A BETA INTEGRAL ASSOCIATED WITH

THE ROOT SYSTEM G_{2}*

F. G. GARVAN†

Abstract. Some conjectures of Askey are proven that have to do with adding roots in the Macdonald-Morris conjecture for G_{2}. This is done by extending Aomoto’s proof of Selberg’s integral. This yields a new proof of the Macdonald-Morris root system conjecture for G_{2} which should extend to other root systems.

1. Introduction. Let

G(x_{1}, x_{2}; a, b) (1.1)

= (1 − x_{1})^{a} (1 − \frac{1}{x_{1}} )^{a} (1 − x_{2})^{a} (1 − \frac{1}{x_{2}} )^{a} (1 − x_{1}x_{2})^{a} (1 − \frac{1}{x_{1}x_{2}} )^{a}

· (1 − \frac{x_{1}}{x_{2}} )^{b} (1 − \frac{x_{2}}{x_{1}} )^{b} (1 − x_{1}^{2}x_{2})^{b} (1 − \frac{1}{x_{1}^{2}x_{2}} )^{b} (1 − x_{2}^{2}x_{1})^{b} (1 − \frac{1}{x_{2}^{2}x_{1}} )^{b} .

Then the Macdonald-Morris root system conjecture for G_{2} is

(1.2)  C.T. G(x_{1}, x_{2}; a, b) = \frac{(3a + 3b)!(3b)!(2a)!(2b)!}{(2a + 3b)!(a + 2b)!(a + b)!a!b!b!} = g(a, b).

Here C.T. means the constant term in the Laurent expansion as a polynomial in x_{1}, x_{1}^{−1}, x_{2}, x_{2}^{−1}. This has been proved independently by Habsieger [5] and Zeilberger [12]. They have also proved the q-analogue of (1.2). Although their proofs are elegant, they are special to G_{2}. Recently Zeilberger [13] has also proved the G2^{∨} case of the Macdonald-Morris conjectures. His proof should extend to other root systems.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 33A15, 33A75.

Key words and phrases. Askey’s G_{2} conjectures, Macdonald-Morris root system conjectures, Aomoto, Selberg’s integral, multidimensional beta integrals.

†Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706. Current address, School of Mathematics, Macquarie University, Sydney, New South Wales 2109, Australia .

2 F. G. GARVAN†

In this paper we give a new proof of (1.2) which is entirely in terms of integrals and which should also extend to other root systems. Our proof is an extension of Aomoto’s [1] proof of Selberg’s [11] integral. See Askey [3] for a good exposition of Aomoto’s proof. We were led to this by considering conjectures of Askey [3] that have to do with adding roots in the Macdonald-Morris root system conjecture for G_{2}. Askey conjectured

C.T. (1 − x_{1}) (1 − \frac{1}{x_{1}} ) G(x_{1}, x_{2}; a, b) = \frac{2(3a + 3b + 1)}{2a + 3b + 1}g(a, b), (1.3)

C.T. (1 − x_{1}) (1 − \frac{1}{x_{1}} ) (1 − x_{2}) (1 − \frac{1}{x_{2}} ) G(x_{1}, x_{2}; a, b) (1.4)

= \frac{2(3a + 3b + 2)(3a + 3b + 1)}{(2a + 3b + 1)(a + 2b + 1)}g(a, b),

C.T. (1 − x_{1}^{2}x_{2}) (1 − \frac{1}{x_{1}^{2}x_{2}} ) G(x_{1}, x_{2}; a, b) = \frac{2(3a + 3b + 1)(3b + 1)}{(2a + 3b + 1)(a + 2b + 1)}g(a, b), (1.5)

C.T. (1 − x_{1}^{2}x_{2}) (1 − \frac{1}{x_{1}^{2}x_{2}} ) (1 − x_{2}^{2}x_{1}) (1 − \frac{1}{x_{2}^{2}x_{1}} ) G(x_{1}, x_{2}; a, b) (1.6)

= \frac{6(3a + 3b + 2)(3a + 3b + 1)(3b + 1)(3b + 2)}{(2a + 3b + 3)(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}g(a, b).

In §2 we give the idea of the proof of (1.2). It is well known that (1.2) may be written as a trigonometric integral formula (see, for example, Morris [10, p.46]). The starting point of our proof is to write (1.2) as

(1.7) \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} \frac{(t_{1} + t_{2})^{2a}(t_{1} − t_{2})^{2b}(t_{2}^{2} + 2t_{1}t_{2} − 1)^{2b}(t_{1}^{2} + 2t_{1}t_{2} − 1)^{2b}}{(1 + t_{1}^{2})^{2a+4b+1}(1 + t_{2}^{2})^{2a+4b+1}} dt_{1}dt_{2}
= g(a, b).

