# Different Riemannian manifolds with the same Riemannian volume form

Let $(X,g_1)$ and $(X,g_2)$ be two Riemannian manifolds over the same space $X$.

My (vague) question is the following : If I know that the two induced Riemannian volume form coincide, what can I say on $g_1$ and $g_2$ ?

A variant of the question : what kind of extra assumption(s) (like curvature, etc) can I add on $X$ and/or $g_1$ and $g_2$ so that the equality of the volume forms implies the equality of the Riemannian metrics (maybe up to identifiable "easy" transformations).

For a concrete example, we can consider for $X$ the $n$-th space of (unordered) configurations on $\mathbb R$, that is $$X=\Big\{(x_1,\ldots,x_n)\in\mathbb R^n : \; x_i\neq x_j \mbox{ for } i\neq j\Big\}/S_n,$$ where $S_n$ stands for the symmetric group over $n$ elements.

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## migrated from mathoverflow.netJul 15 '13 at 22:39

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Consider the linear map $L: T_pX \to T_pX$ given by $L:(g_1)^{-1}\circ g_2$, treating $g_i: T_p X \to T_p^*X$ as the duality map. (1) The requirement that $g_1$ and $g_2$ have the same volume forms is equivalent to saying that $\det L = 1$. (2) The requirement that $g_1$ and $g_2$ are identical is equivalent to saying $L = \mathrm{Id}$. So just on the linear algebra level you see how many degrees of freedom you are missing here. – Willie Wong Jul 15 '13 at 13:09
A mildly related comment: there's a theorem by J. Moser saying that if $M$ is a compact oriented manifold with two volume forms $\Lambda_{0}$ and $\Lambda_{1}$ such that $\int_{M}\Lambda_{0} = \int_{M}\Lambda_{1}$, then there exists a smooth isotopy $f_{t} \in \text{Diff}(M)$ such that $f_{1}^{*}\Lambda_{1} = \Lambda_{0}$. (J.Moser: On the volume elements on a manifold, 1965.) – Oldřich Spáčil Jul 15 '13 at 13:26
A concrete example stemming from Willie Wong's comment. Consider $\mathbb{R}^2$ equipped with the metrics $$g_1=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$g_2=\begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 2\end{bmatrix}.$$ Then the volume forms are $$\text{vol}_1=\text{vol}_2=dx\wedge dy,$$ but clearly the metrics are different. – Giuseppe Negro Jul 15 '13 at 13:47
One last remark: curvature will not help you here. @GiuseppeNegro's example can be modified to work over the torus $\mathbb{T}^2$. The two metrics have both identically vanishing curvature. There is no isometry between the two (consider the length of the shortest closed geodesics). One may also want to think about in general manifolds of the form $\mathbb{S}^{k_1}\times \mathbb{S}^{k_2} \times \cdots\times \mathbb{S}^{k_d}$ with the metric for each factor weighed differently. If you are faced with a specific problem it is better now to look at the specifics for that problem, rather than ... – Willie Wong Jul 15 '13 at 15:12
For any given $g_1$, in each conformal class of $g_2$, there is a unique metric with the same volume form as $g_1$. So I suppose you could require $g_1$ and $g_2$ to lie in the same conformal class. In particular, a necessary condition for them to be equivalent is that they have the same conformal curvature invariants. In 2 dimensions, there are no conformal invariants. But in 3 dimensions, there is the Schouten tensor, and in 4 dimensions, the Weyl curvature. However, this will not be enough to concluded equivalence, for example if the Weyl tensor vanishes, since this is diffeo. invariant. – Agol Jul 15 '13 at 15:58