how can to prove this Differential Geometry Question Let $M$ be a smooth manifold of dimension $n$, and $\sigma$ is a covariant tensor field of order $k$ over $M$. If $(U, (x^{i}))$ and $( ~{U}, (~x^{i}))$ are charts in $M$ with non-empty intersection, we can write
$\sigma= \sigma_{i_{1}...i_{k}}dx^{i_{1}} \oplus \cdot\cdot \cdot \oplus dx^{i_{k}} =  \tilde{\sigma}_{i_{1}...i_{k}}d{\tilde x}^{i_{1}} \oplus \cdot\cdot \cdot \oplus d{\tilde x}^{i_{k}}
$
Find a transformation law that expresses the components $\sigma_{i_{1}...i_{k}} $  in terms of the components $ \tilde{\sigma}_{i_{1}...i_{k}}$.
 A: You should show more of your effort. I don't think anyone here is actually going to give you the full step by step computations since you should do them yourself. But here's the way to go: notice that $\tilde{\sigma}_{i_1 \cdots i_k} = \sigma(\tilde{e}_1, \cdots, \tilde{e_k}) = \sigma_{i_1 \cdots i_k} \ (\mathrm{d}x^{i_1} \otimes \cdots \otimes \mathrm{dx}^{i_k})(\tilde{e}_1, \cdots, \tilde{e_k})$. It's easy to show that if $(x, U), (y, \tilde{U})$ are charts around $p$ such that $U \cap \tilde{U} \neq \emptyset$ and $e_i = \dfrac{\partial}{\partial x^i}, \tilde{e}_i = \dfrac{\partial}{\partial y^i}$, then:
$$
\left.\frac{\partial}{\partial y^{j}}\right|_{q}=\left.\left.\frac{\partial}{\partial y^{j}}\right|_{q}\left(x^{i}\right) \frac{\partial}{\partial x^{i}}\right|_{q}
$$
for every $q \in U \cap \tilde{U}$, where I'm using Einstein notation. If you understand well what $(\mathrm{d}x^{i_1} \otimes \cdots \otimes \mathrm{dx}^{i_k})$ does, it's pretty straightforward to finish it from here.
