I would like to prove that the multilinearity condition for a tensor field is equivalent to the transformation rule for its components in a coordinate basis. One way is straightforward, but I haven't been able to find a proof of the converse. For the sake of simplicity, I consider the case of a (1,1) tensor.
Let $\mathfrak{X}(M)$ be the set of tangent vector fields on $M$ and $\mathfrak{X}(M)^*$ the set of dual vector fields. Let $\mathbf{T}$ be a $\mathbb{R}$-bilinear map from $\mathfrak{X}(M)^* \times \mathfrak{X}(M)$ to $\mathbb{R}^M$. When is $\mathbf{T}$ a tensor field? In other words, when is it $\mathbb{R}^M$-bilinear?
For any two maps $(U,x)$ and $(V,y)$ on $M$ such that $U\cap V\neq \emptyset$, we define $J_{x\rightarrow y}{}^\mu{}_\nu |_P:=\frac{\partial {y^\mu}}{\partial {x^\nu}}|_P \in M_2(\mathbb{R})$, or $J^\mu{}_\nu$ for short. Hence, we have $\partial_{y^\mu}=(J^{-1})^\rho {}_\mu \partial_{x^\rho}$ and $dy^\nu=J^\nu{}_\sigma dx^\sigma$. The components of $\mathbf{T}$ in the coordinate map $(U,x)$ are defined by $T_{(x)\mu}{}^\nu:=\mathbf{T}(\partial_{x^\mu}{},dx^\nu)$.
I would like to show that $\mathbb{R}^M$-bilinearity for $\mathbf{T}$ is $equivalent$ to the transformation rule for $T_\mu{}^\nu$:
$``$Lower indices transform with $J^{-1}$ and upper indices transform with $J"$.
- Proof of $\implies$:
If $\mathbf{T}$ is $\mathbb{R}^M$-bilinear, since $(J^{-1})^\rho {}_\mu$ and $J^\nu{}_\sigma$ are scalar fields, we have:
$T_{(y)\mu}{}^\nu=\mathbf{T}(\partial_{y^\mu}{},dy^\nu)=\mathbf{T}((J^{-1})^\rho {}_\mu \partial_{x^\rho},J^\nu{}_\sigma dx^\sigma)=(J^{-1})^\rho {}_\mu J^\nu{}_\sigma \mathbf{T}(\partial_{x^\rho},dx^\sigma)=(J^{-1})^\rho {}_\mu J^\nu{}_\sigma T_{(x)\rho}{}^\sigma$
Hence, $T_\mu{}^\nu$ satisfies the transformation rule. $\square$
- Proof of $\impliedby$?
Conversely, given $\mathbb{R}$-bilinear $\mathbf{T}$, if $T_\mu{}^\nu$ satisfies the transformation rule for any two maps $(U,x)$ and $(V,y)$, how does one show that $\mathbf{T}$ is $\mathbb{R}^M$-bilinear?
That is, how does one show $\mathbf{T}(a.X,\omega)=\mathbf{T}(X,a.\omega)=a.\mathbf{T}(X,\omega)$ for an arbitrary scalar field $a\in \mathbb{R}^M$, vector field $X$ and dual field $\omega$?
I also posted this question here: https://physics.stackexchange.com/posts/790871/.