Tensor change of coordinates - why one way is inverse matrix of the other?

35 Views Asked by At

I am working on tutorial on connections from Gravity and Light winter school. I have a problem understanding solution from walkthrough on youtube. To calculate connection coefficients in different chart I have to calculate

$(\frac{\partial{x^i}}{\partial{y^j}})_{y^{-1}(r, \phi)} = \partial_j(x^i \circ y^{-1})(r, \phi)$ and $(\frac{\partial{y^i}}{\partial{x^j}})_{x^{-1}(a, b)} = \partial_j(y^i \circ x^{-1})(x(p)) = \partial_j(y^i \circ x^{-1})(x \circ y^{-1} \circ y(p)) = \partial_j(y^i \circ x^{-1})(x \circ y^{-1}(r, \phi))$.

I see how that's correct, what I am missing is what follows: $(\frac{\partial{y^i}}{\partial{x^j}})_{x^{-1}(a, b)} = (\frac{\partial{y^i}}{\partial{x^j}})_{y^{-1}(r, \phi)}$ and based on that it is calculated as matrix inverse of the $(\frac{\partial{x^i}}{\partial{y^j}})_{y^{-1}(r, \phi)}$. Why is it correct? What happens to the chart transition function $x \circ y^{-1}$ from the second last equation?

1

There are 1 best solutions below

1
On BEST ANSWER

Let $V\subset M$ be the domain of the chart $y$ and consider the open set $y(U)\subset\mathbb R^n$. Observe the equality $$(y^i\circ x^{-1})\circ(x\circ y^{-1})=\pi^i$$ where $\pi^i:y(V)\to\mathbb R$ is the projection to the $i$th coordinate. Now use the chain rule on this equality.