Confusion in the order of the differentials in integral using differential forms

Question

My doubt is very general but I'm going to give an example so I can explain why I'm struggling with this.

I have a 2-form $\alpha$ written in Cartesian coordinates

$$ \alpha=\alpha_{ij}\,dx^i\wedge dx^j $$

If I want to integrate this on a 2-sphere $S^2\in\mathbb R^3$, I'm going to change $\alpha$ to spherical coordinates $(x,y,z)\to(r,\theta,\phi)$ and have something like this

$$ \int_{S^2}\alpha= \int \alpha'_{\theta\phi} \,d\theta \wedge d\phi $$

But I know that $d\theta \wedge d\phi=-d\phi \wedge d\theta $, so I also have

$$ \int_{S^2}\alpha= -\int \alpha'_{\theta\phi} \,d\phi \wedge d\theta $$

Which is the correct way to integrate $\alpha$ over $S^2$?

$$ \int_{S^2}\alpha= \int \alpha'_{\theta\phi} \,d\theta \,d\phi \quad\text{or}\quad \int_{S^2}\alpha= -\int \alpha'_{\theta\phi} \,d\theta \,d\phi $$

I can't discern difference between the two integrals, I need a procedure to choose the correct way to integrate. And the thing is worse when I'm trying to integrate things less intuitive and with other manifolds.

I don't find any source either with clear examples and where I can understand

Answer 1

Suppose $M$ is an oriented $m$-dimensional manifold and $\alpha$ is a smooth compactly-supported $m$-form on $M$. Then, we can consider its integral $\int_M\alpha$. Note that this integral depends on the choice of orientation for $M$. For simple examples like balls, ellipsoids, (one half of) hyperboloids, paraboloids or their boundary surfaces, we can usually use a single chart $(U,\psi)$ to cover the entire manifold, up to a "negligible set". If the chart is orientation-preserving, then we can write \begin{align} \int_M\alpha&=\int_U\alpha=\int_{\psi^{-1}[\psi(U)]}\alpha=\int_{\psi(U)}(\psi^{-1})^*\alpha\tag{$*$} \end{align} where on the RHS, $(\psi^{-1})^*\alpha$ is an $m$-form on an open subset $\psi(U)$ of $\Bbb{R}^m$, and thus can be written as $(\psi^{-1})^*\alpha=f\,dx^1\wedge \cdots \wedge dx^m$, where $(x^1,\dots, x^m)$ is the usual cartesian coordinate chart on $\Bbb{R}^m$ (and thus $dx^1\wedge \cdots\wedge dx^m$ is indeed the volume form), and for a unique function $f:\psi(U)\to\Bbb{R}$. So, we have \begin{align} \int_M\alpha&=\int_{\psi(U)}f\,dx^1\wedge \cdots \wedge dx^k:=\int_{\psi(U)}f, \end{align} where the last symbol is a Riemann/Lebesgue integral of a function $f:\psi(U)\to\Bbb{R}$, i.e a standard integral. So far, all I've done is chase definitions.

As a remark, if $\psi$ is an orientation-reversing chart, then we must modify $(*)$ to \begin{align} \int_M\alpha&=\int_U\alpha=\int_{\psi^{-1}[\psi(U)]}\alpha=-\int_{\psi(U)}(\psi^{-1})^*\alpha. \end{align}

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Now, let us recall how we can induce orientations. Suppose $(M,g)$ is an $m$-dimensional oriented Riemannian manifold with Riemannian volume form $\mu$. Let $S$ is an $(m-1)$ dimensional embedded submanifold (i.e a smooth hypersurface) which admits a smooth unit normal vector field $n$ (i.e $n:S\to TM$ is a smooth map such that for each $p\in S$, we have $g_p(n_p,n_p)=1$ and for each $\xi\in T_pS$, $g_p(n_p, \xi)=0$). Then, we have the induced volume form $\mu_S:= \iota_S^*(n \lrcorner \mu)$, i.e the pullback to $S$ of the interior product of $n$ with $\mu$. This means, for each $p\in S$ and $\xi_1,\dots, \xi_{m-1}\in T_pS$, we have \begin{align} (\mu_S)_p(\xi_1,\dots, \xi_{m-1})&:=\mu_p(n_p,\xi_1,\dots, \xi_{m-1}). \end{align} In particular, what this means is $\{\xi_1,\dots, \xi_{m-1}\}$ is a positively-oriented ordered basis of $T_pS$ (i.e the LHS above is $>0$) if and only if $\{n_p,\xi_1,\dots, \xi_{m-1}\}$ is a positively-oriented ordered basis of $T_pM$ (i.e the RHS above is $>0$). (Also, $\mu_S$ is indeed the Riemannian volume form of $S$, relative to the pullback metric $\iota_S^*g$, because if we chose the $\xi$'s to be positively oriented and orthonormal, then the RHS is equal to $1$ by definition of $\mu$ being the Riemannian volume form of $M$).