A BETA INTEGRAL ASSOCIATED WITH THE ROOT SYSTEM G_{2}* 3

The left-hand side of (1.7) is the integral referred to in the title of this paper. This is done via three changes of variables: first, by letting x_{j} = e^{2iθ_{j}} (j = 1, 2) in (1.1) and using the orthogonality of the exponentials on [0, π] to obtain an integral on [0, π]^{2}; second, by linear change of variables to obtain an integral on [−\frac{π}{2} , \frac{π}{2} ]^{2}; and finally by letting t_{j} = tan θ_{j} to obtain the integral over ℝ^{2}. We note that this is the same change of variables that Morris used in transforming his constant terms formula for A_{n} [10, p. 95] into the Cauchy-Selberg integral [10, (6.6)]. The advantage of this integral over the trigonometric integrals is that the integrand is a rational function of t_{1}, t_{2}, which can be easily manipulated using a computer algebra package like REDUCE. In §3 we prove some preliminary results. In §4 we complete the proof of (1.2) and prove (1.3)–(1.5) as well.

Macdonald [8, conjecture(6.1)] has also conjectured generalizations of Mehta’s integral formula for arbitary root systems [9], [8, (4.1)]. The G_{2} case does not seem to be related to our integral given in (1.7). The Macdonald-Mehta integral conjecture involves parameters that are constant on root length. We note that the two parameter case of the G_{2} Macdonald-Mehta integral may be proved in the same way as the one parameter case, which was proved by Macdonald [8, p.1002]. This is done by transforming to polar coordinates. The resulting integral turns out to be the product of a gamma integral and a beta integral.

We should mention that (1.3)–(1.6) and their q-analogues may be proved by other methods. Zeilberger [12] proved (1.2) using the result of Morris, mentioned above, related to A_{n} and Dixon’s [4, §3.1] summation of a well-poised _{3}F_{2}. Equations (1.3)–(1.6) could be proved by trying to generalize Morris’s results and Dixon’s summation. See Kadell [6] for such generalizations of q-analogues of Morris’s results and Askey [2] for some extensions of Dixon’s summation. The author has proved (1.3)–(1.6), as well as other similar results, by these methods. In §4 we state these other results without proof. For the most part we restrict attention to (1.2)–(1.5), preferring to take a more direct approach.

Kadell [7] has found yet another approach to proving (1.2) which should extend to other root systems. This involves working with the function G(x_{1}, x_{2}; a, b) directly rather than writing it as an integral, and using the fact that derivatives have no residues. Finally, (1.2)–(1.6) could be proved by extending Zeilberger’s [13] method for the G2^{∨} case and then letting q → 1.

2. The idea of the proof. We label the roots of G_{2} as in Fig. 1.

FIG. 1

Let

(2.1)  g′(a, b) = C.T. G(x_{1}, x_{2}; a, b).

Our goal is to prove that g′(a, b) = g(a, b) for all a, b ≥ 0. The idea is to proceed by induction on a. However to jump from a to a + 1 in one step would be too much to ask for. Considering (1.3) and (1.4) we break it up into three stages:

Stage 1.  C.T. [α_{1}]G = \frac{2(3a + 3b + 1)}{2a + 3b + 1}g′(a, b).

Stage 2. C.T. [α_{1}][α_{2}]G = \frac{2(3a + 3b + 1)(3a + 3b + 2)}{(2a + 3b + 1)(a + 2b + 1)}g′(a, b).

C.T. [α_{1}][α_{2}][α_{1} + α_{2}]G = \frac{6(3a + 3b + 2)(3a + 3b + 1)(2a + 1)}{(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}g′(a, b) Stage 3.

where

[k_{1}α_{1} + k_{2}α_{2}] = (1 − x_{1}^{k_{1}}x_{2}^{k_{2}})(1 − x_{1}^{−k_{1}}x_{2}^{−k_{2}}),

for k_{1}, k_{2} ∈ ℤ. Each stage corresponds to adding an additional pair of opposite short roots. After Stage 3 all that will remain is to prove the result for a = 0 since

(2.2)  \frac{g(a + 1, b)}{g(a, b)} = \frac{6(3a + 3b + 2)(3a + 3b + 1)(2a + 1)}{(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}.

The case a = 0 is equivalent to b = 0 since we have

Long roots of G_{2} ≅ Short roots of G_{2} ≅ A_{2}.

The result is trivially true for a = b = 0. The case b = 0 follows from (2.2) and Stage 3 by induction.

