Tensor analysis - Solving for $A_q$

Question

To show that if $\bar{A}_p=\frac{\partial x^q}{\partial \bar{x}^p}A_q \implies A_q=\frac{\partial \bar{x}^p}{\partial x^q}\bar{A}^p$ I wonder I can multiply by $\frac{\partial \bar{x}^p}{\partial x^q}$ at both sides of the given expresión for $\bar{A}_p$ but how is this turned into a valid argument? In this case is $\bar{A}_p$ a covariant tensor and $A_q$ a contravariant tensor?

Answer 1

$$\bar A_p=\frac{\partial x^q}{\partial \bar x^p}A_q$$ is the coordinate tranformation law for covectors, that is, it describes what relation exists among the components of the same covector under two different coordinate systems, in a given point.

At the left hand side, if $\bar A_p$ is

$(1)$ the first-order partial derivative of the scalar function $A$ w.r.t. the coordinate $\bar x^p$ at the given point with all other coordinates $\{\bar x^i\}$ held constant, than it is

$(2)$ the value that the linear functional $\operatorname{d}\!A$, the differential of $A$, takes on the basis vector $\partial/\partial \bar x^p$ at the given point,

$(3)$ whose expansion over the linear-function basis (covector basis) $\{\operatorname{d}\!x^q\}$ is given by $\operatorname{d}\!A=A_q\operatorname{d}\!x^q$

$(4)$ where $\operatorname{d}\!x^q$, being the dual of the vector basis $\{\partial/\partial x^q\}$, is the differential of the $q$-coordinate function in the said vector basis at the given point, and so it gives the $q$-component of its vector argument. As remembered at the step $(1)$-$(2)$, the differential of a function, when applied on a basis vector, is the first-order partial derivative of the function w.r.t. the coordinate corresponding to the basis vector itself, holding all other coordinates constant in the same coordinate system.

$$\bar A_p\,\overset{_{(1)}}=\,\frac{\partial A}{\partial\bar x^p}\,\overset{_{(2)}}=\,\operatorname{d}\!A\left(\frac{\partial}{\partial\bar x^p}\right)\,\overset{_{(3)}}=\,A_q\operatorname{d}\!x^q\left(\frac{\partial}{\partial\bar x^p}\right)\,\overset{_{(4)}}=\,A_q\frac{\partial x^q}{\partial\bar x^p}$$

More generally $(1)$, $(2)$, and $(3)$ above transform into the followings.

$\bar A_p$ is

$(1')$ the $p$-th component of a linear functional $\bar A_p\operatorname{d}\!\bar x^p$ in the coordinate system relative to the covector basis $\{\operatorname{d}\!\bar x^p\}$ at the given point and so, being $\{\operatorname{d}\!\bar x^p\}$ dual to the vector basis $\{\partial/\partial\bar x^p\}$, such a component is the value that the linear functional takes on the basis vector $\partial/\partial\bar x^p$ at the given point,

$(2'$-superfluous$)$ let's rewrite the linear functional as $\omega=\bar A_p\operatorname{d}\!\bar x^p$,

$(3')$ whose expansion over the linear-function basis (covector basis) $\{\operatorname{d}\!x^q\}$ is given by $\omega=A_q\operatorname{d}\!x_q$

$$\bar A_p\,\overset{_{(1^{'})}}=\, \bar A_q\operatorname{d}\!\bar x^q\left(\frac{\partial}{\partial\bar x^p}\right)\,\overset{_{(2^{'})}}=\, \omega\left(\frac{\partial}{\partial\bar x^p}\right)\,\overset{_{(3^{'})}}=\, A_q\operatorname{d}\!x^q\left(\frac{\partial}{\partial\bar x^p}\right)\,\overset{_{(4)}}=\, A_q\frac{\partial x^q}{\partial\bar x^p}$$

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In this formula we decided to start from the "barred" basis on the left hand side to the "unbarred" one on right hand side. But you can go just as easily in the opposite direction, thereby getting: $$A_q=\frac{\partial\bar x^p}{\partial x^q}\bar A_p$$

Please note that you cannot use, as you were trying to do, a relation like this one: $$\frac{\partial \bar x^p}{\partial x^q}=\left(\frac{\partial x^q}{\partial \bar x^p}\right)^{-1}$$ because it does not hold true (in general): that symbol does not represent, as it should be clear by now, a ratio (as opposed to, e.g., $\frac{df}{dx})$.