<s>[INST] jacobian times its inverse - should be identity Here's an easy one.  
A Jacobian is $\frac{dx^i}{dy^j}$.
The inverse is $\frac{dy^j}{dx^k}$.  So, in tensor notation,

$\frac{dx^i}{dy^j} \frac{dy^j}{dx^k} = \frac{dx^i}{dx^k} = \delta^i_k$

Now I'll try to this as in matrix form, in two dimensions:

$\left[
  \begin{array}{cc}
  \frac{dx^1}{dy^1}  &  \frac{dx^1}{dy^2}  \\
  \frac{dx^2}{dy^1}  &  \frac{dx^2}{dy^2}  \\
  \end{array}
  \right]
  \cdot
  \left[
  \begin{array}{cc}
  \frac{dy^1}{dx^1}  &  \frac{dy^1}{dx^2}  \\
  \frac{dy^2}{dx^1}  &  \frac{dy^2}{dx^2}  \\
  \end{array}
  \right]
$

The 1,2 element of this product is 

$\frac{dx^1}{dy^1}\frac{dy^1}{dx^2} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^2} = \frac{dx^1}{dx^2}+\frac{dx^1}{dx^2} = 0
$

as required.
But looking at the 1,1 element,

$
  \frac{dx^1}{dy^1}\frac{dy^1}{dx^1} + \frac{dx^1}{dy^2}\frac{dy^2}{dx^1} = 2 \frac{dx^1}{dx^1} = 2
$

which is wrong…  but why?
 [/INST] In two dimensions, we have functions $f_1$ and $f_2$ which are both functions of $x_1$ and $x_2$. The Jacobian matrix is given by,
$$J=\left( 
\begin{matrix}
\partial_1 f_1 & \partial_2 f_1 \\
\partial_1 f_2 & \partial_2 f_2
\end{matrix}
\right)$$
where  $\partial_n := \partial/\partial x_n$. Computing the inverse of the Jacobian using the standard formula yields,
$$J^{-1}=\frac{1}{\partial_1f_1 \partial_2f_2-\partial_2f_1\partial_1f_2}\left( 
\begin{matrix}
\partial_2 f_2 & -\partial_2 f_1 \\
-\partial_1 f_2 & \partial_1 f_1
\end{matrix}
\right)$$
Multiplying both matrices yields the desired result,
$$JJ^{-1}=J^{-1}J=\left( 
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right)$$
You just used the wrong expression for the inverse.
</s>