To prove Stages 1–3 we rewrite each stage as an integral and use an idea of Aomoto. To give the reader a taste of our method we work through the proof of Stage 1. If we let

(2.3)

w( t_{∼}) = w(t_{1}, t_{2}; a, b) = \frac{(t_{1} + t_{2})^{2a}(t_{1} − t_{2})^{2b}(t_{2}^{2} + 2t_{1}t_{2} − 1)^{2b}(t_{1}^{2} + 2t_{1}t_{2} − 1)^{2b}}{(1 + t_{1}^{2})^{2a+4b+1}(1 + t_{2}^{2})^{2a+4b+1}} ,

then we may write Stage 1 as

(2.4)  \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} \frac{4}{(1 + t_{1}^{2})} w( t_{∼}) d t_{∼} = \frac{2(3a + 3b + 1)}{(2a + 3b + 1)} \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} w( t_{∼}) d t_{∼},

using the same change of variables used to derive (1.7). It is here that we use Aomoto’s idea. To get ∫_{ℝ^{2}} \frac{1}{1+t_{1}^{2}} w( t_{∼}) d t_{∼} in terms of ∫_{ℝ^{2}} w( t_{∼}) d t_{∼} we use

(2.5) 0 = ∫_{ℝ^{2}} \frac{∂}{∂t_{1}} t_{1}w( t_{∼}) d t_{∼}

= ∫_{ℝ^{2}} w( t_{∼}) d t_{∼} − (2a + 4b + 1) ∫_{ℝ^{2}} \frac{2t_{1}^{2}}{1 + t_{1}^{2}} w( t_{∼}) d t_{∼} + 2a ∫_{ℝ^{2}} \frac{t_{1}}{t_{1} + t_{2}} w( t_{∼}) d t_{∼}

+ 2b ∫_{ℝ^{2}} \frac{t_{1}}{t_{1} − t_{2}} w( t_{∼}) d t_{∼} + 2b ∫_{ℝ^{2}} \frac{2t_{1}t_{2}}{t_{2}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼}

+ 2b ∫_{ℝ^{2}} \frac{2t_{1}(t_{1} + t_{2})}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼}.

We can make some progress by using the fact that w( t_{∼}) is invariant under the transposition t_{1} ↔ t_{2}.

(2.6)  \frac{t_{1}}{t_{1} ± t_{2}} = 1 ∓ \frac{t_{2}}{t_{1} ± t_{2}} .

It follows that

(2.7)  ∫_{ℝ^{2}} \frac{t_{1}}{t_{1} ± t_{2}} w( t_{∼}) d t_{∼} = \frac{1}{2} ∫_{ℝ^{2}} w( t_{∼}) d t_{∼}.

∫_{ℝ^{2}} \frac{t_{1}t_{2}}{t_{2}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼} + ∫_{ℝ^{2}} \frac{t_{1}(t_{1} + t_{2})}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼} (2.8)

= ∫_{ℝ^{2}} \frac{t_{1}t_{2}}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼} + ∫_{ℝ^{2}} \frac{t_{1}(t_{1} + t_{2})}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼}

(via t_{1} ↔ t_{2} on the first integral)

= ∫_{ℝ^{2}} w( t_{∼}) d t_{∼} + ∫_{ℝ^{2}} \frac{1}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼}.

Then (2.5) becomes

(2.9) 0 = (−3a − 3b − 1) ∫_{ℝ^{2}} w( t_{∼}) d t_{∼} + 2(2a + 4b + 1) ∫_{ℝ^{2}} \frac{1}{1 + t_{1}^{2}} w( t_{∼}) d t_{∼}

+ 4b ∫_{ℝ^{2}} \frac{1}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼}.

The problem that remains is to get ∫_{ℝ^{2}} \frac{1}{t_{1}^{2}+2t_{1}t_{2}−1} w( t_{∼}) d t_{∼} in terms of ∫_{ℝ^{2}} w( t_{∼}) d t_{∼}

and ∫_{ℝ^{2}} \frac{1}{1+t_{1}^{2}} w( t_{∼}) d t_{∼}. The transformation t_{1} ↔ t_{2} is not helpful here. We need another transformation that leaves w( t_{∼}) d t_{∼} invariant. The real reason why w( t_{∼}) is invariant under t_{1} ↔ t_{2} is that the root system G_{2} is invariant under the linear transformation given by

α_{1} ⟼ α_{2}  and  α_{2} ⟼ α_{1},

which is also the reflection through the plane orthogonal to the vector α_{2} − α_{1}. Recall that a root system is invariant under any element of the Weyl group, the group generated by the w_{α}, where α is a root and w_{α} is the reflection through the hyperplane orthogonal to α. The Weyl group for G_{2} is generated by w_{α_{2}−α_{1}} and w_{α_{1}}. The extra integral transformation that we need will correspond to w_{α_{1}}. We study the action of this reflection on the roots:

α_{1} ↔ −α_{1},  α_{2} ↔ α_{1} + α_{2},
α_{2} − α_{1} ↔ 2α_{1} + α_{2},  α_{1} + 2α_{2} ↔ α_{1} + 2α_{2}.

