Reading Rovelli's "General Relativity" - how to calculate a "diad"? So I'm reading Carlo Rovelli's "General Relativity" and there is this notion of "diad field" or "frame field" that I do not really get.
According to Rovelli, given a differentiable manifold $\Sigma$ and a point $p$ on it, one can consider the tangent plane $T_p\Sigma$ and Cartesian coordinates $X^i$ on that plane.
Than one can define "local Cartesian coordinates" $X^i_p$ by orthogonally projecting the $X^i$ onto the manifold.
Let also $x^j$ be general coordinates on $\Sigma$ (here I guess Rovelli means that the $x^j$ are coordinates in an open set of the same dimension of the manifold that are mapped to $\Sigma$ via some diffeomorphism $\varphi$).
Then the frame field, or "diad field" is defined as follows:
\begin{equation}
e^i_j :=\frac{\partial X^i_p(x^j_p)}{\partial x^j}
\end{equation}
where $X^i_p$ maps the general coordinates to the local Cartesian coordinates and $x^j_p$ are the coordinates for the point $p$.
I have some problems understanding this definition, and also calculating the actual diad field in some easy cases- i.e. for the paraboloid $z=x^2+y^2$ at a specific point $p$. Can somebody please explain this concept to me?
 A: If $\Sigma$ is merely a differential manifold with no further structure, then "orthogonally projecting the $X^i$ onto the manifold" doesn't make sense. In fact, even "tangent plane" won't make sense, unless it's embedded into some big $\mathbb R^n$. I guess the author is talking about Riemannian or at least pesudo-Riemannian manifold.
But the rest is just simple coordinate change. If $X$ and $x$ are two local coordinate charts (ignoring where they come from), then we have a frame field, i.e. $n$ linearly independent vector fields $\frac{\partial}{\partial X_i}, i=1, \cdots, n$ (if it's unfamiliar, tangent vectors are defined as partial differential operators in mathematics, and this avoids discussing tangent plane, but only the abstract tangent space), and similarly we have another frame field $\frac{\partial }{\partial x_i}, i=1, \cdots, n$.
They are related by chain rule: $\frac{\partial}{\partial x_j}= \sum_{i=1}^n\frac{\partial X_i}{\partial x_j} \frac{\partial}{\partial X_i} = \sum_{i=1}^n e^i_j \frac{\partial}{\partial X_i}$.
When $X$ is "special", there might be more to say, but basically that's it.
A: Too long for a comment:
As pointed out by @Just a user in the answer, I am afraid these concepts do not make sense unless $\Sigma$ is already embedded in some other Riemannian space. Check that the notion of dyad or frame field is introduced in Section 3.1, which is entirely devoted to embedded surfaces $\Sigma\subset \mathbb{R}^3$. Then the notion of orthogonally projecting the tangent plane onto the surface makes sense and it is kind of related to the exponential construction in Riemannian geometry.
For your requested example, perhaps it is instructive to outline which computations you need to do to start.

*

*Pick a point ($p=(0,0,0)$ will do) in your surface $z=x^2+y^2$

*Calculate $T_p\Sigma$ as an affine plane embedded in $R^3$. You should get $$T_pM = \{(a,b,c)\in R^3| c=0\}$$

*Introduce coordinates in this tangent plane. I can take the basis $\{(1,0,0),(0,1,0)\}$ to write any $v\in T_p\Sigma$ as $v=\alpha (1,0,0) + \beta (0,1,0)$.

*Take points $q=(x',y',z')\in R^3$ near $p$. Find a projection formula of these points onto the plane. This is actually easy because I picked the easiest point with the easiest tangent plane ever ever: $\pi(x',y',z') = (x',y',0)$. This is an orthogonal projection, the metric in $R^3$ is very much needed for this step; there other possible projections unless I specify the metric. Note that in terms of our coordinates $\alpha,\beta$ in the plane we get the projected functions $\alpha(x',y',z')=x',\beta(x',y',z')=y'$ for any $q=(x',y',z') \in R^3$ near $p$.

*Invert the formula locally in $\Sigma$ near $p$ in terms of general coordinates in $\Sigma$. As coordinates in $\Sigma$ near $p$, I will choose... well the two first Cartesian coordinates, $x,y:\Sigma\rightarrow\mathbb{R}$. $X_p^\alpha(x,y) = x$ and $X_p^\beta (x,y) = y$. Yep, very simple coordinates.

*The derivatives are now trivially $e^i_j = \delta^i_j$.

Yes, it is a very trivial example. I suggest you do the same with the same surface at a different point to get a feel of what is happening and get possibly more complex frame fields.
My suggestions: if you are not too familiar with "embedded" differential geometry, I wholeheartedly recommend Do Carmo's book(s). Once mastered, learn to forget "embedded geometry" for the abstract approach, which is the road to differential geometry and general relativity!
