Integration on Manifolds

Integration on manifolds is the coordinate-independent integration of differential forms over oriented domains. A $k$-form is integrated over a $k$-dimensional oriented curve, surface, or manifold:

$$\int_\Sigma \omega, \qquad \deg\omega=\dim\Sigma.$$

The key mechanism is pullback. A parametrization of the domain pulls the form back to an ordinary form on a region of Euclidean space, where the integral becomes a familiar multiple integral with the correct Jacobian and orientation signs already built in.

Helpful background. Review Differential Forms for wedge products and Exterior Derivative for the derivative in Stokes’ theorem.

Forms as geometric integrands

A Berry connection is integrated along a path, while its curvature is integrated over a surface. To calculate either integral, substitute a parametrization into the form. This operation, called pullback, supplies the familiar Jacobian factors and the orientation signs.

Pullback Is the Basic Operation

Let $\omega$ be a $k$-form on a manifold $M$, and let

$$\Phi:D\to M$$

parametrize a $k$-dimensional domain $\Sigma\subset M$, where $D\subset\mathbb R^k$. The integral is defined by

$$\int_\Sigma \omega = \int_D \Phi^*\omega.$$

Here $\Phi^*\omega$ is the pullback of the form to the parameter domain. It inserts the tangent vectors produced by the parametrization and automatically supplies the Jacobian factors.

This is why the coordinate formulas in Differential Forms look determinant-like: the determinant is the form’s response to oriented tangent directions.

One-Forms on Curves

Let

$$\alpha=\alpha_i(x)\,dx^i$$

be a 1-form, and let

$$\gamma:[a,b]\to M, \qquad t\mapsto x(t),$$

be an oriented curve. Pulling back $\alpha$ gives

$$\gamma^*\alpha = \alpha_i(x(t))\dot x^i(t)\,dt.$$

Therefore

$$\int_\gamma \alpha = \int_a^b \alpha_i(x(t))\dot x^i(t)\,dt.$$

Reversing the curve orientation changes the sign. If $\bar\gamma(t)=\gamma(a+b-t)$ is the reversed path, then

$$\int_{\bar\gamma}\alpha = - \int_\gamma\alpha.$$

This is the line-integral structure behind Berry phases and circulation integrals.

Two-Forms on Surfaces

Let

$$\omega = \sum_{i<j} \omega_{ij}(x)\,dx^i\wedge dx^j$$

be a 2-form, and let

$$\Phi:D\to M, \qquad (u,v)\mapsto x(u,v),$$

parametrize an oriented surface. Then

$$\Phi^*\omega = \sum_{i<j} \omega_{ij}(x(u,v)) \left( \frac{\partial x^i}{\partial u} \frac{\partial x^j}{\partial v} - \frac{\partial x^i}{\partial v} \frac{\partial x^j}{\partial u} \right) du\wedge dv.$$

Thus

$$\int_\Sigma\omega = \int_D \Phi^*\omega.$$

Changing the order of the parameters from $(u,v)$ to $(v,u)$ reverses the orientation and changes the sign.

Orientation

An orientation is a consistent choice of positive ordered tangent bases. In two dimensions, saying that $(\partial_u,\partial_v)$ is positive is the same kind of choice as saying that $du\wedge dv$ is the positive area form on the parameter domain.

The sign of a form integral depends on this choice:

$$\int_{-\Sigma}\omega = - \int_\Sigma\omega,$$

where $-\Sigma$ denotes the same domain with the opposite orientation.

For curves, orientation is the direction of traversal. For surfaces, orientation is a choice of which ordered tangent-pair direction is positive. In embedded Euclidean three-space, this is often represented by a normal vector using the right-hand rule, but the form definition does not require an ambient normal.

Boundary Orientation

Stokes theorem requires a convention for orienting the boundary $\partial\Sigma$. The standard convention is:

An ordered basis of $T(\partial\Sigma)$ is positive when the outward normal followed by that basis gives the orientation of $\Sigma$.

For an interval $[a,b]$ with its usual orientation, the boundary orientation is

$$\partial[a,b]=\{b\}-\{a\}.$$

For an oriented region in the plane with the usual $dx\wedge dy$ orientation, the outer boundary is counterclockwise and any boundary around a hole is clockwise. In either case, the region stays on the left as one traverses its boundary. The annulus example illustrates why both boundary components matter.

This sign convention is what makes the fundamental theorem of calculus and Green’s theorem appear with the usual signs.

