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Choose a volume form $\omega$ on $M$, oriented manifold. For every $F\in C^{\infty}_c(M)$, we define $$ \int_M F:=\int_M F\omega $$ where in the right hand term $M$ is taken wit positive orientation w.r.t. $\omega$, that is $$ \omega=\lambda_p dx_1\wedge\dots\wedge dx_n $$ for every $p$, and $\lambda_p>0$.

Let $M$ be a $n$-dimensional compact manifold and let $\omega\in\Omega^n(M)$ be a volume form. Define the volume of M as $$ Vol(M):=\int_M 1=\int_M\omega $$

Now I'm asking myself if the Volume is well-defined, i.e. if it doesn't depend on the choice of $\omega$.

I would say that it depends on the choice of $\omega$.

Infact, I know that $\omega$ would be, locally, something of the kind $$ (\omega)_p=F_p(dx_1\wedge\dots\wedge dx_n)_p $$ where $x_1,\dots,x_n$ are the coordinates induced by a local chart and $F\in C^{\infty}(M)$.

So $\omega$ is not unique.

Probably I haven't learnt well this topic.

Remember that a volume form for a $n$-dimensional manifold $M$ is a line section of the canonical bundle $\Lambda^nT^*M$, i.e. a non-vanishing $n$-form.

 How can I prove that the volume of a compact manifold always positive?
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  • $\begingroup$ Okay. How can I, at least, notice that the volume is non-negative? @ThomasRot $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 9:49
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    $\begingroup$ I don't think so. It is an exercise from my professor to prove that the volume of a compact manifold is always positive... @ThomasRot $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 9:54
  • $\begingroup$ I will add some details... $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 9:57
  • $\begingroup$ See the edit @ThomasRot $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 10:04
  • $\begingroup$ I think, I just need to add the hypothesis of orient ability in the definition of volume of a manifold. $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 10:09

2 Answers 2

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Volume is non-unique. Therefore, the smooth structure of a smooth, compact manifold is not enough for the manifold to have a canonical volume.

What can happen is, that if your manifold has a richer structure, then you can associate a canonical volume to that richer structure. For example, if you have a Riemannian manifold $(M,g)$, then the differential form whose expression in a local chart, $(U,x)$, is $$ \omega=\sqrt{|\det g|}dx^1\wedge...\wedge dx^n $$ is a canonical volume form, and then the volume of $M$ as a Riemannian manifold is $$\int_M \omega. $$

Or if $(M,\omega)$ is a symplectic manifold, with $\omega$ being the symplectic form, then $\Omega=\omega^{\wedge n}$ is a canonical volume form, and the volume of $M$ as a symplectic manifold is $$ \int_M \Omega. $$

However, in either case, $M$ by itself has no meaningful volume. Volume isan extra structure you either choose, or a different, but chosen structure (like a riemannian metric, or a symplectic form) chooses for you.

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  • $\begingroup$ What about positivity of the volume? $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 10:10
  • $\begingroup$ @avati91 Volume is generally taken to be positive. $\endgroup$ Commented Aug 20, 2015 at 10:55
  • $\begingroup$ That's what I have to prove... $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 12:10
  • $\begingroup$ @avati91 I don't think that can be proven, it is just convention. You integrate $fdx^1\wedge...\wedge dx^n$ by "erasing the wedges" (and obv. involving partitions unity as well but those wont change a thing), but as a Riemann-integral, $dx^1...dx^n$ is symmetric. But by the wedge product's skew-symmetry, $fdx^1\wedge dx^2\wedge...\wedge dx^n=-fdx^2\wedge dx^1\wedge...\wedge dx^n$, so integrating the latter would give us an opposite-sign integral. However, if the manifold's orientable, and $x^1...x^n$ is a positive coordinate system, then we use the first integral I wrote by convention. $\endgroup$ Commented Aug 21, 2015 at 7:30
  • $\begingroup$ @avati91, which is positive, if $f$ is positive. But for a Riemannian manifold, for example, $\sqrt{|\det g|}$ is always positive, so the volume will be too, since you are doing $\int_M\sqrt{|\det g|}\ dx^1...dx^n$-type Riemann-integrals strung together by partitions of unity. That is manifestly positive. $\endgroup$ Commented Aug 21, 2015 at 7:32
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Since nobody answered me, I will try myself. I have to prove that if $\omega$ is a volume form, then $$ Vol(M):=\int_M 1=\int_M\omega>0 $$ where the second equality follows by definition as well and $M$ is taken positively oriented w.r.t $\omega$, i.e. for a local chart $\omega=\lambda dx_1\wedge\dots\wedge dx_n$ and $\lambda(p)>0$ for every $p$ in the chart.

Since $M$ is compact, I can cover it by a finite number of charts $(U_i,\phi_i)$, for $i=1,\dots,n$. Let's take a partition of unity $\{\rho_i\}_i$ subordinate to $\{U_i\}_i$ and define $\omega_i:=\rho_i\omega$. We have $supp(\omega_i)\subset U_i$. Then $$ Vol(M)=\int_M\omega=\sum_{i=1}^n\int_M\omega_i $$ Observe that, for every $i=1,\dots,n$ $$ \int_M\omega_i=\int_{\phi_i(U_i)}(\phi_i^{-1})^*\omega_i=\int_{\phi_i(U_i)}(\rho_i\lambda\circ\phi_i^{-1})>0 $$ The strictly positivity follows from the fact that $\rho_i\lambda\circ\phi_i^{-1}$ is a positive function and $\phi_i(U_i)$ is an open set, hence a set with positive measure.

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  • $\begingroup$ What do you mean by $\int_M 1$? It does not make sense. $\endgroup$ Commented Aug 20, 2015 at 15:39
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    $\begingroup$ It makes sense, it is a definition. Have you read the first four lines of the question? $\endgroup$
    – avati91
    Commented Aug 20, 2015 at 15:41
  • $\begingroup$ So you define $\int_M 1$ as $\int_M \omega$? $\endgroup$ Commented Aug 20, 2015 at 18:49
  • $\begingroup$ Exactly! Yes. Now I'm stuck $\endgroup$
    – avati91
    Commented Aug 21, 2015 at 0:18

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