On Jacobi polynomials and fractional spectral functions on compact symmetric spaces

Abstract

The Jacobi coefficients $c_{j}^{\ell }(\alpha ,\beta )$ ($1\le j\le \ell ; \alpha ,\beta >-1$) associated with the normalised Jacobi polynomials ${\mathscr {P}}_k^{(\alpha , \beta )}(\eta ):=P_{k}^{(\alpha ,\beta )}(\eta )/P_{k}^{(\alpha ,\beta )}(1)$ ($k\ge 0; \alpha ,\beta >-1, -1\le \eta \le 1$) describe the Maclaurin heat coefficients appearing in the classical Maclaurin expansion of the heat kernel on any N-dimensional compact rank one symmetric space. These coefficients are computed by transforming the even $(2\ell )$th $(\ell \ge 1)$ derivatives of the Jacobi polynomials ${\mathscr {P}}_{k}^{(\alpha ,\beta )}(\eta )$ into a spectral sum involving the Jacobi operator. In this paper, we generalise this idea by constructing the fractional Taylor heat coefficients (i.e., the coefficients appearing in the fractional Taylor series expansion of the heat kernel) on any rank one symmetric space of compact type. The Riemann–Liouville fractional derivative of normalised Jacobi polynomials ${\mathscr {P}}_k^{(\alpha , \beta )}(\eta )$ is considered and an interesting spectral identity explicitly described by the fractional Jacobi coefficients is established. The analytical and spectral implications of these fractional coefficients are in turn underlined. The first fractional coefficients are explicitly computed. By extension, fractional Jacobi coefficients play a crucial role in the explicit descriptions of constants appearing in the fractional power series expansion of eigenfunctions involving Jacobi polynomials. We also introduce and construct new zeta functions ${\mathcal {Z}}^{(\alpha ,\beta )}_{m,\ell \mu }={\mathcal {Z}}^{(\alpha ,\beta )}_{m,\ell \mu }(s)$ $(1\le \ell \le m, 0<\mu \le 1, s\in {\mathbb {C}})$ associated with these fractional Taylor heat coefficients. It is interesting to see that this new zeta function can be explicitly described by the newly introduced fractional Minakshisundaram-Pleijel zeta function.

Appendices

Appendix A: Explicit calculations of the first coefficients $\mathbb {F}_{j,k}^{m,\mu \ell }(\alpha ,\beta )$

This section presents in explicit form the first fractional Jacobi coefficients ${\mathbb {F}}_{j,k}^{m,\ell \mu }(\alpha ,\beta )$ appearing in the spectral relation (40). Note that the proof of Theorem  5.1 does not reveal in explicit form the fractional Jacobi coefficients. We consider the special cases ${\mathbb {F}}_{j,k}^{m,\ell \mu }(\alpha ,\beta )$ $(1\le j\le \ell \le m\le 4,m\ge 1;\mu =1/2)$. These coefficients generalise those in [17, 18].

For simplicity of notation, let ${\mathbb {D}}_{k, m}^{\mu \ell }(\eta ):={\mathbf {D}}_{m}^{\mu \ell }{\mathscr {P}}_{k}^{(\alpha ,\beta )}(\eta )$.

• (a) $m=1:$ Here we see that

  $$\begin{aligned} {\mathbb {D}}_{k, 1}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathsf {A}}^{1,\ell /2}_{1}{\mathsf {F}}^{1,\ell /2}_{1,k}(\alpha ,\beta ){\mathsf {y}}'(1), \end{aligned}$$

  (A.1)

  where

  $$\begin{aligned}&{\mathsf {F}}^{1,\ell /2}_{1,k}(\alpha ,\beta )=\ _3F_2\left( 1-k,1-\frac{\ell }{2},\alpha +\beta +k+2;3-\frac{\ell }{2},\alpha +2;\frac{1}{2}\right) \nonumber \\&{\mathsf {A}}^{1,\ell /2}_{1}=\frac{1}{\Gamma \left( 3-\frac{\ell }{2}\right) }, \qquad {\mathsf {y}}'(1)=\frac{k(k+\alpha +\beta +1)}{2(\alpha +1)}. \end{aligned}$$

  (A.2)

  Thus

  $$\begin{aligned} {\mathbb {D}}_{k,1}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathbb {F}}^{1,\ell /2}_{1,k}(\alpha ,\beta )k(k+\alpha +\beta +1) =e^{1,\ell /2}_{1,1}(\alpha ,\beta ){\mathsf {F}}^{1,\ell /2}_{1,k}(\alpha ,\beta ) k(k+\alpha +\beta +1), \end{aligned}$$

  (A.3)

  where

  $$\begin{aligned} e^{1,\ell/2}_{1,1}(\alpha,\beta)=\frac{1}{2(\alpha+1)\Gamma \left(3-\frac{\ell}{2}\right)}. \end{aligned}$$

  (A.4)

  • $(\ell =1)$ Here we see that

    $$\begin{aligned} e^{1,1/2}_{1,1}(\alpha ,\beta )=\frac{2}{3 \sqrt{\pi } (\alpha +1)}. \end{aligned}$$