For more details about orientation, you should consult any differential geometry book, for example Spivak or Lee (Chapter 15 of Intro to Smooth Manifolds)

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Now, we come to the case of the sphere $S^2$ in $\Bbb{R}^3$. Note that $\Bbb{R}^3$ has the standard Riemannian metric \begin{align} g=dx\otimes dx+dy\otimes dy+dz\otimes dz \end{align} and volume form \begin{align} \mu&=dx\wedge dy\wedge dz \end{align} Now, introduce spherical coordinates $(r,\theta,\phi)$ on a certain open subset of $\Bbb{R}^3$. Then, a standard calculation shows (equality obviously only on an appropriate open set of $\Bbb{R}^3$) \begin{align} \mu&=r^2\sin\theta\,dr\wedge d\theta\wedge d\phi. \end{align} The sphere obviously admits a unit normal vector field, and by convention, we choose $n=\frac{\partial}{\partial r}$ (restricted to $S^2$) to be the "outward pointing" unit normal. By my remarks above, this induces an orientation on $S^2$ with the volume form (I suppress the pullback to a certain open subset of $S^2$ in the notation) \begin{align} \mu_{S^2}&=\frac{\partial}{\partial r} \lrcorner \,r^2\sin\theta\,dr\wedge d\theta\wedge d\phi\\ &=\sin\theta\,d\theta\wedge d\phi \end{align} (the $r^2$ becomes $1$ since when we pullback to $S^2$, the radius is $1$). The range of $(\theta(\cdot),\phi(\cdot))$ is the open set $A:=(0,\pi)\times(0,2\pi)\subset\Bbb{R}^2$, which means $\sin\theta>0$, and thus the differential form $d\theta\wedge d\phi$ defines the same orientation as $\mu_{S^2}=\sin\theta\,d\theta\wedge d\phi$. Therefore, the spherical coordinate mapping which identifies an open subset of $S^2$ with $A$ preserves orientation.

In short, once you pull everything back to $A$, you deal with $d\theta\wedge d\phi$ (I'm abusing notation slightly here, since throughout I've used $\theta,\phi$ to mean functions on an open subset of $S^2$, so logically speaking I should refer to the corresponding coordinate functions on $A$ as $\tilde{\theta}$ and $\tilde{\phi}$), and as I mentioned in my formula $(*)$ in the beginning, due to the orientation-preserving nature, we don't introduce any rogue minus signs.

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Edit In Response to Comment.

I didn't want to have to do this initially because it would require us to introduce extra notation. One should always be careful of where each object lives. Recall that in general, when we say $(U,\psi)$ is a chart for an $m$-dimensional manifold $M$, it means $U\subset M$ is open and $\psi:U\to \psi[U]\subset \Bbb{R}^m$ is a mapping from the manifold to the cartesian space. Since $\psi$ maps into $\Bbb{R}^m$, we can consider its component functions $\psi(\cdot)=(x^1_{\psi}(\cdot),\dots, x^m_{\psi}(\cdot))$, so $x^i_{\psi}(\cdot)$ is the $i^{th}$ coordinate function of $\psi$ (it's what one might typically write as $\psi^i(\cdot)$, but let us stick to $x^i_{\psi}(\cdot)$).

Next, if $A\subset \Bbb{R}^m$ is an open set, when we speak of "cartesian coordinates on $A$" what we mean is the identity chart $\text{id}_A:A\to A\subset\Bbb{R}^m$. In keeping with the above notation, $\text{id}_A(\cdot)= (x^1_{\text{id}_A}(\cdot),\dots, x^m_{\text{id}_A}(\cdot))$.

