I am reading Nakahara's Geometry, topology and physics and I am a bit confused about the non-coordinate basis (section 7.8). Given a coordinate basis at a point on the manifold $\frac{\partial}{\partial x^\mu}$ we can pick a linear combination of this in order to obtain a new basis $\hat{e}_\alpha=e_\alpha^\mu\frac{\partial}{\partial x^\mu}$ such that the new basis is orthonormal with respect to the metric defined on the manifold i.e. $g(\hat{e}_\alpha,\hat{e}_\beta)=e_\alpha^\mu e_\beta^\nu g_{\mu\nu}=\delta_{\alpha \beta}$ (or $=\eta_{\alpha \beta}$). I am a bit confused about the way we perform this linear transformation of the basis. Is it done globally? This would mean that at each point we turned the metric into the a diagonal metric, which would imply that the manifold is flat, which shouldn't be possible as the change in coordinates shouldn't affect the geometry of the system (i.e. the curvature should stay the same). So does this mean that we perform the transformation such that the metric becomes the flat metric just at one point, while at the others will also change, but without becoming flat (and thus preserving the geometry of the manifold)? Then Nakahara introduces local frame rotations, which are rotations of these new basis at each point, which further confuses me to why would you do that, once you already obtained the flat metric at a given point. So what is the point of these new transformations as long as we perform the first kind in the first place? Sorry for the long post I am just confused.
[Physics] Non-coordinate basis in GR
differential-geometrygeneral-relativitymetric-tensor
Related Solutions
Taking the determinant on both sides, you get:
$$g = -\left|\frac{\partial y(x)^\alpha}{\partial x^\beta}\right|^2$$
where $g = \text{det} (g_{\mu \nu})$ and $\text{det} (\eta_{\mu \nu}) = -1$. On the RHS is the Jacobian (squared) of the coordinate transformation. Can you take it from here?
The simple answer is that the two are different (but related) concepts and you have confused them. The Lie derivative of a vector $\mathbf{u}$ in the direction $\mathbf{v}$ (tangent to the flow $\phi_\epsilon$) is defined as $$\mathcal{L}_{\mathbf{v}} \mathbf{u}= \lim_{\epsilon \to 0} \frac{\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]-\mathbf{u}(x)}{\epsilon}$$
The focus is the term $\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]$. What this means is that we need to take the vector $\mathbf{u}$ at a nearby point $\phi_\epsilon (x)$ further down the flow, which is $\mathbf{u}(\phi_\epsilon (x))$, and then pull it back to our starting point $x$ using $\phi_\epsilon^*$.
To first order $\phi_\epsilon (x) \approx x+\epsilon\mathbf{v}$, so we obtain by Taylor expansion, in components, $$u^\alpha(\phi_\epsilon (x)) \approx u^\alpha(x+\epsilon\mathbf{v})= u^\alpha(x)+\epsilon v^\beta \frac{\partial u^\alpha (x)}{\partial x^\beta}$$
However, we are not done yet. We need to pull it back to our starting point $x$ and this is accomplished by performing an infinitesimal coordinate transformation $x^\mu \to x^\mu -\epsilon v^\mu$ on both of the above terms. Applying the standard tensor transformation law, the first term becomes $$u^\alpha \to u^\alpha-\epsilon u^\beta \frac{\partial v^\alpha}{\partial x^\beta}$$ while the second term gives an additional term of order $\epsilon^2$ which can be ignored. So we end up with $$\left(\phi_\epsilon^*[\mathbf{u}(\phi_\epsilon (x))]\right)^\alpha = u^\alpha + \epsilon v^\beta \frac{\partial u^\alpha}{\partial x^\beta}-\epsilon u^\beta \frac{\partial v^\alpha}{\partial x^\beta}$$
Inserting it into the definition gives us the Lie derivative $$\mathcal{L}_{\mathbf{v}} \mathbf{u}= [\mathbf{v},\mathbf{u}] = \left(v^\beta \frac{\partial u^\alpha}{\partial x^\beta}- u^\beta \frac{\partial v^\alpha}{\partial x^\beta}\right)\mathbf{e}_\alpha$$
The infinitesimal coordinate transformation is only part of the construction of the Lie derivative. If you simply perform an infinitesimal coordinate transformation (which is what you did in your first equation), you will, of course, not obtain the term arising from the Taylor expansion.
Best Answer
This transformation is done locally, i.e. so that $g_{\alpha\beta} = \eta_{\alpha\beta}$ in a neighbourhood rather than a single point. I believe that due to topological effects, we cannot in general do it globally. However, even though the new basis is orthonormal, we have to remember that it is (in general) non-holonomic, i.e that there is no set of functions $y^\alpha$ satisfying $$ e^\mu_\alpha = \frac{\partial x^\mu}{\partial y^\alpha}. $$ This means that the curvature does not (in general) vanish. Indeed, it holds that $$T_{\alpha\beta\ldots\gamma} = e^\mu_\alpha e^\nu_\beta e^\sigma_\gamma T_{\mu\nu\ldots\sigma},$$ for all tensors $T_{\mu\nu\ldots\sigma}$, as is obvious by the linearity of $e^\mu_\alpha$, and thus in particular for $R_{\mu\nu\sigma\tau}$. You may be confused if you have learned that the connection coefficients are given by $$ \Gamma_{\mu\nu\sigma} = \frac{1}{2}\left(g_{\mu\nu,\sigma} + g_{\mu\sigma,\nu} - g_{\nu\sigma,\mu}\right), $$ but this is only valid in a holonomic frame. More generally $$ \Gamma_{\alpha\beta\gamma} = \frac{1}{2}\left(g_{\alpha\beta|\gamma} + g_{\alpha\gamma|\beta} - g_{\beta\gamma|\alpha} + C_{\gamma\alpha\beta} + C_{\beta\alpha\gamma} - C_{\alpha\beta\gamma}\right), $$ where $f_{|\alpha} \equiv e_\alpha(f)$, $[e_\alpha,e_\beta] = C^\gamma{}_{\alpha\beta}e_{\gamma}$, and $C_{\gamma\alpha\beta} \equiv g_{\gamma\delta}C^\delta{}_{\alpha\beta}$. In particular, in an orthonormal frame: $$ \Gamma_{\alpha\beta\gamma} = \frac{1}{2}\left(C_{\gamma\alpha\beta} + C_{\beta\alpha\gamma} - C_{\alpha\beta\gamma}\right). $$ Unless I misunderstand you, the rotations you refer to are local Lorentz transformations, and we are interested in them because each orthonormal frame corresponds to the instantaneous rest frame of an observer (with velocity field equal to $e_0$), and thus these rotations transform between different rest frames.