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The volume form is known to be invariant under a change of coordinates $T$ with $\det(T)>0$, so consequently integral of forms are also invariant. But what happens when the change of coordinates has $\det(T)<0$? Is the integral invariant under transformations with negative determinant? In Calculus of one variable, when the determinant is negative one would just reverse the limit of integration, why we don't do the same when dealing with forms? In Calculus of multivariable, the absolute value of the determinant is taken, so I guess my question is related to why there's an absolute value there (is it put by hand or does it show up naturally?).

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  • $\begingroup$ Whether or not you use the absolute value of the Jacobian depends on what you’re computing. For some integrals, orientation matters, for others it doesn’t. In the language of differential forms, you need to know whether you’re integrating a form (such as $f\,dx\wedge dy$) or a density ($f\,dx\,dy$), which transform differently. $\endgroup$
    – amd
    Sep 1, 2017 at 19:21
  • $\begingroup$ As I understand, the definition of an integral of a form as an integral of a density is only meaningful when det(T)>0, since that's when the volume form is invariant. But it seems to me that the outcome of the integral is always the same either way (for det(T)>0 or det(T)<0). So why not take this fact into account? I mean, why not say that det(T)<0 transformations are fine as long as one also reverse the limit of integration? In symbols this would mean $ \int_Df\omega = \int_{T(D)} T^*f\omega $ for any T. $\endgroup$
    – user219526
    Sep 2, 2017 at 14:54
  • $\begingroup$ $|J(T)|$ is in general not constant throughout the region. Showing that the Riemann sum converges when $|J(T)|\lt0$ in some subregions and $\gt0$ in others becomes difficult, if not impossible. If $T$ is orientation reversing throughout, then one could certainly do what you propose. BTW, the basic pullback identity is $\int_DT^*\omega=\int_{T(D)}\omega$, which is a bit different from what you’ve got. $\endgroup$
    – amd
    Sep 2, 2017 at 18:28
  • $\begingroup$ I see, this answers my question, thanks. Just out of curiosity, is it possible to prove that $|J(T)|$ is constant in the whole domain in the Euclidean space, for any $T$? $\endgroup$
    – user219526
    Sep 3, 2017 at 12:26
  • $\begingroup$ Well, $|J(T)|=\text{const}$ means that $T$ multiplies volumes by the constant factor $|J(T)|$. That’s certainly true of affine transformations, and I’m sure there are others for which $J(T)$ itself isn’t constant, but it’s not true for even the common polar-to-Cartesian transformation, which has $|J(T)|=\rho$. Another useful transformation is the parabolic coordinates $x=uv$, $y=\frac12(u^2-v^2)$, for which $|J(T)|=u^2+v^2$. The transformation is certainly orientation-preserving, but the Jacobian determinant isn’t constant for this one, either. $\endgroup$
    – amd
    Sep 3, 2017 at 19:43

1 Answer 1

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$\newcommand{\di}{\mathrm{d}} \newcommand{\de}{\partial}$ All kinds of coordinate transformations can be considered when integrating forms. The integral of an $n$-form over an $n$-dimensional submanifold is independent of any coordinates, so it's "invariant" no matter which coordinates we use and no matter their orientation. See below.

Sometimes people prefer to use only coordinates with the same orientation in order not to think about absolute values of determinants. Such absolute values appear in the coordinate transformation of a particular kind of forms. Typically, "volume form" are of that kind.

To understand this it's important to distinguish between integration over inner-oriented submanifolds and outer-oriented submanifolds.

Here's an example of inner-oriented 2D submanifold: inner-oriented 2D submanifold

and of outer-oriented 2D submanifold in 3D space: outer-oriented 2D submanifold

The first has a circulation sense on it; the second a flux direction through it.

Here's an example of inner-oriented 3D submanifold: inner-oriented 3D submanifold

and of outer-oriented 3D submanifold in 3D space: enter image description here

The first has a screw-sense; the second just a global "$+$" or "$-$" sign.

(Pictures adapted from Schouten [1989], see refs below.)

Non-orientable manifolds such as the Möbius strip are actually non-inner-orientable, but they are outer-orientable in 2D: we can just give them the sign "$+$" everywhere. Any manifold is outer-orientable when considered as a submanifold of itself.

Many integrals typically considered in the sciences do not require an inner orientation (a screw-sense in the example above), but an outer orientation. For example, if we calculate the total mass in a region of space from the mass density given there, we don't care about any screw-senses in that region.

In the case of the Möbius strip, for example, imagine that a 2D mass density is defined on it – some parts have more mass, some have less mass. It's clear that we can ask what the total mass of the strip is, and we must be able to calculate it by integrating the mass density all over the strip. But wasn't the strip non-orientable? how is such an integral possible, then? It's possible because we're considering it as an outer-oriented manifold, and as such it has a well-defined "$+$" orientation. And we integrate an inner-oriented 2-form over it – the mass density – to get its total mass.


Before saying something more on the two kinds of integration, it's useful to remember: outer-oriented forms are integrated over inner-oriented submanifolds; inner-oriented forms are integrated over outer-oriented submanifolds. There's a sort of complementarity between the two. (I'm sure I'm going to mess up the two terms below, sorry in that case.)

