$\mathbb {C}P^{2S}$ Sigma Models Described Through Hypergeometric Orthogonal Polynomials

Abstract

The main objective of this paper is to establish a new connection between the Hermitian rank-1 projector solutions of the Euclidean $\mathbb {C}P^{2S}$ sigma model in two dimensions and the particular hypergeometric orthogonal polynomials called Krawtchouk polynomials. We show that any Veronese subsequent analytical solutions of the $\mathbb {C}P^{2S}$ model, defined on the Riemann sphere and having a finite action, can be explicitly parametrized in terms of these polynomials. We apply these results to the analysis of surfaces associated with $\mathbb {C}P^{2S}$ models defined using the generalized Weierstrass formula for immersion. We show that these surfaces are homeomorphic to spheres in the $\mathfrak {su}(2s+1)$ algebra and express several other geometrical characteristics in terms of the Krawtchouk polynomials. Finally, a connection between the $\mathfrak {su}(2)$ spin-s representation and the $\mathbb {C}P^{2S}$ model is explored in detail.

Properties of the Krawtchouk Polynomials

In this appendix, we recall and prove useful properties of the Krawtchouk polynomials and of the projectors $P_k$ (3.6) According to R. Koekoek [23], the properties of the Krawtchouk polynomials follow directly from their definition

$$\begin{aligned} K_j(k; p, 2s) = {}_2 F_{1}\left( -j, -k; -2s; 1/p\right) , \qquad 0\le j,k\le 2s, \end{aligned}$$

(A.1)

where we recall that $0<p={\textstyle {\xi _+\xi _-\over 1+\xi _+\xi _-}}<1$. In view of the formula (1.10.6) from [23], the Krawtchouk polynomial forward shift operator is

$$\begin{aligned} K_j(k+1; p, 2s) - K_j(k; p, 2s) = -\, {\textstyle {\displaystyle j\over \displaystyle 2sp}}K_{j-1}(k;p,2s-1). \end{aligned}$$

(A.2)

Lemma A.1

The first derivative with respect to $\xi _\pm$ of the Krawtchouk polynomials are given by

$$\begin{aligned} \partial K_j(k;p,2s)&= {\textstyle {\displaystyle -k\over \displaystyle \xi _+(1+\xi _+\xi _-)}}(K_j(k;p,2s)-K_{j}(k-1;p,2s)), \end{aligned}$$

(A.3)

$$\begin{aligned} \overline{\partial }K_j(k;p,2s)&= {\textstyle {\displaystyle -k\over \displaystyle \xi _-(1+\xi _+\xi _-)}}(K_j(k;p,2s)-K_{j}(k-1;p,2s)). \end{aligned}$$

(A.4)

Proof

We recall the formula for the derivative of the hypergeometric function

$$\begin{aligned} {\textstyle {\partial \over \partial x}} \left( {}_{2}F_{1}(a, b; c; x)\right) = {\textstyle {\displaystyle ab\over \displaystyle c}}\ {}_{2}F_{1}(a+1, b+1; c+1; x). \end{aligned}$$

(A.5)

Hence, from the definition (A.1), we can evaluate the first derivatives with respect to $\xi _+$ of the Krawtchouk polynomial

$$\begin{aligned} \partial K_j\left( k; {\textstyle {\xi _+\xi _-\over (1+\xi _+\xi _-)}}, 2s\right)&= \partial \left( {}_{2}F_{1}\left( -j, -k; -2s; {\textstyle {1+\xi _+\xi _-\over \xi _+\xi _-}}\right) \right) \nonumber \\&= -{\textstyle {\xi _-\over (\xi _+\xi _-)^2}}\left( {\textstyle {-jk\over 2s}}\right) {}_{2}F_{1}\left( -j+1, -k+1; -2s+1; {\textstyle {1+\xi _+\xi _-\over \xi _+\xi _-}}\right) \nonumber \\&={\textstyle {1\over \xi _+^2\xi _-}}{\textstyle {jk\over 2s}}K_{j-1}(k-1; p, 2s-1). \end{aligned}$$

(A.6)

In view of (A.2), we obtain the expression (A.3). Similarly, we obtain the relation (A.4). $\square$

Lemma A.2

The following orthogonality relations hold

$$\begin{aligned}&\sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) (\xi _+\xi _-)^q K_q(k)K_q(\ell ) = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s}\over \displaystyle (\xi _+\xi _-)^{k}\left( {\begin{array}{c}2s\\ k\end{array}}\right) }} \ \delta _{k,\ell }, \end{aligned}$$

(A.7)

$$\begin{aligned}&\sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) (\xi _+\xi _-)^q \,q\, K_q^2 = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-1}\over \displaystyle (\xi _+\xi _-)^{k}\left( {\begin{array}{c}2s\\ k\end{array}}\right) }}(k+(2s-k)\xi _+\xi _-), \end{aligned}$$

(A.8)

$$\begin{aligned}&\sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) (\xi _+\xi _-)^q \,q\, K_qK_q(k-1) = -{\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-1}\over \displaystyle (\xi _+\xi _-)^{k-1}\left( {\begin{array}{c}2s\\ k\end{array}}\right) }}(2s-k+1), \end{aligned}$$