• (b) $m=2:$ Here we have

  $$\begin{aligned} {\mathbb {D}}_{k, 2}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathsf {A}}^{2,\ell /2}_{1}{\mathsf {F}}^{2,\ell /2}_{1,k}(\alpha ,\beta ){\mathsf {y}}'(1) +{\mathsf {A}}^{2,\ell /2}_{2}{\mathsf {F}}^{2,\ell /2}_{2,k}(\alpha ,\beta ){\mathsf {y}}''(1), \end{aligned}$$

  (A.5)

  where

  $$\begin{aligned} {\mathsf {F}}^{2,\ell /2}_{1,k}(\alpha ,\beta )&=\ _3F_2\left( 1-k,2-\frac{\ell }{2},\alpha +\beta +k+2;4-\frac{\ell }{2},\alpha +2;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{2,\ell /2}_{2,k}(\alpha ,\beta )&=\ _3F_2\left( 2-k,2-\frac{\ell }{2},\alpha +\beta +k+3;5-\frac{\ell }{2},\alpha +3;\frac{1}{2}\right) \nonumber \\ {\mathsf {A}}^{2,\ell /2}_{1}&=\frac{2 \Gamma \left( 3-\frac{\ell }{2}\right) }{\Gamma \left( 2-\frac{\ell }{2}\right) \Gamma \left( 4-\frac{\ell }{2}\right) }=\frac{4-\ell }{\Gamma \left( 4-\frac{\ell }{2}\right) }, \quad {\mathsf {A}}^{2,\ell /2}_{2}=\frac{2}{\Gamma \left( 5-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {y}}''(1)&=\frac{[k(k+\alpha +\beta +1)]^{2}}{4(\alpha +1)(\alpha +2)} -\frac{(\alpha +\beta +2)k(k+\alpha +\beta +1)}{4(\alpha +1)(\alpha +2)}. \end{aligned}$$

  (A.6)

  Simplifying further we see that

  $$\begin{aligned} {\mathbb {D}}_{k,2}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathbb {F}}_{1,k}^{2,\ell /2}(\alpha ,\beta )k(k+\alpha +\beta +1) +{\mathbb {F}}_{2,k}^{2,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{2}, \end{aligned}$$

  (A.7)

  where

  $$\begin{aligned} {\mathbb {F}}^{2,\ell /2}_{1,k}(\alpha ,\beta ) & =e^{2,\ell /2}_{1,1}(\alpha ,\beta ){\mathsf {F}}^{2,\ell /2}_{1,k}(\alpha ,\beta ) +e^{2,\ell /2}_{1,2}(\alpha ,\beta ){\mathsf {F}}^{2,\ell /2}_{2,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}^{2,\ell /2}_{2,k}(\alpha ,\beta ) & =e^{2,\ell /2}_{2,2}(\alpha ,\beta ){\mathsf {F}}^{2,\ell /2}_{2,k}(\alpha ,\beta ), \end{aligned}$$

  (A.8)

  with

  $$\begin{aligned} e^{2,\ell /2}_{1,1}(\alpha ,\beta ) & =\frac{ \left( 4-\ell \right) (8-\ell )}{4(\alpha +1)\Gamma \left( 5-\frac{ \ell }{2}\right) },\ \ e^{2,\ell /2}_{1,2}(\alpha ,\beta ) =-\frac{\alpha +\beta +2}{2(\alpha +1)(\alpha +2)\Gamma \left( 5-\frac{\ell }{2}\right) }\nonumber \\ e^{2,\ell /2}_{2,2}(\alpha ,\beta ) & =\frac{1}{2(\alpha +1) (\alpha +2)\Gamma \left( 5-\frac{ \ell }{2}\right) }. \end{aligned}$$

  (A.9)

  • $(\ell =1)$ In this case

    $$\begin{aligned} e^{2,1/2}_{1,1}(\alpha ,\beta ) & =\frac{4}{5 \sqrt{\pi } (\alpha +1)},\ \ e^{2,1/2}_{1,2}(\alpha ,\beta )=-\frac{8 (\alpha +\beta +2)}{105 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{2,1/2}_{2,2}(\alpha ,\beta ) & =\frac{8}{105 \sqrt{\pi } (\alpha +1) (\alpha +2)}. \end{aligned}$$

    (A.10)

  • $(\ell =2)$ We have

    $$\begin{aligned} e^{2,1}_{1,1}(\alpha ,\beta ) & =\frac{1}{2 (\alpha +1)},\ \ e^{2,1}_{1,2}(\alpha ,\beta )=-\frac{\alpha +\beta +2}{12 (\alpha +1) (\alpha +2)}\nonumber \\ e^{2,1}_{2,2}(\alpha ,\beta ) & =\frac{1}{12 (\alpha +1) (\alpha +2)}. \end{aligned}$$

    (A.11)

• (c) $m=3:$ It is seen in this case that

  $$\begin{aligned} {\mathbb {D}}_{k,3}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathsf {A}}^{3,\ell /2}_{1}{\mathsf {F}}^{3,\ell /2}_{1,k}(\alpha ,\beta ){\mathsf {y}}'(1) +{\mathsf {A}}^{3,\ell /2}_{2}{\mathsf {F}}^{3,\ell /2}_{2,k}(\alpha ,\beta ){\mathsf {y}}''(1) +{\mathsf {A}}^{3,\ell /2}_{3}{\mathsf {F}}^{3,\ell /2}_{3,k}(\alpha ,\beta ){\mathsf {y}}'''(1), \end{aligned}$$

