Generalised Kähler structure on $  \mathbb{C}P^2 $ and elliptic functions

Francesco Bonechi; Jian Qiu; Marco Tarlini; Francesco Bonechi; Jian Qiu; Marco Tarlini

doi:10.3934/jgm.2023009

Journal of Geometric Mechanics

2023, Volume 15, Issue 1: 188-223. doi: 10.3934/jgm.2023009

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Research article

Generalised Kähler structure on $\mathbb{C}P^2$ and elliptic functions

1.
INFN Sezione di Firenze, Italy
2.
Matematiska institutionen, Institutionen för fysik och astronomi, Uppsala Universitet, Sweeden

Received: 21 December 2021 Revised: 01 September 2022 Accepted: 04 October 2022 Published: 02 February 2023
Primary: 53D18; Secondary: 53D17, 53C15

We construct a toric generalised Kähler structure on $\mathbb{C}P^2$ and show that the various structures such as the complex structure, metric etc are expressed in terms of certain elliptic functions. We also compute the generalised Kähler potential in terms of integrals of elliptic functions.

Keywords:

Citation: Francesco Bonechi, Jian Qiu, Marco Tarlini. Generalised Kähler structure on $\mathbb{C}P^2$ and elliptic functions[J]. Journal of Geometric Mechanics, 2023, 15(1): 188-223. doi: 10.3934/jgm.2023009

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Abstract

1. Introduction

In ^[18] Ramanujan showed a total of 17 series for $1/\pi$ but he did not indicate how he arrived at these series. The Borwein brothers ^[5] gave rigorous proofs of Ramanujan's series for the first time and also obtained many new series for $1/\pi.$ Till now, many new Ramanujan's-type series for $1/\pi$ have been published, (see, for example, ^[4,6,8]). Chu ^[7], Liu ^[15,16] and Wei et al. ^[21,22] gave many $\pi$ -formula with free parameters by means of gamma functions and hypergeometric series. Guillera ^[10] proved a kind of bilateral semi-terminating series related to Ramanujan-like series for negative powers of $\pi$ . Moreover, Guillera and Zudilin ^[11] outlined an elementary method for proving numerical hypergeometric identities, in particular, Ramanujan-type identities for $1/\pi$ . Recently, $q$ -analogues of Ramanujan-type series for $1/\pi$ have caught the interests of many authors (see, for example, ^{[9,12,13,14,20,21]}).

Although various definitions for gamma functions are used in the literature, we adopt the following definition [23, p.76]

$\frac 1{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n = 1}^\infty\left(1+\frac zn\right)e^{-\frac zn}$

where $\gamma$ is the Euler constant defined as

$\gamma = \lim\limits_{n\rightarrow \infty}\left(1+ \frac 12+\cdots+ \frac 1n-\log n\right).$

It is easy to verify that $\Gamma(1) = 1, \Gamma({ \frac 12}) = \sqrt\pi$ and $\Gamma({z+1}) = z\Gamma(z).$ It follows that for every positive integer $n$ , $\Gamma(n) = (n-1)!.$

For any complex $\alpha$ , we define the general rising shifted factorial by

$\begin{align} (z)_{{\alpha}} = \Gamma({z+{\alpha}})/\Gamma(z). \end{align}$

(1.1)

Obviously, $(z)_0 = 1.$ For every positive integer $n$ , we have

$(z)_n = \Gamma({z+n})/\Gamma(z) = z(z+1)\cdots(z+n-1)$

and

$(z)_{-n} = \Gamma({z-n})/\Gamma(z) = \frac 1{(z-1)(z-2)\ldots(z-n)}.$

For convenience, we use the following compact notations

$(a_1,a_2,\ldots,a_m)_n = (a_1)_n(a_2)_n\ldots(a_m)_n$

and

$(a)_{(n_1,n_2,\ldots,n_m)} = (a)_{n_1}(a)_{n_2}\ldots(a)_{n_m}.$

Following ^[1,3], the hypergeometric series is defined by

$_{{r+1}}F_{s}\left[\begin{array}{cccc}a_0,&a_1,&\ldots,&a_r\\&b_1,&\ldots,&b_s\end{array};z\right] = \sum\limits_{k = 0}^\infty \frac {(a_0, a_1,\ldots,a_r)_k}{(b_1,\ldots,b_s)_k} \frac {z^k}{k!},$

where $a_i, b_j(i = 0, 1, \ldots, r, j = 1, 2, \ldots, s)$ are complex numbers such that no zero factors appear in the denominators of the summand on the right hand side.

