# Is it trivial that the Levi-Civita connection can be pulled back to a isometric manifold?

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...

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