  (A.12)

  where

  $$\begin{aligned} {\mathsf {A}}^{3,\ell /2}_{1}&=\frac{3 \Gamma \left( 4-\frac{\ell }{2}\right) }{\Gamma \left( 2-\frac{\ell }{2}\right) \Gamma \left( 5-\frac{\ell }{2}\right) }=\frac{3 \left( 6-\ell \right) (4-\ell )}{4\Gamma \left( 5-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {A}}^{3,\ell /2}_{2}&=\frac{6 \Gamma \left( 4-\frac{\ell }{2}\right) }{\Gamma \left( 3-\frac{\ell }{2}\right) \Gamma \left( 6-\frac{\ell }{2}\right) }=\frac{3 \left( 6-\ell \right) }{\Gamma \left( 6-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {A}}^{3,\ell /2}_{3}&=\frac{6}{\Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {F}}^{3,\ell /2}_{1,k}(\alpha ,\beta )&=\ _3F_2\left( 1-k,3-\frac{\ell }{2},\alpha +\beta +k+2;5-\frac{\ell }{2},\alpha +2;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{3,\ell /2}_{2,k}(\alpha ,\beta )&=\ _3F_2\left( 2-k,3-\frac{\ell }{2},\alpha +\beta +k+3;6-\frac{\ell }{2},\alpha +3;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{3,\ell /2}_{3,k}(\alpha ,\beta )&=\ _3F_2\left( 3-k,3-\frac{\ell }{2},\alpha +\beta +k+4;7-\frac{\ell }{2},\alpha +4;\frac{1}{2}\right) \end{aligned}$$

  (A.13)
  $$\begin{aligned} {\mathsf {y}}'''(1)&=\frac{[k(k+\alpha +\beta +1)]^{3}}{8(\alpha +1)(\alpha +2)(\alpha +3)}-\frac{(3\alpha +3\beta +8)[k(k+\alpha +\beta +1)]^{2}}{8(\alpha +1)(\alpha +2)(\alpha +3)}\nonumber \\&\quad +\frac{2(\alpha +\beta +3)(\alpha +\beta +2)k(k+\alpha +\beta +1)}{8(\alpha +1)(\alpha +2)(\alpha +3)}. \end{aligned}$$

  (A.14)

  Further simplification gives

  $$\begin{aligned} \mathbb {D}_{k,3}^{\ell /2}(\eta )\bigg |_{\eta =1}&={\mathbb {F}}_{1,k}^{3,\ell /2}(\alpha ,\beta )k(k+\alpha +\beta +1) +{\mathbb {F}}_{2,k}^{3,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{2}\nonumber \\&\quad +{\mathbb {F}}_{3,k}^{3,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{3}, \end{aligned}$$

  (A.15)

  where

  $$\begin{aligned} {\mathbb {F}}_{1,k}^{3,\ell /2}(\alpha ,\beta )&= e^{3,\ell /2}_{1,1}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{1,k}(\alpha ,\beta )+e^{3,\ell /2}_{1,2}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{2,k}(\alpha ,\beta )+e^{3,\ell /2}_{1,3}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{3,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}_{2,k}^{3,\ell /2}(\alpha ,\beta )&=e^{3,\ell /2}_{2,2}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{2,k}(\alpha ,\beta )+e^{3,\ell /2}_{2,3}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{3,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}_{3,k}^{3,\ell /2}(\alpha ,\beta )&=e^{3,\ell /2}_{3,3}(\alpha ,\beta ){\mathsf {F}}^{3,\ell /2}_{3,k}(\alpha ,\beta ), \end{aligned}$$

  (A.16)

  with the coefficients

  $$\begin{aligned} e^{3,\ell /2}_{1,1}(\alpha ,\beta )&= \frac{3 (4-\ell ) (6-\ell ) (10-\ell ) (12-\ell )}{32 (\alpha +1) \Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ e^{3,\ell /2}_{1,2}(\alpha ,\beta )&=-\frac{3 (6-\ell ) (12-\ell ) (\alpha +\beta +2)}{8 (\alpha +1) (\alpha +2) \Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ e^{3,\ell /2}_{1,3}(\alpha ,\beta )&=\frac{3 (\alpha +\beta +2) (\alpha +\beta +3)}{2 (\alpha +1) (\alpha +2) (\alpha +3) \Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ e^{3,\ell /2}_{2,2}(\alpha ,\beta )&=\frac{3 (6-\ell ) (12-\ell )}{8 (\alpha +1) (\alpha +2) \Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ e^{3,\ell /2}_{2,3}(\alpha ,\beta )&=-\frac{3 (3 \alpha +3 \beta +8)}{4 (\alpha +1) (\alpha +2) (\alpha +3) \Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ e^{3,\ell /2}_{3,3}(\alpha ,\beta )&=\frac{3}{4(\alpha +1)(\alpha +2)(\alpha +3)\Gamma \left( 7-\frac{ \ell }{2}\right) }. \end{aligned}$$

  (A.17)