We let $F_{q:s; v}^{p:r; u}\ ({p, q, r, s, u, v\in \Bbb N_0 = \{0, 1, 2, \ldots\}})$ denote a general (Kampé de Fériet's) double hypergeometric function defined by (see ^[2,19])

$\begin{eqnarray*} &&F_{q:s;v}^{p:r;u}\left[\begin{array}{ccc}{\alpha}_1,\ldots,{\alpha}_p:&a_1,\ldots,a_r;&c_1,\ldots,c_u;\\ \beta_1,\ldots,\beta_q:&b_1,\ldots,b_s;&d_1,\ldots,d_v;\end{array}x,y\right]\\&& = \sum\limits_{m,n = 0}^\infty \frac {({\alpha}_1,\ldots,{\alpha}_p)_{m+n} (a_1,\ldots,a_r)_m(c_1,\ldots,c_u)_n}{(\beta_1,\ldots,\beta_q)_{m+n} (b_1,\ldots,b_s)_m(d_1,\ldots,d_v)_n} \frac {x^m}{m!} \frac {y^n}{n!}, \end{eqnarray*}$

where, for convergence of the double hypergeometric series,

$p+r\leq q+s+1{\quad}\text{and}{\quad} p+u\leq q+v+1,$

with equality only when $|x|$ and $|y|$ are appropriately constrained (see, for details, [19,Eq 1.3(29),p.27]).

There exist numerous identities for such series. For example, we have

Theorem 1.1 (See [,(30)] ) If $Re(e-d) > 0$ and $Re(d+e-a-b-c) > 0,$ then

$F_{1:1;1}^{0:3;3}\left[\begin{array}{ccc}-:&a,b,c;&d-a,d-b,d-c;\\d:&e;&d+e-a-b-c;\end{array}1,1\right] = \frac {\Gamma({e})\Gamma({e+d-a-b-c})\Gamma({e-d})}{\Gamma({e-a})\Gamma({e-b})\Gamma({e-c})}.$

In ^[15], Liu used the general rising shifted factorial and the Gauss summation formula to prove the following four-parameter series expansions formula, which implies infinitely many Ramanujan type series for $1/\pi$ and $\pi$ .

Theorem 1.2 For any complex ${\alpha}$ and $Re(c-a-b) > 0,$ we have

$\sum\limits_{n = 0}^\infty \frac {({\alpha})_{a+n}(1-{\alpha})_{b+n}}{n!\Gamma({c+n+1})} = \frac {({\alpha})_a(1-{\alpha})_b\Gamma({c-a-b})} {({\alpha})_{c-b}(1-{\alpha})_{c-a}}\cdot \frac {\sin \pi{\alpha}}\pi.$

Motivated by Liu's work, in this paper we derive the following result from Theorem 1.1 which enables us to give many double series expansions for $1/\pi$ and $\ \pi$ . To the best of our knowledge, most of the results in this paper have not previously appeared.

Theorem 1.3 If $d\in \Bbb N_0, Re(e-d+{\sigma}-{\delta}) > 0$ and $Re(d+e-a-b-c+{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}) > 0,$ then

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\alpha})_{a+m}(\beta)_{b+m}({\gamma})_{c+m}({\delta}-{\alpha})_{d-a+n}({\delta}-\beta)_{d-b+n}({\delta}-{\gamma})_{d-c+n}} {m!n!({\delta}+d)_{m+n}({\sigma})_{e+m}({\delta}+{\sigma}-{\alpha}-\beta-{\gamma})_{d+e-a-b-c+n}}\\ & = \frac {({\alpha})_a(\beta)_b({\gamma})_c({\delta}-{\alpha})_{d-a}({\delta}-\beta)_{d-b}({\delta}-{\gamma})_{d-c}({\sigma}-{\delta})_{e-d}} {({\sigma}-{\alpha})_{e-a}({\sigma}-\beta)_{e-b}({\sigma}-{\gamma})_{e-c}} \cdot \frac {\Gamma({\sigma})\Gamma({{\sigma}-{\delta}})\Gamma({{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}})}{\Gamma({{\sigma}-{\alpha}})\Gamma({{\sigma}-\beta})\Gamma({{\sigma}-{\gamma}})}. \end{align*}$

Several examples of such formulae are

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_m^3( \frac 12)_n^2}{m!n!(m+n)!(m+1)!(2n+1)} = \frac 4{\pi},$

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+3)!( \frac 12)_{m+1}} = \pi,$

and

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^2( \frac 13)_n^3}{m!n!(n+1)!(2-3m)(- \frac 13)_{m+n}} = \frac {\sqrt3\pi}3.$

The remainder of the paper is organized as follows. In section 2 we give the proof of Theorem 1.3. Sections 3 and 4 are devoted to the double series expansions for $1/\pi$ and $\pi$ , respectively.

