J-不変量
J ( 4 ( 5 i ± 1 ) 13 ) = ( 1 − 9 ( 1 − 5 ) 38 2 41 2 ( 7485 − 762 2 − 1479 5 + 3072 10 ± i 5 4 ( 178 − 2221 2 − 3148 5 + 1289 10 ) ) 2 ) 3 J ( 5 ( 4 i ± 1 ) 17 ) = ( 1 + 9 ( 1 − 5 ) 38 2 41 2 ( 7485 + 762 2 − 1479 5 − 3072 10 ± i 5 4 ( 178 + 2221 2 − 3148 5 − 1289 10 ) ) 2 ) 3 {\displaystyle {\begin{aligned}J\left({\tfrac {4\left(5i\pm 1\right)}{13}}\right)=\left(1-{\tfrac {9\left(1-{\sqrt {5}}\right)^{38}}{2^{41}{\sqrt {2}}}}{\Bigl (}7485-762{\sqrt {2}}-1479{\sqrt {5}}+3072{\sqrt {10}}\pm i{\sqrt[{4}]{5}}\left(178-2221{\sqrt {2}}-3148{\sqrt {5}}+1289{\sqrt {10}}\right){\Bigr )}^{2}\right)^{3}\\J\left({\tfrac {5\left(4i\pm 1\right)}{17}}\right)=\left(1+{\tfrac {9\left(1-{\sqrt {5}}\right)^{38}}{2^{41}{\sqrt {2}}}}{\Bigl (}7485+762{\sqrt {2}}-1479{\sqrt {5}}-3072{\sqrt {10}}\pm i{\sqrt[{4}]{5}}\left(178+2221{\sqrt {2}}-3148{\sqrt {5}}-1289{\sqrt {10}}\right){\Bigr )}^{2}\right)^{3}\end{aligned}}}
J ( i ) = J ( 1 + i 2 ) = 1 J ( 2 i ) = ( 5 3 ) 3 J ( 2 i ) = ( 11 2 ) 3 J ( 2 2 i ) = 125 216 ( 19 + 13 2 ) 3 J ( 4 i ) = 1 64 ( 724 + 513 2 ) 3 J ( 1 + 2 i 2 ) = 1 64 ( 724 − 513 2 ) 3 J ( 1 + 2 2 i 3 ) = 125 216 ( 19 − 13 2 ) 3 J ( 3 i ) = 1 27 ( 2 + 3 ) 2 ( 21 + 20 3 ) 3 J ( 2 3 i ) = 125 16 ( 30 + 17 3 ) 3 J ( 1 + 7 3 i 2 ) = − 64000 7 ( 651 + 142 21 ) 3 J ( 1 + 3 11 i 10 ) = 64 27 ( 23 − 4 33 ) 2 ( − 77 + 15 33 ) 3 J ( 21 i ) = 1 128 ( 3 + 7 ) 5 ( 17 + 7 3 + 59 7 + 35 21 ) 3 J ( 30 i 1 ) = 1 4 ( 7 + 5 2 + 3 5 + 2 10 ) 4 ( 55 + 30 2 + 12 5 + 10 10 ) 3 J ( 30 i 2 ) = 1 4 ( 7 + 5 2 − 3 5 − 2 10 ) 4 ( 55 + 30 2 − 12 5 − 10 10 ) 3 J ( 30 i 5 ) = 1 4 ( 7 − 5 2 + 3 5 − 2 10 ) 4 ( 55 − 30 2 + 12 5 − 10 10 ) 3 J ( 30 i 10 ) = 1 4 ( 7 − 5 2 − 3 5 + 2 10 ) 4 ( 55 − 30 2 − 12 5 + 10 10 ) 3 J ( 1 + 31 i 2 ) = ( 1 − ( 1 + 19 2 ( 13 − 93 13 + 93 ⋅ 31 + 27 31 − 27 3 + 13 + 93 13 − 93 ⋅ 31 − 27 31 + 27 3 ) ) 2 ) 3 J ( 5 i ) = ( 1 + 9 4 5 ( 13 + 5 5 ) 2 ) 3 J ( 5 i + 1 2 ) = ( 1 − 9 4 5 ( 13 − 5 