Given a differential form or vector field, this routine returns the ideal generated by the scalar coefficients of such element.
In this example we compute the equations that the scalar coefficients of a closed differential 1-form must satisfy.
i1 : w = newForm(2,1,2,"a")
2 2 2 2
o1 = (a x + a x x + a x + a x x + a x x + a x )dx + (a x + a x x +
0 0 3 0 1 9 1 6 0 2 12 1 2 15 2 0 1 0 4 0 1
------------------------------------------------------------------------
2 2 2 2
a x + a x x + a x x + a x )dx + (a x + a x x + a x + a x x +
10 1 7 0 2 13 1 2 16 2 1 2 0 5 0 1 11 1 8 0 2
------------------------------------------------------------------------
2
a x x + a x )dx
14 1 2 17 2 2
o1 : DiffAlgForm
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i2 : diff w
o2 = ((2a - a )x + (a - 2a )x + (a - a )x )dx dx + ((2a - a )x + (a
1 3 0 4 9 1 7 12 2 0 1 2 6 0 5
------------------------------------------------------------------------
- a )x + (a - 2a )x )dx dx + ((a - a )x + (2a - a )x + (a -
12 1 8 15 2 0 2 5 7 0 11 13 1 14
------------------------------------------------------------------------
2a )x )dx dx
16 2 1 2
o2 : DiffAlgForm
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i3 : moduliIdeal(diff w)
o3 = ideal (2a - a , a - 2a , a - a , 2a - a , a - a , a - 2a , a -
1 3 4 9 7 12 2 6 5 12 8 15 5
------------------------------------------------------------------------
a , 2a - a , a - 2a )
7 11 13 14 16
QQ[i]
o3 : Ideal of ------[][a , a , a , a , a , a , a , a , a , a , a , a , a , a , a , a , a , a ]
2 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
i + 1
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