Top-Degree Forms and Volume

On an $n$-dimensional oriented manifold, an $n$-form can be integrated over $n$-dimensional regions. In local coordinates,

$$\Omega = f(x)\,dx^1\wedge\cdots\wedge dx^n$$

integrates as

$$\int_U \Omega = \int_{\varphi(U)} f(x)\,dx^1\cdots dx^n$$

in an orientation-preserving chart.

A scalar function by itself is not automatically integrable over a manifold in a coordinate-independent way. To integrate a scalar $f$, one needs a chosen volume form or measure, such as the metric volume form on a Riemannian manifold:

$$\int_M f\,dV_g.$$

This distinction prevents a common confusion: forms are integrands; scalar functions become integrands only after choosing the measure or volume form.

Example: Polar Coordinates

The standard area 2-form on the plane is

$$dx\wedge dy.$$

Use polar coordinates

$$x=r\cos\theta, \qquad y=r\sin\theta.$$

Then

$$\begin{aligned} dx&=\cos\theta\,dr-r\sin\theta\,d\theta,\\ dy&=\sin\theta\,dr+r\cos\theta\,d\theta. \end{aligned}$$

Therefore

$$dx\wedge dy = r\,dr\wedge d\theta.$$

The familiar polar-coordinate Jacobian appears because the 2-form has been pulled back:

$$\int_{\text{disk}} dx\wedge dy = \int_0^{2\pi}\int_0^R r\,dr\,d\theta = \pi R^2.$$

No separate mnemonic for the Jacobian is needed; it is the coefficient of the pulled-back form.

Stokes Theorem

Let $\Sigma$ be a compact oriented $k$-dimensional manifold with boundary, and let $\omega$ be a smooth $(k-1)$-form on a neighborhood of $\Sigma$. Stokes theorem says

$$\int_\Sigma d\omega = \int_{\partial\Sigma}\omega.$$

The left side integrates a $k$-form over the $k$-dimensional region. The right side integrates a $(k-1)$-form over the oriented boundary.

This is the master theorem behind several familiar results:

• the fundamental theorem of calculus;
• Green’s theorem in the plane;
• the Kelvin-Stokes curl theorem;
• the divergence theorem.

The power of the form statement is that the same equation works in any dimension and any coordinate system where the hypotheses are satisfied.

Fundamental Theorem of Calculus

Let $f$ be a smooth function on an interval $[a,b]$. Since

$$df=f'(x)\,dx,$$

Stokes theorem gives

$$\int_{[a,b]}df = \int_{\partial[a,b]}f.$$

Using the boundary orientation $\partial[a,b]=\{b\}-\{a\}$,

$$\int_a^b f'(x)\,dx = f(b)-f(a).$$

Thus the fundamental theorem of calculus is the one-dimensional case of Stokes theorem.

Green and Curl Theorems

Let

$$\alpha=P(x,y)\,dx+Q(x,y)\,dy$$

on a region $D$ in the plane. Then

$$d\alpha = \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\wedge dy.$$

Stokes theorem gives

$$\int_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\wedge dy = \int_{\partial D} P\,dx+Q\,dy.$$

This is Green’s theorem. The three-dimensional curl theorem is the same statement with a 1-form integrated around the boundary of an oriented surface.

Divergence Theorem

In Euclidean three-space, represent a vector field $\mathbf B=(B_x,B_y,B_z)$ by the flux 2-form

$$\mathcal B = B_x\,dy\wedge dz + B_y\,dz\wedge dx + B_z\,dx\wedge dy.$$

Then

$$d\mathcal B = (\nabla\cdot\mathbf B)\,dx\wedge dy\wedge dz.$$

Stokes theorem gives

$$\int_V d\mathcal B = \int_{\partial V}\mathcal B,$$

which is the divergence theorem after translating the 2-form flux back into vector notation.

Berry Phase and Curvature

For a Berry connection one-form $A_n$ in a smooth eigenvector frame that is periodic around a closed loop $C$, the Berry phase is, modulo $2\pi$,

$$\gamma_n[C] = \oint_C A_n.$$

For a nonperiodic frame, include the endpoint overlap. If $C=\partial\Sigma$ and a smooth gauge is available on the surface $\Sigma$, Stokes theorem gives

$$\oint_C A_n = \int_\Sigma dA_n = \int_\Sigma F_n.$$

This is the geometric form of the statement that Berry phase can be computed as curvature flux through a spanning surface.