  • $(\ell =1)$ Indeed

    $$\begin{aligned} e^{3,1/2}_{1,1}(\alpha ,\beta )&=\frac{6}{7 \sqrt{\pi } (\alpha +1)},\ e^{3,1/2}_{1,2}(\alpha ,\beta )=-\frac{8 (\alpha +\beta +2)}{63 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,1/2}_{1,3}(\alpha ,\beta )&=\frac{32 (\alpha +\beta +2) (\alpha +\beta +3)}{3465 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)},\ \ e^{3,1/2}_{2,2}(\alpha ,\beta )=\frac{8}{63 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,1/2}_{2,3}(\alpha ,\beta )&=-\frac{16 (3 \alpha +3 \beta +8)}{3465 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{3,1/2}_{3,3}(\alpha ,\beta )&=\frac{16}{3465 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}. \end{aligned}$$

    (A.18)

  • $(\ell =2)$ It is clear in this case that

    $$\begin{aligned} e^{3,1}_{1,1}(\alpha ,\beta )&= \frac{1}{2 (\alpha +1)},\ \ e^{3,1}_{1,2}(\alpha ,\beta )=-\frac{\alpha +\beta +2}{8 (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,1}_{1,3}(\alpha ,\beta )&=\frac{(\alpha +\beta +2) (\alpha +\beta +3)}{80 (\alpha +1) (\alpha +2) (\alpha +3)},\ \ e^{3,1}_{2,2}(\alpha ,\beta )=\frac{1}{8 (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,1}_{2,3}(\alpha ,\beta )&=-\frac{3 \alpha +3 \beta +8}{160 (\alpha +1) (\alpha +2) (\alpha +3)},\ \ e^{3,1}_{3,3}(\alpha ,\beta )=\frac{1}{160 (\alpha +1) (\alpha +2) (\alpha +3)}. \end{aligned}$$

    (A.19)

  • $(\ell =3)$ Here we have

    $$\begin{aligned} e^{3,3/2}_{1,1}(\alpha ,\beta )&=\frac{3}{5 \sqrt{\pi } (\alpha +1)},\ \ e^{3,3/2}_{1,2}(\alpha ,\beta )=-\frac{12 (\alpha +\beta +2)}{35 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,3/2}_{1,3}(\alpha ,\beta )&=\frac{16 (\alpha +\beta +2) (\alpha +\beta +3)}{315 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)},\ \ e^{3,3/2}_{2,2}(\alpha ,\beta )=\frac{12}{35 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{3,3/2}_{2,3}(\alpha ,\beta )&=-\frac{8 (3 \alpha +3 \beta +8)}{315 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{3,3/2}_{3,3}(\alpha ,\beta )&=\frac{8}{315 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}. \end{aligned}$$

    (A.20)

• (d) $m=4:$ Indeed

  $$\begin{aligned} {\mathbb {D}}_{k,4}^{\ell /2}(\eta )\bigg |_{\eta =1}&={\mathsf {A}}^{4,\ell /2}_{1}{\mathsf {F}}^{4,\ell /2}_{1,k}(\alpha ,\beta ){\mathsf {y}}'(1)+{\mathsf {A}}^{4,\ell /2}_{2}{\mathsf {F}}^{4,\ell /2}_{2,k}(\alpha ,\beta ){\mathsf {y}}''(1)\nonumber \\&\quad +{\mathsf {A}}^{4,\ell /2}_{3}{\mathsf {F}}^{4,\ell /2}_{3,k}(\alpha ,\beta ){\mathsf {y}}'''(1)+{\mathsf {A}}^{4,\ell /2}_{4}{\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta ){\mathsf {y}}^{(4)}(1), \end{aligned}$$

  (A.21)

  where

  $$\begin{aligned} {\mathsf {A}}^{4,\ell /2}_{1}= & {} \frac{4 \Gamma \left( 5-\frac{\ell }{2}\right) }{\Gamma \left( 2-\frac{\ell }{2}\right) \Gamma \left( 6-\frac{\ell }{2}\right) }=\frac{(8-\ell ) (6-\ell ) (4-\ell )}{2 \Gamma \left( 6-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {A}}^{4,\ell /2}_{2}= & {} \frac{12 \Gamma \left( 5-\frac{\ell }{2}\right) }{\Gamma \left( 3-\frac{\ell }{2}\right) \Gamma \left( 7-\frac{\ell }{2}\right) }=\frac{3 (8-\ell ) (6-\ell )}{\Gamma \left( 7-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {A}}^{4,\ell /2}_{3}= & {} \frac{24 \Gamma \left( 5-\frac{\ell }{2}\right) }{\Gamma \left( 4-\frac{\ell }{2}\right) \Gamma \left( 8-\frac{\ell }{2}\right) }=\frac{12 (8-\ell )}{\Gamma \left( 8-\frac{\ell }{2}\right) },\ \ {\mathsf {A}}^{4,\ell /2}_{4}=\frac{24}{\Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ {\mathsf {F}}^{4,\ell /2}_{1,k}(\alpha ,\beta )= & {} \ _3F_2\left( 1-k,4-\frac{\ell }{2},\alpha +\beta +k+2;6-\frac{\ell }{2},\alpha +2;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{4,\ell /2}_{2,k}(\alpha ,\beta )= & {} \ _3F_2\left( 2-k,4-\frac{\ell }{2},\alpha +\beta +k+3;7-\frac{\ell }{2},\alpha +3;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{4,\ell /2}_{3,k}(\alpha ,\beta )= & {} \ _3F_2\left( 3-k,4-\frac{\ell }{2},\alpha +\beta +k+4;8-\frac{\ell }{2},\alpha +4;\frac{1}{2}\right) \nonumber \\ {\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta )= & {} \ _3F_2\left( 4-k,4-\frac{\ell }{2},\alpha +\beta +k+5;9-\frac{\ell }{2},\alpha +5;\frac{1}{2}\right) \end{aligned}$$

