Zoran Skoda
gl2 linear realization

The following shows that the Gurevich-Saponov differential calculus for gl(2)gl(2), linear in derivatives is just an example of our framework – the right hand side are the components of matrix ϕ=(ϕ j i( 1,1,, 2,2)) i,j=1,,2\phi = (\phi^i_j(\partial^{1,1},\ldots,\partial^{2,2}))_{i,j = 1,,2} satisfying our differential equation for realizations hence it comes from some isomorphism of coalgebras between U(gl 2)U(gl_2) and S(gl 2)S(gl_2). Here the matrix elements, which are in general formal power series, is linear in partial derivatives.

Let us put h=1h = 1 (it is just a rescale). If we postulate commutative coordinates

X 1=A,X 2=B,X 3=C,X 4=DX_1 = A, X_2 = B, X_3 = C, X_4 = D and set

a=A(1+ a)+B ba = A(1+\partial_a) + B\partial_b
b=A c+B( d+1)b = A \partial_c + B (\partial_d +1)
c=C( a+1)+D bc = C (\partial_a +1) + D \partial_b
d=C c+D( d+1)d = C \partial_c + D (\partial_d+1)

and [ a,A]=1,[ b,B]=1,[ c,C]=1,[ d,D]=1,[\partial_a,A] = 1, [\partial_b, B] = 1, [\partial_c,C] = 1, [\partial_d,D] = 1, then we obtain

1) [a,b]=b,[a,c]=c,[a,d]=0,[b,c]=ad,[b,d]=b,[c,d]=c[a,b] = b, [a,c] = -c, [a,d] = 0, [b,c] = a-d, [b,d] = b, [c,d] = -c

2) a=X iϕ a ia = \sum X_i \phi^i_a

b=X iϕ b ib = \sum X_i \phi^i_b etc. and ϕ\phi satisfies our differential equation

and

3) your relations follow

aaa a= a+1 abb a= c acc a=0 add a=0 baa b= b bbb b= d+1 bcc b=0 bdd b=0 caa c=0 cbb c=0 ccc c= a+1 cdd c= c daa d=0 dbb a=0 dcc d= b ddd d= d+1\array{ \partial_a a - a \partial_a = \partial_a +1\\ \partial_a b - b \partial_a = \partial_c\\ \partial_a c - c \partial_a = 0\\ \partial_a d - d \partial_a = 0\\ \partial_b a - a \partial_b = \partial_b\\ \partial_b b - b \partial_b = \partial_d +1\\ \partial_b c - c \partial_b = 0\\ \partial_b d - d \partial_b = 0\\ \partial_c a - a \partial_c = 0\\ \partial_c b - b \partial_c = 0\\ \partial_c c - c \partial_c = \partial_a + 1\\ \partial_c d - d \partial_c = \partial_c\\ \partial_d a - a \partial_d = 0\\ \partial_d b - b \partial_a = 0\\ \partial_d c - c\partial_d = \partial_b \\ \partial_d d - d \partial_d = \partial_d + 1}

what is just the form [ j,x^ i]=ϕ i j[\partial_j, \hat{x}_i] = \phi^j_i

category: valenciennes

Created on December 30, 2012 at 00:00:34. See the history of this page for a list of all contributions to it.