2. Proof of Theorem 1.3

First of all, by making use of (1.1), Theorem 1.3 can be restated as follows:

$\begin{equation} \begin{split} &\sum\limits_{m,n = 0}^\infty \frac {\Gamma({a+m})\Gamma({b+m})\Gamma({c+m})\Gamma({d-a+n})\Gamma({d-b+n})\Gamma({d-c+n})} {m!n!\Gamma({d+m+n})\Gamma({e+m})\Gamma({d+e-a-b-c+n})}\\ & = \frac {\Gamma(a)\Gamma(b)\Gamma(c)\Gamma({d-a})\Gamma({d-b})\Gamma({d-c})\Gamma({e-d})}{\Gamma(d)\Gamma({e-a})\Gamma({e-b})\Gamma({e-c})}. \end{split} \end{equation}$

(2.1)

From (1.1) it is easy to see that

$\begin{align*} &\Gamma({a+{\alpha}+m}) = ({\alpha})_{a+m}\Gamma({{\alpha}}), \ \Gamma({b+\beta+m}) = (\beta)_{b+m}\Gamma({\beta}),\ \Gamma({c+{\gamma}+m}) = ({\gamma})_{c+m}\Gamma({{\gamma}}),\\ &\Gamma({d-a+{\delta}-{\alpha}+n}) = ({\delta}-{\alpha})_{d-a+n}\Gamma({{\delta}-{\alpha}}),\ \Gamma({d-b+{\delta}-\beta+n}) = ({\delta}-\beta)_{d-b+n}\Gamma({{\delta}-\beta}),\\ &\Gamma({d-c+{\delta}-{\gamma}+n}) = ({\delta}-{\gamma})_{d-c+n}\Gamma({{\delta}-{\gamma}}),\ \Gamma({d+{\delta}+m+n}) = ({\delta})_{d+m+n}\Gamma({{\delta}})\\ &\Gamma({e+m+{\sigma}}) = ({\sigma})_{e+m}\Gamma({\sigma}),\ \Gamma({a+{\alpha}}) = ({\alpha})_a\Gamma({\alpha}),\ \Gamma({b+\beta}) = (\beta)_b\Gamma(\beta),\ \Gamma({c+{\gamma}}) = ({\gamma})_c\Gamma({\gamma}),\\ &\Gamma({d-a+{\delta}-{\alpha}}) = ({\delta}-{\alpha})_{d-a}\Gamma({{\delta}-{\alpha}}),\ \Gamma({d-b+{\delta}-\beta}) = ({\delta}-\beta)_{d-b}\Gamma({{\delta}-\beta}),\\ &\Gamma({d-c+{\delta}-{\gamma}}) = ({\delta}-{\gamma})_{d-c}\Gamma({{\delta}-{\gamma}}),\ \Gamma({e-d+{\sigma}-{\delta}}) = ({\sigma}-{\delta})_{e-d}\Gamma({{\sigma}-{\delta}}),\\ &\Gamma({d+{\delta}}) = ({\delta})_d\Gamma({\delta}),{\quad} \Gamma({e-a+{\sigma}-{\alpha}}) = ({\sigma}-{\alpha})_{e-a}\Gamma({{\sigma}-{\alpha}}),\\ &\Gamma({e-b+{\sigma}-\beta}) = ({\sigma}-\beta)_{e-b}\Gamma({{\sigma}-\beta}),\ \Gamma({e-c+{\sigma}-{\gamma}}) = ({\sigma}-{\gamma})_{e-c}\Gamma({{\sigma}-{\gamma}}),\\ &\Gamma({d+e-a-b-c+{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}})\\& = ({\delta}+{\sigma}-{\alpha}-\beta-{\gamma})_{d+e-a-b-c}\Gamma({{\delta}+{\sigma}-{\alpha}-\beta-{\gamma}}). \end{align*}$

and we realize that $({\delta})_{d+m+n} = ({\delta})_d({\delta}+d)_{m+n}$ when $d\in \Bbb N_0.$ Replacing $(a, b, c, d, e)$ by $(a+{\alpha}, b+\beta, c+{\gamma}, d+{\delta}, e+{\sigma})$ in (2.1) and substituting above identities into the resulting equation, we get the desired result.

