Verify that if \([\vec{\xi}',\vec{\eta}']=[\vec{\xi},\vec{\eta}]A\) for some \(2\times 2\) matrix \(A\text{,}\) then \(\xi_j'\eta_i' - \xi_i'\eta_j'=(\det A)(\xi_j\eta_i - \xi_i\eta_j) \text{,}\) and that if \(\{ \vec{\xi}',\vec{\eta}'\}\) and \(\{ \vec{\xi},\vec{\eta}\}\) are orthonormal, then \(A\) is an orthogonal matrix.
Section 7.1 A Brief Review of the Notion of Curl and Divergence of a Vector Field
In multi-variable calculus the line integral of a vector field \(\vec{X}=(X_{1}(\bx),\cdots, X_{n}(\bx)), n=2 \text{ or } 3,\) along a path (or loop) \(\Gamma: t\in [a, b]\mapsto \vec{r}(t)\text{,}\) is defined as
\begin{equation*}
\int_{\Gamma} \vec{X}\cdot d\vec{r} = \int_{a}^{b}
\left( X_1(\vec{r}(t))r_1'(t)+ \cdots + X_n(\vec{r}(t))r_n'(t)\right)dt,
\end{equation*}
and the flux of \(\vec{X}\) across a surface \(S\text{,}\) is defined as
\begin{equation*}
\int_S \vec{X} \cdot \vec{n}\, dA,
\end{equation*}
where \(\vec{n}(\bx)\) is a (continuous) choice of unit normal vector to \(S\) at \(\bx\text{.}\) In both cases, the integrand is not a given scalar function but a bilinear function of a given vector field and of a tangent vector of the curve or a normal vector of the surface.
One natural question is what infinitesimal quantities measure the "strength" of the circulation of the vector field along a closed loop or flux of the vector field across a closed surface near a point? The answers turn out to be the curl and, respectively, divergence of the vector field \(\vec{X}\) at the point.
For \(n=3\text{,}\) the curl of \(\vec{X}=(X_{1}(\bx),X_{2}(\bx), X_{3}(\bx))\) at \(\bx\) is the vector
\begin{equation*}
\text{curl} \vec{X}(\bx)=
(\frac{\partial X_3}{\partial x_2}-\frac{\partial X_2}{\partial x_3},
\frac{\partial X_1}{\partial x_3}-\frac{\partial X_3}{\partial x_1},
\frac{\partial X_2}{\partial x_1}-\frac{\partial X_1}{\partial x_2}).
\end{equation*}
For \(n=2\text{,}\) we can treat \(\vec{X}\) as a special case of \(n=3\) with \(X_{3}(\bx)=0\) and \(X_{i}(\bx)\) for \(i=1, 2\) depend only on \((x_{1}, x_{2})\text{,}\) so the curl of such a vector field takes the form of
\begin{equation*}
(0, 0, \frac{\partial X_2}{\partial x_1}-\frac{\partial X_1}{\partial x_2}).
\end{equation*}
The divergence of \(\vec{X}=(X_{1}(\bx), X_{2}(\bx), X_{3}(\bx))\) is the scalar function
\begin{equation*}
\text{div}\vec{X}(\bx)= \sum_{j=1}^{3} \frac{\partial X_j(\bx)}{\partial x_j}.
