# Any volume form on a smooth $n$-dimensional manifold is locally a pullback of the standard volume form $dx_1\wedge …\wedge dx_n$ on $\mathbb R^n$?

Let $M$ be an $n$-dimensional smooth manifold equipped with a volume form $\omega$ , and let $\omega_0:=dx_1\wedge ...\wedge dx_n$ be the standard volume form on $\mathbb R^n$ , then is it true that for every $a \in M$ , there exists an open set $U$ containing $a$ in $M$ and a diffeomorphism $g:U \to \mathbb R^n$ such that $\omega = g^*\omega_0$ ?

• Are you familiar with Moser's trick, which is used in the proof of Darboux theorem for symplectic manifolds? – Amitai Yuval Apr 29 '17 at 15:15
• @AmitaiYuval : um no – user228169 Apr 29 '17 at 15:16
• @AmitaiYuval: I don't think this is nearly that difficult. The OP is just asking whether n-manifolds are locally isometric to $\mathbb{R}^n$. – Faraad Armwood Apr 29 '17 at 15:17
• @FaraadArmwood No. The answer to the question you're talking about is negative, of course. However, volume preserving is weaker than being an isometry. – Amitai Yuval Apr 29 '17 at 15:21
• @AmitaiYuval: You are right, I apologize. I'm writing something right now in response to this and it is not very technical yet. – Faraad Armwood Apr 29 '17 at 15:22

The following is a variation of Pedro's answer (which doesn't work in my opinion, as explained in Ted's comment).

Take a coordinate chart around $a$ and write $\omega=fdx^1\wedge\ldots\wedge dx^n$. Try new coordinates as follows: $y_1=gx_1,y_2=x_2,\ldots,y_n=x_n$, where $g$ is some smooth function. Then\begin{align}dy^1&=\left(\frac{\partial g}{\partial x^1}x_1+g\right)dx^1+\frac{\partial(gx_1)}{\partial x^2}dx^2+\ldots+\frac{\partial(gx_1)}{\partial x^n}dx^n,\end{align}and so,\begin{align}dy^1\wedge\ldots\wedge dy^n=\left(\frac{\partial g}{\partial x^1}x_1+g\right)dx^1\wedge\ldots\wedge dx^n.\end{align}Hence, in order for the $y$'s to be good coordinates, we need to solve $$\frac{\partial g}{\partial x^1}x_1+g=f.$$ For convenience, we may assume that $x_1$ does not vanish, and then the above first order linear equation is easy to solve.

Edit: In second thought, the computation is even easier when guessing a coordinate change of the form $y_1=g,y_2=x_2,\ldots,y_n=x_n$, where $g$ is a smooth function (here, $g$ replaces $gx_1$). Now we have $$dy^1\wedge\ldots\wedge dy^n=dg\wedge dx^2\wedge\ldots\wedge dx^n=\frac{\partial g}{\partial x^1}dx^1\wedge\ldots\wedge dx^n,$$and so, the function $g$ only has to satisfy $\frac{\partial g}{\partial x^1}=f.$

It's true. If you choose coordinates $x_1, \ldots, x_n$ in $M$, the volume form will have the form $f dx_1 \wedge \ldots \wedge dx_n$, with $f$ non-vanishing. So just make the coordinate change $y_1 = x_1/f$ and $y_2 = x_2, \ldots, y_n = x_n$. Then in these coordinates the volume form is $dy_1 \wedge \ldots \wedge dy_n$ as you wish.

In fancier language: orientations are really $\text{GL}^{+}$-structures, and these are always integrable (as the argument above shows).

• Ah, I knew there was some messing around you could do once you knew that locally you could write that form as $f dx^1 \wedge \cdots \wedge dx^n$. Thanks for seeing what I couldn't. – Faraad Armwood Apr 29 '17 at 15:53
• @Pedro : um so could you just please point out what is $g$ ? – user228169 Apr 29 '17 at 16:07
• Not so fast, @Pedro. First, I think you meant $y_1=fx_1$. But, regardless, what happened to the $-\dfrac{x_1}{f^2} df\wedge dx_2\wedge\dots \wedge dx_n$ term? Your trick only works if $f=f(x_2,\dots,x_n)$, which is highly unlikely. – Ted Shifrin Apr 29 '17 at 16:16
• This coordinate change only works if $f$ is constant in $x_1$. Otherwise, it doesn't. – Amitai Yuval Apr 29 '17 at 16:17
• @TedShifrin You are right, of course. Would like to hear your opinion on my answer, which is an adjusted version of Pedro's attempt. – Amitai Yuval Apr 29 '17 at 16:55