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Section 3.3 Fields and Forms

This section introduces three kinds of objects that organize vector calculus: vector fields, differential forms, and chains. The goal is practical. We want enough language to describe local expansion, local rotation, circulation, and flux before turning to the integral theorems. For a more theoretical treatment, see [1].

Subsection 3.3.1 Vector Fields

Definition 3.3.1. Vector Field.

Let \(U \subseteq \R^n\) be open. A vector field on \(U\) is a function \(\mathbf{F} \colon U \to \R^n\text{.}\) Writing
\begin{equation*} \mathbf{F}(\vx) = \langle F_1(\vx),\ldots,F_n(\vx) \rangle, \end{equation*}
the functions \(F_1,\ldots,F_n\) are called the component functions of \(\mathbf{F}\text{.}\) A real-valued function \(f \colon U \to \R\) is called a scalar field. When \(n=1\text{,}\) a vector field on \(U \subseteq \R\) can be identified with a scalar field.
A vector field is continuous, differentiable, or \(C^k\) if each of its component functions has the same property.
Addition and scalar multiplication are defined pointwise:
\begin{gather*} (\mathbf{F}+\mathbf{G})(\vx) = \mathbf{F}(\vx)+\mathbf{G}(\vx),\\ (c\mathbf{F})(\vx) = c\,\mathbf{F}(\vx). \end{gather*}
So the familiar algebra of vectors can be carried out at each point of the domain.
The examples below will serve as our basic models in the plane:
\begin{gather*} \mathbf{F}_{\mathrm{src}}(x,y) = \langle x,y \rangle,\\ \mathbf{F}_{\mathrm{snk}}(x,y) = \langle -x,-y \rangle,\\ \mathbf{F}_{\mathrm{ccw}}(x,y) = \langle -y,x \rangle,\\ \mathbf{F}_{\mathrm{cw}}(x,y) = \langle y,-x \rangle. \end{gather*}
The first field points away from the origin, the second points toward the origin, and the last two rotate around the origin in opposite directions.
Four square vector-field plots arranged in a two-by-two grid..
Figure 3.3.2. Four basic planar vector fields: a source, a sink, a counterclockwise rotation, and a clockwise rotation.
To measure how a field behaves near a point, we package the first-order partial derivatives into the formal operator
\begin{equation*} \nabla = \langle D_1,\ldots,D_n \rangle. \end{equation*}
If \(f \colon U \to \R\) is a differentiable scalar field, then \(\nabla f\) is its gradient, as in Definition 2.4.5.

Definition 3.3.3. Divergence and Curl.

Let \(U \subseteq \R^n\) be open, and let \(\mathbf{F}=\langle F_1,\ldots,F_n \rangle \colon U \to \R^n\) be differentiable. The divergence of \(\mathbf{F}\) is the scalar field
\begin{equation*} \nabla \cdot \mathbf{F} = D_1F_1 + \cdots + D_nF_n. \end{equation*}
If \(n=3\) and \(\mathbf{F}=\langle F_1,F_2,F_3 \rangle\text{,}\) then the curl of \(\mathbf{F}\) is the vector field
\begin{equation*} \nabla \times \mathbf{F} = \langle D_2 F_3-D_3 F_2,\; D_3 F_1-D_1 F_3,\; D_1 F_2-D_2 F_1 \rangle. \end{equation*}
In classical notation, if \(\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z) \rangle\text{,}\) then one often writes \(\operatorname{div}\mathbf{F}\) and \(\operatorname{curl}\mathbf{F}\text{,}\) where
\begin{gather*} \operatorname{div}\mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z},\\ \operatorname{curl}\mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle. \end{gather*}
As a memory aid, \(\operatorname{curl}\mathbf{F}\) can be computed formally as the determinant
\begin{equation*} \operatorname{curl}\mathbf{F} = \det \begin{bmatrix} \ui \amp \uj \amp \uk \\ \frac{\partial}{\partial x} \amp \frac{\partial}{\partial y} \amp \frac{\partial}{\partial z} \\ P \amp Q \amp R \end{bmatrix}, \end{equation*}
expanded along the first row.
Informally, divergence measures local expansion or compression, while curl measures local rotation.
For a planar field \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y) \rangle\text{,}\) we often view it as the three-dimensional field \(\langle P(x,y),Q(x,y),0 \rangle\text{.}\) Then
\begin{equation*} \nabla \times \mathbf{F} = \langle 0,0,D_1Q-D_2P \rangle. \end{equation*}
The scalar quantity \(D_1Q-D_2P\) records the signed rotation in the plane: positive means counterclockwise rotation, and negative means clockwise rotation.

