Let $f:X\to Y$ be a submersion between orientable smooth manifolds of respective dimensions $n,p$ and let $j:M=f^{-1}(y)\hookrightarrow X$ denote the inclusion of some fibre of $f$.
My naïve (I am not a differential geometer) question is: what data do we need in order to define a volume form on $M$, given the existence of volume forms on $X$ and $Y$ ?
In the case where $Y=\mathbb R^p$ and $f=(f_1,\cdots,f_p)$, I think that if for some $\omega \in \Omega^{n-p}(X)$ the form $df_1\wedge\cdots\wedge df_p\wedge\omega\in \Omega^n(X)$ is a volume form (i.e. vanishes nowhere), then the form $j^* \omega\in\Omega^{n-p}(M)$ is a volume form for $M$.
In the even more particular case where $X=\mathbb R^n$ and $p=1$ we may take for $\omega$ the $(n-1)$-form $$\omega=\sum_{i=1}^{n} (-1)^{i-1}\partial_i f \wedge dx_1\cdots\wedge \widehat{ dx_i}\wedge\cdots \wedge dx_n \in \Omega^{n-1}(\mathbb R^n)$$ and obtain the required volume form $j^* \omega\in\Omega^{n-p}(M)$.
I once long ago read that last beautiful formula somewhere and the remembrance of that thing past motivates my present question, which is to find generalizations of that formula.
$\begingroup$
$\endgroup$
4
asked May 10, 2021 at 20:18
-
$\begingroup$ 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 $\omega(b_1,\ldots,b_{n-p}) = 1$. $\endgroup$– HugoCommented May 16, 2021 at 18:13
-
$\begingroup$ Obs.: In the above comment, I'm assuming the volume form of $X$ is constructed from a fixed Riemannian metric. $\endgroup$– HugoCommented May 16, 2021 at 18:43
-
$\begingroup$ 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 $Y$. $\endgroup$– HugoCommented May 16, 2021 at 18:48
-
$\begingroup$ Thanks for your comments, @Hugocito. $\endgroup$– Georges ElencwajgCommented May 18, 2021 at 16:01
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Take a unit tangent p-vector $a$ based at $y \in Y$ (you need the volume form on $Y$ to say what a unit p-vector is). At each point $x$ of the fiber $f^{-1}(y)$ consider the contraction of your volume form on $X$ with any p-vector $b$ in $\Lambda^p(T_xX)$ that projects down to $a$ by the linear map $D_xf$. Note that the pullback of this contracted form to the fiber (you only evaluate it on $n-p$ vectors tangent to the fiber) does not depend on the choice of the p-vectors that project to $a$.
This is basically the pushforward construction for forms, which also works for densities.
-
$\begingroup$ Thank you for this interesting answer. $\endgroup$ Commented May 11, 2021 at 17:59