can we evaluate the integral
Z
Ck1,...,kn [0,1]
nY
s=1
ksY
i=1
`
t(s)i
´αs−1`
1 − t(s)i
´βs−1
×
n−1Y
s=1
˛
˛∆`
t(s)
, t(s+1)
´˛
˛
−γn
Y
s=1
˛
˛∆`
t(s)
´˛
˛
2γdt
(1)· · · dt
(n)
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−1 versusA1The root systemAn
An Selbergintegral
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem I
q-BinomialTheorem II
q-BinomialTheorem III
Open Problems
The End