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:
- $F$ 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 $x$).
- $F_{x_0}$ 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.
A possible way to prove this is to remember that the LC connection is the unique torsion free connexion for which the metric tensor is parallel.
The fact that your formula gives a connexion is obvious.
To check that it is torsion free, note that $g_N(\tilde \nabla _X Y, Z)= g(\nabla _X Y, Z)$ for every triple of tangent vector fields on $N$
To check that it preserves the induced tensor metric let $X,Y,Z$ three tangent vector fields on $N$. We can extend these fields on $M$ to compute :
$(\tilde \nabla _X g) (X,Y)= X. g(Y,Z)-g(\tilde \nabla _X Y, Z)- g(Y, \tilde \nabla _XZ)=X. g(Y,Z)-g( \nabla _X Y, Z)- g(Y, \nabla _XZ) $
Best Answer
I'll register a shorter coordinate-free proof. Let the Koszul formula be written as $$2g(\nabla_XY,Z) = A(X,Y,Z) + B(X,Y,Z),$$where $A$ is the part of the Koszul formula containing directional derivatives and $B$ is the part containing Lie brackets. We know that any two connections differ by a tensor, so write $\widetilde{\nabla}_XY = \nabla_XY + T_XY$ -- the goal is to find $T$, and we know that $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = \widetilde{A}(X,Y,Z) + \widetilde{B}(X,Y,Z).$$Clearly $\widetilde{B}(X,Y,Z) = {\rm e}^{2f}B(X,Y,Z)$, while $$ X\widetilde{g}(Y,Z) = X({\rm e}^{2f})g(Y,Z) + {\rm e}^{2f}X(g(Y,Z))$$says that $\widetilde{A}(X,Y,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + {\rm e}^{2f}A(X,Y,Z)$. Thus $$2\widetilde{g}(\widetilde{\nabla}_XY,Z) = X({\rm e}^{2f})g(Y,Z) + Y({\rm e}^{2f})g(X,Z) - Z({\rm e}^{2f})g(X,Y) + 2{\rm e}^{2f}g(\nabla_XY,Z).$$Evaluating $X({\rm e}^{2f}) = 2{\rm e}^{2f}\,X(f)$, etc., and simplyfying $2{\rm e}^{2f}$ on everything, we get $$g(\nabla_XY + T_XY,Z) = X(f)g(Y,Z) + Y(f)g(X,Z) - Z(f)g(X,Y) + g(\nabla_XY,Z).$$Eliminate $\nabla_XY$ from the above and use the definition of $g$-gradient to write the right side in the form $g({\rm something}, Z)$, obtaining $$g(T_XY,Z) = g(X(f)Y + Y(f)X - g(X,Y){\rm grad}(f), Z).$$This means that $$T_XY = X(f)Y + Y(f)X - g(X,Y){\rm grad}(f)$$and hence $$\widetilde{\nabla}_XY = \nabla_XY +X(f)Y + Y(f)X - g(X,Y){\rm grad}(f).$$