Problem Statement
This is an in-text exercise from John Lee's Smooth Manifolds book chapter 14.
If $F: M \rightarrow N$ is a smooth function from manifolds $M$ to $N$, and $\mu$ is a smooth density on $N$, I want to show that $F^*\mu$ is a smooth density on $M$.
I can show it's a density on $M$. There's already a question that deals with part of that here. If I assume that $F$ is a local diffeomorphism, I can prove the statement. I've seen the problem stated with just smoothness in another textbook, so I feel like it should be possible without assuming $F$ is a local diffeomorphism.
Attempt
I've already shown:
- If $f$ is a smooth function on $N$ and $\mu$ is a density, $F^*(f\mu) = (f \circ F)F^*\mu.$
- If $w$ is an n-form on $N$, then $F^*|w| = |F^*w|$.
Write $m$ = dim $M$ and $n$ = dim $N$.
Pick a point $p$ in $M$. Find a neighborhood, $V$, of $F(p)$ where we have a non-vanishing n-form, $w$ ($w$ is a local frame over $V$ for $\Lambda^n(N)$, the n-forms on $N$). Using the fact that $F$ is smooth, we can find a neighborhood $U$ of $p$ where $F(U) \subset V$. Shrinking $U$ if necessary, we have a non-vanishing m-form on $U$, $w'$.
If I show $(F|_U)^*\mu|_V$ is smooth, then since $(F|_U)^*\mu|_V = (F^*\mu)|_U$, I'll have shown the density is locally smooth, and since smoothness is a local property, it's globally smooth.
Below I'll just write $F$ for $F|_U$.
Since $|w|$ is a local frame for the density bundle on $N$, for some smooth function $f: V \rightarrow R$, we have, $$\mu = f|w|$$
Now use our two results above: $$F^*\mu = F^*(f|w|) = (f \circ F)F^*|w| = (f \circ F)|F^*w|$$
We also have, since $w'$ is a frame for the m-forms bundle on $M$, $$F^*w = gw'$$ where $g : U \rightarrow R$, is a smooth function.
We then have $|F^*w| = |gw'|$, and you can check that that's also $|g||w'|$ by evaluating on a local frame of $U$ in the tangent bundle. We then have: $$|F^*w| = |g| |w'|$$
Because $|w'|$ is a local frame on $U$ for the density bundle, $|g|$ is the component function of $|F^*w|$ in that frame, so $|F^*w|$ is smooth iff $|g|$ is smooth as a function. So here's where I have to assume $F$ is a local diffeomorphism.
If $F$ is a local diffeomorphism, then $F^*w$ is non-vanishing, so $|F^*w|$ is a positive density, making $|g|$ non-vanishing on $U$, so $g$ is either always positive or always negative. Since absolute value is smooth away from zero, $|g|$ is smooth.
Once you have that $|F^*w|$ is smooth, just note that $f \circ F$ is smooth so the product is smooth.
Does anybody see some way to tweak that to work with just smoothness of $F$, or did I go wrong somewhere?
Aack! That statement is wrong. I've posted a correction on my web page.
Good catch!