In this discussion a program for computing first terms of the series expansion for \sigma-function associated with a genus two curve is presented. The series is computed for the curve in its canonical form. \sigma-Functions associated with curves in other forms are obtained by a proper transformation of parameters of the curve, and coordinates of the Jacobian varieties.
Canonical curve
Let the canonical form of genus two curves be defined by the equation
\mathcal{C}_{(2,5)}:\quad f(x,y;\lambda) \equiv - y^2 + x^5 + \lambda_4 x^3 + \lambda_6 x^2 + \lambda_8 x + \lambda_{10} = 0.
\tag{1}
The canonical form is the (2,5)-curve defined in 1, which
is the Weierstass canonical form described in 2, \S\S 60–63. Let
\lambda \equiv (\lambda_4, \lambda_6, \lambda_8, \lambda_{10} ) be parameters of the canonical curve (1).
Let first and second kind differentials be defined by
\begin{aligned}
&\mathrm{d} u= \begin{pmatrix} \mathrm{d} u_1 \\
\mathrm{d} u_3 \end{pmatrix} =
\begin{pmatrix} x \\ 1 \end{pmatrix}
\frac{\mathrm{d}x}{\partial_y f(x,y;\lambda)},\\
&\mathrm{d} r = \begin{pmatrix} \mathrm{d} r_1 \\
\mathrm{d} r_3 \end{pmatrix} =
\begin{pmatrix} x^2 \\ 3 x^3 + \lambda_4 x \end{pmatrix}
\frac{\mathrm{d}x}{\partial_y f(x,y;\lambda)}.
\end{aligned}
The \sigma-function is obtained as a solution of the system of heat equations
Q_k \sigma(u_1,u_3;\lambda) = 0
with the initial condition given by the Weierstrass–Schur polynomial u_3 - \frac{1}{3} u_1^3.
Operators Q_k, which annihilate the \sigma-function, have the following form, see arXiv:1509.01490, Sect. 4 p.6, and also arXiv:1711.08395, Sect 6.4,
\begin{aligned}
&Q_0 = - u_1 \partial_{u_1} − 3 u_3 \partial_{u_3} + 3 + L_0,\\
&Q_2 = − \tfrac{1}{2} \partial_{u_1,u_1} + \tfrac{4}{5} \lambda_4 u_3 \partial_{u_1}
− u_1 \partial_{u_3} + \tfrac{3}{10} \lambda_4 u_1^2
− \tfrac{1}{10}(15 \lambda_8 − 4 \lambda_4^2) u_3^2 + L_2,\\
&Q_4 = − \partial_{u_1,u_3} + \tfrac{6}{5} \lambda_6 u_3 \partial_{u_1}
− \lambda_4 u_3 \partial_{u_3} + \tfrac{1}{5} \lambda_6 u_1^2 − \lambda_8 u_1 u_3
− \tfrac{1}{10} (30 \lambda_{10} − 6 \lambda_6 \lambda_4) u_3^2 + \lambda_4 + L_4,\\
&Q_6 = − \frac{1}{2} \partial_{u_3,u_3} + \tfrac{3}{5} \lambda_8 u_3 \partial_{u_1}
+ \tfrac{1}{10} \lambda_8 u_1^2 − 2 \lambda_{10} u_1 u_3
+ \tfrac{3}{10} \lambda_8 \lambda_4 u_3^2
+ \tfrac{1}{2} \lambda_6 + L_6,
\end{aligned}
where the linear differential operators with respect to parameters \lambda are obtained by
\begin{pmatrix} L_0 \\ L_2 \\ L_4 \\ L_6
\end{pmatrix} =
\begin{pmatrix}
4 \lambda_4 & 6 \lambda_6 & 8 \lambda_8 & 10 \lambda_{10} \\
6 \lambda_6 & 8 \lambda_8 − \tfrac{12}{5} \lambda_4^2 &
10 \lambda_{10} − \tfrac{8}{5} \lambda_6 \lambda_4 & − \tfrac{4}{5} \lambda_8 \lambda_4 \\
8 \lambda_8 &10 \lambda_{10} − \tfrac{8}{5} \lambda_6 \lambda_4 &
4\lambda_8 \lambda_4 − \tfrac{12}{5} \lambda_6^2 &
6 \lambda_{10} \lambda_4 − \tfrac{6}{5} \lambda_8 \lambda_6 \\
10 \lambda_{10} & − \tfrac{4}{5} \lambda_8 \lambda_4 &
6 \lambda_{10} \lambda_4 − \tfrac{6}{5} \lambda_8 \lambda_6 &
4\lambda_{10} \lambda_6 − \tfrac{8}{5} \lambda_8^2
\end{pmatrix}
\begin{pmatrix}
\partial_{\lambda_4} \\ \partial_{\lambda_6} \\ \partial_{\lambda_8} \\ \partial_{\lambda_{10}}
\end{pmatrix}.
The system is solved by means of representing the unknown \sigma-function as a series
with terms formed according to the Sato weight, see arXiv:1711.08395, Sect 6.4. Recurrence relations and the corresponding series expansion are implemented below.
