Motivation: I'm trying to work out some calculations on the cotangent bundle of a Lie group and I'd like to use the fact that on Lie groups there are global frames. I'm not sure if what I'm doing is correct, so I was hoping I could get some feedback.
Set-up: Let $M$ be an $n$-dimensional Lie group and $g$ a left-invariant riemannian metric on $M$. Then we can construct a global orthonormal frame $V_1, \cdots, V_n$ for the tangent bundle and write \begin{equation} TM= M \times \mathbb R^n. \quad \quad \quad (1) \end{equation}
Similarly, by taking the dual coframe $W_1, \cdots, W_n$, we get $$ T^*M = M\times \mathbb R^n. \quad \quad \quad (2) $$ Then the tangent bundle of $T^*M$ can be written $$ T(T^*M) = TM \times T\mathbb R^n =T^*M \times \mathbb R^{2n}, $$ where the first equality holds since $T^*M$ is a product manifold and the second equality holds by regrouping or just by $(1)$ since $T^*M$ is also a Lie group because it is the product of two Lie groups.
My question is the following: I'd like to say that $V_1, \cdots, V_n, W_1, \cdots, W_n$ form a global frame for $T(T^*M)$. This seems like it should be true because $T^*M$ is $M\times \mathbb R^n$, and so at any point I can move along $M$ (giving me the $V_1, \cdots, V_n$ ) or I can move within the fibre $\mathbb R^n$ (giving me $W_1, \cdots, W_n$). However, $W_1, \cdots, W_n$ technically arent "tangent" vectors, like $V_1, \cdots, V_n$ are, so I'm wondering if I need to worry about that. On the other hand, because the fibre is a vectorspace it is isomorphic to its tangent space and so maybe this identification is okay?