3
$\begingroup$

Consider an arbitrary connection $\Gamma$, compatible with the metric, in 4 dimensional spacetime: $$\nabla_{\lambda} \, g_{\mu \nu} \equiv \partial_{\lambda} \, g_{\mu \nu} - \Gamma_{\lambda \mu}^{\kappa}\, g_{\kappa \nu} - \Gamma_{\lambda \nu}^{\kappa} \, g_{\mu \kappa} = 0. \tag{1}$$ I don't assume symetry of that connection (torsion may be present): $\Gamma_{\mu \nu}^{\kappa} \ne \Gamma_{\nu \mu}^{\kappa}$.

For a long time, I was believing that it was possible to find a special coordinates system such that we could cancel all the $4 \times 4 \times 4 = 64$ connection components, at any given point $\mathcal{P}$ in spacetime (but only for that point): $$\tilde{\Gamma}_{\mu \nu}^{\lambda}(\tilde{x}_{\mathcal{P}}) = a_{\mu}^{\; \rho} \: a_{\nu}^{\; \sigma} \: a^{\lambda}_{\; \kappa} \: \Gamma_{\rho \sigma}^{\kappa}(x_{\mathcal{P}}) - a_{\mu}^{\; \rho} \: a_{\nu}^{\; \sigma} \: \partial_{\rho} \, a^{\lambda}_{\; \sigma} = 0, \tag{2}$$ where \begin{align} a^{\mu}_{\; \nu} &= \frac{\partial \tilde{x}^{\mu}}{\partial x^{\nu}} \, \bigg|_{\mathcal{P}}, &a^{\; \nu}_{\mu} &= \frac{\partial x^{\nu}}{\partial \tilde{x}^{\mu}} \, \bigg|_{\mathcal{P}}. \tag{3} \end{align} But then, recently I had to re-read my old personnal notes on this subject and I'm now confused (I must have forgotten something important here, which I didn't described in my notes). Consider the following coordinates transformation (for simplicity, I assume cartesian-like coordinates $x^{\mu}$ such that the metric is Minkowskian at point $\mathcal{P}$ of coordinates $x_{\mathcal{P}}^{\mu}$ : $g_{\mu \nu}(x_{\mathcal{P}}) = \eta_{\mu \nu}$): $$\tilde{x}^{\mu} = x_0^{\mu} + \Lambda^{\mu}_{\: \nu} \, x^{\nu} + \frac{1}{2} \, \Lambda^{\mu}_{\: \nu} \: \Gamma_{\rho \sigma}^{\nu}(x_{\mathcal{P}}) (x^{\rho} - x^{\rho}_{\mathcal{P}})(x^{\sigma} - x^{\sigma}_{\mathcal{P}}), \tag{4}$$ where $x_0^{\mu}$ and $\Lambda^{\mu}_{\: \nu}$ are arbitrary constants (translation and Lorentz transformations), such that $a^{\mu}_{\: \nu} = \Lambda^{\mu}_{\: \nu}$. Then, (2) gives $$\tilde{\Gamma}_{\mu \nu}^{\lambda} = \Lambda_{\mu}^{\; \rho} \, \Lambda_{\nu}^{\; \sigma} \, \Lambda^{\lambda}_{\; \kappa} \, (\, \Gamma_{\rho \sigma}^{\kappa} - \Gamma_{\sigma \rho}^{\kappa}) = 0. \tag{5}$$ This equation then implies that the connection need to be symetric (no torsion), which implies 40 components instead of 64.

So I have two questions:

  1. Is there another - more general - coordinates transformation (in place of (4) above) that could cancel all 64 components of the arbitrary connection (evaluated at a given point)? Or is that impossible to cancel all of the 64 components (evaluated at a point) with any coordinates transformation (4 functions)?
  2. In the special case of the symetric connection (AKA Levi-Civita connection), how is it possible that transformation (4) of four coordinates is able to cancel at once all 40 components of the symetric Levi-Civita connection, evaluated at a given point?
$\endgroup$
3
$\begingroup$

Both questions are easily answered once we realize that $\partial_\mu a^\lambda_{\ \nu}=\partial_\mu \partial_\nu \tilde{x}^\lambda = \partial_\nu \partial_\mu \tilde{x}^\lambda = \partial_\nu a^\lambda_{\ \mu}$.