We need a transformation

(2.10)  f : ℝ^{2} ⟶ ℝ^{2},  (t_{1}, t_{2}) ⟼ (f_{1}(t_{1}, t_{2}), f_{2}(t_{1}, t_{2}))

with the following action:

\frac{1}{1 + t_{1}^{2}} ↔ \frac{1}{1 + t_{1}^{2}} , \frac{1}{1 + t_{2}^{2}} ↔ \frac{(t_{1} + t_{2})^{2}}{(1 + t_{1}^{2})(1 + t_{2})^{2}} ,

(2.11)

\frac{(t_{1} − t_{2})^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})} ↔ \frac{(t_{1}^{2} + 2t_{1}t_{2} − 1)^{2}}{(1 + t_{1}^{2})^{2}(1 + t_{2}^{2})} , \frac{(t_{2}^{2} + 2t_{1}t_{2} − 1)^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})^{2}} ↔ \frac{(t_{2}^{2} + 2t_{1}t_{2} − 1)^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})^{2}} .

The transformation that does the job is

(2.12)  f_{1}(t_{1}, t_{2}) = t_{1},  f_{2}(t_{1}, t_{2}) = \frac{1 − t_{1}t_{2}}{t_{1} + t_{2}}  (t_{1} ≠ −t_{2}).

In §3 we show that

(2.13)  ∫_{ℝ^{2}} g( t_{∼})w( t_{∼}) d t_{∼} = ∫_{ℝ^{2}} g(f( t_{∼}))w( t_{∼}) d t_{∼},

for a certain restricted class of functions g.

Now we return to the problem of evaluating ∫_{ℝ^{2}} \frac{1}{t_{1}^{2}+2t_{1}t_{2}−1} w( t_{∼}) d t_{∼}. A routine

calculation shows that

\frac{1}{t_{1}^{2} + 2t_{1}t_{2} − 1} ⟶^{f} \frac{(t_{1} + t_{2})}{(1 + t_{1}^{2})(t_{1} − t_{2})}, (2.14)

\frac{(t_{1} + t_{2})}{(1 + t_{1}^{2})(t_{1} − t_{2})} = −\frac{(t_{1} + t_{2})^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})} + \frac{(t_{1} + t_{2})}{(1 + t_{2}^{2})(t_{1} − t_{2})}. (2.15)

It follows that

∫_{ℝ^{2}} \frac{1}{t_{1}^{2} + 2t_{1}t_{2} − 1} w( t_{∼}) d t_{∼} = ∫_{ℝ^{2}} \frac{(t_{1} + t_{2})}{(1 + t_{1}^{2})(t_{1} − t_{2})} w( t_{∼}) d t_{∼} (2.16)

(by (2.13) and (2.14))

= −\frac{1}{2} ∫_{ℝ^{2}} \frac{(t_{1} + t_{2})^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})} w( t_{∼}) d t_{∼}

(by applying t_{1} ↔ t_{2} and using (2.15))

= −\frac{1}{2} ∫_{ℝ^{2}} \frac{1}{1 + t_{1}^{2}} w( t_{∼}) d t_{∼}

(by applying f, then t_{1} ↔ t_{2}).

Substituting this into (2.9) gives

(3a + 3b + 1) ∫_{ℝ^{2}} w( t_{∼}) d t_{∼} = 2(2a + 3b + 1) ∫_{ℝ^{2}} \frac{1}{1 + t_{1}^{2}} w( t_{∼}) d t_{∼}

and Stage 1 follows. The proof of Stages 2–3 is analogous and will be given in §4.

3. Preliminary results. The main result of this section is Lemma 3.6. It contains a list of integrals that we will need in the proof of Stages 2–3. In the proof of this lemma we will use the transformation formula (2.13) for f. A more formal statement of this formula is given in Lemma 3.2. The idea of the proof of Lemma 3.2 is to write both sides of (2.13) as an integral over [0, π]^{2}, using the same change of variables mentioned after (1.7) in the introduction, use the transformation T (θ_{1}, θ_{2}) = (−θ_{1}, θ_{1} + θ_{2}) and apply Lemma 3.1. The proofs of Lemmas 3.1 and 3.2 are omitted. In the proof of Lemma 3.6 we will also need to calculate the image of certain rational functions in t_{1}, t_{2} under f. This was done using the computer algebra package REDUCE.

