I am trying to describe the Jacobian variety $J(C)$ of Fermat quartic curve $C$:
$1+x_{10}^4+x_{20}^4 = 0$ on affine $U_0$,
$x_{01}^4+1+x_{21}^4 = 0$ on affine $U_1$,
$x_{02}^4+x_{12}^4+1 = 0$ on affine $U_2$, related by $x_{01}=1/x_{10}, x_{21}=x_{20}/x_{10}$ etc.
Points on $C$ can be described by $[X:Y:Z]$ satisfying $X^4+Y^4+Z^4=0$, corresponding to
$(x_{10},x_{20})=(Y/X,Z/X)$ on $U_0$,
$(x_{01},x_{21})=(X/Y,Z/Y)$ on $U_1$,
$(x_{02},x_{12})=(X/Z,Y/Z)$ on $U_2$.
I found $3$ independent holomorphic differentials:
$\omega_1 = dx_{10}/x_{20}^3 = -dx_{20}/x_{10}^3 = -x_{01}dx_{01}/x_{21}^3 = dx_{21}/x_{01}^2 = x_{02}dx_{02}/x_{12}^3 = -dx_{12}/x_{02}^2,$
$\omega_2 = x_{10}dx_{10}/x_{20}^3 = -dx_{20}/x_{10}^2 = -dx_{01}/x_{21}^3 = dx_{21}/x_{01}^3 = dx_{02}/x_{12}^2 = -x_{12}dx_{12}/x_{02}^3,$
$\omega_3 = dx_{10}/x_{20}^2 = -x_{20}dx_{20}/x_{10}^3 = -dx_{01}/x_{21}^2 = x_{21}dx_{21}/x_{01}^3 = dx_{02}/x_{12}^3 = -dx_{12}/x_{02}^3,$
I know that
$\Lambda = \{ \int_\gamma\omega:= (\int_\gamma \omega_1, \int_\gamma \omega_2, \int_\gamma \omega_3)\ | \gamma$: cycle in $C \}$ forms a lattice in $\mathbb{C}^3$,
and I want to determine the generators for this lattice $\Lambda$.
To describe explicitly, I use following notation of points:
$P_{a} = [0:1:\zeta^a], Q_{a}= [\zeta^{a}:0:1], R_{a} = [1:\zeta^a:0]$,
where $a \in \{1,3,5,7\}, \zeta = (1+i)/\sqrt2$.
First I calculated the integral for each segments as follow:
$\int_{Q_aR_b} \omega_1 = \int_0^{\zeta^b} dx_{10}/ (\zeta^{-a} \sqrt[4]{1+x_{10}^4} )^3
= \zeta^{3a+b}\int_0^1 dt/ ( \sqrt[4]{1-t^4} )^3 = \zeta^{3a+b}k\sqrt2$,
$\int_{Q_aR_b} \omega_2 = \int_0^{\zeta^b} x_{10}dx_{10}/ (\zeta^{-a} \sqrt[4]{1+x_{10}^4} )^3
= \zeta^{3a+2b}\int_0^1 tdt/ ( \sqrt[4]{1-t^4} )^3 = \zeta^{3a+2b}k$,
$\int_{Q_aR_b} \omega_3 = \int_0^{\zeta^b} dx_{10}/ (\zeta^{-a} \sqrt[4]{1+x_{10}^4} )^2
= \zeta^{2a+b}\int_0^1 dt/ \sqrt{1-t^4} = \zeta^{2a+b}k$.
Similarly,
$\int_{R_aP_b}(\omega_1,\omega_2,\omega_3) = (\zeta^{2a+b}k, \zeta^{3a+b}k\sqrt2, \zeta^{3a+2b}k), $
$\int_{P_aQ_b}(\omega_1,\omega_2,\omega_3) = (\zeta^{3a+2b}k, \zeta^{2a+b}k, \zeta^{3a+b}k\sqrt2).$
EDIT: I found mistake and tried again. I used computer for calculation.
For $P_aQ_bR_cP_a$-type cycles,
$(a,b,c)=(1,1,1)$ gives $(-4,-4,-4)k/\sqrt2$.
For $P_1Q_aP_bQ_cP_1$-type cycles,
$(a,b,c)=(1,5,5)$ gives $(0,0,-8)k/\sqrt2$, and $(a,b,c)=(1,7,5)$ gives $(0,4i-4,4i-4)k/\sqrt2$.
From these, I guessed the generators of $\Lambda$ as:
$v_1=(0,0,8)k/\sqrt2$,
$v_2=(0,0,8i)k/\sqrt2$,
$v_3=(0,4+4i,4+4i)k/\sqrt2$,
$v_4=(0,4-4i,4+4i)k/\sqrt2$,
$v_5=(4+4i,0,4+4i)k/\sqrt2$,
$v_6=(4,4,4)k/\sqrt2$.
I roughly observed that integrals for every $P_aQ_bR_cP_a$-type and $P_1Q_aP_bQ_cP_1$-type cycles are actually generated by these, but I am still not sure that I am not missing anything coming from other type cycles.