  (A.22)

  and

  $$\begin{aligned} {\mathsf {y}}^{(4)}(1)= & {} \frac{[k(k+\alpha +\beta +1)]^{4} - 2(3\alpha +3\beta +10) [k(k+\alpha +\beta +1)]^{3}}{16(\alpha +1)(\alpha +2)(\alpha +3)(\alpha +4)}\nonumber \\&\quad +\frac{\left( 11\alpha ^{2}+11\beta ^{2}+22\alpha \beta +70\alpha +70\beta +108\right) [k(k+\alpha +\beta +1)]^{2}}{16(\alpha +1)(\alpha +2)(\alpha +3)(\alpha +4)}\nonumber \\&\quad -\frac{3(\alpha +\beta +2)(\alpha +\beta +3)(\alpha +\beta +4)k(k+\alpha +\beta +1)}{8(\alpha +1)(\alpha +2)(\alpha +3)(\alpha +4)}. \end{aligned}$$

  (A.23)

  Simplifying further we see that

  $$\begin{aligned} {\mathbb {D}}_{k,4}^{\ell /2}(\eta )\bigg |_{\eta =1} & ={\mathbb {F}}_{1,k}^{4,\ell /2}(\alpha ,\beta )k(k+\alpha +\beta +1)+{\mathbb {F}}_{2,k}^{4,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{2}\nonumber \\&+{\mathbb {F}}_{3,k}^{4,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{3}+{\mathbb {F}}_{4,k}^{4,\ell /2}(\alpha ,\beta )[k(k+\alpha +\beta +1)]^{4}, \end{aligned}$$

  (A.24)

  where

  $$\begin{aligned} {\mathbb {F}}_{1,k}^{4,\ell /2}(\alpha ,\beta )&=e^{4,\ell /2}_{1,1}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{1,k}(\alpha ,\beta )+e^{4,\ell /2}_{1,2}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{2,k}(\alpha ,\beta )\nonumber \\&\quad +e^{4,\ell /2}_{1,3}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{3,k}(\alpha ,\beta )+e^{4,\ell /2}_{1,4}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}_{2}^{4,\ell /2}(\alpha ,\beta )&=e^{4,\ell /2}_{2,2}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{2,k}(\alpha ,\beta )+e^{4,\ell /2}_{2,3}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{3,k}(\alpha ,\beta )+e^{4,\ell /2}_{2,4}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}_{3,k}^{4,\ell /2}(\alpha ,\beta )&=e^{4,\ell /2}_{3,3}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{3,k}(\alpha ,\beta )+e^{4,\ell /2}_{3,4}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta )\nonumber \\ {\mathbb {F}}_{4,k}^{4,\ell /2}(\alpha ,\beta )&=e^{4,\ell /2}_{4,4}(\alpha ,\beta ){\mathsf {F}}^{4,\ell /2}_{4,k}(\alpha ,\beta ) \end{aligned}$$

  (A.25)

  with the fractional coefficients

  $$\begin{aligned} e^{4,\ell /2}_{1,1}(\alpha ,\beta )&=\frac{(4-\ell ) (6-\ell ) (8-\ell ) (12-\ell ) (14-\ell ) (16-\ell ) }{32 (\alpha +1) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{1,2}(\alpha ,\beta )&=-\frac{3 (6-\ell ) (8-\ell ) (14-\ell ) (16-\ell )(\alpha +\beta +2)}{16 (\alpha +1) (\alpha +2) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{1,3}(\alpha ,\beta )&=\frac{3 (8-\ell ) (16-\ell ) (\alpha +\beta +2) (\alpha +\beta +3)}{2 (\alpha +1) (\alpha +2) (\alpha +3) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{1,4}(\alpha ,\beta )&=-\frac{9 (\alpha +\beta +2) (\alpha +\beta +3) (\alpha +\beta +4)}{(\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{2,2}(\alpha ,\beta )&=\frac{3 (6-\ell ) (8-\ell ) (14-\ell ) (16-\ell ) }{16 (\alpha +1) (\alpha +2) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{2,3}(\alpha ,\beta )&=-\frac{3 (8-\ell ) (16-\ell ) (3 \alpha +3 \beta +8)}{4 (\alpha +1) (\alpha +2) (\alpha +3) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{2,4}(\alpha ,\beta )&=\frac{3 \left( 11 \left( \alpha ^2+\beta ^2\right) +70 (\alpha +\beta )+22 \alpha \beta +108\right) }{2 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{3,3}(\alpha ,\beta )&=\frac{3 (8-\ell ) (16-\ell ) }{4 (\alpha +1) (\alpha +2) (\alpha +3) \Gamma \left( 9-\frac{\ell }{2}\right) } \end{aligned}$$

