In the book Riemannian Geometry, by Mandredo do Carmo, he supposes that $M$ is a riemannian oriented manifold and then defines the volume of a region $R$ contained in some image $\boldsymbol x(U)$ of a positive parametrization $\boldsymbol x:U\subset \Bbb R^n\to M$ as $$vol(R)=\int_{\boldsymbol x^{-1}(R)}\sqrt{\det (g_{ij})}\,dx_1\dots dx_n,$$ and then he argues that this definition is independent of the parametrization, using the change of variables theorem. At some moment, he simply says that $$\tag{1}\sqrt{\det(g_{ij})}(p)=J\sqrt{\det(h_{ij})}(p),$$ where $J=\det (d(\boldsymbol y^{-1}\circ \boldsymbol x)_p)$ and $(h_{ij})$ is the metric for other parametrization $\boldsymbol y:V\to M$, $R\subset \boldsymbol y(V)$. Said this, he emphasizes the importance of $\boldsymbol x$ and $\boldsymbol y$ being positive parametrizations (to the orientation of $M$), since then $J>0$ and $|J|=J$ and we can use the change of variables theorem to conclude the referred independence.

Then, I've tried to prove (1). After some computation, I've proved (if nothing is wrong) that $$\tag{2} G(p)=A^T(p) H(\boldsymbol y^{-1}\circ \boldsymbol x(p))A(p),$$ where $G=(g_{ij})$, $H=(h_{ij})$ and $A(p)=d(\boldsymbol y^{-1}\circ \boldsymbol x)_p$, with $p\in U$. And then $$\int_{ x^{-1}(R)}\sqrt{\det G(p)}\,dx_1\dots dx_n=\int_{\boldsymbol x^{-1}(R)}\sqrt{(\det A(p))^2\det H(\boldsymbol y^{-1}\circ \boldsymbol x(p))}\,dx_1\dots dx_n\\ =\int_{\boldsymbol x^{-1}(R)}\sqrt{\det H(\boldsymbol y^{-1}\circ \boldsymbol x(p))}|J|\,dx_1\dots dx_n=\int_{\boldsymbol y^{-1}(R)}\sqrt{\det H(p)}\,dy_1\dots dy_n,$$ using only the change of variables theorem.

My questions are:

(1) How did the author get equation (1)?

(2) If my computation is right, why is the orientability required necessary?

(3) Furthermore: Why is $\det(g_{ij})\geq 0$?? We were writing $\sqrt{\det (g_{ij})}$ all the time...

  • $\begingroup$ @ChristianBlatter So this is indeed my doubt: why does the author requires orientability? It seems that without that the sign of the volume is not well-defined. Which is indeed somewhat frustrating... $\endgroup$ – Derso Mar 3 '18 at 12:47

You don't need orientability to define the volume of a parametrized region. Your value $vol(R)$ is completely well-defined in terms of a parametrization, and doesn't even depend on the parametrization if we take the modulus of its value, which is arguably the sensible thing to do if we are talking about "volumes". Therefore, I think that this discussion about orientability when you are restricting yourself to volumes of parametrizations can be a little uninteresting.

However, you need orientability to define a volume form on the entire manifold, so that global things like $$\int_M \omega_{vol}$$ make sense, and even so that $\omega_{vol}$ be well-defined on $M$. This is the point where orientability is important: when we talk about global phenomena. Integration in a chart is only a matter of sign choice. However, if we try to clump up the work we do in charts in order to make something in the whole manifold, these sign choices need to be chosen in such a way that things will be well-defined when we mix them up. Orientability tells us that we can do that. It is not only a matter of sign choice of the end result anymore, it is a matter of possibility, due to sign choices of building blocks, to have an end result that makes sense. If it is possible (i.e., if the manifold is orientable), then it is only a matter of sign choice of the end result.

There is already another answer dealing with the computations directly, but it may be fruitful to know that the volume form is given at each point $p$ by taking an orthogonal basis $E_1,\cdots,E_n$ with respect to $g_p$ with the correct orientation given by the orientation of the manifold, and then considering the dual basis of $T_pM^*$: $\delta_1,\cdots,\delta_n$. Then, $$(\omega_{vol})_p:=\delta_1\wedge \cdots \wedge \delta_n.$$ The fact that this is well-defined is due to the fact that changing bases will change it by a factor of a determinant. Since we defined it in terms of orthogonal basis with the same orientation, this determinant will be $1$, and thus the form is well-defined.