3. Double series expansions for $1/\pi$

In this section we will use Theorem 1.3 to prove the following double series expansion formula for $1/\pi$ .

Theorem 3.1 If $d\in \Bbb N_0, Re(e-d+1) > 0$ and $Re(d+e-a-b-c+ \frac 32) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(a+m,b+m,c+m,d-a+n,d-b+n,d-c+n)}} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e-a-b-c+n}} = \frac {( \frac 12)_{(a,b,c,d-a,d-b,d-c)}(1)_{e-d}} {( \frac 32)_{(e-a,e-b,e-c)}}\cdot \frac 4{\pi}.$

Proof. Let $({\alpha}, \beta, {\gamma}, {\delta}, {\sigma}) = ( \frac 12, \frac 12, \frac 12, 1, 2)$ in Theorem 1.3. We find that

$\begin{align} \sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(a+m,b+m,c+m,d-a+n,d-b+n,d-c+n)}} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e-a-b-c+n}} = \frac {( \frac 12)_{(a,b,c,d-a,d-b,d-c)}(1)_{e-d}} {( \frac 32)_{(e-a,e-b,e-c)}}\cdot \frac {\Gamma(2)\Gamma(1)\Gamma({ \frac 32})}{\Gamma^3( \frac 32)}. \end{align}$

(3.1)

Substituting $\Gamma({ \frac 32}) = \frac {\sqrt{\pi}}2$ into (3.1) we obtain the result immediately. Putting $(a, b, c) = (0, 0, 0)$ in Theorem 3.1 we get the following general double summation formula for $1/\pi$ with two free parameters.

Corollary 3.2 If $d\in \Bbb N_0, Re(e-d+1) > 0$ and $Re(d+e+ \frac 32) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,d+n)}^3} {m!n!(d+1)_{m+n}(2)_{e+m}( \frac 32)_{d+e+n}} = \frac {4( \frac 12)_d^3(1)_{e-d}}{\pi( \frac 32)_e^3}.$

Setting $d = 0$ and $e = k\in\Bbb N_0$ in Corallary 3.2 we have the following result.

Proposition 3.3 Let $k$ be a nonnegative integer. Then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,n)}^3}{m!n!(m+n)!(m+k+1)!( \frac 32+k)_n} = \frac {4k!}{\pi( \frac 32)_k^2}.$

Example 3.1 ( $k = 0$ in Proposition 3.3).

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_m^3( \frac 12)_n^2}{m!n!(m+n)!(m+1)!(2n+1)} = \frac 4{\pi}.$

If $d = e = k\in\Bbb N_0$ in Corollary 3.2 we achieve

Proposition 3.4 Let $k$ be a nonnegative integer. Then

$\sum\limits_{m,n = 0}^\infty \frac {( \frac 12)_{(m,n+k)}^3} {m!n!(k+1)_{m+n}(m+k+1)!( \frac 32)_{n+2k}} = \frac 4{\pi(2k+1)^3}.$

If we put $k = 0$ into Proposition 3.4, then we can also get Example 3.1.

4. Double series expansions for $\pi$

In this section we will prove the following theorem, which allows us to derive infinitely double series expansions for $\pi$ .

Theorem 4.1 If $d\in \Bbb N_0, Re(e-d-{\sigma}+1) > 0$ and $Re(d+e-a-b-c+2) > 0,$ then

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_{(a+m,b+m,c+m)}({\sigma})_{(d-a+n,d-b+n,d-c+n)}} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e-a-b-c+n}}\\ & = \frac {({\sigma}-1)_{(a,b,c)}({\sigma})_{(d-a,d-b,d-c)}(1-{\sigma})_{e-d}}{(1)_{(e-a,e-b,e-c)}} \cdot \frac {\pi}{\sin{{\sigma}\pi}}. \end{align*}$

Proof. Let $({\alpha}, \beta, {\gamma}, {\delta}) = ({\sigma}-1, {\sigma}-1, {\sigma}-1, 2{\sigma}-1)$ in Theorem 1.3. We obtain that

$\begin{align*} &\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_{(a+m,b+m,c+m)}({\sigma})_{(d-a+n,d-b+n,d-c+n)}} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e-a-b-c+n}}\\ & = \frac {({\sigma}-1)_{(a,b,c)}({\sigma})_{(d-a,d-b,d-c)}(1-{\sigma})_{e-d}} {(1)_{(e-a,e-b,e-c)}} \cdot \frac {\Gamma({\sigma})\Gamma({1-{\sigma}})\Gamma({2})}{\Gamma^3(1)}. \end{align*}$

(4.1)

Combining $\Gamma({{\sigma}})\Gamma({1-{\sigma}}) = \frac \pi{\sin{\sigma}\pi}$ with (4.1) we get the desired result immediately. Putting $a = b = c = 0$ in Theorem 4.1 we obtain the following equation.

