Let $M, N$ be Riemannian Manifolds and $\phi: M \to N$ a isometric diffeomorphism. We know that we have unique Levi-Civita Connections $\nabla^M, \nabla^N$ on $M, N$ respectively. One can check that $\tilde\nabla^N_V W = \phi^*\nabla^N_{\phi_* V}\phi_* W$ satisfies the properties of a Levi-Civita connection and hence (by uniqueness) $\tilde\nabla^N = \nabla^M$.

However it seems to me that this should be trivial, since $M$ and $N$ are "the same" (when identified via $\phi$) as far as their differentiable structures and metrics are concerned. Since the properties of a Levi-Civita connection are expressed using only "the language of Riemannian manifolds" (i.e. smooth functions, vector fields, the metric, etc.) it seems that it should be obvious that $\tilde\nabla^N$ is a Levi-Civita connection on $M$.

To me this seems analogous to other trivial facts like

  1. The tangent bundles of diffeomorphic manifolds are isomorphic (as vector bundles)
  2. The centers of isomorphic groups are isomorphic

(Number 1 can actually be argued to follow from the functoriality of taking tangent spaces, but AFAIK there is no suitable functor that works for 2).

So is it a valid argument to say that the Levi-Civita connection can be pulled back to $M$ because the two manifolds are "the same in every relevant aspect"?

Also is there any way to generalize this kind of argument?

EDIT: It feels to me like this question (and the other two facts I stated) are an example of the following principle:

Since $\phi$ preserves all structures on $M$ and $N$ (that is a Topological space, a smooth structure and a metric) and The Levi-Civita connections only depends on these structures they are equivalent (using $\phi$ to "translate" them).

For the center of a group you would argue analogously, saying that it only depends on the group structure, which is preserved by group isomorphisms. Therefore the centers must be isomorphic (with the isomorphism given by the isomorphism on the groups). And so on...

  • $\begingroup$ In my opinion, your argument is perfectly legit. $\endgroup$ – Amitai Yuval Dec 10 '18 at 18:34

One can certainly argue that an isometry preserves the Levi-Civita connection insofar as it preserves the metric and commutes with the evaluation and Lie bracket of vector fields. However, this isn't some kind of magical contentless proof, and in particular it's not clear to me what kind of argument you're trying to generalize.

Your examples are examples of functors (in the case of the center, it's only a functor on the category of groups and isomorphisms,) and functors preserve isomorphisms, but at some point you still have to actually check that your construction is functorial. That said, if this is of interest to you, then you could certainly argue that the Levi-Civita connection gives a functor from the category of Riemannian manifolds with isometries to the category of smooth manifolds equipped with a connection on their tangent bundle and diffeomorphisms mapping the connections into each other.

  • $\begingroup$ If you want to view the center as a "mapping" from Groups with isomorphisms to abelian groups isn't the functoriality equivalent to proving that isomorphic groups have isomorphic centers? So that wouldn't yield a proof of the latter fact. $\endgroup$ – 0x539 Dec 10 '18 at 20:04
  • $\begingroup$ @0x539 Well, yes, that's kind of my point. I don't see any meaningful way in which these claims are "truly" trivial, or purely formal-they're just things you have to prove which are pretty easy. $\endgroup$ – Kevin Carlson Dec 11 '18 at 0:29

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