Now, when we speak of spherical coordinates on $S^2$, we're talking about a specific chart $(U,\psi)$. It is rather cumbersome to give the definition of $\psi$ (because inverting the trig functions is a little annoying). So, it is convenient to first define $A=(0,\pi)\times (0,2\pi)$ and consider the mapping $A\subset\Bbb{R}^2\to \Bbb{R}^3$, \begin{align} (\theta,\phi)\mapsto (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \end{align} The image of this mapping is an open set $U\subset S^2$, and the inverse of this mapping is called $\psi$. This is the definition of the chart $(U,\psi)$. What I've claimed above is that \begin{align} \mu_{S^2}&=\sin(x_{\psi}^1(\cdot))\,dx_{\psi}^1\wedge dx_{\psi}^2 \end{align} Notice that $\sin(x_{\psi}^1(\cdot))$ is a function $U\to\Bbb{R}$, and $dx_{\psi}^1$ is the exterior derivative of a function on $U$, and thus is a $1$-form on $U$ and so on. Thus, the RHS is properly a $1$-form on $U$.

To check that $\psi:U\to A$ is indeed an orientation preserving chart it means I have to take the volume form on $\mu_A$ and then show that $\psi^*(\mu_A)=(\text{strictly positive function})\mu_{S^2}$.

Now, $\mu_A=dx_{\text{id}_A}^1\wedge dx_{\text{id}_A}^2$ by definition. Let us now calculate the pullback of $\mu_A$ under $\psi$. For this, we need to recall that pullback commutes with wedge products and exterior derivatives: \begin{align} \psi^*(\mu_A)&=\psi^*(dx_{\text{id}_A}^1\wedge dx_{\text{id}_A}^2)\\ &=d(\psi^*x^1_{\text{id}_A})\wedge d(\psi^*x^2_{\text{id}_A})\\ &=d(x^1_{\text{id}_A}\circ \psi)\wedge d(x^2_{\text{id}_A}\circ \psi)\\ &=d(x^1_{\psi})\wedge d(x^2_{\psi})\\ &=\frac{1}{\sin(x^1_{\psi}(\cdot))}\,\mu_{S^2} \end{align} Since the sin is strictly positive here, we have indeed shown that $\psi^*(\mu_A)$ is a positive multiple of $\mu_{S^2}$, and thus $(U,\psi)$ is an orientation-preserving chart.

Prior to this edit, I called $x^1_{\psi},x_{\psi}^2$ as $\theta,\phi$ respectively, and $x^1_{\text{id}_A},x^2_{\text{id}_A}$ as $\tilde{\theta},\tilde{\phi}$. So, $\theta=\tilde{\theta}\circ \psi$ and likewise for $\phi$ (this is what I mean by non-tilde are functions on $U$ and tilde are the associated functions on $A$). So, \begin{align} \psi^*(\mu_A)&=\psi^*(d\tilde{\theta}\wedge d\tilde{\phi})\\ &=d(\tilde{\theta}\circ \psi)\wedge d(\tilde{\phi}\circ\psi)\\ &=d\theta\wedge d\phi\\ &=\frac{1}{\sin\theta}\mu_{S^2} \end{align} and this differs by a positive multiple from $\mu_{S^2}$, hence $\psi$ is orientation preserving.

Note that because the pullback operation is so natural and plays nicely with wedge products and exterior derivatives, we usually don't like to introduce excessive notation to distinguish between functions on $U$ vs functions on $A$ obtained by composing with the chart map $\psi$.

Answer 2

I thought I'd expand on my comment about the difference between $d\phi\wedge d\theta$ and $d\phi d\theta$, since I consider this point a rather pervasive source of confusion. All of what I say is consistent with peek-a-boo's excellent answer, but I want to place the emphasis elsewhere: on the difference between integrating differential forms like $d\phi\wedge d\theta$ and integrating densities like $d\phi d\theta$ (the differential geometric analogues of measures).

The central difference between the two kinds of integrals is how they transform under a change of variables. Differential forms transform via a factor of the Jacobian, whereas densities (and measures) transform via a factor of the absolute value of the Jacobian. It's the fact that the differential forms version of the change of variables theorem lacks the absolute value sign that causes all the trouble with sign book-keeping and orientations.

