Let $(M,g_{ij})$ be a smooth Riemannian manifold. We introduce a new metric $\tilde{g}_{ij}: = (e^{\phi} g)_{ij}$ where $\phi:M \to \mathbb{R}$ is a smooth function. Now we want to calculate the volume form w.r.t. the new metric in terms of the old volume form. All of this in local coordinates. I found two ways to do this:

Let $(U,x_1,...,x_n)$ be the local coordinates on M, such that the frame $(\partial x_1,..., \partial x_n)$ is orthonormal w.r.t. $g_{ij}$.

the "old" volume form is then just $vol = dx_1 \wedge dx_2 \wedge ... \wedge dx_n$.

  1. an orthonormal frame $(f_1,...,f_n)$ w.r.t. $\tilde{g}_{ij}$ is then given by $f_i := e^{-\frac{\phi}{2}} \partial{x_i}$. So that we get the new volume form: $vol_{new} = e^{-\frac{\phi}{2}} d{x_1} \wedge ... \wedge e^{-\frac{\phi}{2}}d{x_n} = e^{-n\frac{\phi}{2}} vol$

  2. $vol_{new} = \sqrt{det(\tilde{g}_{ij}})dx_1 \wedge ... \wedge dx_n = e^{n \frac{\phi}{2}}\sqrt{(det(g_{ij})}dx_1 \wedge...\wedge dx_n = e^{n \frac{\phi}{2}} vol$.

Now why don't I get the same result ?

  • 1
    $\begingroup$ A coordinate system such that associated frame is orthonormal may not exist: this would imply that the metric is flat. Also, there is no reason for $e^{-\frac{\phi}{2}}\partial_i$ to be tangent to a coordinate system $\endgroup$
    – Didier
    Oct 16, 2022 at 12:26
  • $\begingroup$ Let's check what happens in one dimension: Let $\partial$ denote the constant vector field $1$ and $dx$ the dual coordinate $1$-form. If $e^{-\phi/2}\,\partial$ is a unit vector in a scaled metric, then the dual $1$-form is...? $\endgroup$ Oct 16, 2022 at 12:39

1 Answer 1


In your first attempt, forget about using coordinates: there may not exist any coordinates system such that the associated frame is orthonormal. Although, there is something to do with your idea of using dual forms to certain vector fields: these vector fields will just not be given by a coordinates system.

Let $\{E_1,\ldots,E_n\}$ be a local orthonormal frame with respect to the metric $g$, with direct orientation. Let $\theta^i=g(\cdot,E_i)$ be the associated $1$-forms. Then the Riemannian volume form $v_g$ is locally given by $$ v_g = \theta^1\wedge \cdots \wedge \theta^n. $$ Indeed, this is a volume form that is equal to $1$ at the orthonormal frame $\{E_1,\ldots,E_n\}$, and hence, at any orthonormal frame.

Now, consider the metric $\tilde{g}=e^{\phi}g$. The frame $\{\tilde{E}_1,\ldots,\tilde{E}_n\}$ given by $\tilde{E}_i = e^{-\frac{\phi}{2}}E_i$ is orthonormal with respect to the new metric $\tilde{g}$. It follows that the associated Riemannian volume form $v_{\tilde{g}}$ is locally given by $$ v_{\tilde{g}} = \tilde{\theta}^1\wedge \cdots \wedge \tilde{\theta}^n, $$ with $\tilde{\theta}^i = \tilde{g}(\cdot,\tilde{E}_i) = e^{\phi}g(\cdot,e^{-\frac{\phi}{2}}E_i)=e^{\frac{\phi}{2}}\theta^i$. It finally follows that $$ v_{\tilde{g}}= e^{\frac{n}{2}\phi}\theta^1\wedge\cdots\wedge\theta^n = e^{\frac{n}{2}\phi}v_g. $$ This gives you the same result that your second attempt.


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