Remark 2

Suppose that G is finite-dimensional. The mapping

of $U_s(G) \times G$ into $T^{(s)}(G)$ is an isomorphism of analytic vector bundles; for this mapping is bijective, linear on each fibre and analytic (no. 5); on the other hand, let $\phi : T^{(s)}(G) \to U_s(G) \times G$ be the inverse bijection; if $t \in T^{(s)}_g(G)$, then $\phi(t) = (t * \varepsilon_{g^{-1}}, g)$ and hence $\phi$ is analytic. The isomorphism $\phi$ is called the right trivialization of $T^{(s)}(G)$. Similarly, consider the mapping $(t, g) \mapsto (\mathbf{L}_t)_g = \varepsilon_g * t$ of $U_g(G) \times G$ into $T^{(s)}(G)$; the inverse isomorphism is called the left trivialization of $T^{(s)}(G)$. By restriction we recover the right and left trivializations of $T(G)$ (§ 2, no. 2).