I often read the statement, that the Christoffel symbols aren't tensors. But then, under which representation do they transform?


2 Answers 2


First, they do not transform in an actual "representation" in the sense of a linear representation of the group of coordinate transformation since their behaviour under a coordinate transformations $x\mapsto y(x)$ is given as $$ {\Gamma^\alpha}_{\beta\gamma} \overset{y(x)}\mapsto \frac{\partial x^\mu}{\partial y^\beta}\frac{\partial x^\nu}{\partial y^\gamma}{\Gamma^\sigma}_{\mu\nu}\frac{\partial y^\alpha}{\partial x^\sigma} + \frac{\partial y^\alpha}{\partial x^\sigma}\frac{\partial^2 y^\sigma}{\partial x^\beta \partial x^\gamma}\tag{1}$$

Nevertheless, their transformation is not "random", since they have that form to make the covariant derivative transform as a proper tensor. They transform as an object in the jet bundle of the frame bundle of the spacetime $\mathcal{M}$. Now, what does this mean?

A coordinate transformation $y(x)$ determines at every point $p\in\mathcal{M}$ an invertible linear map $$ y'_p : T_p\mathcal{M}\to T_p\mathcal{M}, \frac{\partial}{\partial x^\mu}\mapsto\frac{\partial x^\nu}{\partial y^\mu}\frac{\partial}{\partial x^\nu} $$ i.e. $y'$ is a diffeomorphism $T\mathcal{M}\to T\mathcal{M}$ of the tangent bundle that's compatible with the projection onto the base.

To the tangent bundle, there is the frame bundle $F\mathcal{M}$ of all ordered bases of $T_p\mathcal{M}$ at every point. "Ordered base" means you just take the $n$ basis vectors and write them into an $n\times n$-matrix. Since it is a base, this matrix is invertible - the space that the frame bundle associates to every point is that of $\mathrm{GL}(n)$, and it is hence a $\mathrm{GL}(n)$-principal bundle.

The "Christoffel symbols" are now just the components of a principal connection on that bundle, where a "connection form" is better known to physicists as a gauge field, this one taking values in $\mathfrak{gl}(n)$, i.e. the Christoffel form is a 1-form $\Gamma : T\mathcal{M}\to\mathfrak{gl}(n)$. Its gauge transformations are given by $y_\text{gauge} : \mathcal{M}\to\mathrm{GL}(n), p \mapsto y'_p = \frac{\partial x^\nu}{\partial y^\mu}\rvert_p$ for a coordinate transformation $y$, and like every gauge field, it transforms as $$ y_\text{gauge} \Gamma y_\text{gauge}^{-1} + y_\text{gauge}\mathrm{d}y_\text{gauge}^{-1}$$ The coordinate expression is $(1)$ is reobtained from this by writing $\Gamma = {\Gamma^\mu}_{\nu\sigma}{T^\nu}_\mu\mathrm{d}x^\sigma$ for a basis ${T^a}_b$ of the $n\times n$-matrices $\mathfrak{gl}(n)$.

  • $\begingroup$ Highly useful and concise explanation utilizing the terminology of fiber bundles in the context of "standard" phrasing $\endgroup$
    – R. Rankin
    Feb 24, 2020 at 20:16

The Christoffel symbols do not transform under any representation. The reason for this is that they do not transform linearly, which puts them out of the game altogether. The transformation law is

$$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde x^\nu}{\partial x^\gamma \over \partial \tilde x^\kappa} + {\partial ^2 x^\alpha \over \partial \tilde x^\nu \partial \tilde x^\kappa} \right ]$$

(for a proof see e.g. this math.se question). As you can see, its possible for $\tilde \Gamma^{\mu}_{\nu\kappa}$ to be nonzero even if $\Gamma^{\mu}_{\nu\kappa}$ vanishes identically, which is fatal for linearity.

The reason this is important is that a representation is a mapping from the symmetry group $G$ of spacetime into the group of linear transformations on some given vector space $V$. Thus: no linearity, no representation.


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