# Smoothness of local trivialisation of a vector bundle

I'm building a vector bundle structure for the space $L_{alt}^k(TM) = \bigcup_{p \in M}L_{alt}^k(T_pM)$, the bundle of alternating k-multilinear maps in $T_pM$.

$L_{alt}^k(T_pM)$ is a vector space of dimension $n \choose k$ and has $\{dx^{i_1} \wedge \dotsc \wedge dx^{i_k}:1 \leq i_1 < \dotsc < i_k \leq n \}$ as a basis.

So, if $\alpha \in L_{alt}^k(T_pM)$, then $\alpha = \sum a_{i_1\dotsc i_k} dx^{i_1} \wedge \dotsc \wedge dx^{i_k}$, the coefficients uniquely determined by $\alpha$ and $\{dx^{i_1} \wedge \dotsc \wedge dx^{i_k}:1 \leq i_1 < \dotsc < i_k \leq n \}$.

This defines a linear isomorphism $\Phi_p:L_{alt}^k(T_pM) \to \mathbb{R}^d$ where $d = {n \choose k}$ for each $p \in M$.

Given a chart $(U_\alpha, x_\alpha)$ of $M$, the collection of maps $\{\Phi_p\}_{p \in U_\alpha}$ is my candidate for the principal part $\Phi_\alpha$ of the local trivialisation $\phi_\alpha=(\pi,\Phi_\alpha)$ for that chart.

How can I show that $\Phi_\alpha$ is smooth and $\Phi_p$ is a diffeomorphism for each $p \in U_\alpha$??

• By brute force, using the definitions. Only perserverance is required. Oct 7 '15 at 22:30
• @Mariano Could you elaborate on that? I only know smoothness for maps between smooth manifolds... Oct 7 '15 at 22:55
• You know the definition of manifolds and you have candidate for charts. Compute the transition functions and observe that they are smooth. Oct 7 '15 at 23:02
• @MarianoSuárez-Alvarez edited, is this argument correct? Oct 8 '15 at 0:20
• I suggest you review the construction of the tangent bundle itself and its dual. Oct 8 '15 at 0:29

Finally solved it by revisiting the construction of the cotangent bundle. We have that given two charts that overlap $(U,x), (V,y)$
$$dx^{i_k} = \sum_{j_k}\frac{\partial x^{i_k}}{\partial y^{j_k}}\cdot dy^{j_k}$$ so after some tedious but straightforward calculations we get that $$a_{i_1\cdots i_k} = \sum_{\overset{\to}{L}}a_{i_1\cdots i_k}\frac {\partial(x^{i_1},\ldots,x^{i_k})}{\partial(y^{l_1},\ldots,y^{l_k})}:= \sum_{\overset{\to}{L}}a_{i_1\cdots i_k}\frac{\partial x^I}{\partial x^L}$$ gives the transformation of local coordinates for each coefficient, so given $a \in \mathbb{R}^d$ we get that each $a_i$ goes to $$a_i \overset{proy_i \circ \Phi_x^{-1}}{\longmapsto} \alpha_i = a_{i_1\dotsc i_k} dx^{i_1} \wedge \dotsc \wedge dx^{i_k} \overset{proy_i \circ \Phi_y}{\longmapsto}\sum_{\overset{\to}{L}}a_{i_1\cdots i_k}\frac{\partial x^I}{\partial x^L}$$ which is clearly smooth. A similar procedure gives the inverse transformation and since $\Phi$ is biyective we get the result I was looking for.