\end{equation*}
The Stokes Theorem says that if \(\Gamma: t\in [a, b]\mapsto \vec{r}(t)\) is a differentiable closed loop (meaning \(\Gamma(a)=\Gamma(b)\)) in \(\bbR^{3}\) and spans a differentiable surface \(S\text{,}\) then
\begin{equation*}
\int_{\Gamma} \vec{X}\cdot d\vec{r} = \int_S \text{curl}\vec{X}(\bx) \cdot \vec{n}(\bx)\, dA,
\end{equation*}
where \(\vec{n}(\bx)\) is an appropriately chosen unit normal vector field to \(S\text{.}\)
If we accept Stokes Theorem, it can be used to give some geometric interpretation for \(\text{curl}\vec{X}\text{.}\) Fix some \(\bx_{0}\) and a plane \(\Pi\) containing \(\bx_{0}\) with a unit normal vector \(\vec{n}\text{.}\) Take \(S\) to be the disc in the plane \(\Pi\) of radius \(\epsilon >0\) centered at \(\bx_{0}\) and \(\Gamma\) to be the boundary of this disc. Then
\begin{equation*}
\lim_{\epsilon \to 0} \frac{1}{\pi \epsilon^{2}} \int_{\Gamma} \vec{X}\cdot d\vec{r}=
\lim_{\epsilon \to 0} \frac{1}{\pi \epsilon^{2}} \int_S \text{curl}\vec{X}(\bx) \cdot \vec{n}(\bx)\, dA
=\text{curl}\vec{X}(\bx_{0}) \cdot \vec{n}.
\end{equation*}
Thus \(\text{curl}\vec{X}(\bx_{0}) \cdot \vec{n}\) is the infinitesimal rate of circulation\(\mbox{---}\)circulation per unit area\(\mbox{---}\)of \(\vec{X}\) along closed loops surrounding \(\bx_{0}\) in the plane through \(\bx_{0}\) with unit normal \(\vec{n}\text{.}\)
In fact, we can see why \(\text{curl}\vec{X}\) is defined this way by examining
\begin{equation*}
\lim_{\epsilon \to 0} \frac{1}{\pi \epsilon^{2}} \int_{\Gamma} \vec{X}\cdot d\vec{r}
\end{equation*}
without the knowledge of Stokes Theorem. We will see that, if the vector field \(\vec{X}\) is differentiable at \(\bx_{0}\text{,}\) then this limit exists and is given by the expression above.
In the simple setting of the two dimensional plane, if \(\vec{X}\) is continuously differentiable in a neighborhood of \({\mathbf x}_{0}=(x_{0}, y_{0})\text{,}\) and we take \(\Gamma\) to be a small square loop of side length \(2\epsilon\) around \({\mathbf x}_{0}\text{,}\) then
\begin{align*}
\amp \int_{\Gamma} \vec{X}\cdot d\vec{r} \\
= \amp
\int_{x_{0}-\epsilon}^{x_{0}+\epsilon} \left\{X_{1}(x, y_{0}-\epsilon)- X_{1}(x, y_{0}+\epsilon)\right\}\, dx\\
\amp + \int_{y_{0}-\epsilon}^{y_{0}+\epsilon} \left\{X_{2}(x_{0}+\epsilon, y)-X_{2}(x_{0}-\epsilon, y)\right\}\, dy\\
= \amp
\int_{x_{0}-\epsilon}^{x_{0}+\epsilon} \int_{y_{0}-\epsilon}^{y_{0}+\epsilon}
- \frac{\partial X_{1}(x, y)}{\partial y} \, dy dx +
\int_{y_{0}-\epsilon}^{y_{0}+\epsilon} \int_{x_{0}-\epsilon}^{x_{0}+\epsilon}
\frac{\partial X_{2}(x, y)}{\partial x} \, dx dy\\
= \amp \int_{[x_{0}-\epsilon, x_{0}+\epsilon] \times [y_{0}-\epsilon, y_{0}+\epsilon]}
\left\{ \frac{\partial X_{2}(x, y)}{\partial x}- \frac{\partial X_{1}(x, y)}{\partial y}\right\}\, dx dy.
\end{align*}
This is the simplest form of Green’s theorem and motivates the definition of the curl of a two dimensional vector field \((X_{1}(x, y), X_{2}(x, y))\) as \(\frac{\partial X_{2}(x, y)}{\partial x}- \frac{\partial X_{1}(x, y)}{\partial y}\text{.}\) Note that at any point \({\mathbf x}_{0}=(x_{0}, y_{0})\text{,}\) its value is determined as
\begin{equation*}
\lim_{\epsilon\to 0}\frac{1}{\text{Area enclosed by } \Gamma} \int_{\Gamma} \vec{X}\cdot d\vec{r}.
