Computing $\left(\frac{\partial x^{i}}{\partial y^{j}}\right)_{p}$ for $\mathbb{R}$ valued component functions.

Question

Consider the manifold $(\mathbb{R}^{2}, \mathcal{O})$ (where $\mathcal{O}$ is the standard topology on $\mathbb{R}^{2}$) be equipped with the atlas $\mathcal{A} = \{(\mathbb{R}^{2}, x), (\mathbb{R}^{2}, y)\}$ such that $x: \mathbb{R}^{2} \to \mathbb{R}^{2}$, where $x(a,b) := (a,b)$, and $y: \mathbb{R}^{2} \to \mathbb{R}^{2}$ where $y(a,b) : =(a,b + a^{3})$.

We are asked to compute the $(\text{dim}\mathbb{R}^{2})^{2} = 4$ objects $\left(\frac{\partial x^{i}}{\partial y^{j}}\right)_{p} := \partial_{j}(x^{i} \ \circ \ y^{-1})(y(p))$.

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I found the inverse of $y$ to be defined as $y^{-1}(u,v) = (u, v-u^{3})$ and so we have $x^{i} \ \circ \ y^{-1} : \mathbb{R^{2}} \to \mathbb{R}$ where using the chart map $x$ we have $(x \ \circ \ y^{-1})(u,v) = (u, v-u^{3})$. In the solution for $\left(\frac{\partial x^{2}}{\partial y^{1}}\right)_{p}$, for example, the author says to differentiate $x^{2}$ (the second component function of $x$) with respect to the first entry (which in this case would be the symbol $u$) which he claims would give us

$$\partial_{1}(x^{2} \ \circ \ y^{-1})(a, b+a^{3}) = -3u^{2} = -3a^{2}. (*)$$

However, in the video, there's a lot of steps skipped and I can't seem to understand how he arrived at this solution. Do we have to perform the multidimensional chain rule for each $\left(\frac{\partial x^{i}}{\partial y^{j}}\right)_{p}$? I.e., maybe for the example above, if $$h(a, b+a^{3}) := x^{2}(y^{-1}(a, b+a^{3})) $$ then (writing something like the following, where $u=a$, $v = b+a^{3}$) $$Dh(u,v) = Dx^{2}(y^{-1}(u,v)))Dy^{-1}(u,v)?$$

If someone could please walk me through these steps or show me that I'm completely overthinking it would be greatly appreciated.

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$(*)$ Another thought: It seems like, since $x$ is the identity function on $\mathbb{R}^{2}$, we don't need to use the chain rule explicitly and can just take the partial derivative of $x^{i}$ with respect to either $u$ or $v$.

Answer 1

There is no need for the chain rule. As you showed$$ x \circ y^{-1} : \mathbb R^2 \to \mathbb R^2 : (u,v) \mapsto(u,v-u^3).$$

Therefore define a function $$ g : \mathbb R^2 \to \mathbb R : (u,v) \mapsto v-u^3 $$ notice that this function is exactly $g = x^2\circ y^{-1}.$

Can you compute the partial derivative of the above function with respect to the first component $u$? Yes ! $$ \frac{\partial g}{\partial u} = \frac{\partial }{\partial u}(v-u^3) = -3u^2 = \partial_1(x^2\circ y^{-1})(u,v)$$ Now fix a specific point $p = (a,b)$ and evaluate the above partial derivatives at $y(p) = (a,b+a^3)$ and you find that

$$\partial_1(x^2\circ y^{-1})(a,b+a^3) = -3a^2$$

Answer 2

This is much simpler than it sounds and as the comment above implies, it's due to poor notation. You've got two charts, each with a function from $\mathbb{R^2}$ to $\mathbb{R^2}$ and denoted by $x$ and $y$. Since these are vector-valued functions of two variables, this can be improved by bolding them $\mathbb{x}$ and $\mathbb{y}$. Then the notation for the 4 partial derivatives using $x$ and $y$ does not seem right to me since $x$ and $y$ are not being used as dependent and independent variables as the usual partials notation assumes. In other words, just express the transition function $x \circ y^{-1}$ as a function of $u$ and $v$ as you did and the 4 objects you want form the standard Jacobian, namely $1, 0, -3u^2$ and $1$ with some care to get each into the right spot in the resulting 2 x 2 matrix. Using $a$ and $b$ would just mean putting in an actual point. Added: above excellent answer appeared while I was writing this.