nrussoniello

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These are replies submitted by nrussoniello

@vv I appreciate the clarification. Of course I understand if we achieve maximal rank, we can call it a day immediately.

The examples I'll be computing will typically be either maximal rank or #rows-1, so I'll have to be a little careful, but it should work out within one or two iterations most of the time.

Thanks again!

@vv I'm having a hard time understanding what this actually does. Does this replace each instance of "s#" with a random number, or does it replace each instance of "s" with a random number?

Also, the symbols in CI and CO represent basis elements, so there is no relationship between "s1" and "s2," for example. If we are simply assigning random numbers to these basis elements, isn't it possible to conjure relationships (common divisor, etc.) which would affect the rank? Is the "fix" for this to run your code multiple times to obtain the correct rank?

@dharr Your help has been very much appreciated. Thank you so much.

@dharr This appears to be working, though not quite as easily as I originally hoped. I'm still trying to fully understand what is going on, but I'm getting there. However, while I have you, is there an easy way to append a row which contains 0 as the (1,1) entry, then lists s1,...,sn as entries ((1,2),...,(1,n+1)), while maintaining skew-symmetry? (See the CO matrix in my last worksheet for clarification).

 

I deeply appreciate all your help on this.

@dharr I haven't done the computation for the rank of CO; I was more referencing CI in this particular case. That is, the value Index should be greater than 0 (in particular, 1, which I know from a theorem related to Index). You can see this by eliminating row 2 by adding row 1 to it. This would give rank less than or equal to 3, meaning Index is at least 1.

@dharr Apologies. I thought this issue was fully resolved, but upon trying more complicated examples, I noticed the rank is incorrect. It appears any time e_i and e_j (i not equal to j) appear in the same column, it returns rank equal to the dimension of the matrix, which is often not correct.

 

Any idea as to what could solve this?
 

with(DifferentialGeometry); with(LieAlgebras)

``

``

M := [Matrix([[1, 0, 0], [0, 0, 0], [0, 0, -1]]), Matrix([[0, 0, 0], [0, 1, 0], [0, 0, -1]]), Matrix([[0, 0, 0], [1, 0, 0], [0, 0, 0]]), Matrix([[0, 1, 0], [0, 0, 0], [0, 0, 0]])]

[Matrix(%id = 18446744078258831598), Matrix(%id = 18446744078258831718), Matrix(%id = 18446744078258831838), Matrix(%id = 18446744078258831958)]

(1)

L := LieAlgebraData(M, Alg1)

_DG([["LieAlgebra", Alg1, [4, table( [ ] )]], [[[1, 3, 3], -1], [[1, 4, 4], 1], [[2, 3, 3], 1], [[2, 4, 4], -1], [[3, 4, 1], -1], [[3, 4, 2], 1]]])

(2)

DGsetup(L)

`Lie algebra: Alg1`

(3)

T := MultiplicationTable("LieTable")

Matrix(%id = 18446744078223965230)

(4)

CI := T[3 .. (), 3 .. ()]

Matrix(%id = 18446744078245951718)

(5)

Index := LinearAlgebra:-RowDimension(CI)-LinearAlgebra:-Rank(CI)

0

(6)

``CO := T[[1, 3 .. ()], [1, 3 .. ()]]

Matrix(%id = 18446744078223999814)

(7)

CO[1 .. (), 1] := Vector([0, -e1, -e2, -e3, -e4])

Vector[column](%id = 18446744078223989206)

(8)

CO

Matrix(%id = 18446744078223999814)

(9)

LinearAlgebra[Rank](CO)

5

(10)

``

``


 

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@dharr That's embarrassing...I appreciate your patience.

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