# finding the volume form under a change of metric

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 ?

• 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 Commented Oct 16, 2022 at 12:26
• 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...? Commented Oct 16, 2022 at 12:39

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.