  1. The antisymmetric part of $\Gamma^{\lambda}_{\mu \nu}$, which we denote as $\Gamma^\lambda_{[\mu\nu]}$ (where $[\mu,\nu]\equiv \frac{1}{2}(\mu\nu-\nu\mu)$), transforms as a tensor. The part of the transformation of $\Gamma^{\lambda}_{\mu\nu}$ which does not transform as a tensor is symmetric under the exchange of $\mu$ and $\nu$, which can be shown using the property above. Therefore, it is not possible to find coordinates where $\Gamma^\lambda_{\mu\nu}=0$ in general. If it were possible, then $\Gamma^\lambda_{[\mu\nu]}$ would vanish, and then by properties of tensors, $\Gamma^\lambda_{[\mu\nu]}=0$ would be zero in every coordinate system. Indeed, the antisymmetric part of $\Gamma^\lambda_{\mu \nu}$ is a way to measure the torsion -- so coordinate invariance by itself is not enough to set $\Gamma^\lambda_{[\mu\nu]}=0$, you need an additional assumption that the torsion vanishes. This assumption is made in GR, but logically speaking this is an extra assumption you need to make.

  2. As we established above, $\partial_\mu a^\lambda_{\ \nu}$ is symmetric under exchanging $\mu$ and $\nu$. It therefore has 40 components (4 possibilities $\lambda$ times 10 possibilities for the symmetric pair ($\mu,\nu$). This gives enough freedom to cancel the freedom in the symmetric part of $\Gamma^{\lambda}_{\mu \nu}$.

$\endgroup$
11
  • $\begingroup$ Appears to be clear, thanks. I will medit on this. $\endgroup$ – Cham Jul 17 '20 at 3:44
  • $\begingroup$ @Andrew I am not the world's expert on this, but this does not seem to be correct. $\Gamma$ can indeed be made to vanish at one point, by choosing geodesic coordinates. This does not imply it vanishes in every coordinate system, because it does not transform as a tensor. so your statement "by properties of tensors $\Gamma _{\mu\nu}$ would be zero in every coordinates system" is just completely wrong. $\endgroup$ – joseph f. johnson Dec 7 '20 at 21:29
  • $\begingroup$ @josephf.johnson $\Gamma^{\lambda}_{[\mu\nu]}$ (where $[\mu\nu]=\frac{1}{2}(\mu\nu-\nu\mu)$) is a tensor. So if $\Gamma^{\lambda}_{[\mu\nu]}=0$ in one coordinate system, then it is zero in all coordinate systems. If $\Gamma^{\lambda}_{[\mu\nu]}=0$ in one coordinate system, then $\Gamma^{\lambda}_{[\mu\nu]}=0$ in all coordinate systems. However this is not guaranteed to be the case, and is an extra assumption that the geometry has no torsion. I edited the answer to be more clear on the logic. $\endgroup$ – Andrew Dec 7 '20 at 21:49
  • $\begingroup$ @Andrew You seem to be making general statements. For simplicity, let's focus on whether your statements are even true for connexions associated to the metric and without torsion, clearly a special case. Your statements are false for this special case, e.g., you can pick geodesic coordinates, then all first derivatives of the metric vanish, so all Christoffel symbols vanish. But the space need not be flat, so the Christoffel symbols do not vanish in every coordinate system. $\endgroup$ – joseph f. johnson Dec 7 '20 at 22:23
  • 2
    $\begingroup$ And the diffeerence with the metric compatible, torsion free connection is that $\nabla_{a}v^{b} = \partial_{a}v^{b} + \Gamma_{ac}{}^{b}v^{c}$ has two terms on the left hand side that are not tensors, but whose combination becomes a tensor, so a coodinate transformation CAN make one of them zero, because neither term transforms like a tensor.You can't make the torsion vanish locally with a coordinate change for the exact same reasons you can't make the riemann tensor vanish locally with a coordinate transformation. $\endgroup$ – Jerry Schirmer Dec 8 '20 at 0:31
1
$\begingroup$

