Suppose I have $M^2$ a closed (i.e. compact with no boundary) Riemannian manifold, and
$$T: \text{Sym}^2(TM) \to \mathbb{R}$$
Coordinatize an open set $U \subseteq M^2$ with $\{(x,y)\}$. Suppose I know that
$$ T(\partial_x, \partial_x), T(\partial_x, \partial_y), T(\partial_y, \partial_y)$$
are constant on $U$. This holds for any coordinate system on $U$, as well as any $U$ which admits a chart map. Can I conclude that $T$ is the $0$ two-tensor? Or can I conclude anything in general about $T$?
With functions, one would just paste together the value across all charts, but this doesn't make sense for tensors since the tangent space is different for every point. Maybe $M$ is parallelizable?
For convenience let's assume $T_{xx} = c \ne 0$ and that $x > 0$ on $U.$ If we make the change of variables $(x,y) \to (z,y)$ such that $x = z^2,$ then $$T_{zz} = \left(\frac{\partial x}{\partial z}\right)^2T_{xx} = 4z^2T_{xx} = 4cz^2$$ which is not constant. Thus the only tensor that is constant in all coordinates is the zero tensor.