  (A.26)
  $$\begin{aligned} e^{4,\ell /2}_{3,4}(\alpha ,\beta )&=-\frac{3 (3 \alpha +3 \beta +10)}{(\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4) \Gamma \left( 9-\frac{\ell }{2}\right) }\nonumber \\ e^{4,\ell /2}_{4,4}(\alpha ,\beta )&=\frac{3 }{2 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4) \Gamma \left( 9-\frac{\ell }{2}\right) }. \end{aligned}$$

  (A.27)

  • $(\ell =1)$ Clearly we have

    $$\begin{aligned} e^{4,1/2}_{1,1}(\alpha ,\beta )&=\frac{8}{9 \sqrt{\pi } (\alpha +1)},\ \ e^{4,1/2}_{1,2}(\alpha ,\beta )=-\frac{16 (\alpha +\beta +2)}{99 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{4,1/2}_{1,3}(\alpha ,\beta )&=\frac{128 (\alpha +\beta +2) (\alpha +\beta +3)}{6435 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1/2}_{1,4}(\alpha ,\beta )&=-\frac{256 (\alpha +\beta +2) (\alpha +\beta +3) (\alpha +\beta +4)}{225225 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)} \end{aligned}$$

    (A.28)
    $$\begin{aligned} e^{4,1/2}_{2,2}(\alpha ,\beta )&=\frac{16}{99 \sqrt{\pi } (\alpha +1) (\alpha +2)},\ \ e^{4,1/2}_{2,3}(\alpha ,\beta )=-\frac{64 (3 \alpha +3 \beta +8)}{6435 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1/2}_{2,4}(\alpha ,\beta )&=\frac{128 \left( 11 \left( \alpha ^2+\beta ^2\right) +70 (\alpha +\beta )+22 \alpha \beta +108\right) }{675675 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,1/2}_{3,3}(\alpha ,\beta )&=\frac{64}{6435 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1/2}_{3,4}(\alpha ,\beta )&=-\frac{256 (3 \alpha +3 \beta +10)}{675675 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,1/2}_{4,4}(\alpha ,\beta )&=\frac{128}{675675 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}. \end{aligned}$$

    (A.29)

  • $(\ell =2)$ Here we see that

    $$\begin{aligned} e^{4,1}_{1,1}(\alpha ,\beta )&=\frac{1}{2 (\alpha +1)},\ \ e^{4,1}_{1,2}(\alpha ,\beta )=-\frac{3 (\alpha +\beta +2)}{20 (\alpha +1) (\alpha +2)}\nonumber \\ e^{4,1}_{1,3}(\alpha ,\beta )&=\frac{(\alpha +\beta +2) (\alpha +\beta +3)}{40 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1}_{1,4}(\alpha ,\beta )&=-\frac{(\alpha +\beta +2) (\alpha +\beta +3) (\alpha +\beta +4)}{560 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)} \end{aligned}$$

    (A.30)
    $$\begin{aligned} e^{4,1}_{2,2}(\alpha ,\beta )&=\frac{3}{20 (\alpha +1) (\alpha +2)},\ e^{4,1}_{2,3}(\alpha ,\beta )=-\frac{3 \alpha +3 \beta +8}{80 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1}_{2,4}(\alpha ,\beta )&=\frac{11 \left( \alpha ^2+\beta ^2\right) +70 (\alpha +\beta )+22 \alpha \beta +108}{3360 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,1}_{3,3}(\alpha ,\beta )&=\frac{1}{80 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,1}_{3,4}(\alpha ,\beta )&=-\frac{3 \alpha +3 \beta +10}{1680 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,1}_{4,4}(\alpha ,\beta )&=\frac{1}{3360 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}. \end{aligned}$$

    (A.31)

  • $(\ell =3)$ In this case we have

    $$\begin{aligned} e^{4,3/2}_{1,1}(\alpha ,\beta )&=\frac{4}{7 \sqrt{\pi } (\alpha +1)},\ e^{4,3/2}_{1,2}(\alpha ,\beta )=-\frac{8 (\alpha +\beta +2)}{21 \sqrt{\pi } (\alpha +1) (\alpha +2)}\nonumber \\ e^{4,3/2}_{1,3}(\alpha ,\beta )&=\frac{64 (\alpha +\beta +2) (\alpha +\beta +3)}{693 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,3/2}_{1,4}(\alpha ,\beta )&=-\frac{128 (\alpha +\beta +2) (\alpha +\beta +3) (\alpha +\beta +4)}{15015 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)} \end{aligned}$$

    (A.32)
    $$\begin{aligned} e^{4,3/2}_{2,2}(\alpha ,\beta )&=\frac{8}{21 \sqrt{\pi } (\alpha +1) (\alpha +2)},\ \ e^{4,3/2}_{2,3}(\alpha ,\beta )=-\frac{32 (3 \alpha +3 \beta +8)}{693 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,3/2}_{2,4}(\alpha ,\beta )&=\frac{64 \left( 11 \left( \alpha ^2+\beta ^2\right) +70 (\alpha +\beta )+22 \alpha \beta +108\right) }{45045 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,3/2}_{3,3}(\alpha ,\beta )&=\frac{32}{693 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,3/2}_{3,4}(\alpha ,\beta )&=-\frac{128 (3 \alpha +3 \beta +10)}{45045 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,3/2}_{4,4}(\alpha ,\beta )&=\frac{64}{45045 \sqrt{\pi } (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}. \end{aligned}$$

