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I am learning the frame formalism in differential geometry and I am trying to reconcile this with applications in general relativity, especially in contexts like the tetrad formalism.

Consider a vector field $X$ on a differentiable manifold $\mathcal{M}$. Then, in some coordinate patch we can expand the vector in a basis $\partial_\mu$ as, $X = X^{\mu}(x)~\partial_\mu$ where $X^\mu(x)$ are the components of the vector field. If instead, I use coordinates $y$, then the components transform as, $$X'^\mu(y) = \frac{\partial y^\mu}{\partial x^\nu}~X^{\nu}(x)~.$$

An alternative way to describe these objects is through the introduction of frame fields which are a collection of vector fields $E_a$ with $a = 1,2,3,...,n$ for an $n$ dimensional manifold, which are orthonormal $\langle E_a, E_b\rangle_G = \delta_{ab}$ with respect to some metric $g_{\mu \nu}$ on the manifold. Then, these $n$ vector fields act as a coordinate system at each point $p \in \mathcal{M}$. Since these are vector fields I still have $E_a = E_a^{~\mu}~\partial_\mu$ and similarly I can define a dual frame $e^a = e^a_{~\mu}~dx^\mu$ which are one-forms such that $\langle E_a, e^b\rangle = \delta_{a}^{~b}$. Now, given the same vector field $X$, I can project this vector field into my frame to obtain

$$X^a = \langle e^a, X\rangle = e^a_{~\mu}X^{\mu}, $$

which is a scalar under coordinate transformations. However, the frame does transform under $O(n)$ rotations. So, for $M^a_{~b} \in O(n)$, $e'^a = M^a_{~b}~e^b$ and hence, $X^a \to X'^a = M^a_{~b}X^b$ under $O(n)$. Therefore, $X^a$ is a scalar under coordinate transformations but an $O(n)$ vector.

This is my confusion. In special relativity for example, one can make a Lorentz transformation (where we apply the above formalism but with $O(1,n-1)$) and move between different inertial frames. However, these transformations are also considered coordinate transformations, since I have an invertible coordinate map between the coordinates in the new frame and the old one.

  1. Therefore, what really is the difference between a coordinate transformation and a frame change?

  2. As a corollary, I understand that going between different frames via an $O(n)$ rotation, is also like making a gauge transformation. But I have heard the statement that "diffeomorphisms are much like gauge transformations." How is this relevant in this context, since coordinate invariance is promoted to diffeomorphism invariance in GR?

newtothis
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  • The author of https://arxiv.org/abs/1711.09503 writes that tetrad indices are transformed under local Lorentz transformations. To me it seems that those transformations are sloppily called "rotations" even though they are something more general. – Kurt G. Jul 12 '23 at 16:18
  • I mean, even in special relativity before introducing any additional geometry, Lorentz transformations are treated as rotations of space $\textit{and}$ time. So I'm not sure I understand what you mean by them not being rotations – newtothis Jul 12 '23 at 16:36
  • Here is a detailed description what a Lorentz transformation is and how those (sub) groups are denoted. $O(n)$ is not one of them, and usually we denote rotations by $SO(n),.$ – Kurt G. Jul 12 '23 at 16:43
  • @KurtG., I understand why $O(n)$ is not a Lorentz transformation. My point is that even if we exclude reflections and stick to $SO(n)$, then I still have the same confusion because it doesn't change the coordinate maps between different $SO(n)$ frames. – newtothis Jul 14 '23 at 05:47
  • I believe the equivalent to your $e'^a={M^a}b=e^b$ is the author's equation (107) at the bottom of p. 18: $$\tilde\theta'^{\alpha}(p)={\Lambda^\alpha}\beta(p),\tilde\theta^\beta(p),.$$ Dos Santos calls this a "local rotation" but knows as much as we do that it is a local Lorentz transformation. Where does your idea come from that these "rotations" are $SO(n),?$ – Kurt G. Jul 14 '23 at 06:28
  • I think I am confused about two things. Firstly, regarding eqn 107, I understand that the local rotation the author defines is itself in terms of coordinates, since the transformation depends on the point $p \in \mathcal{M}$. Therefore, explicitly writing out the equation for $\tilde{\theta}'^{\alpha}$ and the corresponding equation for $e'_{\alpha}$ will simply give map for the basis vectors before and after rotations, in terms of the coordinates. – newtothis Jul 14 '23 at 07:04
  • Therefore, as a trivial case, consider the map $x' = x\cos{\theta} - y\sin{\theta}$ and $y' = x\sin{\theta} + y\cos{\theta} $ gives how basis vectors are rotated in the XY plane, if my initial basis vectors where $e_{\alpha} \equiv {e_1,e_2} = {x,y}$ and $e'_{\alpha} \equiv {e'_1,e'_2} = {x',y'}$ – newtothis Jul 14 '23 at 07:08
  • My second confusion has to do with the fact that moving between orthonormal frames uses an $O(n)$ transformation, and Lorentz transformations (which are a special case of physical interest) are $SO(n)$. What I had understood earlier (possibly incorrectly) are that these are independent of coordinate transformations. As with my OP, I am following the book "Lectures on Advanced Mathematical Methods for Physicists" by Sunil Mukhi (https://www.worldscientific.com/doi/abs/10.1142/9789814299749_others01). This is given in eqn 4.17 in this book once he defines vielbeins. – newtothis Jul 14 '23 at 07:13
  • I am reading again that Lorentz transformations are $\color{red}{SO(n)},.$ That's wrong. If anything they are $SO(1,n-1),.$ I recommend to patiently read a source about non coordinate bases, tetrads, moving frames that you like best. The subject is too important to rely on a single one. In my case Sean Carroll Spacetime and Geometry was a good start but not the end. – Kurt G. Jul 14 '23 at 07:28
  • Assuming the metric is of Lorentzian signature, the orthonormality condition reads $g_{\mu\nu}=\eta_{ab}e^a_\mu e^b_\nu$. The choice of frame is of course not unique, since any other (co)frame $e' = \Lambda e$ with $\Lambda\in \mathrm{SO}(1,D-1)$ satisfies the orthonormality condition due the defining property of the Lorentz group. Hence, this is a sort of redundancy in the system, i.e., a gauge freedom. If the metric is of Euclidean signature, you just replace $\eta$ with $\delta$ and the gauge group is now SO(D). – Stuckelberator Jul 14 '23 at 10:58

