Tangent and cotangent bundle as an associated bundle

Question

In a book I read the isomorphisms below were mentioned without any explanation. Is there any intuitive way to see these identities hold?

Let $M$ be a smooth $n$-manifold, $F(M)$ a frame bundle of $M$, and $\lambda_{0}$ the standard action of $\operatorname{GL}(n, \mathbb{R})$ on $\mathbb{R}^{n}$. Then

$$\begin{align} F(M) \times_{\lambda_{0}} \mathbb{R}^{n} &\cong TM \\ F(M) \times_{\lambda^{*}_{0}} \mathbb{R}^{n} &\cong T^{*}M \end{align} $$ where $\lambda^{*}_{0}$ denotes the dual representation to $\lambda_{0}$.

Answer 1

Take an $n$-dimensional real vector space $V$. The group $\operatorname{GL}(V)$ acts on the left of $V$ in the usual way $g \cdot v = gv$, and on the left of the dual $V^*$ by $g \cdot \varphi = \varphi g^{-1}$ for a linear functional $\varphi \colon V \to \mathbb{R}$.

Now let $FV$ be the set of frames in $V$, i.e. $$ FV = \{(v_1, \ldots, v_n) \mid \text{The } v_i \text{ are a basis of } V\}.$$ The set $FV$ is a left $\operatorname{GL}(V)$-space under the action $g \cdot (v_1, \ldots, v_n) = (gv_1, \ldots, gv_n)$, and a right $\operatorname{GL}_n$-space under the action $(v_1, \ldots, v_n) \cdot A = [v_1 \cdots v_n] A$ (i.e. consider $(v_1, \ldots, v_n)$ to be a row vector, and right-multiply by the matrix $A$). Note that these two actions commute. We want to "use up" the right action by forming an associated bundle, while keeping the left action.

There is a natural map $FV \times \mathbb{R}^n \to V$, given by $$((v_1, \ldots, v_n), (w_1, \ldots, w_n)) \mapsto w_1 v_1 + \cdots + w_n v_n.$$ Furthermore, this map is $\operatorname{GL}(V)$-equivariant on the left. You can check that this map factors through the equivalence relation $(fA, w) \sim (f, Aw)$ for $A \in \operatorname{GL}_n$, and hence we get a $\operatorname{GL}(V)$-equivariant homomorphism $FV \times_\lambda \mathbb{R}^n \to V$, where $\lambda \colon \operatorname{GL}_n \times \mathbb{R}^n \to \mathbb{R}^n$ is the standard action $\lambda(A, w) = Aw$.

Aside: A quick way to check that this map respects the equivalence relation is to view a frame $(v_1, \ldots, v_n)$ as an injective linear map $v \colon \mathbb{R}^n \to V$, where $v_i = v(e_i)$. The right $\operatorname{GL}_n$-action on a frame is $v \cdot A = vA$, i.e. precomposition by the matrix $A$. Given a vector $w = (w_1, \ldots, w_n) \in \mathbb{R}^n$, the linear combination $w_1 v_1 + \cdots + w_n v_n$ is equal to $v(w)$. Therefore we are simply checking that $(vA)(w) = v(Aw)$.

There is also a natural map $\sigma \colon FV \times \mathbb{R}^n \to V^*$, given by $$((v_1, \ldots, v_n), (w_1, \ldots, w_n)) \mapsto w_1 v_1^* + \cdots + w_n v_n^*,$$ where $(v_1^*, \ldots, v_n^*)$ is the basis dual to $(v_1, \ldots, v_n)$. In view of the aside above, treating a frame as an injective map $v \colon \mathbb{R}^n \to V$, the dual frame is the inverse map $v^* \colon V \to \mathbb{R}^n$ where $v_i^* = e_i^* \circ v^{-1}$, and the linear combination $w_1 v_1^* + \cdots + w_n v_n^* = w^T \circ v^*$. Now we can check what equivariance we can expect in the middle: $$ \sigma (vA, w) = w^T \circ (vA)^{-1} = w^T A^{-1} v^{-1} = ((A^{-1})^T w)^T v^{-1}) = \sigma (v, (A^{-1})^T w),$$ and so $\sigma$ descends to a homomorphism $\pi: FV \times_{\lambda^*} \mathbb{R}^n \to V^*$ where $\lambda^*$ is the left action $\lambda^*(A, w) = (A^{-1})^T w$ of $\operatorname{GL}_n$ on $\mathbb{R}^n$. It just remains to check that this map is $\operatorname{GL}(V)$-equivariant on the left: $$ \pi(g \cdot [v, w]) = \pi([gv, w]) = w^T (gv)^{-1} = w^T v^{-1} g^{-1} = \pi([v, w]) g^{-1},$$ and indeed it is.