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Can I reconstruct a Riemannian metric out of its Levi-Civita connection? In other words: Given two Riemannian metrics $g$ and $h$ on a manifold $M$ with the same Levi-Civita connection, can I conclude that $g=h$ up to scalars?

If not, what can I say about the relationship between $g$ and $h$? How rigid is the Levi-Civita-Connection?

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This is something I have been thinking about recently, allow me to complete Mariano Suárez-Alvarez' answer.

First just an observation: "pick any Riemannian manifold with trivial holonomy at each point: for example, a space form of curvature zero": actually you have no choice, a connection with trivial holonomy has vanishing curvature, so the only candidate metrics are Euclidean. In the language of geometric structures, a flat torsion-free connection is equivalent to an affine structure.

Now to the point. As we are going to see, the answer to your question is "generically, yes". First note that given a connection $\nabla$ (let's permanently assume that $\nabla$ is torsion-free otherwise there's no chance of it being a Riemanniann connection), by definition a metric $g$ has Levi-Civita connection $\nabla$ if and only if $g$ is $\nabla$-parallel: $\nabla g = 0$.

Let us make a general observation about parallel tensor fields with respect to a given connection. If $F$ is a parallel tensor field, then $F$ is preserved by parallel transport. In particular:

  1. $F$ is completely determined by what it is at some point $x_0 \in M$ (to find $F_x$, parallel transport $F_{x_0}$ along some path from $x_0$ to $x$).
  2. $F_{x_0}$ must be invariant under the holonomy group $\operatorname{Hol}(\nabla, x_0)$.

Conversely, given a tensor $F_{x_0}$ in some tangent space $T_{x_0} M$ such that $F_{x_0}$ is invariant under $\operatorname{Hol}(\nabla, x_0)$, there is a unique parallel tensor field $F$ on $M$ extending $F_{x_0}$ (obtained by parallel-transporting $F_{x_0}$).

So you have the answer to your question in the following form:

There are as many Riemannian metrics having Levi-Civita connection $\nabla$ as there are inner products $g$ in $T_{x_0} M$ preserved by $\operatorname{Hol}(\nabla, x_0)$.

Now you may want to push the analysis further: how many is that? The answer is provided by analysing the action of the restricted holonomy group $\operatorname{Hol}_0(\nabla, x_0)$ on $T_{x_0}M$. Now this is just linear algebra: let's just call $G = \operatorname{Hol}_0(\nabla, x_0)$ and $V = T_{x_0} M$. Let $g$ and $h$ be two inner products in $V$ that are preserved by $G$, in other words $G \subset O(g)$ and $G \subset O(h)$. If $G$ acts irreducibly on $V$, i.e. there are no $G$-stable subspaces $\{0\} \subsetneq W \subsetneq V$, then a little exercise that I am leaving to you shows that $g$ and $h$ must be proportional. So in the generic case where $\nabla$ is irreducible, the answer to your question is yes:

If $\nabla$ is irreducible, all Riemannian metrics with connection $\nabla$ must be equal up to positive scalars.

NB: note that there might not be any such metrics if $G$ does not preserve any inner product on $V$, in other words $G$ must be conjugated to a subgroup of $O(n)$.

On the opposite side of the spectrum, if $G$ is trivial, i.e. $\nabla$ is flat, then $g$ and $h$ can be anything, there are no restrictions:

If $\nabla$ is flat, there are as many Riemannian metrics with connection $\nabla$ as there are inner products in a $\dim M$-dimensional vector space, they are the Euclidean metrics on $M$.

In the "general" case where $\nabla$ is reducible, I hope I am not mistaken (I won't write the details) in saying that you can derive from the de Rham decomposition theorem that the situation is a mix of the two previous "extreme" cases:

If $\nabla$ is reducible, locally one can write $M = M_0 \times N$, such that both $g$ and $h$ split as products. The components of $g$ and $h$ on $M_0$ are both Euclidean and their components on $N$ are equal up to a scalar.

If this is correct, I believe your question is answered completely.

NB: In this paper (see also this one), Richard Atkins addresses this question. I haven't really looked but since it seems to me that there is not much more to say than what I've written, I have no idea what he's really doing in there.

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  • $\begingroup$ ''if ∇ is flat, there are as many Riemannian metrics with connection ∇ as there are inner products in a M-dimensional vector space, they are the Euclidean metrics on M''. Isn't there just one Euclidean metric, why many? $\endgroup$ – Shadumu Nov 6 '18 at 10:37
  • $\begingroup$ @Shadumu There is more than one inner product in a vector space. Given a basis, any positive definite matrix defines an inner product. $\endgroup$ – Seub Nov 6 '18 at 11:49
  • $\begingroup$ so they are all Euclidean metrics? I thought it is just $\delta_{ij}$ is called so $\endgroup$ – Shadumu Nov 6 '18 at 12:02
  • $\begingroup$ The inner product $\delta_{ij}$ depends on a choice of basis. If you pick different a different basis, you get a different inner product (unless the new basis is orthonormal). $\endgroup$ – Seub Nov 6 '18 at 12:04
  • $\begingroup$ I guess my confusion is that isn't Euclidean metric just the dot product/pythagorean theorem? why there are many Euclidean metrics? $\endgroup$ – Shadumu Nov 6 '18 at 14:10
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No. If $g$ is a metric, $2g$ is also a metric and they both have the same L-C connection.

More interestingly: In general, a connection is metric (that is, comes from a metric) if its holonomy at each point is contained in the orthogonal subgroup (this was discussed at https://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita), and any metric whose value at each point is preserved by the holonomy is one whose L-C connection is the one we started with. To get examples, pick any Riemannian manifold with trivial holonomy at each point: for example, a space form of curvature zero (the Euclidean plane or a flat torus, say,to keep things simple): there are many metrics which induce the same connection (and not multiples of each other), which can be constructed by picking any inner product in the tangent space at one point and transport it to the rest of the manifold parallelly w.r.t the connection of the flat metric we started with.

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  • $\begingroup$ Okay, for sure. I mean up to scalars. I will edit the question. $\endgroup$ – archipelago Mar 24 '13 at 11:20

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