# Do Christoffel symbols commute?

Do Christoffel symbols commute? For example, does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$?

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They are just numbers; so yes, they commute. –  user10001 Oct 22 '12 at 17:15
no they don't, he showed you what he meant by commute with the pattern of index contraction. –  Ron Maimon Oct 22 '12 at 17:15
Yes, to clarify, I am specifically referring to the 'commutator' above. –  user12345 Oct 22 '12 at 17:17
You didn't do anything with the indices above rather than swap the order of the terms. If you take the indices as abstract indices, then yes, they commute by the commutation of ordinary numbers. –  Jerry Schirmer Oct 22 '12 at 17:19
Do Christoffel symbols commute? As matrices they do not. See @Ron's answer. does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$? Yes, because they are just numbers. –  user10001 Oct 22 '12 at 18:22

## 3 Answers

In classical theory, all observables commute. The components $\Gamma^a_{bc}$ are just real numbers so of course that they commute.

In quantum theory, they don't commute. It's probably a bit laborious to calculate the commutator.

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Sorry, @Ron, but $\Gamma^a_{bc}$ is no "linear transformation" or any other transformation. It's a dynamical variable. Classically, it's a pure number. Quantum mechanically, it's a $q$-number, an operator. What you may have computed is the "commutator" in which the indices $b,c$ of $\Gamma^a_{bc}$ are treated as two indices of a matrix and one multiplies two Gammas as matrices. Then they don't commute. But in the products, the indices would be written differently. The original question writes a clear formula which proves that the OP is asking whether the components commute, and they do. –  Luboš Motl Oct 25 '12 at 7:58
The two matrix indices are a and c, not b and c, and treating them as a matrix is the definition of connection. You are doing rhetoric now, not honest science. I will delete my comment above yours. –  Ron Maimon Oct 25 '12 at 12:58
At any rate, your answer is wrong and my answer is right. –  Luboš Motl Oct 25 '12 at 21:07
yes, in the technical sense that I interpreted "commutator of connection coefficients" in a non-vacuous way. You were right about the OP's confusion, so +1. That's more rhetoric by the way, since we all agree on what happened. By the way, Lubos, you might be interested in Thomas Gold's book "The Deep Hot Biosphere", which makes the case (completely persuasively IMO) that we will never run out of oil, since oil is abiogenic and produced deep inside the Earth. While, unlike you, I believe in global warming and think this is a bad thing, it is still good science by Gold. –  Ron Maimon Oct 25 '12 at 23:21

No they don't commute in this sense, except you screwed up the commutator. It should be

$$\Gamma^{e}_{db}\Gamma^c_{ea} - \Gamma^c_{eb}\Gamma^{e}_{da}$$

Which is

$$\Gamma_b \Gamma_a - \Gamma_a \Gamma_b$$

in a matrix form, where I have taken one upper and one lower index and suppressed them in the Christoffel symbol (which lower index doesn't matter, becuase of the symmetry on the lower indices). This has the interpretation I give below.

The Christoffel symbols are the coordinate form of the infinitesimal rotations associated with parallel transporting a vector a ways along a short distance. If all the frames were orthonormal, the parallel transport would be SO rotations, and the infinitesimal form would be a bunch of stuff in the Lie Algebra of SO, and these are antisymmetric matrices (or the upper index representation of antisymmetric forms for the Lorentzian case). Their commutators tell you when the rotations corresponding to moving in a certain direction is noncommutative with the rotation corresponding to moving in another direction.

For a coordinate basis for the tangent space, the basis vectors are not orthonormal, so the connection coefficients don't obey the Lie algebra conditions, but they still stay noncommutative. You can easily verify this in a generic example by direct computation, but it is also obvious from those cases where you happen to choose coordinates where the coordinate basis is orthogonal, like spherical coordinates, and think about the different coordinate rotations associated with parallel transport in different infinitesimal directions.

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Look at the index contractions he's made. This isn't the curvature term. –  Jerry Schirmer Oct 22 '12 at 17:24
@JerrySchirmer: Yeah, I didn't see is screw up. I think he meant what I say above--- it's easy to screw up index stuff like this. –  Ron Maimon Oct 22 '12 at 17:26
No, it wasn't a screw up. I meant exactly what I'd written in the question. –  user12345 Oct 22 '12 at 17:35
Yet, my answer is still correct, this is the proper interpretation of non-commutativity of Christoffel symbols, and this is what people mean when they usually say it. I believe the answer is superior to others, and does not deserve downvotes. It doesn't matter that I misunderstood what user16307's screw up was, the question is best answered as above. –  Ron Maimon Oct 22 '12 at 18:07
Ron's answer has quite useful information, and moreover its not irrelevant to OP's question. +1 –  user10001 Oct 22 '12 at 18:09

This might be overkill, but here it goes:

Let $\pi:E\rightarrow B$ be an arbitrary fiber bundle with typical fiber $F$ and $\Phi=\tau\times\varphi$ be a local trivialization over $U\subset B$ $$\Phi:\pi^{-1}(U)\rightarrow U\times F$$

Geometrically, the Christoffel symbol $\Gamma$ of a connection is an element $$\Gamma\in\mathrm{Hom}(\tau^*(\mathrm TU),\varphi^*(\mathrm TF))$$ which is a fancy way of saying that for each $e\in E$, there is a linear map $$\Gamma(e):\mathrm T_{\tau(e)}U\rightarrow\mathrm T_{\varphi(e)}F$$

In general, talking about the composition $$\Gamma(e)\circ\Gamma(e')$$ makes no sense as domain and codomain don't agree.

In relativity however, $E=\mathrm TB$ which allows us to identify these spaces and we end up with $$\Gamma(V):\mathrm T_bU\rightarrow\mathrm T_bU$$ where $V\in\mathrm T_bU$ is a tangent vector, eg a 4-velocity.

As $\Gamma(V)$ is linear, it can be expressed in local coordinates via matrix multiplication $$W^i\mapsto\Gamma(V)^i{}_j W^j$$ In case of the Levi-Civita connection, the map $$V\mapsto\Gamma(V)$$ is linear as well and we arrive at $$W^i\mapsto(\Gamma_k V^k)^i{}_j W^j = \Gamma_k{}^i{}_j V^k W^j$$

The compositions read $$(\Gamma(V)\circ\Gamma(W))^i{}_j = \Gamma_k{}^i{}_a \Gamma_l{}^a{}_j V^k W^l \\ (\Gamma(W)\circ\Gamma(V))^i{}_j = \Gamma_l{}^i{}_a \Gamma_k{}^a{}_j V^k W^l$$ which in general do not commute.

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hopefully correct –  Christoph Oct 22 '12 at 23:52
Correct, but I said the same thing without obfuscation. –  Ron Maimon Oct 23 '12 at 2:40
@Ron: which is why I wasn't sure if I should add this answer; in the end, I decided that explicitly mentioning the domains of the map and showing where the indices come from had sufficient surplus value –  Christoph Oct 23 '12 at 6:03
I see, yes, it does add value, +1. –  Ron Maimon Oct 23 '12 at 15:23