Lemma 3.1. Let h : [0, π]^{2} ⟶ ℝ be continuous; then

∫_{[0,π]^{2}} h(θ_{1}, θ_{2}) dθ_{1}dθ_{2} = ∫_{[0,π]^{2}} h^{∗}(θ_{1}, θ_{2}) dθ_{1}dθ_{2}

where

h^{∗}(θ_{1}, θ_{2}) = \{ \begin{matrix}h(π − θ_{1}, θ_{1} + θ_{2}) & if 0 ≤ θ_{1} + θ_{2} ≤ π, \\ h(π − θ_{1}, θ_{1} + θ_{2} − π) & if π < θ_{1} + θ_{2} ≤ 2π. \end{matrix}

Let

w_{0}( t_{∼}) = (1 + t_{1}^{2})(1 + t_{2}^{2})w( t_{∼}).

Lemma 3.2. Suppose g : ℝ^{2} ⟶ ℝ is a continuous function that satisfies (i) gw_{0} is bounded on ℝ^{2}, (ii) (g ∘ f)w_{0} can be extended to a continuous function on ℝ^{2},

where f is defined in (2.10) and (2.12). Then

(3.3)  ∫_{ℝ^{2}} g( t_{∼}) w( t_{∼}) d t_{∼} = ∫_{ℝ^{2}} g(f( t_{∼})) w( t_{∼}) d t_{∼}.

For notational convenience we let

(3.4) <α_{1} >= \frac{1}{1 + t_{1}^{2}} , <α_{2} >= \frac{1}{1 + t_{2}^{2}} , <α_{1} + α_{2} >= \frac{(t_{1} + t_{2})^{2}}{(1 + t_{1}^{2})(1 + t_{2}^{2})}.

A BETA INTEGRAL ASSOCIATED WITH THE ROOT SYSTEM G_{2}* 9

This notation is related to the notation introduced in Stages 1–3. Using the same change of variables used to derive (1.7) we have

(3.5) C.T.[α_{1}]^{k_{1}}[α_{2}]^{k_{2}}[α_{1} + α_{2}]^{k_{3}}G

= \frac{4^{3a+3b+k_{1}+k_{2}+k_{3}}}{π^{2}} ∫_{ℝ^{2}} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}}<α_{1} + α_{2} >^{k_{3}} w( t_{∼}) d t_{∼},

where w( t_{∼}) is defined in (2.3) and k_{1}, k_{2}, k_{3} are nonnegative integers.

Lemma 3.6.

(3.7)  ∫_{ℝ^{2}} (t_{2}^{2} + 2t_{1}t_{2} − 1) <α_{1} > w( t_{∼}) d t_{∼} = 0,

10 F. G. GARVAN†

∫_{ℝ^{2}} \frac{t_{1}}{t_{1} + t_{2}} <α_{2} > w( t_{∼}) d t_{∼} = \frac{5}{4} ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼} (3.8)

− ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼},

∫_{ℝ^{2}} \frac{t_{1}}{t_{1} − t_{2}} <α_{2} > w( t_{∼}) d t_{∼} = \frac{3}{4} ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼}, (3.9)

∫_{ℝ^{2}} \frac{t_{2}(t_{1} + t_{2})}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} > w( t_{∼}) d t_{∼} = \frac{1}{2} ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼}, (3.10)

∫_{ℝ^{2}} \frac{1 − t_{1}t_{2}}{t_{1}^{2} + 2t_{1}t_{2} − 1} <α_{1} > w( t_{∼}) d t_{∼} = −\frac{3}{4} ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼}, (3.11)

∫_{ℝ^{2}} \frac{1}{t_{1}^{2} + 2t_{1}t_{2} − 1} <α_{1} > w( t_{∼}) d t_{∼} = − ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼}, (3.12)

∫_{ℝ^{2}} <α_{1} ><α_{2} ><α_{1} + α_{2} > w( t_{∼}) d t_{∼} = −\frac{3}{4} ∫_{ℝ^{2}} <α_{1} >^{2} w( t_{∼}) d t_{∼} (3.13)

+\frac{3}{2} ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼},

∫_{ℝ^{2}} \frac{1}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} > w( t_{∼}) d t_{∼} = −\frac{1}{2} ∫_{ℝ^{2}} <α_{1} >^{2} w( t_{∼}) d t_{∼}, (3.14)

∫_{ℝ^{2}} \frac{t_{1}t_{2}}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} > w( t_{∼}) d t_{∼} = \frac{1}{2} ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼} (3.15)

−\frac{1}{2} ∫_{ℝ^{2}} <α_{1} >^{2} w( t_{∼}) d t_{∼}.

Proof. Equation (3.7) follows from (3.3) and

(t_{2}^{2} + 2t_{1}t_{2} − 1) <α_{1} ><α_{2} > ⟶^{f} − (t_{2}^{2} + 2t_{1}t_{2} − 1) <α_{1} ><α_{2} > .

