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.

  • $\begingroup$ You're so used to working with coordinate bases that when you see the metric $\delta_{\alpha \beta}$, your brain screams 'flat space!' at you. But this is not so. The Christoffel symbols in an arbitrary basis contain terms proportional to the derivatives of the metric, which you're familiar with, as well as terms proportional to the commutators of the basis vectors, which vanish for coordinate bases. Hence, even if we write $g_{\alpha \beta} = \delta_{\alpha \beta}$ everywhere in some patch, we can still find a non-zero Riemann tensor if our basis vectors don't commute. $\endgroup$ – gj255 Nov 21 '17 at 16:54

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.


Note that the Transformation you mentioned

$g_{\mu \nu} \mapsto g(\hat{e_\alpha},\hat{e_\beta})$

by the function $e_\alpha^{\mu}$ (also called the tetrad) is only a Change of coordinates if it holds:

$e_\alpha^{\mu} = \frac{\partial u^\mu}{\partial x^\alpha}$.

Here, $u^\mu(x)$ is a vector function and you can choose this function such that at some Point $x_0$ you have a flat metric. If you have done a suitable choice of $u^\mu$, metric tensors in a neighborhood of $x_0$ will not be flat in General. The Transformation

$g_{\mu \nu} \mapsto g(\hat{e_\alpha},\hat{e_\beta}):=\frac{\partial u^\mu}{\partial x^\alpha}\frac{\partial u^\nu}{\partial x^\beta}g_{\mu \nu} (*)$

will preserve all geometric Tensors like curvature Tensor, Torsion Tensor, etc. up to a Change of coordinate Basis. More precisely, if $X_{\mu_1 \mu_2 \dots \mu_n}$ is a geometric Tensor like the Riemann Tensor, it will Change under the Transformation (*) as

$X_{\mu_1 \mu_2 \dots \mu_n} \mapsto \frac{\partial u^{\mu_1}}{\partial x^{\nu_1}} \frac{\partial u^{\mu_2}}{\partial x^{\nu_2}} \dots \frac{\partial u^{\mu_n}}{\partial x^{\nu_n}}X_{\mu_1 \mu_2 \dots \mu_n}=X'_{\nu_1 \dots \nu_n}$

like a Tensor should transform. Another Transformation that does not Change geometric quantities are local rotations $\Lambda^{\mu}_{\nu}(x)$. In this case you have $e^{\mu}_{\nu} = \Lambda^{\mu}_{\nu}$. Such rotation matrices have the property that its inverse matrices is simply the transpose of it (from which it follows that $det(\Lambda) = \pm 1$). A geometric Tensor will transform similary:

$X_{\mu_1 \mu_2 \dots \mu_n} \mapsto \Lambda^{\mu_1}_{\nu_1} \dots \Lambda^{\mu_n}_{ \nu_n} X_{\mu_1 \mu_2 \dots \mu_n}=X''_{\nu_1 \dots \nu_n}$

Other forms of the Tensors $e_\alpha^\mu$ (not coordinate transformations or rotations) change the intrinsic geometry!

  • $\begingroup$ I see, so the first kind of transformation mentioned in the book, transforms the metric into the flat metric just for a given point, not over the whole manifold, but the transformation applied at that point is applied to all the other point, right? And for your last statement, why would one make a transformation that changes the geometry of the manifold? Doing this you lose all the information you have about your space. $\endgroup$ – Silviu Nov 21 '17 at 13:41
  • $\begingroup$ @Silviu It is not true that more general linear transformations change the intrinsic geometry. Additionally, the transformations referred to in the book are almost guarantueed to be intended as to make $g_{\alpha\beta} = \eta_{\alpha\beta}$ locally (in a neighbourhood, not in a point). $\endgroup$ – Erik Jörgenfelt Nov 21 '17 at 16:35

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