Independent of Choice of Local Coordinates/Global Tensor Field

Question

Let $M$ be a smooth manifold, and let $x_ 1 , . . . , x_n$ be a local coordinate system defined on an open set $U ⊆ M$. Consider the $(1, 1)$-tensor field $C$ defined on $U$ in local coordinates by $C = \sum_{i=1}^n dx_i\otimes \frac{\partial}{\partial x^i}$.

Show that $C$ is independent of the choice of local coordinates and hence defines a smooth global tensor field on $M$.

I'm really stuck on how to do this problem. I'm practicing for quals and don't know where to really start. I think I should use change of coordinates given two local coordinate systems, but I don't know how.

Answer 1

The physicist's way: use standard formulas (BEST!)
If $x=x(y)$ compute:
$$\sum_i dx^i\otimes\frac{\partial}{\partial x^i}=\sum_i (\sum_j \frac{\partial x_i}{\partial y^j} dy^j \otimes \sum_k\frac{\partial y^k}{\partial x^i} \frac{\partial}{\partial y^k})=\sum_i (\sum_{j,k}\frac{\partial x_i}{\partial y^j}\frac{\partial y^k}{\partial x^i}dy^j\otimes \frac{\partial}{\partial y^k})\\=\sum_{j,k} \delta ^k_jdy^j\otimes \frac{\partial}{\partial y^k}=\sum_j dy^j\otimes \frac{\partial}{\partial y^j}$$ and notice that you end up with the same expression in the $y$ coordinates.

The pure mathematician's way (VERY CHIC!)
We have a canonical isomorphism of vector bundles on $M$: $$HOM(T_M,T_M) \stackrel \sim \to T^*_M\otimes T_M $$ Taking global sections we obtain an isomorphism of $ C^\infty (M)$-modules $$\varphi:\Gamma(M, HOM(T_M,T_M))=Hom(T_M,T_M)\stackrel \sim \to \Gamma(M, T^*_M\otimes T_M)$$ The identity morphism on the left $[Id:T_M\stackrel = \to T_M]\in Hom(T_M,T_M)$ is sent to the tensor field $\varphi (Id)=C\in \Gamma(M, T^*_M\otimes T_M)$ on the right.
And this $C$ is your mysterious tensor field: its expression in a local chart $(U,x)$ with domain $U\subset M$ is $$C\vert U=\sum_i dx^i\otimes\frac{\partial}{\partial x^i}\in \Gamma(U, T^*_M\otimes T_M)$$

A personal point of view
As an algebraic geometer used to the abstract tools of scheme theory and as a Bourbaki fan, my natural tendency is to prefer the second point of view.
However in order to keep myself honest I try to occasionally compute the way the physicists (for example the general relativists) do.
Of course Spivak's fantastic treatise convinced me once and for all that one can give a completely rigorous explanation for the sort of computation in the first answer, so that now I do these computations purely algorithmatically, with my brain completely disengaged...

Answer 2

I second Georges's answer, and I'll add one more approach:

A tensor of type $(1,1)$ at a point $p$ can be viewed as a bilinear map from $T_pM\times T_p^*M$ to $\mathbb R$. What does $C$ do? If you apply it to a pair $(V,\omega)$, where $V = \sum_j V^j \partial/\partial x^j$ and $\omega = \sum_k \omega_k dx^k$, you get $C(V,\omega) = V^i\omega_i = \omega(V)$. This last expression is manifestly independent of coordinates.