# Are two metrics with the same Levi-Civita connection and the same volume form identical?

Given a (smooth, orientable) n-dimensional manifold $$M$$ with two (pseudo-)Riemannian metrics $$g_{1}$$ and $$g_{2}$$ of the same signature that induce the same Levi-Civita connection and satisfy $$\sqrt{|\det{g_1}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n} = \sqrt{|\det{g_2}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n}$$,

is $$g_{1}=g_{2}$$?

• From the metric we can derive several quantities (eg. curvature tensor). A question I'd like to ask is: which derived quantities determine the metric? – Student Jun 5 at 13:42
• The right statement is: Suppose that $(M,g)$ does not split locally as Riemannian product. Then the LC connection determines the metric up to a constant factor. – Moishe Kohan Jun 5 at 15:45

Take two inner products on $$\mathbb{R}^n$$ given by symmetric matrices $$(g_{ij})_{1\leq i,j\leq n}$$, $$(h_{ij})_{1\leq i,j\leq n}$$ such that $$\det g_{ij} =\det h_{ij}=1$$. They have the same volume forms and Levi-Civita connections though they are different if the matrices are different.
Consider a Riemannian manifold admitting a parallel (with respect to the Levi-Civita connection $$\nabla^g$$) symmetric bilinear form $$\beta$$ which is not a multiple of the metric $$g$$. Then, for an open set of constants $$(a,b)\in\mathbb R^2$$ $$g_{a,b}:=a g+b\beta$$ is a Riemannian metric, i.e., positive definite. The Levi-Civita connection of $$g_{a,b}$$ is $$\nabla^g$$. Moreover, the volume form of $$g_{a,b}$$ is a constant multiple (depending on $$a,b$$) of the volume form of $$g$$, as the volume forms $$vol_{a,b}$$ are parallel with respect to $$\nabla^g$$. Clearly, at $$(a,b)=(1,0)$$ we have $$g_{a,b}=g$$ and $$\frac{\partial vol_{a,b}}{\partial a}\neq0.$$ Hence, you always find a curve of Riemannian metrics with the same Levi-Civita connection and the same volume form under the above condition. Of course, explicit examples are provided by the answers of Ben and Liviu, but also by product metrics. Analogous arguments apply to the pseudo-Riemann case.