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The Christoffel symbols of the Levi-Civita connection are calculated through the metric, but does that necessarily mean that its existence depends on whether or not we have a metric?

Specifically, the Levi-Civita connection offers a specific way to parallel transport a vector along a manifold, so if we don't have a metric on a manifold, does this particular way to parallel transport a vector along the manifold gets "lost"? I mean, of course we need the metric to determine the Levi-Civita connection, but as a geometrical concept, it seems intuitive to me that as a way to transport vectors, its existence should not depend on whether on not we defined a metric on our manifold.

Note that this question is motivated by the fact that we define connections before even talking about a metric. But in my Riemannian geometry course, we defined the Levi-Civita connection through the metric, but I wanted to know if this necessarily means that it can't be defined(as a concept) without a metric.

Thanks in advance.

TheQuantumMan
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    I hope you don't mind if I try and give you some advice. Consider concrete examples and include them in your reasoning. This kind of question is fueled by the lack of examples. If you had computed the Christoffel symbols of a couple of Levi-Civita connections, it would have been clear to you that these are a function of the metric tensor (and derivatives of its components). It does not make sense at all to even speak about Levi-Civita connections without metric tensors. – Giuseppe Negro Feb 01 '18 at 20:29
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    I am not trying to patronize; I also learned mathematics in an abstract way, mostly. That was not a good idea, and I recognize myself in the wonderful advice given by Georges Elencwajg to budding mathematicians. – Giuseppe Negro Feb 01 '18 at 20:32

2 Answers2

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I think maybe you're misinterpreting the phrase "the Levi-Civita connection." Despite the definite article, there isn't just one such connection -- every Riemannian metric has its own Levi-Civita connection, uniquely determined by the metric. So the question in your title doesn't make sense -- without a specific choice of metric, "the Levi-Civita connection" has no meaning. It's like asking if "the derivative" exists without specifying a particular function to take the derivative of.

Jack Lee
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  • Would it make sense if I specified that the question refers to a single metric? Then there is a unique Levi-Civita connection but we can also find other(not Levi-Civita) connections. Basically, my question is about this. I just ommited the crucial information that I am referring to a single choice of metric. – TheQuantumMan Oct 10 '17 at 20:09
  • @TheQuantumMan: I'm not sure what your question is, in that case. If you've chosen a specific metric, then its Levi-Civita connection is uniquely determined -- that's the fundamental theorem of Riemannian geometry. But there are many, many other connections besides the Levi-Civita connection -- in fact, an infinite-dimensional space of them. They all determine different ways of parallel-transporting vectors. The reason the Levi-Civita connection is interesting is because its version of parallel transport is intimately connected with the metric in specific ways. – Jack Lee Oct 10 '17 at 20:46
  • Say we choose a metric for our smooth manifold. That metric will have a unique Levi Civita connection as you say. With this connection, we can parallel transport a vector along a particular curve in a particular way(just visualizing it being transported in a particular way). If we choose another metric, we get another Levi Civita connection. Since the two metrics are connected through a change of coordinates transformation(if I am allowed to talk a bit loosely here), the same goes for the Levi Civita connections. So, as far as the geometry** – TheQuantumMan Oct 10 '17 at 21:37
  • ** of the parallel transport is concerned they both transport the vector along the curve in the same way. So, I was thinking that since this way of transporting the vector is independent on the metric used(again, I have a geometric view of this), maybe this way of transporting the vector exists without even defining a metric BUT we do need a metric in order to determine the Levi Civita connection that corresponds to it(i.e. calculate the metric's Christoffel symbols). – TheQuantumMan Oct 10 '17 at 21:38
  • @TheQuantumMan: Why do you think both connections "transport the vector along the curve in the same way"? This is not true in general, because they are different connections. – Jack Lee Oct 10 '17 at 22:16
  • Correct me if I am wrong here but if the two metrics and their respective Levi Civita connections are connected via coordinate transformations, shouldn't they describe the same things(again, sorry for the loose tong)? – TheQuantumMan Oct 10 '17 at 23:38
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    @TheQuantumMan: I'm not sure exactly what you mean by "connected via coordinate transformations." Not all metrics can be transformed into each other by changes of coordinates, and not all connections define the samae parallel transport. – Jack Lee Oct 11 '17 at 04:43
  • Oh, so if that's the case, then with all the above comments, you've answered my question! Thanks for your patience. – TheQuantumMan Oct 11 '17 at 07:57
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There is a more general notion of Koszul derivative which allows to define parallel transport.

https://en.wikipedia.org/wiki/Connection_(vector_bundle)

https://en.wikipedia.org/wiki/Covariant_derivative#Formal_definition

Tsemo Aristide
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