\end{equation*}
In three dimension or higher, the simplest loops are planar ones, namely, a loop contained in the plane spanned by a pair of orthonormal vectors \(\vec{\xi}\) and \(\vec{\eta}\) --- we take \({\mathbf x}_{0}\) to be the origin for simplicity, and \(\Gamma: t\mapsto \vec{\gamma}(t)\) be a closed loop near and enclosing \({\mathbf x}_{0}\) in the plane spanned by \(\vec{\xi}\) and \(\vec{\eta}\) given as
\begin{equation*}
\vec{\gamma}(t)=x(t)\vec{\xi} + y(t)\vec{\eta}.
\end{equation*}
Assuming \(\vec{X}\) to have the necessary differentiability, then Taylor expansion
\begin{equation*}
\vec{X}(\vec{\gamma}(t))= \vec{X}({\mathbf x}_{0})+
\left[\frac{\partial \vec{X}}{\partial {\mathbf x}}({\mathbf x}_{0})\right]( \vec{\gamma}(t)- {\mathbf x}_{0})+
h.o.t.\ (\Vert \vec{\gamma}(t)- {\mathbf x}_{0}\Vert )
\end{equation*}
gives the leading order term of \(\int_{\Gamma} \vec{X}\cdot d\vec{r}\) to be
\begin{align*}
\amp \sum_{i,j} \frac{\partial X_i({\mathbf x}_0)}{\partial x_j} \int_{\Gamma} \left(x(t)\xi_j + y(t)\eta_j\right)
\left(x'(t)\xi_i +y'(t)\eta_i\right)dt \\
= \amp \sum_{i,j} \frac{\partial X_i({\mathbf x}_0)}{\partial x_j} \int_{\Gamma} \left(x(t)x'(t)\xi_i\xi_j +
y(t)y'(t)\eta_i\eta_j + x(t)y'(t)\xi_j\eta_i +y(t)x'(t)\xi_i\eta_j\right)dt\\
= \amp \sum_{i,j} \frac{\partial X_i({\mathbf x}_0)}{\partial x_j}\left(\xi_j\eta_i \int_{\Gamma} x(t)y'(t)dt +
\xi_i\eta_j \int_{\Gamma} y(t)x'(t) dt\right)
\end{align*}
where we have used that along any closed loop
\begin{equation*}
\int_{\Gamma} \vec{X}({\mathbf x}_{0})\cdot \vec{\gamma}'(t)\, dt=
\int_{\Gamma} \left[\frac{\partial \vec{X}}{\partial {\mathbf x}}({\mathbf x}_{0})\right] {\mathbf x}_{0}\cdot
\vec{\gamma}'(t)\, dt=0,
\end{equation*}
and
\begin{equation*}
\int_{\Gamma} x(t)x'(t) dt = \int_{\Gamma} y(t)y'(t) dt =0,
\end{equation*}
but
\begin{equation*}
\int_{\Gamma} x(t)y'(t)dt = - \int_{\Gamma} y(t)x'(t) dt
\end{equation*}
equals the area enclosed by \(\Gamma\) --- one may take \(\Gamma\) to be a square or circle loop to see this, so the leading order term of \(\int_{\Gamma} \vec{X}\cdot d\vec{r}\) is
\begin{equation*}
\sum_{i,j} \frac{\partial X_i({\mathbf x}_{0})}{\partial x_j} \left(\xi_j\eta_i - \xi_i\eta_j \right)\cdot
\left(\mbox{Area enclosed by } \Gamma \right).
\end{equation*}
In other words,
\begin{equation*}
\lim_{\epsilon\to 0}\frac{1}{\text{Area enclosed by } \Gamma} \int_{\Gamma} \vec{X}\cdot d\vec{r}
=\sum_{i,j} \frac{\partial X_i({\mathbf x}_{0})}{\partial x_j} \left(\xi_j\eta_i - \xi_i\eta_j \right).
