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Infinite-dimensional Lie algebras with generators and relations

asked 2022-12-20 10:24:59 +0100

updated 2022-12-20 10:25:30 +0100

I need to work with the polynomial current Lie algebra of the general linear Lie algebra, that is $\mathfrak{gl}_N[x]$, which I consider as a complex Lie algebra. In fact I would like to work with its universal enveloping algebra. But the Lie algebra in question is infinite-dimensional. How can I define it by specifying the structure constants?

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answered 2022-12-22 21:18:46 +0100

I found out that my computations boil down to the ‘’truncated’’ current Lie algebra $\mathfrak{gl}_N(\mathbb{C})\otimes\mathbb{C}[x]/(x^n)$, which is finite dimensional over $\mathbb{C}$, and the following code does what I need

d=3 #the size of matrices
n=12 #the ''depth'' of truncated polynomials, i.e. \mathbb{C}[x]/(x^n)
basis=''
for a in range(0,n):
    for i in range(1,d+1):
        for j in range(1,d+1):
            basis=basis+',e'+'_'+str(i)+'_'+str(j)+'_'+str(a)
basis=basis[1:]
def rel(i,j,a,k,l,b):
    if a+b>=n: return {}
    elif k==j and i==l and i!=j: return {'e'+'_'+str(j)+'_'+str(j)+'_'+str(a+b):-1,'e'+'_'+str(i)+'_'+str(i)+'_'+str(a+b):1}
    elif k==j and i==l and i==j: return {}
    elif k==j and i!=l: return {'e'+'_'+str(i)+'_'+str(l)+'_'+str(a+b):1}
    elif k!=j and i==l: return {'e'+'_'+str(k)+'_'+str(j)+'_'+str(a+b):-1}
    elif k!=j and i!=l: return {}

relations={('e'+'_'+str(i)+'_'+str(j)+'_'+str(a),'e'+'_'+str(k)+'_'+str(l)+'_'+str(b)): rel(i,j,a,k,l,b) for i in range(1,d+1) for j in range(1,d+1) for k in range(1,d+1) for l in range(1,d+1) for a in range(0,n) for b in range(0,n)}

L = LieAlgebra(QQ, relations, names=basis)
ee=list(L.lie_algebra_generators())
def e(i,j,a):
    if a<n: return ee[d*(i-1)+j+a*d^2-1]
    else: return 0
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Comments

You should accept your own answer in order to mark it as solved for the benefit of future ask.sagemath.org (per-)users...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2022-12-23 12:43:12 +0100 )edit

I haven’t got enough points to accept my own answer=)

Nikita_Safonkin gravatar imageNikita_Safonkin ( 2022-12-23 13:11:24 +0100 )edit

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Asked: 2022-12-20 10:24:59 +0100

Seen: 98 times

Last updated: Dec 22 '22