Concept

Representation theory of diffeomorphism groups

Summary
In mathematics, a source for the representation theory of the group of diffeomorphisms of a smooth manifold M is the initial observation that (for M connected) that group acts transitively on M. A survey paper from 1975 of the subject by Anatoly Vershik, Israel Gelfand and M. I. Graev attributes the original interest in the topic to research in theoretical physics of the local current algebra, in the preceding years. Research on the finite configuration representations was in papers of R. S. Ismagilov (1971), and A. A. Kirillov (1974). The representations of interest in physics are described as a cross product C∞(M)·Diff(M). Let therefore M be a n-dimensional connected differentiable manifold, and x be any point on it. Let Diff(M) be the orientation-preserving diffeomorphism group of M (only the identity component of mappings homotopic to the identity diffeomorphism if you wish) and Diffx1(M) the stabilizer of x. Then, M is identified as a homogeneous space Diff(M)/Diffx1(M). From the algebraic point of view instead, is the algebra of smooth functions over M and is the ideal of smooth functions vanishing at x. Let be the ideal of smooth functions which vanish up to the n-1th partial derivative at x. is invariant under the group Diffx1(M) of diffeomorphisms fixing x. For n > 0 the group Diffxn(M) is defined as the subgroup of Diffx1(M) which acts as the identity on . So, we have a descending chain Diff(M) ⊃ Diffx1(M) ⊃ ... ⊃ Diffxn(M) ⊃ ... Here Diffxn(M) is a normal subgroup of Diffx1(M), which means we can look at the quotient group Diffx1(M)/Diffxn(M). Using harmonic analysis, a real- or complex-valued function (with some sufficiently nice topological properties) on the diffeomorphism group can be decomposed into Diffx1(M) representation-valued functions over M. So what are the representations of Diffx1(M)? Let's use the fact that if we have a group homomorphism φ:G → H, then if we have a H-representation, we can obtain a restricted G-representation. So, if we have a rep of Diffx1(M)/Diffxn(M), we can obtain a rep of Diffx1(M).
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