Also, to close the loop, let's prove this by controposition. Let's assume that, for some torsion-having metric, we HAVE produced a transformation that makes all $\Gamma$ zero at some point. Then, at this point, we have:

$$\nabla_{a}v^{b} = \partial_{a}v^{b}$$

Now, transform to some other coordinate system. This will require that we multiply $v^{b}$ by the transformation matrix $P^{a}{}_{b} = \frac{\partial y^{a}}{\partial{x^{b}}}$

Now, we have, in this new coordinate system (suppressing multiplication on the left by two factors of $P$ on both sides to transform the basis components:

$$\nabla_{a}v^{b} = \partial_{a}\left(P^{b}{}_{c}v^{c}\right) = P^{b}{}_{c}\partial_{a}v^{c} + v^{c}\partial_{a}P^{b}{}_{c}$$

The first term becomes just the partial derivative term again, because the factors of P that we suppressed satisfy the identity $P^{a}_{b}P^{b}{}_{c} = P^{a}{}_{c}$ by the chain rule. That second term must be the Christoffel symbol in our new frame with nonzero Christoffel symbols, so:

$$\Gamma_{ab}{}^{c} = \partial_{a}P^{c}{}_{b} = \frac{\partial^{2}y^{c}}{\partial x^{a} \partial x^{b}}$$

This cannot have any torsion associated with it, because it is symmetric in a and b.

$\endgroup$
1
$\begingroup$

If the connection has torsion, the answer is negative. Indeed, if you found a coordinate system around a point where all connection coefficients vanish at that point, then also their antisymmetric part would vanish, and this is not possible because the torsion is a tensor so that, if it vanishes in a coordinate system, then it has to vanish in every coordinate system.

Hence, let us assume that the connection is symmetric without assuming that it is metric. What follows does not need a metric, just an affine symmetric connection on a manifold with generic dimension $n$.

Consider a point $p$. In that case you can define the local diffeomorphism known as the exponential map at $p$. The map identifies an open neighborhood of the tangent space at $p$ with a neighborhood of $p$.

$$\exp_p{v} = \gamma(1,p, v)$$ Where $t \mapsto \gamma(t,p,v)$ is the geodesic with initial point $p$ and initial tangent vector $v$.

(Actually there are a few mathematical subtleties. First, one has to prove that it is always possible to include $1$ in every maximal parameter domain of all geodetics emanated from $p$ if $v$ stays in a suitable small neighborhood of the origin of $T_pM$. Secondly, one has to prove that the map written above is in fact a diffeomorphism, i.e., smooth, bijective with inverse smooth, by possibly shrinking the said neighborhood. All that can be fixed with a little effort.)

Fixing a basis in the tangent space at $p$, the components of $v$ are coordinates of the corresponding point on the manifold. Such a coordinate system is known as a normal coordinates system centered on $p$.

The geodesics emanated by $p$ are straight lines exiting the origin in that coordinate system, $$v^a(s) = sv_0^a\tag{1}$$ This is an immediate consequence of the identity arising from the geodesic equation (and the uniqueness theorem for that differential equation) $$\gamma(s,p,v_0) = \gamma(1,p, sv_0)$$ Eventually, (1) easily implies from the geodesic equation that the connection coefficients must vanish at $p$ in the said coordinate system. In fact, it must hold at $s=0$ $$\Gamma(0)_{ab}^cv^av^b=- \frac{d^2sv^c}{dt^2}|_0=0$$ for every choice of the coefficients $v^a$. Using the fact that the connection is symmetric, replacing $v$ for $u\pm w$, you find $$\Gamma(0)_{ab}^c=0.$$

If the connection is metric (Levi-Civita), we also have for free that $$\frac{\partial g_{ab}}{\partial x^c}|_{p}=0$$ in the said coordinate system. It arises by just writing the above derivative in terms of connection coefficients (using the metricity condition).