Example 3.3.4.

Compute the divergence and curl of the four fields in Figure 3.3.2.
Solution.
For the source field,
\begin{align*} \nabla \cdot \mathbf{F}_{\mathrm{src}} \amp = D_1(x)+D_2(y) = 1+1 = 2,\\ \nabla \times \mathbf{F}_{\mathrm{src}} \amp = \langle 0,0,D_1(y)-D_2(x) \rangle = \vz. \end{align*}
For the sink field,
\begin{align*} \nabla \cdot \mathbf{F}_{\mathrm{snk}} \amp = D_1(-x)+D_2(-y) = -2,\\ \nabla \times \mathbf{F}_{\mathrm{snk}} \amp = \vz. \end{align*}
For the counterclockwise rotation field,
\begin{align*} \nabla \cdot \mathbf{F}_{\mathrm{ccw}} \amp = D_1(-y)+D_2(x) = 0,\\ \nabla \times \mathbf{F}_{\mathrm{ccw}} \amp = \langle 0,0,D_1(x)-D_2(-y) \rangle = \langle 0,0,2 \rangle. \end{align*}
For the clockwise rotation field,
\begin{align*} \nabla \cdot \mathbf{F}_{\mathrm{cw}} \amp = D_1(y)+D_2(-x) = 0,\\ \nabla \times \mathbf{F}_{\mathrm{cw}} \amp = \langle 0,0,D_1(-x)-D_2(y) \rangle = \langle 0,0,-2 \rangle. \end{align*}
These four examples show positive, negative, and zero divergence, and also positive, negative, and zero curl.

Example 3.3.5.

Show that \(\mathbf{F}_{\mathrm{ccw}}(x,y)=\langle -y,x \rangle\) is not the gradient of any \(C^2\) scalar field on \(\R^2\text{.}\)
Solution.
If \(\mathbf{F}_{\mathrm{ccw}}=\nabla f\text{,}\) then \(f_x=-y\) and \(f_y=x\text{.}\) Differentiating gives
\begin{equation*} f_{xy}=-1, \qquad f_{yx}=1. \end{equation*}
But a \(C^2\) function satisfies \(f_{xy}=f_{yx}\text{,}\) which is impossible here. Therefore \(\mathbf{F}_{\mathrm{ccw}}\) is not a gradient field.

Subsection 3.3.2 Differential Forms

For our purposes, it is enough to define differential forms on open subsets of \(\R^3\text{.}\) We emphasize the computational point of view needed later for line, surface, and volume integrals. For the general theory, see again [1].

Definition 3.3.6. Differential Forms on \(\R^3\).

Let \(U \subseteq \R^3\) be open. For each \(0 \le p \le 3\text{,}\) the collection of smooth \(p\)-forms on \(U\) is denoted by \(\Omega^p(U)\text{.}\)
A \(0\)-form is just a smooth scalar field \(f \colon U \to \R\text{,}\) so \(\Omega^0(U)=C^{\infty}(U)\text{.}\)
A \(1\)-form is an expression of the form
\begin{equation*} \omega = P\,dx + Q\,dy + R\,dz, \end{equation*}
where \(P,Q,R \in \Omega^0(U)\text{.}\)
A \(2\)-form is an expression of the form
\begin{equation*} \eta = A\,dy \wedge dz + B\,dz \wedge dx + C\,dx \wedge dy, \end{equation*}
where \(A,B,C \in \Omega^0(U)\text{.}\)
A \(3\)-form is an expression of the form
\begin{equation*} \mu = F\,dx \wedge dy \wedge dz, \end{equation*}
where \(F \in \Omega^0(U)\text{.}\)
The symbols \(dx\text{,}\) \(dy\text{,}\) and \(dz\) behave like basis elements, while smooth functions supply the coefficients. We add forms of the same degree coefficientwise, multiply a form by a smooth function coefficientwise, and combine forms of different degrees using the wedge product from Definition 1.3.1.
Because the wedge product is alternating,
\begin{gather*} dx \wedge dx = dy \wedge dy = dz \wedge dz = 0,\\ dx \wedge dy = -dy \wedge dx, \qquad dy \wedge dz = -dz \wedge dy, \qquad dz \wedge dx = -dx \wedge dz. \end{gather*}
So the order of the factors matters. For example,
\begin{equation*} P\,dx \wedge dy \wedge dz = -P\,dx \wedge dz \wedge dy = P\,dz \wedge dx \wedge dy. \end{equation*}

