A Lie group whose Lie algebra is equal to (the Lie algebra? of )all functions with fibrewise polynomial growth

Question

Let $M$ be a Riemannian manifold. We denote by $\mathfrak{g}$ the space of all smooth function $f:TM\to \mathbb{R}$ with fibre wise polynomial growth. Is it a Lie algebra wrt the Poisson bracket on symplectic manifold $TM$? What is a precise infinite dimensional Lie group whose Lie algebra is the above $\mathfrak{g}$? Is the isomorphism class of Lie algebra mentioned above independent of choosing Riemanian metric?

Answer 1

The following will only deal with the Lie algebra, the question about the Lie group is far beyond my capabilities.

The symplectic structure is (I guess) the one coming from the musical isomorphism of the Riemannian metric and the canonical one on $T^*M$. Since the musical isomorphism is a vector bundle isomorphism (over the identity), the Poisson bracket you get on $TM$ has similar homogeneity properties as the canonical one on $T^*M$. One has \begin{equation} \{\operatorname{Pol}^k(TM), \operatorname{Pol}^\ell(TM)\}_g \subseteq \operatorname{Pol}^{k+\ell-1}(TM) \end{equation} where $\operatorname{Pol}^k(TM)$ denotes those smooth functions which are homogeneous polynomials of degree $k \in \mathbb{N}_0$ in fiber direction.

Since all these Poisson brackets (resp. symplectic forms) on $TM$ originate from the canonical one on $T^*M$ they are all isomorphic as Poisson algebras (resp. symplectomorphic). Since the musical isomorphism is a vector bundle isomorphism it preseves polynomial degrees of functions, too. This is in fact an equivalent characterization of a vector bundle morphism.

So: the first answer I can give is that the polynomial functions $\operatorname{Pol}^\bullet(TM)$ form a Lie subalgebra of all smooth functions with respect to the above Poisson bracket $\{.,.\}_g$. Any two metrics $g$ and $g'$ lead to isomorphic such subalgebras.

As a hint on the "Lie group" one notes that there is a subalgebra inside: $\operatorname{Pol}^0(TM) \oplus \operatorname{Pol}^1(TM)$ is a Lie subalgebra, though not a Poisson subalgebra. On $T^*M$ the characterization becomes more clear: The constant functions $\operatorname{Pol}^0(T^*M) \cong C^\infty(M)$ generate fiber translations by exact one-forms on $T^*M$, the linear functions $\operatorname{Pol}^1(T^*M) \cong \Gamma^\infty(TM)$ generate (by the flows of the corresponding vector fields) point transformations of $T^*M$. One can now transfer this back to $TM$ by the musical isomorphism and gets a similar description of the flows (and hence the group they generate). It is isomorphic to a semidirect product of the (small) diffeomorphisms $\operatorname{Diffeo}(M)$ with the vector space of exact one-forms and action given by the usual pull-back.

The Hamiltonian flows of all the polynomial functions generate way more general symplectomorphism. In fact, I guess that not much intersting things can be said here, this is very close to group of all (small) symplectomorphisms.

That being said, let us try to answer the actual question (at last):

• If you mean by polynomial growth the above polynomial functions, then the above discussion should give you all you need.

• However, if polynomial growth really means smooth functions which can be estimated by polynomials in fiber directions, then things really become way more complicated: first, one most probably wants to have locally uniform such estimates in $M$-direction. Second, and this is crucial, one needs to estimate also derivatives: otherwise the Poisson bracket of two functions, which are say bounded but oscillate very wildly, will not have polynomial growth anymore. So without conditions on the derivatives polynomial growth will not lead to a Lie subalgebra at all.

This will ultimately lead to symbol spaces on $TM$, which are precisely designed such that one has good control on the behaviour in fiber direction including all derivatives as well. There are many competing ways define symbol spaces, so perhaps you should clarify first which one is useful for you.

In any case, the symbol spaces are intrinsic to the vector bundle structure. Since the above Poisson brackets are compatible with the vector bundle structure in a way which does not depend on the chosen metric, I guess that one can get isomorphic Lie subalgebras once one has guaranteed that the symbol spaces in question yield Lie subalgebras at all.

Again, the integrating Lie group is most probably very complicated. From a technical point of view, it will suffice to consider $T^*M$ instead. Then works of Hörmander, Duistermaat, Guillemin, Sternberg and many more on Fourier integral operators and symbol spaces might give you an impression what can be done.