ADDENDUM. Actually the exponential map is defined also when the connection has torsion (I did not use that hypothesis in defining the exponential map). In that case, the reasoning above proves that the symmetric part of the connection coefficients vanishes at the center of a normal system of coordinates. The antisymmetric part cannot be cancelled in any cases as said at the beginning of this answer.

$\endgroup$
2
  • $\begingroup$ It seems as though the next-to-last equation is true even for non-symmetric connections, since the geodesics only depend on the symmetric part of the connexions coefficients. $\endgroup$ – joseph f. johnson Dec 8 '20 at 4:29
  • $\begingroup$ Yes it is, but it does not matter here because we have already excluded that case. We can always cancel the symmetric part of a connection at a point by choosing the coordinate system. $\endgroup$ – Valter Moretti Dec 8 '20 at 8:11
-1
$\begingroup$

The OP is not the only person who has wondered about this. And even a textbook can be mistaken, or at least misleading, about this.

I have in my possession a textbook which assigns this as an exercise. It says that for any point $P$ of a manifold and any connexion on any neighbourhood of it $U$, there exist coordinates (in a neighbourhood of $P$ which is possibly smaller than $U$) such that all $\Gamma_{ij}^k$ are zero at $P$. Part of the exercise is to construct the coordinates using the geodesics of $\Gamma$ which begin at $P$.

Not every textbook is error-free, and Doubrovine, Novikov, and Fomenko, is certainly not. But there it is, on p. 285, of Geometrie Contemporaine, Methodes et Applications, Premier Partie, Geometrie des surfaces, des groupes de transformations et des champs, Moscow, 1982, tranlated from the Russian edition of 1979. I understand that the new revised Russian edition is freely available on the web. Also, I have read this theorem elsewhere for the special case of connexions which are torsion free and compatible with the metric. But neither of those assumptions are made here.

One can not trust one exercise posed in one textbook, and Moretti has solved their exercise under the hypothesis that there is no torsion.

So the same question that the OP wanted clarified is actually the subject of a mistake in a reputable textbook.

$\endgroup$
8
  • $\begingroup$ @Andrew Excercise 4 to section 19 of Chapter 4, so it can be found in any edition. I am living out of a suitcase and all the libraries are closed, so I have only four or five texts on differential geometry at hand. $\endgroup$ – joseph f. johnson Dec 8 '20 at 0:02
  • $\begingroup$ This is only true in the absence of torsion. (and metric compatibility). $\endgroup$ – Jerry Schirmer Dec 8 '20 at 0:11
  • 1
    $\begingroup$ ...you do have enough coordinate freedom to set the symmetric part of $\Gamma^\lambda_{\mu\nu}$ to zero. On the other hand it is no problem that you don't have enough freedom to set the antisymmetric part to zero, because the antisymmetric part is a tensor. If, like in GR, the torsion is zero in one coordinate system, it is guaranteed to be zero in all other coordinate systems. So you can consistently assume the torsion is zero; you don't need to set it to zero. I was trying to get at the logical distinction between these two parts of $\Gamma^\lambda_{\mu\nu}$ in my answer. $\endgroup$ – Andrew Dec 8 '20 at 0:23
  • 1
    $\begingroup$ Thanks for the feedback, I do appreciate chance to improve my communication. It's not always easy to write a clear answer; questions like this help me find better ways to express the ideas. $\endgroup$ – Andrew Dec 8 '20 at 0:39
  • 1
    $\begingroup$ "It is still false to say... then they vanish in every coordinate system." I agree, but I don't think I am saying this. The words "in general" in my answer carry a lot of weight. I mean "If you allow for torsion," then you can't find a coordinate system where every component $\Gamma^\lambda_{\mu\nu}=0$. Because, if there is torsion, then $\Gamma^\lambda_{[\mu\nu]}$ is non zero, and this quantity is a tensor. This is backwards from the way I'd explain this if I was writing a review article or book, but my answer was meant to respond to the OP's question. $\endgroup$ – Andrew Dec 8 '20 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.