5 ) 2 ) 3 J ( 6 i ) = 1 216 ( 2 + 3 ) 10 ( 231 + 380 3 + ( 204 + 158 3 ) 12 4 ) 3 J ( 70 i ) = ( 1 + 9 4 ( 303 + 220 2 + 139 5 + 96 10 ) 2 ) 3 J ( 94 i ) = ( 1 + 9 8192 ( 3 + 2 2 + 9 + 8 2 ) 8 ( 8 + ( − 1 − 2 + 9 + 8 2 ) ( − 2 + 2 2 + 3 + 4 2 + 3 9 + 8 2 ) ) 2 ) 3 J ( 7 i ) = ( 1 + 9 32 28 4 ( 3 + 7 ) 3 ( 13 + 3 7 + ( 6 + 7 ) 28 4 ) 2 ) 3 J ( 8 i ) = ( 1 + 9 4 2 4 ( 1 + 2 ) ( 123 + 104 2 4 + 88 2 + 73 8 4 ) 2 ) 3 J ( 10 i ) = ( 1 + 9 8 ( 2402 + 1607 5 4 + 1074 25 4 + 719 125 4 ) 2 ) 3 J ( 5 i 2 ) = ( 1 + 9 8 ( 2402 − 1607 5 4 + 1074 25 4 − 719 125 4 ) 2 ) 3 J ( 130 i ) = ( 1 + 9 4 ( 7392 + 3289 5 + 2040 13 + 917 65 ) 2 ) 3 J ( 190 i ) = ( 1 + 18 ( 31570 + 22323 2 + 14139 5 + 9998 10 ) 2 ) 3 J ( 2 58 i ) = ( 1 + 9 256 ( 1 + 2 ) 5 ( 5 + 29 ) 5 ( 793 + 907 2 + 237 29 + 103 58 ) 2 ) 3 J ( 1 + 1435 i 2 ) = ( 1 − 9 ( 9892538 + 4424079 5 + 1544955 41 + 690925 205 ) 2 ) 3 J ( 1 + 1555 i 2 ) = ( 1 − 9 ( 22297077 + 9971556 5 + ( 3571365 + 1597163 5 ) 31 + 21 5 2 ) 2 ) 3 {\displaystyle {\begin{aligned}J(i)&=J\left({\tfrac {1+i}{2}}\right)=1\\J\left({\sqrt {2}}i\right)&={\big (}{\tfrac {5}{3}}{\big )}^{3}\\J(2i)&={\big (}{\tfrac {11}{2}}{\big )}^{3}\\J\left(2{\sqrt {2}}i\right)&={\tfrac {125}{216}}\left(19+13{\sqrt {2}}\right)^{3}\\J(4i)&={\tfrac {1}{64}}\left(724+513{\sqrt {2}}\right)^{3}\\J\left({\tfrac {1+2i}{2}}\right)&={\tfrac {1}{64}}\left(724-513{\sqrt {2}}\right)^{3}\\J\left({\tfrac {1+2{\sqrt {2}}i}{3}}\right)&={\tfrac {125}{216}}\left(19-13{\sqrt {2}}\right)^{3}\\J(3i)&={\tfrac {1}{27}}\left(2+{\sqrt {3}}\right)^{2}\left(21+20{\sqrt {3}}\right)^{3}\\J\left(2{\sqrt {3}}i\right)&={\tfrac {125}{16}}\left(30+17{\sqrt {3}}\right)^{3}\\J\left({\tfrac {1+7{\sqrt {3}}i}{2}}\right)&=-{\tfrac {64000}{7}}\left(651+142{\sqrt {21}}\right)^{3}\\J\left({\tfrac {1+3{\sqrt {11}}i}{10}}\right)&={\tfrac {64}{27}}\left(23-4{\sqrt {33}}\right)^{2}\left(-77+15{\sqrt {33}}\right)^{3}\\J\left({\sqrt {21}}i\right)&={\tfrac {1}{128}}\left(3+{\sqrt {7}}\right)^{5}\left(17+7{\sqrt {3}}+59{\sqrt {7}}+35{\sqrt {21}}\right)^{3}\\J\left({\tfrac {{\sqrt {30}}i}{1}}\right)&={\tfrac {1}{4}}\left(7+5{\sqrt {2}}+3{\sqrt {5}}+2{\sqrt {10}}\right)^{4}\left(55+30{\sqrt {2}}+12{\sqrt {5}}+10{\sqrt {10}}\right)^{3}\\J\left({\tfrac {{\sqrt {30}}i}{2}}\right)&={\tfrac {1}{4}}\left(7+5{\sqrt {2}}-3{\sqrt {5}}-2{\sqrt {10}}\right)^{4}\left(55+30{\sqrt {2}}-12{\sqrt {5}}-10{\sqrt {10}}\right)^{3}\\J\left({\tfrac {{\sqrt {30}}i}{5}}\right)&={\tfrac {1}{4}}\left(7-5{\sqrt {2}}+3{\sqrt {5}}-2{\sqrt {10}}\right)^{4}\left(55-30{\sqrt {2}}+12{\sqrt {5}}-10{\sqrt {10}}\right)^{3}\\J\left({\tfrac {{\sqrt {30}}i}{10}}\right)&={\tfrac {1}{4}}\left(7-5{\sqrt {2}}-3{\sqrt {5}}+2{\sqrt {10}}\right)^{4}\left(55-30{\sqrt {2}}-12{\sqrt {5}}+10{\sqrt {10}}\right)^{3}\\J\left({\tfrac {1+{\sqrt {31}}i}{2}}\right)&=\left(1-\left(1+{\frac {\sqrt {19}}{2}}\left({\sqrt {\tfrac {13-{\sqrt {93}}}{13+{\sqrt {93}}}}}\cdot {\sqrt[{3}]{\tfrac {{\sqrt {31}}+{\sqrt {27}}}{{\sqrt {31}}-{\sqrt {27}}}}}+{\sqrt {\tfrac {13+{\sqrt {93}}}{13-{\sqrt {93}}}}}\cdot {\sqrt[{3}]{\tfrac {{\sqrt {31}}-{\sqrt {27}}}{{\sqrt {31}}+{\sqrt {27}}}}}\right)\right)^{2}\right)^{3}\\J(5i)&=\left(1+{\tfrac {9}{4}}{\sqrt {5}}\left(13+5{\sqrt {5}}\right)^{2}\right)^{3}\\J\left({\tfrac {5i+1}{2}}\right)&=\left(1-{\tfrac {9}{4}}{\sqrt {5}}\left(13-5{\sqrt {5}}\right)^{2}\right)^{3}\\J(6i)&={\tfrac {1}{216}}\left(2+{\sqrt {3}}\right)^{10}\left(231+380{\sqrt {3}}+\left(204+158{\sqrt {3}}\right){\sqrt[{4}]{12}}\right)^{3}\\J({\sqrt {70}}i)&=\left(1+{\tfrac {9}{4}}\left(303+220{\sqrt {2}}+139{\sqrt {5}}+96{\sqrt {10}}\right)^{2}\right)^{3}\\J({\sqrt {94}}i)&=\left(1+{\tfrac {9}{8192}}\left(3+2{\sqrt {2}}+{\sqrt {9+8{\sqrt {2}}}}\right)^{8}\left(8+\left(-1-{\sqrt {2}}+{\sqrt {9+8{\sqrt {2}}}}\right)\left(-2+2{\sqrt {2}}+{\sqrt {3+4{\sqrt {2}}+3{\sqrt {9+8{\sqrt {2}}}}}}\right)\right)^{2}\right)^{3}\\J(7i)&=\left(1+{\tfrac {9}{32}}{\sqrt[{4}]{28}}\left(3+{\sqrt {7}}\right)^{3}\left(13+3{\sqrt {7}}+\left(6+{\sqrt {7}}\right){\sqrt[{4}]{28}}\right)^{2}\right)^{3}\\J(8i)&=\left(1+{\tfrac {9}{4}}{\sqrt[{4}]{2}}\left(1+{\sqrt {2}}\right)\left(123+104{\sqrt[{4}]{2}}+88{\sqrt {2}}+73{\sqrt[{4}]{8}}\right)^{2}\right)^{3}\\J(10i)&=\left(1+{\tfrac {9}{8}}\left(2402+1607{\sqrt[{4}]{5}}+1074{\sqrt[{4}]{25}}+719{\sqrt[{4}]{125}}\right)^{2}\right)^{3}\\J\left({\tfrac {5i}{2}}\right)&=\left(1+{\tfrac {9}{8}}\left(2402-1607{\sqrt[{4}]{5}}+1074{\sqrt[{4}]{25}}-719{\sqrt[{4}]{125}}\right)^{2}\right)^{3}\\J({\sqrt {130}}i)&=\left(1+{\tfrac {9}{4}}\left(7392+3289{\sqrt {5}}+2040{\sqrt {13}}+917{\sqrt {65}}\right)^{2}\right)^{3}\\J({\sqrt {190}}i)&=\left(1+18\left(31570+22323{\sqrt {2}}+14139{\sqrt {5}}+9998{\sqrt {10}}\right)^{2}\right)^{3}\\J(2{\sqrt {58}}i)&=\left(1+{\tfrac {9}{256}}\left(1+{\sqrt {2}}\right)^{5}\left(5+{\sqrt {29}}\right)^{5}\left(793+907{\sqrt {2}}+237{\sqrt {29}}+103{\sqrt {58}}\right)^{2}\right)^{3}\\J\left({\tfrac {1+{\sqrt {1435}}i}{2}}\right)&=\left(1-9\left(9892538+4424079{\sqrt {5}}+1544955{\sqrt {41}}+690925{\sqrt {205}}\right)^{2}\right)^{3}\\J\left({\tfrac {1+{\sqrt {1555}}i}{2}}\right)&=\left(1-9\left(22297077+9971556{\sqrt {5}}+\left(3571365+1597163{\sqrt {5}}\right){\sqrt {\tfrac {31+21{\sqrt {5}}}{2}}}\right)^{2}\right)^{3}\\\end{aligned}}}
j-不変量はある無限和(下記の g2, g3 を参照)で純粋に定義することができるが、これらは楕円曲線の同型類を考えることが動機となる。C 上のすべての楕円曲線 E は複素トーラスであるので、ランク 2 の格子、つまり C の 2 次元格子と同一視できる。格子の互いに平行な反対側の辺を同一視することで、そのようにみなすことができる。複素数を格子に掛けることは格子の回転やスケーリングに対応し、これらは楕円曲線の同型類を保存することがわかり、このことから、格子を 1 と上半平面 H のある元 τ によって生成されると考えてよい。逆に、
g 2 = 60 ∑ ( m , n ) ≠ ( 0 , 0 ) ( m + n τ ) − 4 , {\displaystyle g_{2}=60\sum _{(m,n)\neq (0,0)}(m+n\tau )^{-4},}
g 3 = 140 ∑ ( m , n ) ≠ ( 0 , 0 ) ( m + n τ ) − 6 , {\displaystyle g_{3}=140\sum _{(m,n)\neq (0,0)}(m+n\tau )^{-6},}