The hypotheses matter. If no single smooth gauge is available on $\Sigma$, or if the surface crosses a degeneracy where the eigenstate is not defined, one must use patches or remove singular points. The physical discussion is in Berry Phase, while the connection and curvature objects are defined in Connections and Curvature; this page explains the integration theorem.

Manifolds with Several Charts

If a domain is not covered by one coordinate chart, integration is defined by covering it with charts, using a partition of unity, integrating in each chart, and summing the results. The coordinate changes agree because differential forms transform with the correct Jacobian factors.

For most quantum-mechanics calculations, the practical lesson is simpler: when a single chart has a coordinate singularity, do not automatically conclude that the integral or form is singular. Check whether the form is globally defined, whether patches are needed, and whether the singular point is actually part of the manifold.

Common Mistakes

• Integrating a $k$-form over a domain of the wrong dimension.
• Ignoring orientation and losing a sign.
• Treating $du\,dv$ as coordinate-independent without pulling back a form.
• Forgetting that a scalar function needs a volume form or measure before it can be integrated invariantly.
• Applying Stokes theorem when the form is not smooth on the region.
• Using a single gauge or chart across a surface where it is not defined.
• Forgetting the induced boundary orientation.
• Confusing a coordinate singularity with a genuine singularity of the integrand.

Exercises

1. Use Stokes theorem to compute $\int_C x\,dy-y\,dx$ for the counterclockwise unit circle.

Solution

Let

$$\alpha=x\,dy-y\,dx.$$

Then

$$d\alpha = 2\,dx\wedge dy.$$

If $D$ is the unit disk and $C=\partial D$ with counterclockwise orientation, Stokes theorem gives

$$\int_C\alpha = \int_D d\alpha = \int_D 2\,dx\wedge dy = 2\pi.$$

2. Let $f(x)=x^3$ on $[0,2]$. Verify the one-dimensional Stokes theorem.

Solution

The exterior derivative is

$$df=3x^2\,dx.$$

Thus

$$\int_{[0,2]}df = \int_0^2 3x^2\,dx = 8.$$

The boundary orientation is $\partial[0,2]=\{2\}-\{0\}$, so

$$\int_{\partial[0,2]}f = f(2)-f(0) = 8.$$

3. Derive the polar-coordinate area element by pulling back $dx\wedge dy$.

Solution

For $x=r\cos\theta$ and $y=r\sin\theta$,

$$dx=\cos\theta\,dr-r\sin\theta\,d\theta, \qquad dy=\sin\theta\,dr+r\cos\theta\,d\theta.$$

Wedge them:

$$\begin{aligned} dx\wedge dy &= (\cos\theta\,dr-r\sin\theta\,d\theta) \wedge (\sin\theta\,dr+r\cos\theta\,d\theta)\\ &= r(\cos^2\theta+\sin^2\theta)\,dr\wedge d\theta\\ &= r\,dr\wedge d\theta. \end{aligned}$$

4. Let $R=[0,L_x]\times[0,L_y]$ with the standard orientation. Compute $\int_{\partial R}x\,dy$.

Solution

Let $\alpha=x\,dy$. Then

$$d\alpha=dx\wedge dy.$$

By Stokes theorem,

$$\int_{\partial R}x\,dy = \int_R dx\wedge dy = L_xL_y.$$

5. Suppose $F=F_0\,dR^1\wedge dR^2$ is a constant Berry-curvature 2-form on a rectangular patch $0\le R^1\le a$, $0\le R^2\le b$. If $A$ is a smooth connection on the patch with $dA=F$, what is $\oint_{\partial\Sigma}A$?

Solution

By Stokes theorem,

$$\oint_{\partial\Sigma}A = \int_\Sigma dA = \int_\Sigma F.$$

Since $F$ is constant,

$$\int_\Sigma F = F_0ab.$$

6. Why can Stokes theorem not be applied blindly across a singular gauge patch?

Solution

Stokes theorem assumes the form is smooth on the region being integrated. If a gauge potential is not defined or not smooth somewhere on the spanning surface, the hypothesis fails. One must use multiple patches, remove the singular point and account for the new boundary, or reformulate the calculation in terms of globally defined objects.

References

• T. Frankel, The Geometry of Physics, 3rd ed., Cambridge University Press, 2011.
• J. M. Lee, Introduction to Smooth Manifolds, 2nd ed., Springer, 2013.
• L. W. Tu, An Introduction to Manifolds, 2nd ed., Springer, 2011.