As an example, \sigma-function is computed with different accuracy: up to the weights -13, and -25 with respect to coordinates u_1, u_3 of the Jacobian variety of \mathcal{C}_{(2,5)}. Fundamental cubic relations and the identities for 4-index \wp-functions are used for verification.
A series representation is computed only for the \sigma-function associated with the canonical curve \mathcal{C}_{(2,5)}. This is enough to obtain the \sigma-function associated with any curve of genus two. A transformation of the \sigma-function along with the corresponding transformation between curves is illustrated below by two examples: the case of a more general (2,5)-curve with term \mu_2 x^4, and the case of a (2,6)-curve.
(2,5)-Curve with additional term \mu_2 x^4
Let a genus two curve be defined by the equation
\tilde{\mathcal{C}}_{(2,5)}:\quad
\tilde{f}(\tilde{x},\tilde{y};\mu) \equiv - \tilde{y}^2 + \tilde{x}^5 + \mu_2 \tilde{x}^4+ \mu_4 \tilde{x}^3 + \mu_6 \tilde{x}^2 + \mu_8 \tilde{x} + \mu_{10} = 0.
\tag{2}
The curve (1) is obtained from (2) by the transformation \tilde{x} = x - \tfrac{1}{5} \mu_2, \tilde{y} = y. The backward transformation
x = \tilde{x} + \tfrac{1}{5} \mu_2, \qquad y = \tilde{y}. \tag{3}
along with the transformation of parameters \lambda into \mu
\begin{aligned}
&\lambda_{10} = \tfrac{4}{5^5}\mu _2^5 - \tfrac{1}{5^3} \mu _4 \mu_2^3
+\tfrac{1}{5^2} \mu_6 \mu_2^2-\tfrac{1}{5} \mu_8 \mu_2+\mu_{10}\\
&\lambda_8 = -\tfrac{3}{5^3}\mu_2^4 + \tfrac{3}{5^2} \mu_4 \mu_2^2 - \tfrac{2}{5}\mu_6 \mu_2 + \mu_8\\
&\lambda_6 = \tfrac{4}{5^2} \mu_2^3 -\tfrac{3}{5}\mu_4 \mu_2 + \mu_6\\
&\lambda_4 = \mu_4-\tfrac{2}{5} \mu_2^2,
\end{aligned} \tag{4}.
induces the transformation of first and second kind differentials
\begin{pmatrix} \mathrm{d} u \\ \mathrm{d} r \end{pmatrix} = \begin{pmatrix} A & B \\ C & D
\end{pmatrix} \begin{pmatrix} \mathrm{d} \tilde{u} \\ \mathrm{d} \tilde{r} \end{pmatrix}.
\tag{5}
where
\begin{aligned}
&A = \begin{pmatrix} 1 & \tfrac{1}{5} \mu_2 \\ 0 & 1 \end{pmatrix},& &B = 0,& \\
&C = \begin{pmatrix} \tfrac{2}{5} \mu_2 & \tfrac{1}{25} \mu_2^2 \\
- \tfrac{1}{25} \mu_2^2 & \tfrac{1}{5} \mu_2 (\mu_4 - \tfrac{7}{25} \mu_2^2) \end{pmatrix},&
&D = \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{5} \mu_2 & 1 \end{pmatrix}.&
\end{aligned}
Then the \sigma-function associated with \tilde{\mathcal{C}}_{(2,5)} is computed by
\tilde{\sigma} (\tilde{u}; \mu) = \exp \Big( \tfrac{1}{2} \tilde{u}^t T \tilde{u} \Big)
\sigma \big(A \tilde{u}; \lambda(\mu)\big),
where T is a symmetric matrix,
T = D^{-1} C =
\begin{pmatrix} \tfrac{2}{5} \mu_2 & \tfrac{1}{25} \mu_2^2 \\
\tfrac{1}{25} \mu_2^2 & \tfrac{1}{5} \mu_2 (\mu_4 - \tfrac{6}{25} \mu_2^2) \end{pmatrix},
and \lambda(\mu) is given by (4). This series coincides with the one given in
arXiv:1711.08395, p.48, and also in 3, p.89.
(2,6)-Curve
Now, we consider a genus two curve defined by the equation
\begin{split}
\bar{\mathcal{C}}_{(2,6)}:\quad
\bar{f}(\bar{x},\bar{y};\mu) \equiv & - \bar{y}^2 + P(\bar{x}) \\
&- \bar{y}^2 + \nu_0 \bar{x}^6 + \nu_1 \bar{x}^5
+ \nu_2 \bar{x}^4 + \nu_3 \bar{x}^3 + \nu_4 \bar{x}^2 + \nu_5 \bar{x} + \nu_6 = 0.
\end{split}
\tag{6}
Let e_0, such that P(e_0)=0, maps to infinity on \tilde{\mathcal{C}}_{(2,5)}.
