Zoran Skoda
coproduct from exponentials

Preliminaries on realizations

Let x^ ix^ ϕ:= j=1 nx jϕ i j( 1,, n)\hat{x}_i \mapsto \hat{x}^\phi :=\sum_{j=1}^n x_j\phi^j_i(\partial^1,\ldots,\partial^n) be the formula (on generators) for a realization of an enveloping algebra of a finite dimensional Lie algebra LL over a field kk in characteristic zero with basis x^ 1,,x^ n\hat{x}_1,\ldots,\hat{x}_n via formal differential operators with linear coefficients; here x 1,,x nx_1,\ldots,x_n are the commutative coordinates i=/x i\partial^i = \partial/\partial x_i and ϕ i j\phi^j_i is a formal power series of the form δ i j+higherorder\delta^j_i + higher order. The realization f^f^ ϕ\hat{f}\mapsto \hat{f}^\phi is a homomorphism of associative algebras U(L)A^ n,kU(L)\to \hat{A}_{n,k} where A^ n,k\hat{A}_{n,k} is the nn-the order Weyl algebra over kk completed with respect to the filtration given by the order of differential operator. This Weyl algebra has the Fock representation on S(L)S(L) where the unit element will be denote by |0|0\rangle and called the (undeformed) vacuum.

Define ξ ϕ 1:U(L)S(L)\xi^{-1}_\phi : U(L)\to S(L) by u^u^ ϕ|0\hat{u} \mapsto \hat{u}^\phi |0\rangle; it appears that this map is an isomorphism of kk-coalgebras and its inverse is denoted by ξ:S(L)U(L)\xi : S(L)\to U(L). The star product on S(L)S(L) is obtained by the transport via ξ\xi of the noncommutative product on U(L)U(L) i.e. ξ(f*g)=ξ(f) U(L)ξ(g)\xi(f\ast g) = \xi(f)\cdot_{U(L)}\xi(g).

The formal power series in L *L^* form a ring S^(L *)\hat{S}(L^*) with an isomorphism of kk-vector spaces S^(L *)S(L) *\hat{S}(L^*)\cong S(L)^* where the linear dual S(L) *=Hom(L,k)S(L)^*= Hom(L,k) is on the right hand side. The dual variables to x ix_i are i\partial^i and the isomorphism with the linear dual is determining by considering polynomials in \partial-s as the constant coefficient differential operator and defining the pairing with xx-s by applying the formal differential operator and evaluating the result at zero,

The star product (which depends on ϕ\phi) on S(L)S(L) is dualized to S(L) *S(L)^* gives a topological coassociative coproduct which is transferred via the isomorphism S(L) *S^(L *)S(L)^*\cong \hat{S}(L^*). Over there, several alternative definitions of the coproduct, depending on ϕ\phi, are possible. One of them is via deformed Leibniz rules i.e. by asking for

m *Δ(P())(fg)=P()(f*g)foreveryP()S(L *). m_\ast \Delta(P(\partial))(f\otimes g) = P(\partial) (f\ast g) for every P(\partial)\in S(L^*).

where m *(fg):=f*gm_\ast(f\otimes g) :=f\ast g and =( 1,, n)\partial = (\partial^1,\ldots,\partial^n).

Coproduct via exponentials

Though the star product is well defined on polynomials and not on general formal power series, it is still well defined on some subspace of the space of formal power series, and one can include the exponential of linear expressions in x^ i\hat{x}_i among allowed power series. Exponentials are dense and hence if there is a candidate for Δ()\Delta(\partial) which gives the correct Leibniz rule on all star products of exponentials (and if the formula is continuous in each variable) then it holds for all ff and gg.

Thus it is sufficients to calculate i(e ikxe iqy)\partial_i(e^{i k x}\star e^{i q y}) and extend to series in i\partial^i by multiplicativity of coproduct. Here ikxi k x denoted the scalar product ik jx ji\sum k^j x_j for k=(k 1,,k n)k = (k^1,\ldots,k^n) and i=1i =\sqrt{-1} (one can work without ii in the formulas, if 1\sqrt{-1} is not in the field).

Define the vector K(p,q)K(p,q) by

e ipx^ ϕ(e iqx):=e iK(p,q)x e^{i p\hat{x}^\phi}(e^{i q x}) := e^{i K(p,q)x}

(it is not difficult to show that the left hand side can be written in the form of the right hand side for some K(k,q)K(k,q), see the article Meljanac, Škoda, Svrtan). Let K 0(k)=K(k,0)K_0(k) = K(k,0), then exp(ipx^ ϕ)|0=exp(iK 0(p)x)exp(i p\hat{x}^\phi)|0\rangle = exp(i K_0(p)x) and ξ(exp(ipx))=exp(iK 0 1(p)x^)\xi(exp(i p x)) = exp(i K_0^{-1}(p)\hat{x}) where K 0 1K_0^{-1} is the inverse of K 0K_0.

By the definition of star product *\ast and of ξ 1\xi^{-1},

e ikx*e iqx=e iK 0 1(k)x^ ϕ(e iqx) e^{ikx}\ast e^{iqx} = e^{iK_0^{-1}(k)\hat{x}^\phi} (e^{iqx})

and we calculate

j(e ikx*e iqx) = j(e iK 0 1(k)x^ ϕ(e iqx)) = iK j(K 0 1(k),q)e iK(K 0 1(k),q)x = im *K j(K 0 1(k1),1q)(e ikxe iqx) = im *K j(K 0 1(i1),1(i))(e ikxe iqx).\array{ \partial^j(e^{ikx}\ast e^{iqx}) &=&\partial^j( e^{i K_0^{-1}(k)\hat{x}^\phi} (e^{iqx})) \\&=& i K^j(K_0^{-1}(k),q)e^{i K(K_0^{-1}(k),q)x} \\&=& i m_\ast K^j(K_0^{-1}(k\otimes 1),1\otimes q)(e^{i k x}\otimes e^{i q x}) \\&=& i m_\ast K^j(K_0^{-1}(-i\partial\otimes 1),1\otimes (-i\partial))(e^{i k x}\otimes e^{i q x}). }

Hence Δ( j)=iK j(K 0 1(i 11,,i n1),i1 1,,i1 n).\Delta (\partial^j) = i K^j(K_0^{-1}(-i\partial^1\otimes 1,\ldots,-i\partial^n\otimes 1),-i1\otimes \partial^1,\ldots,-i1\otimes\partial^n).

In particular if ϕ\phi is the universal formula, ξ\xi is the symmetrization map, K(k,q)=k+qK(k,q) = k+q and Δ( j)= j1+1 j\Delta(\partial^j) = \partial^j\otimes 1+1\otimes \partial^j. Some physics literature denotes K(k,q)K(k,q) in general by kqk\oplus q. The main innovation in our approach is to systematically use realizations of the form above. For KK we can write differential equations as in the article below.

This is used in coproduct for Ugln dual.

Literature

  • S. Meljanac, D. Svrtan, Z. Škoda, Exponential formulas and Lie algebra type star products, arxiv/1006.0478

Created on July 28, 2011 at 12:46:39. See the history of this page for a list of all contributions to it.