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 bracet 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?

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?

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? 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?

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 bracet 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?