2 Answers2

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The difference between a coordinate transformation and a "frame change" is the following.

If you think of the rotation as a coordinate transformation, a diffeomorphism, you are rotating both the frame ($e^a \to M^a_{\ \ b} e^b$) and the vectors ($X^a \to M^a_{\ \ b} X^b$), so that the projection of the rotated vector on the rotated frame $X^a = \langle e^a, X \rangle$ is invariant, as you mentionned.

However, if you only rotate the frame ($e^a \to M^a_{\ \ b} e^b$, $X^a \to X^a$), you are only rotating the coordinate system, so that the projection of the initial vector on the rotated frame is behaving like an $O(n)$ vector. Equivalently, you can fix the frame and only rotate the vectors ($e^a \to e^a$, $X^a \to M^a_{\ \ b} X^b$).

How to relate this to physical theories ? Let us consider a physical differential equation $$ D(x) \phi(x) = 0 $$ where $D(x)$ is a differential operator written in $x$ coordinates. When we claim that the theory is invariant under rotations, we mean that if $\phi(x)$ is a solution, the rotated solution $\phi(M x)$ is also a solution in the same coordinate system, of the same differential operator $D(x)$ $$ \text{Rotationally invariant theory : } D(x) \phi(x) = 0 \implies D(x) \phi(M x) = 0 $$

This is the concrete "rotational invariance", rotating the solution without changing the coordinate system is also a solution of physics. Note that we could equivalently check $D(x) = D(M x)$. This is the standard verification that the differential operator is invariant under rotation.

This is to be contrasted to diffeomorphism invariance : $$ \text{Diffeomorphic invariant theory : } D(x) \phi(x) = 0 \implies D(M x) \phi(M x) = 0 $$ This is satisfied by design, because differential equations are invariant under coordinate reparametrization $x = M x'$. I think this post may be useful for details on diffeomorphism invariance, Diffeomorphism Invariance of General Relativity

Adrien Martina
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Geometrically, you have a tangent space at a point on the manifold, and a vector in that space.

When you add coordinates to that, in the tetrad/frame formalism, you have two different bases for the tangent space: one derived from the coordinate patch on the manifold and indexed by Greek letters, and another that is orthonormal and indexed by Roman letters. You can write the vector as a linear combination of basis vectors from either set. If you express it in terms of the Roman basis then that expression won't be affected by a change of Greek basis, and vice versa. No change of either basis changes what the vector is.

You could say that the vector expressed in one basis is a scalar with respect to changes of the other basis, but I would rather say that it's a vector, because geometrically it is one.

Gauge transformations are physically meaningless. What it means to say that a theory has a gauge symmetry is that the theory can be written as a quotient of a certain mathematical structure by a certain symmetry group. The physical theory is the result of taking that quotient; neither of the objects in the quotient is the theory, and one could imagine writing the theory in a way that doesn't involve such a quotient at all—although it tends to be rather difficult. General relativity is the quotient of a coordinate-laden formulation of general relativity by the group of coordinate changes. Since the coordinates are fictions anyway, nothing prevents you from introducing two different coordinate systems and dividing both of them away, and that's effectively what happens in the tetrad/frame formalism. You can think of both of them as gauge symmetries.

benrg
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