Fordham University

Computer calculations for "Some singular curves in Mukai's model of \(\overline{M}_7\)", Section 3

Code 3.4: Algorithm 3.2, Steps 1-4

We continue the session begun in Code 3.3. For Step 1 in Algorithm 3.2, we compute the dimension of the intersection \( W_{p_0}^{\perp} \cap U_{\infty}.\) 
i17 : intersectionDimension(Wpperp(points_0,I2),Uinfty)

o17 = 0
Since it is even, we do not need to swap \(U_0\) and \(U_\infty\).

For Steps 2 and 3 in Algorithm 3.2, we compute the half spinors of the points \(p_i\), and the hyperplanes that vanish on these half spinor, and an echelonized basis of their rowspace. 

i18 : Spinors = matrix apply(points, k -> spinorOfPoint(k,I2,U0,Uinfty,Q))

o18 = | 1 3/35       0 -1/14     -1/35    1/21      0 3/35       3/7        -3/7       -4/7        1/30 -4/105        -3/70        -3/70        1/105      |
      | 1 321/560    0 515/224   97/70    -97/42    0 261/140    261/28     201/35     268/35      1/30 769/8400      2307/22400   -321/1120    -103/336   |
      | 1 261/140    0 97/28     103/112  -515/336  0 321/560    321/112    2307/2240  769/560     1/30 268/525       201/350      -261/280     -97/210    |
      | 1 10937/8505 0 2917/378  1313/189 -6565/567 0 4379/315   4379/63    2193/35    2924/35     1/30 52492/382725  13123/85050  -10937/17010 -2917/2835 |
      | 1 4379/315   0 6565/378  2917/945 -2917/567 0 10937/8505 10937/1701 13123/8505 52492/25515 1/30 2924/525      2193/350     -4379/630    -1313/567  |
      | 1 -251/140   0 95/28     -509/560 509/336   0 319/560    319/112    -2301/2240 -767/560    1/30 244/525       183/350      251/280      -19/42     |
      | 1 319/560    0 -509/224  19/14    -95/42    0 -251/140   -251/28    183/35     244/35      1/30 -767/8400     -2301/22400  -319/1120    509/1680   |
      | 1 10933/8505 0 -2915/378 6557/945 -6557/567 0 -4369/315  -4369/63   2181/35    2908/35     1/30 -52484/382725 -13121/85050 -10933/17010 583/567    |

               8        16
o18 : Matrix QQ  <--- QQ

i19 : KerSpinors = transpose gens ker(Spinors)

o19 = | 0     0   1 0    0   0 0 0  0 0    0 0 0    0 0 0 |
      | 0     0   0 0    5/3 1 0 0  0 0    0 0 0    0 0 0 |
      | 0     0   0 0    0   0 1 0  0 0    0 0 0    0 0 0 |
      | 0     0   0 0    0   0 0 -5 1 0    0 0 0    0 0 0 |
      | 0     0   0 0    0   0 0 0  0 -4/3 1 0 0    0 0 0 |
      | -1/30 0   0 0    0   0 0 0  0 0    0 1 0    0 0 0 |
      | 0     0   0 0    0   0 0 0  0 0    0 0 -9/8 1 0 0 |
      | 0     1/2 0 0    0   0 0 0  0 0    0 0 0    0 1 0 |
      | 0     0   0 2/15 0   0 0 0  0 0    0 0 0    0 0 1 |

               9        16
o19 : Matrix QQ  <--- QQ

i20 : transpose gens ker KerSpinors

o20 = | 0  0  0 0     -3/5 1 0 0   0 0   0 0 0   0 0 0 |
      | 0  0  0 0     0    0 0 1/5 1 0   0 0 0   0 0 0 |
      | 0  0  0 0     0    0 0 0   0 3/4 1 0 0   0 0 0 |
      | 30 0  0 0     0    0 0 0   0 0   0 1 0   0 0 0 |
      | 0  0  0 0     0    0 0 0   0 0   0 0 8/9 1 0 0 |
      | 0  -2 0 0     0    0 0 0   0 0   0 0 0   0 1 0 |
      | 0  0  0 -15/2 0    0 0 0   0 0   0 0 0   0 0 1 |

               7        16
o20 : Matrix QQ  <--- QQ
Since this has rank 7, we will continue to Step 5 of Algorithm 3.2 in Code 3.5.