Corollary 4.2 If $d\in \Bbb N_0, Re(e-d-{\sigma}+1) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {({\sigma}-1)_m^3({\sigma})_{d+n}^3} {m!n!(2{\sigma}+d-1)_{m+n}({\sigma})_{e+m}(2)_{d+e+n}} = \frac {({\sigma})_d^3(1-{\sigma})_{e-d}}{(1)_e^3} \cdot \frac {\pi}{\sin{{\sigma}\pi}}.$

4.1. $\sigma=\frac{1}{2}$ in Corollary 4.2

Letting $\sigma=\frac{1}{2}$ in Corollary 4.2, we get the following proposition.

Proposition 4.3 If $d\in \Bbb N_0, Re(e-d+ \frac 12) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 12)_{d+n}^3}{m!n!(d)_{m+n}( \frac 12)_{e+m} (2)_{d+e+n}} = \frac {( \frac 12)_d^3( \frac 12)_{e-d}}{(1)_{e}^3} \pi.$

When we set $d = 1$ and $e = k\in\Bbb N = \{1, 2, 3\ldots\}$ in Proposition 4.3 we obtain

Proposition 4.4 If $k$ is a positive integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+k+2)!( \frac 12)_{m+k}} = \frac {\pi( \frac 12)_{k-1}}{(k!)^3} .$

Example 4.1 ( $k = 1$ in Proposition 4.4).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 12)_m^3( \frac 32)_n^3}{m!n!(m+n)!(n+3)!( \frac 12)_{m+1}} = \pi.$

4.2. $\sigma=\frac{1}{3}$ in Corollary 4.2

Putting $\sigma=\frac{1}{3}$ in Corollary 4.2, we get the following proposition.

Proposition 4.5 If $d\in \Bbb N_0, Re(e-d+ \frac 23) > 0$ and $Re(d+e+2) > 0,$ then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13)_{d+n}^3}{m!n!(d- \frac 13)_{m+n}( \frac 13)_{e+m} (2)_{d+e+n}} = \frac {2\sqrt3\pi( \frac 13)_d^3( \frac 23)_{e-d}}{3(1)_{e}^3}.$

When we set $d = 0$ and $e = k\in\Bbb N_0$ in Proposition 4.5 we obtain

Proposition 4.6 If $k$ is a nonnegative integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13)_n^3}{m!n!(- \frac 13)_{m+n}( \frac 13)_{m+k}(n+k+1)!} = \frac {2\sqrt3\pi( \frac 23)_k}{3k!^3}.$

Example 4.2 ( $k = 0$ in Proposition 4.6).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^2( \frac 13)_n^3}{m!n!(n+1)!(2-3m)(- \frac 13)_{m+n}} = \frac {\sqrt3\pi}3.$

Setting $d = e = k\in\Bbb N_0$ in Proposition 4.5, we get

Proposition 4.7 If $k$ is a nonnegative integer, then

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 13+k)_n^3}{m!n!(n+2k+1)!(k- \frac 13)_{m+n}( \frac 13)_{m+k} } = \frac {2\sqrt3\pi}{3k!^3}.$

Therefore, Example 4.2 can also be deduced by fixing $k = 0$ in the above equation.

Example 4.3 ( $k = 1$ in Proposition 4.7).

$\sum\limits_{m,n = 0}^\infty \frac {(- \frac 23)_m^3( \frac 43)_n^3}{m!n!(n+3)!( \frac 23)_{m+n} ( \frac 43)_m} = \frac {2\sqrt3\pi}9.$

5. Conclusions

Double series expansions for $1/\pi$ and $\pi$ with several free parameters are established and many interesting formulas are obtained. A point that should be stressed is that there is an important connection between the summation formulas for double hypergeometric functions and double series expansions for the powers of $\pi$ .

Acknowledgements

The author was partially supported by the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (grant 19KJB110006).

Conflicts of interest

The author declares that there is no conflict of interest in this paper.

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