We note that (t_{2}^{2} + 2t_{1}t_{2} − 1) < α_{1} >< α_{2} > satisfies the conditions of Lemma 3.2 since it is equal to < α_{1} + α_{2} > − < α_{2} >. Whenever we apply f to a rational function in the remainder of the proof we leave it to the reader to verify that the function involved satisfies the conditions of Lemma 3.2.

A BETA INTEGRAL ASSOCIATED WITH THE ROOT SYSTEM G_{2}* 11

Equation (3.8) follows from (3.7) and

\frac{t_{1}}{t_{1} + t_{2}} <α_{2} > +\frac{t_{2}}{t_{1} + t_{2}} <α_{1} >

= \frac{1}{2} <α_{1} ><α_{2} > \{−(t_{2}^{2} + 2t_{1}t_{2} − 1) + 3(1 + t_{2}^{2}) + 2(1 + t_{1}^{2}) − 4\}.

Similarly (3.9) follows from

\frac{t_{1}}{t_{1} − t_{2}} <α_{2} > +\frac{t_{2}}{t_{2} − t_{1}} <α_{1} >

= \frac{1}{2} <α_{1} ><α_{2} > \{(t_{2}^{2} + 2t_{1}t_{2} − 1) + (1 + t_{2}^{2}) + 2(1 + t_{1}^{2})\}.

We have

\frac{t_{2}(t_{1} + t_{2})}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} > = <α_{1} > +\frac{(1 − t_{1}t_{2})}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} >, (3.16)

\frac{(1 − t_{1}t_{2})}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} > ⟶^{f} − \frac{t_{2}(t_{1} + t_{2})}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} >

and (3.10) follows.

\frac{(1 − t_{1}t_{2})}{t_{1}^{2} + 2t_{1}t_{2} − 1} <α_{1} > ⟶^{f} ⟶^{t_{1}↔t_{2}} − \frac{t_{1}}{t_{1} − t_{2}} <α_{2} >

and (3.11) follows by (3.9).

\frac{<α_{1} >}{t_{1}^{2} + 2t_{1}t_{2} − 1}+ <α_{1} ><α_{2} >

= \frac{<α_{1} + α_{2} >}{t_{1}^{2} + 2t_{1}t_{2} − 1} ⟶^{t_{1}↔t_{2}} ⟶^{f} ⟶^{t_{1}↔t_{2}} − \frac{<α_{1} + α_{2} >}{t_{1}^{2} + 2t_{1}t_{2} − 1}

and (3.12) follows.

We observe that (3.13) is equivalent to

∫_{ℝ^{2}} A( t_{∼})w( t_{∼}) d t_{∼} = 0,

where

A( t_{∼}) = 4I( t_{∼}) + 3 <α_{1} + α_{2} >^{2} −6 <α_{1} ><α_{1} + α_{2} >,

I( t_{∼}) =<α_{1} ><α_{2} ><α_{1} + α_{2} >,

since

<α_{1} >^{2} ⟶^{t_{1}↔t_{2}} ⟶^{f} <α_{1} + α_{2} >^{2},

<α_{1} ><α_{2} > ⟶^{f} <α_{1} ><α_{1} + α_{2} > .

A( t_{∼}) = I( t_{∼})(−2 + 3t_{1}^{2} + 6t_{1}t_{2} − 3t_{2}^{2})

12 F. G. GARVAN†

so that

∫_{ℝ^{2}} A( t_{∼})w( t_{∼}) d t_{∼} = −2 ∫_{ℝ^{2}} I( t_{∼})(1 − 3t_{1}t_{2})w( t_{∼}) d t_{∼},

since I( t_{∼}) is invariant under t_{1} ↔ t_{2}.

(1 − t_{1}t_{2}) ⟶^{f} \frac{t_{2}(1 + t_{1}^{2})}{(t_{1} + t_{2})},

\frac{t_{2}(1 + t_{1}^{2})}{(t_{1} + t_{2})} + \frac{t_{1}(1 + t_{2}^{2})}{(t_{1} + t_{2})} = (1 + t_{1}t_{2}).

It follows that

2 ∫_{ℝ^{2}} I( t_{∼})(1 − t_{1}t_{2})w( t_{∼}) d t_{∼} = ∫_{ℝ^{2}} I( t_{∼})(1 + t_{1}t_{2})w( t_{∼}) d t_{∼},

since I( t_{∼}) is invariant under f and t_{1} ↔ t_{2}. Therefore,

∫_{ℝ^{2}} I( t_{∼})(1 − 3t_{1}t_{2})w( t_{∼}) d t_{∼} = 0,

as required. This completes the proof of (3.13).

Equation (3.14) follows from

\frac{<α_{1} >}{t_{2}^{2} + 2t_{1}t_{2} − 1} ⟶^{f} − \frac{<α_{1} >}{t_{2}^{2} + 2t_{1}t_{2} − 1}− <α_{1} >^{2} .

