Tensor fields and scalar function pullbacks

Question

For convenience, $(p,q)$ tensor fields on a differentiable manifold $M$ is defined to be the entire scalar field.
On the other hand, in my textbook, the pullback of the $(p,q)$ tensor $T(x)$ at $x \in M$ by the diffeomorphism map $\sigma:M \rightarrow M$ is defined as follows:

$$(\sigma^*T)(x):=\sigma_x^*T(\sigma(x))\qquad\qquad(1) $$

This definition is correct in case that $p \geq 1 $ or $q \geq 1$ but if $p$ and $q$ are both $0$, then I think the pullback needs to be defined separately. Namely, for a scalar field $f$

$$(\sigma^*f)(x):= (f \circ \sigma)(x)\qquad\qquad(2)$$

Question : Is definition (2) derived from definition (1)?

Answer 1

You can use the same definition:

$$(\sigma^\ast f)(x)=\sigma_x^\ast(f(\sigma(x)).$$

In fact $f(\sigma(x))$ is a $(0,0)$-tensor (i.e. a real number) and the pullback of a $(0,0)$-tensor is simply the $(0,0)$-tensor itself (by definition).

(Or you can simply define that particular case separately as you did. Math is full of this kind of situations).

Answer 2

As we know, a $(1,1)$-tensor is a linear combination $$ \mathbf{T}={T^i}_j\,\frac{\partial}{\partial y_i}\otimes dy^j $$ where the components ${T^i}_j$ are functions of $y_1,...,y_n\,.$ To pullback $\mathbf{T}$ by $\sigma:x\mapsto y$ you can separately define the pullbacks of its building blocks. I prefer to use a bit of sloppy notation that is reminscent of the (guess what?) chain-rule that is so ubiquitous in DG: \begin{align} \sigma^*\Big(\frac{\partial}{\partial y_i}\Big)&=\frac{\partial x_j}{\partial y_i}\frac{\partial}{\partial x_j}\,, &\sigma^*(dy^i)&=\frac{\partial y_i}{\partial x_j}\,dx^j\,,&\sigma^*({T^i}_j)&={T^i}_j\circ\sigma\,. \end{align} Most of the literature on that subject is simply intimidating. The generalization of the above to $(p,q)$ is trivial.