\end{equation*}
This derivation works for any dimension \(n\text{.}\) We observe that this "infinitesimal strength" of circulation of \(\vec{X}\) along loops near \({\mathbf x}_{0}\) depends on the above specific combinations of derivatives of \(\vec{X}\) as well as on the plane in terms of a choice of an orthonormal basis; and the dependence on these derivatives of \(\vec{X}\) and on \(\vec{\xi}\) and \(\vec{\eta}\) is linear in each when the remaining variables are held as constant\(\mbox{ ---}\)this is the notion of a multi-linear function; here, we momentarily relax the condition that \(\vec{\xi}\) and \(\vec{\eta}\) are orthonormal and allow them to be any pair of vectors. Furthermore, the dependence on \((\vec{\xi}, \vec{\eta})\) is antisymmetrical in \((\vec{\xi}, \vec{\eta})\text{,}\) and if \(\vec{\xi}'\) and \(\vec{\eta}'\) are another pair of vectors such that \(\text{Span}\{ \vec{\xi}',\vec{\eta}'\} = \text{Span}\{ \vec{\xi},\vec{\eta}\}\text{,}\) then the quantities \(\xi_j'\eta_i' - \xi_i'\eta_j' \) and \(\xi_j\eta_i - \xi_i\eta_j \) have a common proportionality constant equal to the determinant of the matrix that relates the two pairs of bases and equal \(\pm 1\) when both bases are orthonormal.
Exercise 7.1.1.
In the case of dimension \(3\text{,}\) if we take
\begin{equation*}
\vec{n}=\vec{\xi}\wedge\vec{\eta}=(\xi_2\eta_3-\xi_3\eta_2, \xi_3\eta_1-\xi_1\eta_3, \xi_1\eta_2-\xi_2\eta_1),
\end{equation*}
then it is a unit normal to the plane spanned by \(\vec{\xi}\) and \(\vec{\eta}\text{,}\) and
\begin{align*}
\amp \sum_{i,j} \frac{\partial X_i}{\partial x_j} ({\mathbf x}_{0})
\left(\xi_j\eta_i - \xi_i\eta_j \right)\\
=\amp (\frac{\partial X_3}{\partial x_2}-\frac{\partial X_2}{\partial x_3},
\frac{\partial X_1}{\partial x_3}-\frac{\partial X_3}{\partial x_1},
\frac{\partial X_2}{\partial x_1}-\frac{\partial X_1}{\partial x_2})\cdot \vec{n},
\end{align*}
thus producing the concept of the curl of a vector field in dimension \(3\text{.}\)
In the general dimension, using exterior algebra motivated by the above discussion and to be introduced soon, we see that \(\xi_j\eta_i - \xi_i\eta_j\) are simply the components of \(\vec{\xi}\wedge \vec{\eta}\text{,}\) and
\begin{equation*}
(\vec{\xi}, \vec{\eta}) \mapsto \sum_{i,j} \frac{\partial X_i}{\partial x_j}({\mathbf x}_{0})
\left(\xi_j\eta_i - \xi_i\eta_j \right)
\end{equation*}
is a bilinear antisymmetric function on \((\vec{\xi}, \vec{\eta})\) (we may remove the orthonormal condition on \(\vec{\xi}, \vec{\eta}\) now).
For any \(i \lt j\text{,}\) the coefficient of \((\xi_i\eta_j - \xi_j\eta_i)\) above is \(\frac{\partial X_j}{\partial x_i}- \frac{\partial X_i}{\partial x_j}\text{.}\) Thus the natural generalization of the notion of the curl of a vector field \(\vec{X}\) in \(\bbR^{n}, n>3\text{,}\) is not a vector field, but an object that acts on any pair of vectors in a bilinear and antisymmetric fashion. This is a heuristic reason for the notion of a tensor.
Another important question is how do vector fields and their curls transform under a change of variables? This is particularly important in the theory of manifolds, where there is usually no canonical coordinates to work with.