Densities / measures are the objects we are more familiar with from ordinary integration theory, before we start doing vector calculus and Stokes's theorem. When we integrate a function over a set in the plane, say (whether Riemann or Lebesgue), we do not require the set to have an orientation. It turns out this notion extends naturally to the setting of manifolds with a Riemannian metric. That is, on any Riemannian manifold, whether orientable or not, it is possible to integrate scalar functions with respect to volume over measurable subsets of the manifold. The Möbius strip, for example, is a two-dimensional non-orientable manifold, but it still makes sense to speak of its area (the integral of the constant function $1$ with respect to the strip's area element). The reason is that we can use the Riemannian metric to define a Riemannian density in a canonical way: that is the "area" (or more generally "volume") "element" we integrate functions with respect to. Roughly speaking, the metric tells us how to compute the volume of infinitesimal parallelepipeds; more precisely, the density at a particular point $p$ of an $n$-manifold $M$ is a nonlinear function of $n$ tangent vectors in the tangent space $T_pM$ that behaves like the absolute value of the determinant on the tangent space. It's because we take the absolute value that the integral of a density remains intrinsically defined, even under an orientation-reversing change of local coordinates. (The word "density" is helpfully suggestive, by the way. Suppose you build a Möbius strip with non-uniform mass density per unit area and you wish to find the total mass of the whole strip. To do so, you just integrate the mass density over the whole strip.)

Now, what about integrating differential forms? The trouble all stems from losing the absolute value sign. Consider two charts, $(U,\phi_U)$ and $(V,\phi_V)$, around a point $p$ in a manifold $M$ (not necessarily oriented). Note that the transition map $\phi_V\circ\phi_U^{-1}$ is a diffeomorphism of $\phi_U(U\cap V)$ onto $\phi_V(U\cap V)$.

Say we want to integrate the form $\omega$ over $U\cap V$. The natural thing to try is to just pull everything back to Euclidean space:

$$\int_{U\cap V}\omega:=\int_{\phi_U(U\cap V)}(\phi_U^{-1})^*\omega$$

where the integral on the right is just ordinary integration of a differential form in $\mathbb{R}^n$. The first trouble is that this definition is explicitly coordinate dependent. So we try to check that it doesn't matter whether we use $\phi_V$ for the pullback on $U\cap V$ (or any other coordinate chart around $p$, for that matter). That is, we need to compare

$$\color{blue}{\int_{\phi_U(U\cap V)}(\phi_U^{-1})^*\omega}\,\,\,\,\,\text{and}\,\,\,\,\,\color{red}{\int_{\phi_V(U\cap V)}(\phi_V^{-1})^*\omega}$$

Proposition. If $U\cap V$ is connected, then the red and blue integrals agree up to sign.

Proof. Observe that

$$\color{blue}{\int_{\phi_U(U\cap V)}(\phi_U^{-1})^*\omega}=\int_{\phi_U(U\cap V)}(\phi_V^{-1}\circ\phi_V\circ\phi_U^{-1})^*\omega=\color{green}{\int_{\phi_U(U\cap V)}(\phi_V\circ\phi_U^{-1})^*(\phi_V^{-1})^*\omega}$$

where I've used $(f\circ g)^*\omega=g^*f^*\omega$.

Now, if $U\cap V$ is connected, so is $\phi_U(U\cap V)$, since $\phi_U$ is continuous. Thus $\det D(\phi_V\circ\phi_U^{-1})$ has constant sign on $\phi_U(U\cap V)$.

If the determinant is positive, the change of variables theorem for differential forms tells us that the green integral equals the red integral. On the other hand, if the determinant is negative, the change of variables theorem implies that the green integral is the opposite of the red integral. QED.

As a consequence of this argument, $\int_{U\cap V}\omega$ will be well-defined so long as we work only with charts whose transition maps have positive Jacobian. An oriented atlas on a manifold is precisely an atlas consisting of such charts.

I hope this concrete calculation clarifies why differential forms cannot be integrated on a manifold unless we pick an orientation (which is a maximal oriented atlas). Picking an orientation eliminates the obstruction to well-definedness under a change of coordinates.