    (A.33)

  • $(\ell =4)$ Here we see that

    $$\begin{aligned} e^{4,2}_{1,1}(\alpha ,\beta )&=0,\ \ e^{4,2}_{1,2}(\alpha ,\beta )=-\frac{\alpha +\beta +2}{4 (\alpha +1) (\alpha +2)}\nonumber \\ e^{4,2}_{1,3}(\alpha ,\beta )&=\frac{(\alpha +\beta +2) (\alpha +\beta +3)}{10 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,2}_{1,4}(\alpha ,\beta )&=-\frac{(\alpha +\beta +2) (\alpha +\beta +3) (\alpha +\beta +4)}{80 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)} \end{aligned}$$

    (A.34)
    $$\begin{aligned} e^{4,2}_{2,2}(\alpha ,\beta )&=\frac{1}{4 (\alpha +1) (\alpha +2)},\ \ e^{4,2}_{2,3}(\alpha ,\beta )=-\frac{3 \alpha +3 \beta +8}{20 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,2}_{2,4}(\alpha ,\beta )&=\frac{11 \left( \alpha ^2+\beta ^2\right) +70 (\alpha +\beta )+22 \alpha \beta +108}{480 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,2}_{3,3}(\alpha ,\beta )&=\frac{1}{20 (\alpha +1) (\alpha +2) (\alpha +3)}\nonumber \\ e^{4,2}_{3,4}(\alpha ,\beta )&=-\frac{3 \alpha +3 \beta +10}{240 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}\nonumber \\ e^{4,2}_{4,4}(\alpha ,\beta )&=\frac{1}{480 (\alpha +1) (\alpha +2) (\alpha +3) (\alpha +4)}. \end{aligned}$$

    (A.35)

The fractional coefficients for other cases $0<\mu <1, 1\le j\le \ell \le m$ can be computed similarly.

Appendix B: The multiplicity $M_{k}({\mathscr {X}})$ and associated coefficients

This section presents polynomial representations of the multiplicity $M_{k}({\mathscr {X}})$ and their corresponding coefficients.

The Sphere ${\mathbb {S}}^{n}$ The polynomial transformation of the multiplicity $M_{k}({\mathbb {S}}^{n})$ according to whether n is odd or even is obtained as follows.

• (a) Odd $n\ge 3$. We have³

  $$\begin{aligned} M_{k}({\mathbb {S}}^{n})&=\frac{(k+n-2)!}{(n-1)!k!}(n+2k-1)=\frac{2k+n-1}{(n-1)!}\prod _{j=1}^{n-2} (k+j)\nonumber \\&= \frac{2}{(n-1)!} \prod \limits _{j=0}^{\frac{n-3}{2}} \left[ \left( k+\frac{n-1}{2} \right) ^{2}-j^{2} \right] \nonumber \\&= \frac{2}{(n-1)!} \sum \limits _{m=0}^{\frac{n-3}{2}}{} \texttt {E}_{m,n}\left( k+\frac{n-1}{2} \right) ^{2m+2} \end{aligned}$$

  (B.1)

  with the first coefficients $\left( \texttt {E}_{m,n}:0\le m\le (n-3)/2\right)$ given in Table 2.

  Table 2 The coefficients $\left( \texttt {E}_{m,n}:0\le m\le (n-3)/2\right)$

• (b) Even $n\ge 2$. Indeed we have⁴

  $$\begin{aligned} M_{k}({\mathbb {S}}^{n})&=\frac{(k+n-2)!}{(n-1)!k!}(n+2k-1)=\frac{2k+n-1}{n-1}\prod _{j=1}^{n-2}\frac{k+j}{j} \nonumber \\&=\frac{2\left( k+\frac{n-1}{2}\right) }{(n-1)!}\prod \limits _{j=1/2}^{\frac{n-3}{2}}\left[ \left( k+\frac{n-1}{2} \right) ^{2}-j^{2} \right] \nonumber \\&=\frac{2}{(n-1)!}\sum \limits _{m=0}^{\frac{n-2}{2}}{} \texttt {F}_{m,n}\left( k+\frac{n-1}{2} \right) ^{2m+1}. \end{aligned}$$

  (B.2)

  Here again the first coefficients $(\texttt {F}_{m,n}:0\le m\le (n-2)/2)$ are given in Table 3.

  Table 3 The coefficients $(\texttt {F}_{m,n}:0\le m\le (n-2)/2)$

The Complex Projective Space ${\mathbf {P}}^{n}({\mathbb {C}})$. The following polynomial representations of the multiplicity $M_{k}\left( {\mathbf {P}}^{n}({\mathbb {C}})\right)$ hold.