Finally, (3.15) follows easily from (3.10), (3.14), and (3.16). □

4. Proof of Stages 2–3 and (1.2)–(1.5). In this section we use Lemma 3.6 to prove Stages 2–3, thus completing the proof of (1.2)–(1.4). Finally, we show how

A BETA INTEGRAL ASSOCIATED WITH THE ROOT SYSTEM G_{2}* 13

(1.5) follows from (1.3) and (1.4). For k_{1}, k_{2} ≥ 0 we have

(4.1)  0 = ∫_{ℝ^{2}} \frac{∂}{∂t_{1}} t_{1} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

= −(4a + 8b + 2k_{1} + 1) ∫_{ℝ^{2}} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

+ 2(2a + 4b + k_{1} + 1) ∫_{ℝ^{2}} <α_{1} >^{k_{1}+1}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

+ 2a ∫_{ℝ^{2}} \frac{t_{1}}{t_{1} + t_{2}} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

+ 2b ∫_{ℝ^{2}} \frac{t_{1}}{t_{1} − t_{2}} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

+ 4b ∫_{ℝ^{2}} \frac{t_{1}(t_{1} + t_{2})}{t_{1}^{2} + 2t_{1}t_{2} − 1} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}

+ 4b ∫_{ℝ^{2}} \frac{t_{1}t_{2}}{t_{2}^{2} + 2t_{1}t_{2} − 1} <α_{1} >^{k_{1}}<α_{2} >^{k_{2}} w( t_{∼}) d t_{∼}.

Letting k_{1} = 0, k_{2} = 1 in (4.1) and using (3.8)–(3.12), we find that

(4.2)  ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼} = \frac{(3a + 3b + 2)}{4(a + 2b + 1)} ∫_{ℝ^{2}} <α_{2} > w( t_{∼}) d t_{∼}.

Hence, by (3.5) we have

C.T. [α_{1}][α_{2}]G = \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 16 <α_{1} ><α_{2} > w( t_{∼}) d t_{∼}

= \frac{(3a + 3b + 2)}{(a + 2b + 1)} \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 4 <α_{1} > w( t_{∼}) d t_{∼}

= \frac{(3a + 3b + 2)}{(a + 2b + 1)} C.T. [α_{1}]G

= \frac{2(3a + 3b + 1)(3a + 3b + 2)}{(2a + 3b + 1)(a + 2b + 1)}g′(a, b) (by Stage 1).

This completes the proof of Stage 2.

We cannot prove Stage 3 directly. Instead we use (4.1) to get ∫_{ℝ^{2}} <α_{1} >^{2} w( t_{∼}) d t_{∼}

in terms of ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼} and ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼}. Then we show how

Stage 3 will follow from Stages 1 and 2 using (3.13). Letting k_{1} = 1, k_{2} = 0 in (4.1) and using (3.8), (3.9), (3.11), and (3.15) we find that

(4.3) 2(2a + 3b + 2) ∫_{ℝ^{2}} <α_{1} >^{2} w( t_{∼}) d t_{∼}

= \frac{3}{2}(3a + 3b + 2) ∫_{ℝ^{2}} <α_{1} > w( t_{∼}) d t_{∼} − 2a ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼}

= 2(2a + 6b + 3) ∫_{ℝ^{2}} <α_{1} ><α_{2} > w( t_{∼}) d t_{∼} (by (4.2)).

We have

C.T. [α_{1}][α_{2}][α_{1} + α_{2}]G

= \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 64 <α_{1} ><α_{2} ><α_{1} + α_{2} > w( t_{∼}) d t_{∼}

= −3 · \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 16 <α_{1} >^{2} w( t_{∼}) d t_{∼}

+ 6 · \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 16 <α_{1} ><α_{2} > w( t_{∼}) d t_{∼} (by (3.13))

= \frac{(−3(2a + 6b + 3) + 6(2a + 3b + 2))}{(2a + 3b + 2)} \frac{4^{3a+3b}}{π^{2}} ∫_{ℝ^{2}} 16 <α_{1} ><α_{2} > w( t_{∼}) d t_{∼}

(by (4.3))

= \frac{3(2a + 1)}{(2a + 3b + 2)}C.T. [α_{1}][α_{2}]G

= \frac{6(3a + 3b + 2)(3a + 3b + 1)(2a + 1)}{(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}g′(a, b) (by Stage 2).

This completes the proof of Stage 3.