Example 3.3.7.

Compute a sum, a scalar multiple, and a wedge product of differential forms.
Solution.
Let
\begin{gather*} \omega = x\,dx + y\,dy,\\ \eta = z\,dx + (x+y)\,dz,\\ f(x,y,z)=x^2. \end{gather*}
Then
\begin{gather*} \omega+\eta = (x+z)\,dx + y\,dy + (x+y)\,dz,\\ f\omega = x^3\,dx + x^2y\,dy. \end{gather*}
For the wedge product, expand and use the alternating property:
\begin{align*} \omega \wedge \eta \amp = (x\,dx+y\,dy)\wedge(z\,dx+(x+y)\,dz)\\ \amp = x(x+y)\,dx \wedge dz + yz\,dy \wedge dx + y(x+y)\,dy \wedge dz. \end{align*}
Since \(dy \wedge dx = -dx \wedge dy\text{,}\) this can also be written as
\begin{equation*} \omega \wedge \eta = -yz\,dx \wedge dy + y(x+y)\,dy \wedge dz + x(x+y)\,dx \wedge dz. \end{equation*}
The basic differential operator on forms is the exterior derivative. It raises degree by one.

Definition 3.3.8. Exterior Derivative.

Let \(\omega = \sum f_{i_1,\ldots,i_p}\,dx_{i_1}\wedge\cdots\wedge dx_{i_p}\) be a smooth \(p\)-form on an open set \(U \subseteq \R^3\text{,}\) written with increasing indices. The exterior derivative of \(\omega\) is the \((p+1)\)-form
\begin{equation*} d\omega = \sum df_{i_1,\ldots,i_p}\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_p}. \end{equation*}
In particular, if \(f \in \Omega^0(U)\text{,}\) then
\begin{equation*} df = f_x\,dx + f_y\,dy + f_z\,dz. \end{equation*}

Example 3.3.9.

Compute the exterior derivative of a general \(1\)-form.
Solution.
Let \(\omega = P\,dx + Q\,dy + R\,dz\text{.}\) Then
\begin{equation*} d\omega = dP \wedge dx + dQ \wedge dy + dR \wedge dz. \end{equation*}
Writing \(dP=P_x\,dx+P_y\,dy+P_z\,dz\text{,}\) and similarly for \(dQ\) and \(dR\text{,}\) we get
\begin{equation*} d\omega = (Q_x-P_y)\,dx \wedge dy + (R_y-Q_z)\,dy \wedge dz + (P_z-R_x)\,dz \wedge dx. \end{equation*}
The coefficients are exactly the components of \(\nabla \times \langle P,Q,R \rangle\text{.}\)

Example 3.3.10.

Compute the exterior derivative of a general \(2\)-form.
Solution.
Let \(\eta = A\,dy \wedge dz + B\,dz \wedge dx + C\,dx \wedge dy\text{.}\) Then
\begin{equation*} d\eta = dA \wedge dy \wedge dz + dB \wedge dz \wedge dx + dC \wedge dx \wedge dy. \end{equation*}
All terms containing repeated factors vanish, so only one term survives from each coefficient differential:
\begin{equation*} d\eta = (A_x+B_y+C_z)\,dx \wedge dy \wedge dz. \end{equation*}
Thus the coefficient of \(d\eta\) is the divergence of the vector field \(\langle A,B,C \rangle\text{.}\)

Example 3.3.11.

