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I've been reading about connections on fibre bundles recently and it's made me think about the exact nature of the Christoffel symbols in GR.

If we have a vector bundle $E$ over $M$ and put a connection $D$ on it, in some local trivialisation we can choose a fixed connection $D^0$ and a vector potential $A\in\Gamma(\textrm{End}(E)\otimes T^*M)$ such that $D=D^0+A$. (Typically $D^0$ is the flat connection where allowed.) Then, if we have a section $s\in\Gamma(E)$ written in our local trivialisation as $s^ie_i$, and similarly $A=A_{\mu j}^ie_i\otimes e^j\otimes dx^\mu$, the covariant derivative of $s$ is given by $$D_\mu s=(D_\mu^0s^i+A_{\mu j}^is^j)e_i$$ Now this will immediately recall the covariant derivative of Riemannian geometry/GR, which, on a vector field $v=v^\mu\partial_\mu$, is given by $$D_\mu v=(\partial_\mu v^\nu+\Gamma^\nu_{\mu\lambda}v^\lambda)\partial_\nu$$ Indeed vector fields are just sections of the tangent bundle so this is to be expected. The Christoffels symbol play the role of the vector potential for general bundles.

Now my question is the following: given $E$, $A$ is a bona fide section of the tensor product bundle $\textrm{End}(E)\otimes T^*M$ - but this seems to suggest that $\Gamma$ should be a bona fide section of the bundle $$\textrm{End}(TM)\otimes T^*M\cong TM\otimes T^*M\otimes T^*M.$$ But then should it not be a bona fide type-$(1,2)$ tensor? However it is well-known that it is not a tensor. The only thing I could think of was that $A$ can in general only be defined locally (i.e. on a local trivialisation); but the familiar calculation in GR that shows the Christoffels are not tensors computes their transformation properties, which should be local, so I don't think that the explicit locality of the vector potential should be a problem. In any case, trivial tangent bundles are possible, on which the vector potential can be defined globally. What am I missing?

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What you're missing is that the action of a Jacobian on $\mathrm{End}(TM)$ in this case is that of a gauge transformation, and not the standard action on $TM\otimes T^\ast M$. The isomorphism you've written down does not preserve the transformation behaviour, it holds purely for the underlying manifolds.

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