Integration of outer-oriented forms over inner-oriented submanifolds

An outer-oriented $n$-form $\omega$ on an $n$-dimensional manifold in a coordinate system $(x_i)$ has just one component $\omega_{1\dotso n}$, and it's represented in that coordinate system by $$\omega(x_i) = \omega_{1\dotso n}(x_i) \;\di x_1 \land \dotsb \land \di x_n \;.$$

This is the standard kind of form discussed in most differential-geometry textbooks. It's also called an "even" form.

Its outer orientation at a point is the same as that determined by the coordinates if $\omega_{1\dotso n}>0$, and opposite otherwise.

Under a change of coordinates to $(x'_i)$ its new component is given by $$\omega'_{1\dotso n} = \omega_{1\dotso n}\; \det\tfrac{\de x_i}{\de x'_j}$$ without any absolute values for the determinant. From this formula it's clear that the outer orientation of $\omega$ at a point doesn't depend on the coordinate system. If we change to coordinates $(x'_{i})$ with orientation opposite to $(x_{i})$, we'll have $\det\tfrac{\de x_i}{\de x'_j}<0$, and this determinant in the above formula will automatically flip the sign of $\omega'_{1\dotso n}$, so that the outer orientation of $\omega$ is the same.  

An inner-oriented $n$D submanifold $C$ is represented as a map $F$ from a subset $U \subset \mathbf{R}^n$ to our manifold, such that $F(U)=C$. This map also defines the inner orientation of $C$: it's the orientation determined by the coordinates $(u_i)$ as mapped onto $C$. In other words, it's the inner orientation of $U$ mapped onto $C$. In the coordinate system $(x_{i})$ the map $F$ is represented by the functions $$ x_{i} = F_{i}(u_{1}, \dotsc, u_{n})\;.$$

In these coordinates, where $\omega$ has component $\omega_{1\dotso n}$, the integral of $\omega$ over $C$ is given by $$ \int_C\omega \equiv \int_U \omega_{1\dotso n}[F_i(u_j)]\; \det\tfrac{\de F_i}{\de u_j} \; \di u_1 \dotsm \di u_n \;. $$ Note that no absolute value is taken for the determinant.

In fact the sign of the determinant in the expression above automatically takes care of the orientation of $C$. If in the integral above we had taken the absolute value, the integral would have been for the domain with orientation induced not by the map $F\colon U \to C$, but by the coordinates $(x_{i})$ -- an orientation we don't care about. If the two orientations coincided, that is if $\det\tfrac{\de F_i}{\de u_j}>0$, there'd be no problem. But if the two were opposite, $\det\tfrac{\de F_i}{\de u_j}<0$, we would need to put a minus sign in front of the integral if the determinant were taken in absolute value.

Integration of inner-oriented forms over outer-oriented submanifolds

Inner-oriented forms are also called "odd" or "twisted" (see refs below), and they are related to tensor densities. Note that "volume elements" are typically inner-oriented forms, not standard, outer-oriented ones as those described above; see Marsden & Ratiu [2007] and de Rham [1984].

An inner-oriented $n$-form $\omega$ on an $n$-dimensional manifold in a coordinate system $(x_i)$ has just one component $\omega_{1\dotso n}$, and it's represented in that coordinate system by $$\omega(x_i) = \omega_{1\dotso n}(x_i) \;\lvert\di x_1 \land \dotsb \land \di x_n\rvert \;.$$ The "absolute-value" expression indicates that the coordinate $n$-form must be considered as inner-oriented with sign "$+$".

The inner orientation of $\omega$ at a point is given by the sign of $\omega_{1\dotso n}$.

Under a change of coordinates to $(x'_i)$ its new component is given by $$\omega'_{1\dotso n} = \omega_{1\dotso n}\; \biggl\lvert\det\tfrac{\de x_i}{\de x'_j}\biggr\rvert$$ taking the absolute value of the determinant.

From this formula it's clear that the inner orientation of $\omega$ at a point doesn't depend on the coordinate system, because the original sign of $\omega_{1\dotso n}$ is preserved.

An outer-oriented $n$D submanifold $C$ is again represented as a map $F$ from a subset $U \subset \mathbf{R}^n$ to our manifold, such that $F(U)=C$. Its outer orientation has sign $\epsilon$, taken as "$+$" or "$-$".

In the coordinate system $(x_{i})$ the map $F$ is again represented by the functions $$ x_{i} = F_{i}(u_{1}, \dotsc, u_{n})\;.$$

In these coordinates, where the inner-oriented $\omega$ has component $\omega_{1\dotso n}$, the integral of $\omega$ over $C$ is given by $$ \int_C\omega \equiv \epsilon\int_U \omega_{1\dotso n}[F_i(u_j)]\; \biggl\lvert\det\tfrac{\de F_i}{\de u_j}\biggr\rvert \; \di u_1 \dotsm \di u_n \;. $$ There is the absolute value of the determinant in this case, and note the $\epsilon$ in front of the integral. (Typically in such region integrals the sign of the submanifold is taken as "$+$".)

The absolute value in the expression above takes care of the fact that the component of $\omega$ has a different transformation, and that the outer orientation of $C$ has nothing to do with the coordinates $(u_i)$ (which define an inner orientation instead). You can easily verify that changing coordinates gives the same result.


de Rham's book [1984] discusses these matters very simply in Chapter II.

References

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