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The fundamental theorem of Riemannian geometry says that for a manifold with a given metric, there is a unique torsion-free connection.

Suppose instead that we are given a connection. According to answers to this question, a metric exists that induces that connection. How much non-uniqueness is there in the metric? It seems that there is at least an ambiguity up to a constant factor, because if the connection is metric-compatible with the metric $g$ (i.e., $\nabla g=0$), then it's also compatible with $cg$.

Does any of this change in the semi-Riemannian case?

  • To clarify, your question is: Given a smooth manifold with a torsion-free connection $\nabla$, can we describe the class of metrics $g$ whose Levi-Civita connection is $\nabla$? Is that right? – Jesse Madnick Jul 30 '17 at 01:34
  • @JesseMadnick: Yes, that's right. –  Jul 30 '17 at 01:34
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    Relevant: https://mathoverflow.net/questions/54434/when-can-a-connection-induce-a-riemannian-metric-for-which-it-is-the-levi-civita – Jesse Madnick Jul 30 '17 at 01:35
  • @JesseMadnick: Nice. I guess that question is about existence, while mine is about uniqueness. I've edited the question to make it more specifically about uniqueness. –  Jul 30 '17 at 01:37
  • @JesseMadnick: Are the restrictions on $a$ and $b$ only for the Riemannian case, and not needed for semi-Riemannian? –  Jul 30 '17 at 01:47
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    related: https://mathoverflow.net/questions/140438/can-an-einstein-metric-have-the-same-levi-civita-connection-with-a-non-einstein –  Aug 02 '17 at 15:24

1 Answers1

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This is a partial answer. We'll reduce the question to one of representation theory (of Lie groups), and give an answer in two "extreme" cases.

Suppose $\nabla$ is the Levi-Civita connection of some Riemannian metric $g_1$. Since $\nabla$ is torsion-free, we know that $\nabla$ will be the Levi-Civita connection of a metric $g_2$ if and only if $\nabla g_2 = 0$. This means that we have to understand which (positive-definite) symmetric $2$-tensor fields are $\nabla$-parallel. Understanding which tensor fields are $\nabla$-parallel can be accomplished via:

The Holonomy Principle: Let $\nabla$ be a connection on a connected smooth manifold $M$. Let $\text{Hol}_x \leq \text{GL}(T_xM)$ denote the holonomy group (really, holonomy representation) of $\nabla$ at $x \in M$.

(a) If $T \in \Gamma(TM^{\otimes r} \otimes T^*M^{\otimes s})$ is a parallel tensor field on $M$, then $T|_x$ is fixed by the $\text{Hol}_x$-action on $T_xM^{\otimes r} \otimes T_x^*M^{\otimes s}$.

(b) Conversely: If $T_0$ is a tensor at $x$ fixed by the $\text{Hol}_x$-action on $T_xM^{\otimes r} \otimes T_x^*M^{\otimes s}$, then there exists a unique parallel tensor field $T$ on $M$ with $T|_x = T_0$.

Since our connection $\nabla$ is the Levi-Civita connection of some (let's say Riemannian) metric $g_1$, we have $\text{Hol}_x \leq \text{SO}(T_xM, g_1) \cong \text{SO}(n)$. The question is now: What is the space of (positive-definite) symmetric $2$-tensors at $x$ which are fixed by the $\text{Hol}_x$-action on $\text{Sym}^2(T^*_xM) \subset T_x^*M^{\otimes 2}$. This is a question of representation theory.

Example: (The trivial example) Suppose $\nabla$ is the Levi-Civita connection of a flat metric $g_1$, and suppose $M$ is connected and simply-connected. Then $\text{Hol}_x = 0$ is the identity group, so every element of $\text{Sym}^2(T_x^*M)$ is fixed.

Concretely: If $g_0$ is any (positive-definite) symmetric $2$-tensor at $x$, then (by the Holonomy Principle) we can extend $g_0$ uniquely to a $\nabla$-parallel tensor field $g$ on all of $M$. The upshot is that, for the sort of connections $\nabla$ in this example, we essentially have an $\binom{n+1}{2}$-dimensional space of compatible metrics.

Note that the dimension $\binom{n+1}{2}$ is the largest possible. For the Levi-Civita connection of a "generic" Riemannian metric $g_1$, the holonomy group will be all of $\text{SO}(n)$, and the space of compatible metrics will be $1$-dimensional.

I do not know about the intermediate cases -- i.e., when $\nabla$ is not generic and not flat. (My representation theory needs some work!)

Jesse Madnick
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  • Nice. I lack the background to understand some of this, but I think I get the general idea. We can rule a plane with, say, a set of evenly spaced, parallel red lines, and another, similar set of blue lines, and call that the "graph paper" of the plane. But on the sphere we have less freedom. A patch of graph paper can be parallel transported around a closed loop, and at the end of that process it may have been rotated. This is inconsistent unless the graph paper satisfies certain constraints, which essentially make it consist of squares. –  Jul 30 '17 at 03:40