Show that the exterior derivative of every \(3\)-form on \(U \subseteq \R^3\) is zero.
Solution.
Let \(\mu = F\,dx \wedge dy \wedge dz\text{.}\) Then
\begin{equation*} d\mu = dF \wedge dx \wedge dy \wedge dz. \end{equation*}
But \(dF = F_x\,dx + F_y\,dy + F_z\,dz\text{,}\) so every term in this wedge product contains a repeated factor such as \(dx \wedge dx\text{.}\) Hence every term vanishes, and therefore
\begin{equation*} d\mu = 0. \end{equation*}

Subsection 3.3.3 Pullbacks

Pullbacks let us rewrite a differential form in new coordinates. This is the differential-form version of the Jacobian factor that appeared in Theorem 3.2.2.

Definition 3.3.12. Pullback of a Differential Form.

Let \(U \subseteq \R^n\) and \(V \subseteq \R^m\) be open, and let \(\Phi=\langle \Phi_1,\ldots,\Phi_m \rangle \colon U \to V\) be a smooth map. Using coordinates \(u_1,\ldots,u_n\) on \(U\) and \(x_1,\ldots,x_m\) on \(V\text{,}\) the pullback by \(\Phi\) is the map \(\Phi^* \colon \Omega^p(V) \to \Omega^p(U)\) defined by
\begin{gather*} \Phi^*f = f \circ \Phi \qquad \text{for } f \in \Omega^0(V),\\ \Phi^*(dx_i) = d(\Phi_i),\\ \Phi^*(\omega+\eta)=\Phi^*\omega+\Phi^*\eta,\\ \Phi^*(\omega \wedge \eta)=\Phi^*\omega \wedge \Phi^*\eta. \end{gather*}
Equivalently, if \(\omega = \sum f_{i_1,\ldots,i_p}\,dx_{i_1}\wedge\cdots\wedge dx_{i_p}\text{,}\) then
\begin{equation*} \Phi^*\omega = \sum (f_{i_1,\ldots,i_p}\circ\Phi)\, d\Phi_{i_1}\wedge\cdots\wedge d\Phi_{i_p}. \end{equation*}
In practice, computing \(\Phi^*\omega\) amounts to two substitutions. First replace each coefficient function by its composition with \(\Phi\text{.}\) Then replace each coordinate differential \(dx_i\) by \(d\Phi_i\text{,}\) expand, and simplify using bilinearity and the alternating property of the wedge product.

Example 3.3.13.

Compute the pullback of the \(1\)-form \(\omega = x\,dy-y\,dx\) under the polar-coordinate map
\begin{equation*} \Phi(r,\theta)=\langle r\cos\theta,\; r\sin\theta \rangle. \end{equation*}
Solution.
Since \(\Phi^*(x)=r\cos\theta\text{,}\) \(\Phi^*(y)=r\sin\theta\text{,}\) \(\Phi^*(dx)=\cos\theta\,dr-r\sin\theta\,d\theta\text{,}\) and \(\Phi^*(dy)=\sin\theta\,dr+r\cos\theta\,d\theta\text{,}\) we have
\begin{equation*} \Phi^*\omega = (r\cos\theta)\Phi^*(dy)-(r\sin\theta)\Phi^*(dx). \end{equation*}
Expanding gives
\begin{equation*} \Phi^*\omega = r\cos\theta(\sin\theta\,dr+r\cos\theta\,d\theta) - r\sin\theta(\cos\theta\,dr-r\sin\theta\,d\theta) = r^2\,d\theta. \end{equation*}
So the form \(x\,dy-y\,dx\) becomes \(r^2\,d\theta\) in polar coordinates.
In particular, the standard area form in the plane is \(dA = dx \wedge dy\text{,}\) and the standard volume form in space is \(dV = dx \wedge dy \wedge dz\text{.}\)

Example 3.3.14.

