Timeline for Finding a volume form on a fibre of a submersion between oriented manifolds

6 events

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May 18, 2021 at 16:01	comment	added	Georges Elencwajg		Thanks for your comments, @Hugocito.
May 16, 2021 at 18:48	comment	added	Hugo		In the general case, I believe the construction is similar. Consider a local frame $b_1,\ldots,b_{n-p}$ on $M$ and extend it to a local frame $b_1,\ldots, b_n$ of X. Define $$\omega_M(b_1,\ldots,b_{n-p}) := \frac{\omega_X(b_1,\ldots,b_n)}{\omega_Y(df(b_{n-p+1}),\ldots,df(b_n))},$$ where $\omega_X$ and $\omega_Y$ are the volume forms of $X$ and $Y$.
May 16, 2021 at 18:43	comment	added	Hugo		Obs.: In the above comment, I'm assuming the volume form of $X$ is constructed from a fixed Riemannian metric.
May 16, 2021 at 18:13	comment	added	Hugo		One way of constructing the volume form on $M$ is the following: Consider a local orthonormal frame $b_1,\ldots,b_{n-p}$ tangent to $M$. We can complete the previous frame to an orthonormal frame on $X$ (by using Gram–Schmidt process). So, we have $b_1,\ldots,b_{n-p},b_{n-p+1},\ldots,b_n$. We can assume $df(b_{n-p+1}),\ldots, df(b_n)$ is compatible with the orientation on $Y$, and, rearranging $b_1,\ldots,b_{n-p}$, we can assume $b_1,\ldots,b_n$ is compatible with the orientation of $X$. The volume form on $X$ is defined by the formula $\omega(b_1,\ldots,b_{n-p}) = 1$.
May 11, 2021 at 13:46	answer	added	alvarezpaiva		timeline score: 2
May 10, 2021 at 20:18	history	asked	Georges Elencwajg	CC BY-SA 4.0