A straight-forward calculation shows that $\omega_{vol}$ is given by the formula you have when expressed in terms of the dual forms of an arbitrary positively-oriented chart (i.e, the $dx_i$'s).

EDIT: We can, in the non-orientable case, define $$\mu:=| \delta_1\wedge \cdots \wedge \delta_n| ,$$ with the $\delta_i$'s as before, but for an arbitrary orthogonal base $E_1,\cdots,E_n$ of $T_pM$. Since an orthogonal transformation will make a factor of $+1/-1$ appear, the modulus will kill it and it is thus well-defined as a density. Note however that $\mu$ is not a form, and by the observation of Manfredo, he has integrations of forms in mind.

  • $\begingroup$ Aloizio Macedo: I am still confused with a detail: Since $\det (g_{ij})>0$, as observed @user43687, wouldn't any integral $\displaystyle \int\limits_{\boldsymbol x^{-1}(R)}\sqrt{\det(g_{ij})}\,dx_1\dots dx_n\geq 0$, no matter which is the parametrization $\boldsymbol x:U\to M$? How is the change of sign possible then? $\endgroup$ – Derso Mar 4 '18 at 17:33
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    $\begingroup$ @AndersonFelipeViveiros The change of sign happens if you look at the form $\sqrt{det(g_{ij}} dx_1 \wedge \cdots \wedge dx_n$, which is what the other answer is doing. If you are looking at the integral in $\mathbb{R}^n$ only and the $dx_i$ are mute integration symbols, then you are correct: there is no change of sign. The point here is essentially that $\int dx_1 \wedge dx_2 \wedge dx_3 \wedge \cdots \wedge dx_n=- \int dx_2 \wedge dx_1\wedge dx_3 \wedge \cdots dx_n$. $\endgroup$ – Aloizio Macedo Mar 4 '18 at 18:47
  • $\begingroup$ So, considering the $dx_i$'s as "mute" integration symbols, do I not need to worry about orientability, even in the global case? $\endgroup$ – Derso Mar 4 '18 at 19:22
  • $\begingroup$ No, in the global case you need to worry. Both because such a volume form is not well-defined in the non-orientable case, and because you need orientability to even define the integral of a form globally (check out the definition of integrals of forms in manifolds). You should check out other things we can integrate in manifolds which do not depend on orientability, if you are interested (but the integration of a form depends). $\endgroup$ – Aloizio Macedo Mar 4 '18 at 19:48
  • $\begingroup$ Aloizio: So, I'm confused again: even considering the $dx_i$'s as mute symbols of integration, aren't the volumes well defined globally? You've said there wasn't changes of sign in this case... $\endgroup$ – Derso Mar 4 '18 at 19:54

First off, one of the axioms for a Riemannian metric is that the bilinear form $g_{ij}$ is positive definite and hence $\det(g_{ij})\geq 0$.

your calculation is correct -- we indeed have $$ \sqrt{\det(H(y^{-1}\circ x))}\vert J \vert =\sqrt{\det(G)}, $$ but also notice that by direct calculation, we have \begin{eqnarray*} \sqrt{\det(H(y^{-1}\circ x))} d (y^{-1}\circ x)_1 ... d (y^{-1}\circ x)_n &=& \int \sqrt{H(y^{-1}\circ x)} J dx_1 ... dx_n \\ &=& \pm \int \sqrt{H(y^{-1}\circ x)} \vert J\vert dx_1 ... dx_n \\ &=& \pm \int \sqrt{\det(G)}dx_1...dx_n\;. \end{eqnarray*} So that means that if the transformation $y^{-1}\circ x$ does not preserve orientation, we have some ambiguity in the way we define volume. On the one hand, change of variables tells us that our volumes ought to agree after transforming the metric via the transition maps. On the other hand, if we simply compute the volume after transforming the coordinates, our result differs by a sign.

  • $\begingroup$ You mean, 'our result may differ by a sign' ,right? $\endgroup$ – Mojojojo Feb 13 at 14:29

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