Show $x_{l,L}X_{L,k}=\delta_{kl}$

Question

In lectures we had $$x_{l,L}=\frac{\partial x_l}{\partial X_L}$$ and $$X_{L,k}=\frac{\partial X_L}{\partial x_k}$$

So this is what I THOUGHT it would be but i was told that can't cancel the partial big $X$s... $$x_{l,L}X_{L,k}=\frac{\partial x_l}{\partial X_L}\frac{\partial X_L}{\partial x_k}=\frac{\partial x_l}{\partial x_k}=\delta_{kl}$$

Please help.

Answer 1

In fact, the $X$'s do not cancel but the following is valid

$${{\partial {x_l}} \over {\partial {X_L}}}{{\partial {X_L}} \over {\partial {x_k}}} = {{\partial {x_l}} \over {\partial {x_k}}}\tag{1}$$

Why? OK, let me explain in a better notation

$$\left\{ \matrix{ {\bf{x}} = {\bf{x}}({\bf{X}}) \hfill \cr {\bf{X}} = {\bf{X}}({\bf{x}}) \hfill \cr} \right.\,\,\, \to \,\,\,{\bf{x}} = {\bf{x}}\left( {{\bf{X}}({\bf{x}})} \right)\tag{2}$$

Now, just differentiate with respect to ${\bf{x}}$ and use the chain-rule to obtain $(1)$. I will leave it to you.