At the beginning of §2 we showed how Stage 3 implies (1.2). We remark that (1.3) and (1.4) follow from Stages 1 and 2 together with (1.2). We have been unable to find a proof of (1.5) or (1.6) in terms of integrals. However (1.5) follows easily from (1.3) and (1.4). In order to show this we need to recall how the Weyl group acts on polynomials. For α = k_{1}α_{1} + k_{2}α_{2}, where k_{1}, k_{2} ∈ ℤ and α_{1}, α_{2} are the roots from the root system G_{2} as in Fig. 1, we let

x^{α} = x_{1}^{k_{1}}x_{2}^{k_{2}}.

The elements w of the Weyl group W act on monomials by

w(x^{α}) = x^{w(α)},

and by linearity on Laurent polynomials that are linear combinations of the x^{α}. For w in the Weyl group W we have

(4.4)  C.T. x^{α}G = C.T. x^{w(α)}G,

since G is symmetric with respect to the Weyl group and w does not change the constant term. Utilizing (4.4) we find that the left-hand sides of (1.3)–(1.5) can be written as

C.T. [α_{1}]G = 2C.T.(1 − x_{1})G, (4.5)

C.T. [α_{1}][α_{2}]G = 2C.T. (2 − 3x_{1} + \frac{x_{1}}{x_{2}} ) G, (4.6)

C.T. [2α_{1} + α_{2}]G = 2C.T. (1 − \frac{x_{1}}{x_{2}} ) G. (4.7)

Hence,

C.T. [2α_{1} + α_{2}]G = 3C.T. [α_{1}]G − C.T. [α_{1}][α_{2}]G

= \{\frac{6(3a + 3b + 1)}{(2a + 3b + 1)} − \frac{2(3a + 3b + 2)(3a + 3b + 1)}{(2a + 3b + 1)(a + 2b + 1)}\} g(a, b)

(by (1.3), (1.4))

= \frac{2(3a + 3b + 1)(3b + 1)}{(2a + 3b + 1)(a + 2b + 1)}g(a, b),

which is (1.5).

5. Other results. In this section we state other results that are similar to (1.3)– (1.6). These may be proved by extending Zeilberger’s [12] proof of the ordinary G_{2} case as mentioned in §1.

C.T. [α_{1}][α_{1} − α_{2}]G = \frac{4(3a + 3b + 2)(3a + 3b + 1)(3b + 1)}{(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}g(a, b), (5.1)

C.T. [α_{1}][α_{1} + α_{2}][α_{1} − α_{2}]G (5.2)

= \frac{6(4a + 3b + 4)(3a + 3b + 2)(3a + 3b + 1)(3b + 1)}{(2a + 3b + 3)(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 1)}g(a, b),

C.T. [α_{1}][α_{2}][α_{1} − α_{2}]G (5.3)

= 6 \{\frac{(3a + 3b + 2)(3a + 3b + 1)(3b + 1)}{(2a + 3b + 1)(a + 2b + 2)(a + 2b + 1)}

−\frac{b(4a + 3b + 4)(3a + 3b + 2)(3a + 3b + 1)(3b + 1)}{(2a + 3b + 3)(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 2)(a + 2b + 1)}\} g(a, b),

C.T. [α_{1}][α_{1} − α_{2}][α_{1} + 2α_{2}]G (5.4)

= \frac{6(3a + 4b + 4)(3a + 3b + 2)(3a + 3b + 1)(3b + 2)(3b + 1)}{(2a + 3b + 3)(2a + 3b + 2)(2a + 3b + 1)(a + 2b + 2)(a + 2b + 1)}g(a, b),

C.T. [α_{1}][α_{1} − α_{2}][2α_{1} + α_{2}]G (5.5)

= \frac{6(6a^{2} + 20ab + 23a + 12b^{2} + 28b + 16)}{(2a + 3b + 4)(2a + 3b + 3)(2a + 3b + 2)}

· \frac{(3a + 3b + 2)(3a + 3b + 1)(3b + 2)(3b + 1)}{(2a + 3b + 1)(a + 2b + 2)(a + 2b + 1)}g(a, b),

C.T.[α_{1}][α_{1} + α_{2}][α_{1} − α_{2}][α_{1} + 2α_{2}]G (5.6)

= \frac{18(3a + 3b + 4)(3a + 3b + 2)(3a + 3b + 1)}{(2a + 3b + 4)(2a + 3b + 3)(2a + 3b + 2)}

· \frac{(2a + 2b + 3)(3b + 2)(3b + 1)}{(2a + 3b + 1)(a + 2b + 2)(a + 2b + 1)}g(a, b).

Acknowledgments. I would like to thank many people: Richard Askey for introducing me to the Macdonald conjectures, for his conjectures on G_{2}, and for help and encouragement; Murad Ozaydin for helpful discussions on root systems and Weyl groups; William Long for getting me started on REDUCE; and Dennis Stanton for his suggestions in writing this paper.

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