# How do I get the correct coordinate transition functions for the dual bundle?

Suppose $E$ is a vector bundle over $X$, and $X$ is equipped with an atlas of charts $U_{i}$ that is locally finite. Now assume up to possible refinement we have local trivilization $f_{i}: E_{U_{i}}\rightarrow U_{i}\times \mathbb{R}^{n}$, where $\mathbb{R}^{n}$ is the fibre over $U_{i}$. If $E\rightarrow X$ is well defined we should have transitional functions between two charts $U_{i}\cap U_{j}\not=\emptyset$ such that in $U_{i}\cap U_{j}$ we can change from one chart to the other by taking $g_{ij}=f_{j}f_{i}^{-1}$. It is clear that $g_{ij}\circ g_{jk}=g_{ik}$ and $g_{ij},g_{jk},g_{ik}\in GL(n,\mathbb{R})$.

My question is, for the same vector bundle $E$ if we take its dualization $E^{*}$ with the fibre to be $\mathbb{R^{*}}^{n}$, what will be the dual transitional map $g^{*}_{ij}$, etc? The book give the answer to be $g^{*}_{ij}={g^{-1}_{ij}}^{T}$, which means taking the inverse and then do the transpose. Why this is true?

Let $x\in X$ and $l\in E^{*}_{x}$. We should note that the local trivializations for $E^{*}$ at $U_{i}$ is the set of all linear maps in $\mathbb{R^{*}}^{n}$ such that its value at $f(x,e)$(where $x\in X, e\in E_{x})$ is equal to that of $l(f(x,e))$. But this definition is difficult to work with. I do not know how to simplify the expression $$h(f(x,e))=l(f(x,e),h\in \mathbb{R^{*}}^{n}$$ into some nice local form in a canonical basis. Therefore I do not know how to get $g^{*}_{ij}=h_{j}h_{i}^{-1}$ and prove it is equal to ${g_{ij}^{*}}^{T}$.

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For a fixed $x \in U_i \cap U_j$, $g_{ij}(x)$ is just a linear map from ${\mathbb R}^n$ to ${\mathbb R}^n$. Denote $g = g_{ij}(x) \in GL(n, {\mathbb R})$. Let $l\in ({\mathbb R}^n)^*$. Then $$\langle g^*l, g u \rangle = \langle (f_i^{-1})^*l, (f_i^{-1}) u \rangle =\langle l, u \rangle.$$ Thus for every $v\in {\mathbb R}^n$, we have $$\langle g^*l, v\rangle = \langle l, g^{-1}v\rangle = \langle (g^{-1})^T l, v\rangle.$$ That is, $g^* = (g^{-1})^T$.
This is clear. Thanks. I would only remark $g_{ij}=f_{j}f_{i}^{-1}$ in literature. –  Bombyx mori Sep 9 '12 at 13:33