If \({\mathbf y}=T({\mathbf x})\) is a continuously differentiable change of variables, then any continuously differentiable curve \({\mathbf x}=\vec{\gamma}(t)\) is mapped to a continuously differentiable curve \({\mathbf y}=T(\vec{\gamma}(t))\text{,}\) with
\begin{equation*}
{\mathbf y}'(t) =\left[\frac{\partial T_{i} ({\mathbf x})}{\partial x_{j}}\right]\Big|_{{\mathbf x}
=\vec{\gamma}(t)}
{\mathbf x}'(t).
\end{equation*}
Since the value of a vector field at any point is naturally identified to be the tangent vector of a continuously differentiable curve passing through that point, if \(Y^{i}(T({\mathbf x}))\) are the components of the vector field at \(T({\mathbf x})\) in the \(\mathbf y\) coordinates that is transformed from \(X({\mathbf x})\) by \(T\text{,}\) we expect
\begin{equation*}
Y(T({\mathbf x}))=\begin{bmatrix} Y^{1}(T({\mathbf x}))\\ \vdots \\ Y^{n}(T({\mathbf x}))\end{bmatrix}
= \left[\frac{\partial T_{i} ({\mathbf x})}{\partial x_{j}}\right]\left.\right|_{{\mathbf x}=\vec{\gamma}(t)}
\begin{bmatrix} X^{1}({\mathbf x}) \\ \vdots \\ X^{n}({\mathbf x})\end{bmatrix}.
\end{equation*}
But the quantity
\begin{equation*}
(X^{1}({\mathbf x}), \ldots , X^{n}({\mathbf x}))\cdot
(x_{1}'(t),\ldots, x_{n}'(t))\text{,}
\end{equation*}
which is the integrand in \(\int_{\Gamma} \vec{X}\cdot d \vec{r}\text{,}\) would then transform to
\begin{equation*}
Y(T({\mathbf x}))^{\rm T}\left( \left[\frac{\partial T_{i} ({\mathbf x})}{\partial x_{j}}\right]^{-1}\right)^{\rm T}
\left[\frac{\partial T_{i} ({\mathbf x})}{\partial x_{j}}\right]^{-1} {\mathbf y}'(t)
\end{equation*}
in the \(\by\) coordinates. This is because \((X^{1}({\mathbf x}), \ldots , X^{n}({\mathbf x}))\cdot
(x_{1}'(t),\ldots, x_{n}'(t))\) encodes the Euclidean inner product between vectors in the \(\bx\) coordinates, while the representation of this inner product in the \(\by\) coordinates may no longer have a simple form. This transformation is not only complicated, but makes it even harder to keep track of the relation between the components of the curl computed in the \(\mathbf y\) and \(\mathbf x\) coordinates.
It turns out that we get a much simpler resolution of this issue if we do not treat \(X(\mathbf x)\) as a vector field, but as a field of covectors, namely, at each point \(\mathbf x\text{,}\) \(\mathbf u \mapsto \langle X(\mathbf x), \mathbf u\rangle :=X(\mathbf x)\cdot \mathbf u\) is a linear function on the vector space of tangent vectors at that point. The simplification is due to the transformation laws of vectors and covectors, to be introduced next. It is such consideration that makes differential forms, instead of vector-fields, the natural objects for integration on surfaces.
Exercise 7.1.2.
Determine
\begin{equation*}
\lim_{\epsilon\to 0}\frac{1}{\text{Area enclosed by } \Gamma} \int_{\Gamma} \vec{X}\cdot d\vec{r},
\end{equation*}
where \(\Gamma\) is the boundary of the disc of radius \(\epsilon\) centered at \(\mathbf 0 \in \bbR^{4}\) of the plane
\begin{equation*}
\begin{cases} x_{1} \hphantom{x_{2}}-a x_{3}-c x_{4} \amp =0 \\
\hphantom{x_{2}} x_{2} -b x_{3} - d x_{4} \amp =0\\
\end{cases}\text{.}
\end{equation*}
Hint.
One could avoid finding explicitly an orthonormal basis for the plane by using Exercise 7.1.1.
