[GAP Forum] Lie Algebra Cohomology

[Willem Adriaan De Graaf]

Dear Alan,

You can compute the quotient of the cocycle space modulo the coboundary
space using GAP's linear algebra functionality.
Example:

gap> L:= SimpleLieAlgebra("A",2,Rationals);;
gap> K:= Subalgebra( L, [ L.1, L.2 ] );;
gap> V:= FaithfulModule( K );;
gap> Dimension(V);
3
gap> B2:= Coboundaries( V, 2 );
<vector space over Rationals, with 9 generators>
gap> C2:=Cocycles( V, 2 );
<vector space of dimension 7 over Rationals>
gap> f:= NaturalHomomorphismBySubspace( C2, B2 );
<linear mapping by matrix, <vector space of dimension
7 over Rationals> -> ( Rationals^2 )>
gap> Q2:= Range(f);
( Rationals^2 )
gap> u1:= PreImagesRepresentative( f, Basis(Q2)[1] );
<2-cochain>
gap> u2:= PreImagesRepresentative( f, Basis(Q2)[2] );
<2-cochain>

So here u1, u2 span the second cohomology space H^2( K, V ). You can get
the internal representation of an s-cochain by the function ExtRepOfObj.
This returns a rather complicated structure of lists, which is explained at
the beginning of Section 64.12 of the manual. Example:

gap> ExtRepOfObj(u2);
[ [  ], [  ], [ [ [ 1, 3 ], 1 ] ] ]

Best wishes,

Willem








On Mon, 2 Aug 2021 at 15:56, Alan Hylton <agh314 at lehigh.edu> wrote:

> Howdy,
>
> For low-dimensional Lie algebras I find myself asking GAP if they have any
> cohomology using the calls I learned on this forum:
> Dimension(Cocycles(my_lie_algebra, n)) -
> Dimension(Coboundaries(my_lie_algebra, n)). Then to get a representative of
> any non-trivial cohomology class I wind up doing all the calculations by
> hand.
>
> I'd like to move into higher dimensions, but I cannot quite figure out how
> to compute cohomology in GAP. Referencing the page
> https://www.gap-system.org/Manuals/doc/ref/chap64.html, I have not seen
> any
> mention of computing cocycles mod coboundaries. Can anyone point me in the
> right direction?
>
> I also find that I do not understand some of the data types; is there a way
> to view the internal representation of a cochain?
>
> Many thanks!
> Alan
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