Compute the pullback of the area form under the polar-coordinate map
\begin{equation*} \Phi(r,\theta)=\langle r\cos\theta,\; r\sin\theta \rangle. \end{equation*}
Solution.
First compute the pullbacks of the coordinate \(1\)-forms:
\begin{gather*} \Phi^*(dx) = d(r\cos\theta) = \cos\theta\,dr-r\sin\theta\,d\theta,\\ \Phi^*(dy) = d(r\sin\theta) = \sin\theta\,dr+r\cos\theta\,d\theta. \end{gather*}
Therefore
\begin{equation*} \Phi^*(dA) = \Phi^*(dx \wedge dy) = \Phi^*(dx)\wedge\Phi^*(dy) = r\,dr \wedge d\theta. \end{equation*}
This is the form version of the familiar identity \(dA = r\,dr\,d\theta\text{.}\)

Example 3.3.15.

Compute the pullback of the volume form under the cylindrical-coordinate map
\begin{equation*} \Phi(r,\theta,z)=\langle r\cos\theta,\; r\sin\theta,\; z \rangle. \end{equation*}
Solution.
We combine the polar-coordinate computation with the identity \(\Phi^*(dz)=dz\text{.}\) This gives
\begin{equation*} \Phi^*(dV) = \Phi^*(dx \wedge dy \wedge dz) = \Phi^*(dx \wedge dy)\wedge\Phi^*(dz) = r\,dr \wedge d\theta \wedge dz. \end{equation*}
So in cylindrical coordinates the volume form becomes \(r\,dr \wedge d\theta \wedge dz\text{.}\)
This property is one reason pullbacks are so useful. It means we can either differentiate first and then change coordinates, or change coordinates first and then differentiate.

Example 3.3.17.

Verify Proposition 3.3.16 for \(\omega = x\,dy-y\,dx\) and the polar-coordinate map.
Solution.
From the previous example, \(\Phi^*\omega=r^2\,d\theta\text{,}\) so
\begin{equation*} d(\Phi^*\omega) = d(r^2)\wedge d\theta = 2r\,dr \wedge d\theta. \end{equation*}
On the other hand,
\begin{equation*} d\omega = d(x\,dy-y\,dx) = dx \wedge dy - dy \wedge dx = 2\,dx \wedge dy. \end{equation*}
Pulling this back gives
\begin{equation*} \Phi^*(d\omega) = 2\,\Phi^*(dx \wedge dy) = 2r\,dr \wedge d\theta. \end{equation*}
Therefore \(\Phi^*(d\omega)=d(\Phi^*\omega)\text{,}\) exactly as the proposition predicts.

Example 3.3.19.

Compute the pullback of the volume form under the spherical-coordinate map
\begin{equation*} \Phi(\rho,\theta,\phi) = \langle \rho\sin\phi\cos\theta,\; \rho\sin\phi\sin\theta,\; \rho\cos\phi \rangle. \end{equation*}
Solution.
A direct determinant computation gives
\begin{equation*} \det(D\Phi(\rho,\theta,\phi)) = -\rho^2\sin\phi. \end{equation*}
Therefore
\begin{equation*} \Phi^*(dV) = -\rho^2\sin\phi\,d\rho \wedge d\theta \wedge d\phi. \end{equation*}
The negative sign comes from the chosen order \((\rho,\theta,\phi)\text{.}\) For volume integrals we use the absolute value of the Jacobian determinant, so the scalar volume element is \(\rho^2\sin\phi\,d\rho\,d\theta\,d\phi\text{,}\) as in Subsection 3.2.3.

Subsection 3.3.4 Closed and Exact Forms

Two classes of differential forms appear constantly in vector calculus. They are defined using the exterior derivative.

Definition 3.3.20. Closed and Exact Forms.

Let \(U \subseteq \R^n\) be open, and let \(\omega \in \Omega^p(U)\text{.}\)
We say that \(\omega\) is exact if there exists \(\eta \in \Omega^{p-1}(U)\) such that \(\omega=d\eta\text{.}\)
We say that \(\omega\) is closed if \(d\omega=0\text{.}\)
Every exact form is closed. Indeed, if \(\omega=d\eta\text{,}\) then \(d\omega=d(d\eta)=0\text{.}\)
For \(1\)-forms in the plane, these conditions are especially concrete. If \(\omega=P\,dx+Q\,dy\text{,}\) then
\begin{equation*} d\omega = (Q_x-P_y)\,dx \wedge dy. \end{equation*}
So \(\omega\) is closed exactly when \(Q_x=P_y\text{.}\) It is exact exactly when there exists a scalar field \(f\) such that \(\omega=df\text{,}\) or equivalently,
\begin{equation*} P=f_x, \qquad Q=f_y. \end{equation*}
Thus exact \(1\)-forms correspond to gradient fields, while closed \(1\)-forms correspond to vector fields with zero planar curl.