• (a) Odd $n\ge 1$. Note that ${\mathbf {P}}^1({\mathbb {C}})$ is the sphere ${\mathbb {S}}^{2}$. We see that

  $$\begin{aligned} M_{k}\left( {\mathbf {P}}^{n}({\mathbb {C}})\right)&=\frac{2k+n}{n}\left[ \frac{\Gamma (k+n)}{k!\Gamma (n)}\right] ^{2}=\frac{2k+n}{n}\prod \limits _{j=1}^{n-1}\left( \frac{k+j}{j}\right) ^{2}\nonumber \\&=\frac{2\left( k+\frac{n}{2}\right) }{n!(n-1)!}\prod \limits _{j=1/2}^{\frac{n-2}{2}}\left[ \left( k+\frac{n}{2} \right) ^{2}-j^{2} \right] ^{2}\nonumber \\&=\frac{2}{n!(n-1)!}\sum \limits _{m=0}^{n-1}{} \texttt {G}_{m,n}\left( k+\frac{n}{2} \right) ^{2m+1}, \end{aligned}$$

  (B.3)

  where the first coefficients $\left( \texttt {G}_{m,n}:0\le m\le n-1\right)$ are illustrated in Table 4.

  Table 4 The coefficients $\left( \texttt {G}_{m,n}:0\le m\le n-1\right)$

• (b) Even $n\ge 2$. It is seen here that

  $$\begin{aligned} M_{k}\left( {\mathbf {P}}^{n}({\mathbb {C}})\right)&=\frac{2k+n}{n}\left[ \frac{\Gamma (k+n)}{k!\Gamma (n)}\right] ^{2}=\frac{2k+n}{n}\prod \limits _{j=1}^{n-1}\left( \frac{k+j}{j}\right) ^{2}\nonumber \\&=\frac{2\left( k+\frac{n}{2}\right) ^{3}}{n!(n-1)!}\prod \limits _{j=1}^{\frac{n-2}{2}}\left[ \left( k+\frac{n}{2} \right) ^{2}-j^{2} \right] ^{2}\ \ (\text{ product } \text{ omitted } \text{ for } n=2)\nonumber \\&=\frac{2}{n!(n-1)!}\sum \limits _{m=0}^{n-2}{} \texttt {H}_{m,n}\left( k+\frac{n}{2} \right) ^{2m+3}, \end{aligned}$$

  (B.4)

  where the first coefficients $\left( \texttt {H}_{m,n}:0\le m\le n-2\right)$ are illustrated in Table 5.

  Table 5 The coefficients $\left( \texttt {H}_{m,n}: 0\le m\le n-2\right)$

The Quaternionic Projective Space ${\mathbf {P}}^{n}({\mathbb {H}})$. It is clear that

$$\begin{aligned} M_{k}\left( {\mathbf {P}}^{n}({\mathbb {H}})\right)&= \frac{(2k+2n+1)\Gamma (k+2n)\Gamma (k+2n+1)}{\Gamma (2n+2)\Gamma (2n)\Gamma (k+1)\Gamma (k+2)}\nonumber \\&=\frac{(k+2n)(2k+2n+1)}{2n(2n+1)(k+1)}\prod \limits _{j=1}^{2n-1}\left( \frac{k+j}{j}\right) ^{2}\nonumber \\&=\frac{2\left( k+\frac{2n+1}{2}\right) \left[ \left( k+\frac{2n+1}{2}\right) ^{2}-\left( \frac{2n-1}{2}\right) ^{2} \right] }{(2n+1)!(2n-1)!}\prod \limits _{j=1/2}^{\frac{2n-3}{2}}\left[ \left( k+\frac{2n+1}{2} \right) ^{2}-j^{2} \right] ^{2}\nonumber \\&=\frac{2}{(2n+1)!(2n-1)!}\sum \limits _{m=0}^{2n-1}{} \texttt {I}_{m,n}\left( k+\frac{2n+1}{2} \right) ^{2m+1}. \end{aligned}$$

(B.5)

The first coefficients $\left( \texttt {I}_{m,n}:0\le m\le 2n-1\right)$ are illustrated in Table 6.

Table 6 The coefficients $\left( \texttt {I}_{m,n}:0\le m\le 2n-1\right)$

The Cayley Projective Plane ${\mathbf {P}}^{2}(\mathrm {Cay})$. It is seen in this case that

$$\begin{aligned} M_{k}\left( {\mathbf {P}}^{2}(\mathrm {Cay})\right)&= 6(2k+11)\frac{\Gamma (k+8)\Gamma (k+11)}{7!11!k!\Gamma (k+4)}\nonumber \\&=\frac{(2k+11)}{11}\prod _{j=1}^{3}\frac{k+j}{j}\prod \limits _{l=4}^{7}\left( \frac{k+l}{l}\right) ^{2}\prod _{p=8}^{10}\frac{k+p}{p}\nonumber \\&=\frac{ 2k+11}{840\cdot 11!}\prod \limits _{j=1/2}^{3/2}\left[ \left( k+\frac{11}{2}\right) ^{2}-j^{2} \right] ^{2} \prod \limits _{j=5/2}^{9/2}\left[ \left( k+\frac{11}{2}\right) ^{2}-j^{2} \right] \nonumber \\&=\frac{ 1}{420\cdot 11!}\sum \limits _{m=0}^{7}{} \texttt {J}_{m,2}\left( k+\frac{11}{2} \right) ^{2m+1}. \end{aligned}$$

(B.6)

The coefficients $\left( \texttt {J}_{m,2}:0\le m\le 7\right)$ are illustrated in Table 7.

Table 7 The coefficients $\left( \texttt {J}_{m,2}:0\le m\le 7\right)$