(A.9)

$$\begin{aligned}&\sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) (\xi _+\xi _-)^q \,q^2\, K_q^2 = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-2}\over \displaystyle (\xi _+\xi _-)^{k}\left( {\begin{array}{c}2s\\ k\end{array}}\right) }} \left( (\xi _+\xi _-)^2(k-2s)^2\right. \nonumber \\&\qquad \qquad \quad \qquad \qquad \qquad \left. +\,2\xi _+\xi _-(4sk-2k^2+s)+k^2\right) . \end{aligned}$$

(A.10)

Proof

Relation (A.7) is directly obtained from [23]. Let us now differentiate the orthogonality relation (A.7) (for $\ell =k$) with respect to $\xi _+$. By using (A.3), one gets

$$\begin{aligned}&\sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) q\xi _+^{q-1}\xi _-^{q}K_q^2 - {\textstyle {\displaystyle 2k\over \displaystyle \xi _+(1+\xi _+\xi _-)}} \sum _{q=0}^{2s}\left( {\begin{array}{c}2s\\ q\end{array}}\right) (\xi _+\xi _-)^qK_q(K_q-K_q(k-1))\nonumber \\&\quad = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-1}\left[ 2s\xi _+\xi _- - k(1+\xi _+\xi _-)\right] \over \displaystyle \xi _+(\xi _+\xi _-)^{k}\left( {\begin{array}{c}2s\\ k\end{array}}\right) }}. \end{aligned}$$

(A.11)

Therefore, by again using (A.7), we get (A.8). Similarly, by differentiating the orthogonality relation (A.7) with respect to $\xi _+$ (for $\ell =k-1$), one gets (A.9). Finally, we differentiate relation (A.8) to get relation (A.10). $\square$

Let us recall that the Krawtchouk polynomials are self-dual i.e., they satisfy

$$\begin{aligned} K_j(k)=K_k(j). \end{aligned}$$

(A.12)

Therefore, from Lemma A.2, one gets the other orthogonality properties

$$\begin{aligned}&\sum _{k=0}^{2s}\left( {\begin{array}{c}2s\\ k\end{array}}\right) (\xi _+\xi _-)^k K_j(k)K_\ell (k) = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s}\over \displaystyle (\xi _+\xi _-)^{j}\left( {\begin{array}{c}2s\\ j\end{array}}\right) }}\ \ \delta _{j,\ell }, \end{aligned}$$

(A.13)

$$\begin{aligned}&\sum _{k=0}^{2s}\left( {\begin{array}{c}2s\\ k\end{array}}\right) (\xi _+\xi _-)^k \,k\, K_j^2(k) = {\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-1}\over \displaystyle (\xi _+\xi _-)^{j}\left( {\begin{array}{c}2s\\ j\end{array}}\right) }}(j+(2s-j)\xi _+\xi _-), \end{aligned}$$

(A.14)

$$\begin{aligned}&\sum _{k=0}^{2s}\left( {\begin{array}{c}2s\\ k\end{array}}\right) (\xi _+\xi _-)^k \,k\, K_j(k)K_{j-1}(k) = -{\textstyle {\displaystyle (1+\xi _+\xi _-)^{2s-1}\over \displaystyle (\xi _+\xi _-)^{j-1}\left( {\begin{array}{c}2s\\ j\end{array}}\right) }}(2s-j+1). \end{aligned}$$

(A.15)

Let us also remark that

$$\begin{aligned} \sum _{k=0}^{2s}\left( {\begin{array}{c}2s\\ k\end{array}}\right) (\xi _+\xi _-)^k \,k\, K_j(k)K_{\ell }(k) = 0 \qquad \text {if}\quad \ell \ne k,k\pm 1\ . \end{aligned}$$

(A.16)

In our notation, the difference equation satisfied by the Krawtchouk polynomials (see (1.10.5) of [23]) read

$$\begin{aligned} -p(2s-k)K_j(k+1)+(k-j+2p(s-k))K_j-k(1-p)K_j(k-1)=0\;.\nonumber \\ \end{aligned}$$

(A.17)

Lemma A.3

The following relation between the Krawtchouk polynomials holds

$$\begin{aligned} {\textstyle {\displaystyle 1\over \displaystyle 1+\xi _+\xi _-}}\left[ 2(s-j)K_j+\xi _+\xi _-(2s-j)K_{j+1}-{\textstyle {\displaystyle j\over \displaystyle \xi _+\xi _-}}K_{j-1} \right] =(2s-k)K_j(k+1).\nonumber \\ \end{aligned}$$

(A.18)

Proof

By using the following properties of the hypergeometric functions [35]

$$\begin{aligned}&(b-c){}_{2}F_{1}\left( a, b-1; c; x\right) +(c-a-b){}_{2}F_{1}\left( a, b; c; x\right) \nonumber \\&\quad = a(x-1){}_{2}F_{1}\left( a+1, b; c; x\right) . \end{aligned}$$

(A.19)

one deduces that

$$\begin{aligned} K_j(k+1)={\textstyle {\displaystyle 1\over \displaystyle 2s-k}}\left[ (2s-j-k)K_j-{\textstyle {\displaystyle j\over \displaystyle \xi _+\xi _-}}K_{j-1} \right] . \end{aligned}$$

(A.20)

Then, by replacing $K_j(k+1)$ in (A.18) in the above formula, we see that (A.18) is equivalent to the recurrence relation of the Krawtchouk polynomial (see (1.10.3) of [23]) which concludes the proof. $\square$