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Assume we have a Riemannian manifold $(M,g)$. Why do we assume that it is always endowed with the Levi-Civita connection. Is it somehow interesting to change the connection and study how geometric properties change? Do you know some books that deal with this topic? Thank you.

Giulio Binosi
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    The metric $g$ is the primary object of interest, while a connection is a tool. If you change your tool, properties of $g$ do not change, of course. – Moishe Kohan Jan 06 '20 at 13:25
  • Thanks, but what does geometrically means for a connection, not to be torsion-free? Nothing? I understand it's just a way to connect different tangent spaces to the manifold, but doesn't it affect the geometry in some way? – Giulio Binosi Jan 07 '20 at 13:28
  • See these MO and MSE answers. Do they answer your question? If not, you should clarify your question. – Moishe Kohan Jan 07 '20 at 16:50
  • Do you know some articles in the literature that deal with this topic? – Giulio Binosi Jan 07 '20 at 17:29
  • You can take a look here. However, as I said, first read the linked Mathoverflow and MSE discussions. – Moishe Kohan Jan 07 '20 at 17:40
  • Thank you very much – Giulio Binosi Jan 07 '20 at 17:41
  • Are you interested in general connections, or just connections that preserve a given metric $g$? – Travis Willse Jan 08 '20 at 03:27
  • @TravisWillse what do you mean with general connection? Isn't it true that every connection that preserves a given metric is Torsion-free? – Giulio Binosi Jan 08 '20 at 14:07
  • No: Given a metric $g$, in any coordinates the Christoffel symbols $\Gamma_{ij}^k$ are symmetric in $i, j$. Then, for any functions $\alpha_{ij}^k$ satisfying $\alpha_{ji} = -\alpha_{ij}$, the connection $\nabla'$ with Christoffel symbols $\Gamma_{ij}^k + \alpha_{ij}^k$ still preserves $g$ but (provided the $\alpha_{ij}^k$ are not all identically zero), the torsion of $\nabla'$ is nonzero. The Fundamental Theorem of Riemannian Geometry says that for any metric $g$ there is precisely one torsion-free connection that preserves $g$, namely the Levi-Civita connection. – Travis Willse Jan 09 '20 at 01:20
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    In fact more (or rather, less) is true: In a sense that can be made precise, for most connections $\nabla$ (including some torsion-free ones) there is no metric $g$ such that for which $\nabla g = 0$. – Travis Willse Jan 09 '20 at 01:23

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Some basic geometric discussion of (not necessarily torsion-free) affine connections (say, when two connections have the same set of geodesics) you can find in Spivak's book,

"Comprehensive Introduction to Differential Geometry" (Publish Or Perish, 2000), volume 2, chapter 6.

See also

F. Hehl, Y. Obukhov, Élie Cartan's torsion in geometry and in field theory, an essay. Ann. Fond. Louis de Broglie 32 (2007), no. 2-3, 157–194.

Moishe Kohan
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