Example 3.3.21.

Give a closed \(1\)-form on the punctured plane that is not exact.
Solution.
On \(U=\R^2 \setminus \{(0,0)\}\text{,}\) consider
\begin{equation*} \omega = \frac{-y}{x^2+y^2}\,dx + \frac{x}{x^2+y^2}\,dy. \end{equation*}
This form is often written \(d\theta\text{,}\) since in polar coordinates it is the differential of the angle function. More precisely, on any region where a continuous choice of angle \(\theta\) exists, we have \(\omega=d\theta\text{.}\)
The form is closed because
\begin{equation*} Q_x = \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right) = \frac{y^2-x^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right) = P_y. \end{equation*}
So \(d\omega=0\text{.}\)
However, \(\omega\) is not exact on all of \(U\text{.}\) Let \(\gamma(t)=\langle \cos t,\sin t \rangle\) for \(0 \le t \le 2\pi\text{.}\) Then
\begin{equation*} \gamma^*\omega = (-\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt) = dt. \end{equation*}
If \(\omega=df\) on \(U\text{,}\) then by Proposition 3.3.16,
\begin{equation*} \gamma^*\omega = \gamma^*(df) = d(f\circ\gamma). \end{equation*}
But \(d(f\circ\gamma)=dt\) implies \(f(\gamma(t))=t+C\text{.}\) This is impossible because \(\gamma(0)=\gamma(2\pi)\) but \(0+C \ne 2\pi+C\text{.}\) Therefore \(\omega\) is closed but not exact.

Proof.
Since \(U\) is star-shaped with respect to the origin, the whole line segment from \((0,0)\) to \((x,y)\) lies in \(U\) for every \((x,y) \in U\text{.}\) Define
\begin{equation*} f(x,y) := \int_0^1 \left(P(tx,ty)x+Q(tx,ty)y\right)\,dt. \end{equation*}
We claim that \(df=\omega\text{.}\) Because \(P\) and \(Q\) are smooth, we may differentiate under the integral sign. First,
\begin{equation*} f_x(x,y) = \int_0^1 \left(P(tx,ty)+txP_x(tx,ty)+tyQ_x(tx,ty)\right)\,dt. \end{equation*}
Since \(\omega\) is closed, \(Q_x=P_y\text{.}\) Therefore
\begin{equation*} f_x(x,y) = \int_0^1 \left(P(tx,ty)+txP_x(tx,ty)+tyP_y(tx,ty)\right)\,dt. \end{equation*}
The integrand is exactly the derivative of \(tP(tx,ty)\) with respect to \(t\text{,}\) so
\begin{equation*} f_x(x,y) = \int_0^1 \frac{d}{dt}\left(tP(tx,ty)\right)\,dt = P(x,y). \end{equation*}
Similarly,
\begin{equation*} f_y(x,y) = \int_0^1 \left(txP_y(tx,ty)+Q(tx,ty)+tyQ_y(tx,ty)\right)\,dt. \end{equation*}
Again using \(P_y=Q_x\text{,}\) we obtain
\begin{align*} f_y(x,y) \amp = \int_0^1 \left(txQ_x(tx,ty)+Q(tx,ty)+tyQ_y(tx,ty)\right)\,dt\\ \amp = \int_0^1 \frac{d}{dt}\left(tQ(tx,ty)\right)\,dt\\ \amp = Q(x,y). \end{align*}
Thus \(df=f_x\,dx+f_y\,dy=P\,dx+Q\,dy=\omega\text{,}\) so \(\omega\) is exact.
By translating coordinates, the same argument works for any open set that is star-shaped with respect to some point, as in Definition A.0.7.
The punctured plane \(\R^2 \setminus \{(0,0)\}\) is not star-shaped, so the example above does not contradict Theorem 3.3.22.
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