Then the transformation from \bar{\mathcal{C}}_{(2,6)} to \tilde{\mathcal{C}}_{(2,5)} has the form
\bar{x} = e_0 + \frac{P'(e_0)}{\tilde{x}},\qquad \bar{y} = \frac{P'(e_0) \tilde{y}}{\tilde{x}^3},
The backward transformation from \tilde{\mathcal{C}}_{(2,5)} to \bar{\mathcal{C}}_{(2,6)}
\tilde{x} = \frac{P'(e_0)}{\bar{x}-e_0},\quad \tilde{y}=\frac{P'(e_0)^2 \bar{y}}{(\bar{x}-e_0)^3},
\qquad
\mu_{2k} = \frac{P'(e_0)^{k-1} P^{(k+1)}(e_0)}{(k+1)!}
\tag{7}
induces the transformation
\begin{pmatrix} \mathrm{d} \tilde{u} \\ \mathrm{d} \tilde{r} \end{pmatrix} =
\begin{pmatrix} \tilde{A} & \tilde{B} \\ \tilde{C} & \tilde{D}
\end{pmatrix} \begin{pmatrix} \mathrm{d} \bar{u} \\ \mathrm{d} \bar{r} \end{pmatrix}.
\tag{8}
where
\begin{aligned}
&\tilde{A} = - \begin{pmatrix} 0 & 1 \\ P'(e_0)^{-1} & -e_0 P'(e_0)^{-1} \end{pmatrix},&
&\tilde{B} = 0,& \\
&\tilde{C} = \frac{h}{c^2} \begin{pmatrix} 2 \nu_0 e_0^3 + \nu_1 e_0^2 &
4 \nu_0 e_0^4 + 3 \nu_1 e_0^3 + 2 \nu_2 e_0^2 + \nu_3 e_0 \\
(6 \nu_0 e^2 + 2 \nu_1 e_0)P'(e_0) & - (4 \nu_0 e_0^3 + \nu_1 e_0^2)P'(e_0) \end{pmatrix},&
&\tilde{D} = - \frac{h}{c^2} \begin{pmatrix} e_0 & 1 \\ P'(e_0) & 0 \end{pmatrix}.&
\end{aligned}
The \sigma-function associated with \bar{\mathcal{C}}_{(2,6)} is computed by
\bar{\sigma} (\bar{u}; \nu ) = \exp \Big( \tfrac{1}{2} \bar{u}^t \tilde{T} \bar{u} \Big)
\tilde{\sigma} \big(\tilde{A} \bar{u}; \mu(\nu)\big),
where \tilde{T} is a symmetric matrix,
\tilde{T} = \tilde{D}^{-1} \tilde{C} =
\begin{pmatrix} -6\nu_0 e_0^2 - 2 \nu_1 e_0 & 4 \nu_0 e_0^3 + \nu_1 e_0^2 \\
4 \nu_0 e_0^3 + \nu_1 e_0^2 &
-8\nu_0 e_0^4 - 4 \nu_1 e_0^3 - 2 \nu_2 e_0^2 - \nu_3 e_0
\end{pmatrix},
and \mu(\nu) is given by (7).
In order to find the expression for \bar{\sigma} through \sigma, we combine the transformation from \tilde{\mathcal{C}}_{(2,5)} to \bar{\mathcal{C}}_{(2,6)} with transformation from \tilde{\mathcal{C}}_{(2,5)} to \mathcal{C}_{(2,5)}.
Then, \lambda(\nu) is obtained by combining (4) with \mu(\nu) given by (7), and
\begin{pmatrix} \mathrm{d} u \\
\mathrm{d} r \end{pmatrix} =
\begin{pmatrix} A \tilde{A} + B \tilde{C}
& A \tilde{B} + B \tilde{D} \\
C \tilde{A} + D \tilde{C}
& C \tilde{B} + D \tilde{D}
\end{pmatrix}
\begin{pmatrix} \mathrm{d} \bar{u} \\
\mathrm{d} \bar{r} \end{pmatrix}.
\tag{9}
and so, taking into account that B=0, and \tilde{B}=0, we have
\bar{\sigma} (\bar{u}; \nu ) = \exp \Big( \tfrac{1}{2} \bar{u}^t \bar{T} \bar{u} \Big)
\sigma \big(A \tilde{A} \bar{u}; \lambda(\nu)\big),
where \bar{T} = (D \tilde{D})^{-1} (C \tilde{A} + D \tilde{C} ) = \tilde{D}^{-1} T \tilde{A} + \tilde{T}.
References
1 Buchstaber,V. M., Enolskii,V.Z., Leikin,D.V., Rational analogs of abelian functions,
Functional Analysis and Its Applications, 33:2 (1999) 83–94.
2 Baker, H.F., Abel’s theorem and the allied theory of theta functions,
Cambridge Univ. Press, Cambridge, 1897.
3 Baker, H.F., Multiply periodic functions,
Cambridge Univ. Press, Cambridge, 1907.
4 Baker, H.F., On a system of differential equations leading to periodic functions, Acta Mathematica, 27 (1903) pp. 135–156.
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