Commutativity of the invariant differential operators on a symmetric space
Commutativity of the invariant differential operators on a symmetric space
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对称空间上不变微分算子的交换性
DOI:
10.1090/s0002-9939-1968-0218494-5
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发表时间:
1968
影响因子:
0.9
通讯作者:
W. Smoke
中科院分区:
文献类型:
--
作者:
W. Smoke
It is known (see Helgason [1]) that the algebra of invariant differential operators on a Riemannian symmetric space is commutative. The algebra may be defined algebraically. We give here an algebraic proof of its commutativity. Let g be a Lie algebra over a field of characteristic zero and let f be a subalgebra of g. Extend the adjoint representation of f on g to the universal enveloping algebra U(g) of g so that ad x, xrf, acts as a derivation of U(g). Then (ad x) (u) = xu ux for u E U(g). For this is the case if u belongs to g, and g generates U(g). It follows from the formula that the invariant elements of U(g)-those annihilated by the action of f-are the elements of U(g) which commute with the elements of f. Let U(g) be the subalgebra of invariant elements. Now let U(g)f be the left ideal generated by f in U(g). This left ideal is preserved by the action of f. Moreover, [U(g)flt= U(g)f n U(g)f is a two-sided ideal in U(g)f. Let U(g, f) be the quotient algebra. If g is the Lie algebra of a Lie group G and f that of a connected subgroup K then U(g, f) is the algebra of invariant differential operators on the homogeneous space G/K (Smoke [2]). If K is compact and the fixed point set of an involutive automorphism of G then G/K is a Riemannian symmetric space and the algebra U(g, f) is commutative.