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I'm a bit struggling to understand one part of how to actually transform the Christoffel symbols, in the case where the manifold is embedded in a higher dimensional space.

As an example consider the $\mathbb{R}^2$ as the standard manifold with the usual Euclidean metric and connection. Define an injective function $f:\mathbb{R}^2 \to \mathbb{R}^d$ which embeds the plane into a higher dimensional in an arbitrary non-linear way. According to the usual rules for change of variables of the Christoffel symbols we have that $$ \Gamma^k_{i j} = \frac{\partial f^k}{\partial x^m} \frac{\partial^2 x^m}{\partial f^i \partial f^j} + \frac{\partial f^k}{\partial x^m} \frac{\partial x^n}{\partial f^i} \frac{\partial x^p}{\partial f^j} \Gamma^m_{n p} $$ Since we are using the standard Euclidian connection in the coordinate system $x$ we have that $\Gamma$ vanishes everywhere so: $$ \Gamma^k_{i j} = \frac{\partial f^k}{\partial x^m} \frac{\partial^2 x^m}{\partial f^i \partial f^j} $$

My main question is how exactly is defined the expression $$ \frac{\partial^2 x^m}{\partial f^i \partial f^j} $$ in order to be able to calculate the Christoffel symbols?

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    $\begingroup$ Above expression is just second partial derivatives of inverse coordinates transformation $(f^i) \to (x^i)$. $\endgroup$ – Si Kucing Nov 29 '19 at 8:38
  • $\begingroup$ Yeah but that must have some specific definition in terms of the forward one. For instance the same way you define the derivative of the inverse, such that the contraction on f gives you the identity - df^k/dx^i dx^j_df^k = delta^j_i, there must be some way of defining this. As an example consider that f : (x,y) -> (x, y, g(x,y)) - e.g. parametric surface. What exactly is dx^2/df^2. I think the point is there is no exact inverse? $\endgroup$ – Alex Botev Nov 29 '19 at 12:04
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You're not really doing a change of variables in this case since your map $\mathbb{R}^2 \to \mathbb{R}^d$ is not invertible (it won't be surjective), so you'll just have to compute Christoffel symbols the old fashioned way (if it were invertible then you would have a map $\mathbb{R}^d \to \mathbb{R}^d$ and the differential of this map would give you a $d$ by $d$ matrix. If you wanted to actually compute, e.g., $\frac{\partial x^m}{\partial f^i \partial f^j}$ you would have to write the differential at each point as a matrix (the Jacobian), invert this matrix, look at the $m,i$ entry and then differentiate that entry with respect to the $j$th variable).

Your map $\mathbb{R}^2 \to \mathbb{R}^d$ is giving you a parametrization for some surface in $\mathbb{R}^d$. We can then write the metric in these coordinates. We have $f(x, y) = (f_1(x, y), ..., f_d(x, y))$. In these coordinates our metric is a matrix of the form $g(x, y) = \begin{bmatrix} \sum_i (\frac{\partial f_i}{\partial x}) ^2 & \sum_i \frac{\partial f_i}{\partial x} \frac{\partial f_i}{\partial y} \\ \sum_i \frac{\partial f_i}{\partial x} \frac{\partial f_i}{\partial y} & \sum_i (\frac{\partial f_i}{\partial y}) ^2 \end{bmatrix}$ (I'm taking the image of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ under $f$ and taking their dot products and recording it in a matrix). If the differential of $f$ has rank 2, then the matrix $g$ is everywhere invertible. Physicists would often write the matrix $g(x,y)$ as $g_{ij}$ and the inverse matrix as $g^{ij}$. Then the formula for Christoffel symbols using Einstein notation is $\Gamma_{k \ell}^i = \frac{1}{2} g^{im} (g_{mk,\ell} +g_{m \ell, k} - g_{k\ell,m})$ where the comma means partial derivative with respect to that variable.

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