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We can parametrize the upper half of the unit circle by \[ x \mapsto \left( \begin{matrix} x \\ \sqrt{1 - {x}^{2}} \end{matrix}\right) , - 1 \leq x \leq 1,\;\text{ or by }\;t \mapsto \left( \begin{matrix} \cos t \\ \sin t \end{matrix}\right) ,0 \leq t \leq \pi .\;\text{ 5.3.17 } \] | We use equation 5.3.4 to compute its length. The first parametrization gives \[ {\int }_{\left\lbrack -1,1\right\rbrack }\left| \left\lbrack \begin{matrix} 1 \\ \frac{-x}{\sqrt{1 - {x}^{2}}} \end{matrix}\right\rbrack \right| \left| {dx}\right| = {\int }_{\left\lbrack -1,1\right\rbrack }\sqrt{1 + \frac{{x}^{2}}{1 - {x}^... | Yes |
The graph of a \( {C}^{1} \) function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is parametried by \( x \mapsto \left( \begin{matrix} x \\ f\left( x\right) \end{matrix}\right) \) ; hence its arc length is given by the integral | \[ {\int }_{\left\lbrack a, b\right\rbrack }\left| \left\lbrack \begin{matrix} 1 \\ {f}^{\prime }\left( x\right) \end{matrix}\right\rbrack \right| \left| {dx}\right| = {\int }_{a}^{b}\sqrt{1 + {\left( {f}^{\prime }\left( x\right) \right) }^{2}}{dx}. \] | Yes |
Let \( p, q \) be two integers, and consider the curve in \( {\mathbb{R}}^{4} \) parametrized by\n\n\[ \n\gamma \left( t\right) = \left( \begin{matrix} \cos {pt} \\ \sin {pt} \\ \cos {qt} \\ \sin {qt} \end{matrix}\right) ,\;0 \leq t \leq {2\pi }\n\] | Its length is given by\n\n\[ \n{\int }_{0}^{2\pi }\sqrt{{\left( -p\sin pt\right) }^{2} + {\left( p\cos pt\right) }^{2} + {\left( -q\sin qt\right) }^{2} + {\left( q\cos qt\right) }^{2}}{dt}\n\]\n\n\[ \n= {2\pi }\sqrt{{p}^{2} + {q}^{2}}.\;\bigtriangleup\n\] | No |
Example 5.3.8 (The total curvature of a closed curve). Recall that the curvature \( \kappa \) of a curve is defined in Definition 3.9.1. Let \( C \) be the intersection of the surfaces of equations \( x = {y}^{2} - 2 \) and \( x = 2 - {z}^{2} \) . We will integrate \( \kappa \) over this curve to get the total curvatur... | Setting \( {y}^{2} - 2 = 2 - {z}^{2} \) gives \( {y}^{2} + {z}^{2} = 4 \), which suggests setting \( y = 2\cos \theta \) , \( z = 2\sin \theta \), and \( x = 4{\cos }^{2}\theta - 2 = 2\left( {2{\cos }^{2}\theta - 1}\right) = 2\cos {2\theta } \) . That is, we parametrize \( C \) by \( \gamma \left( \theta \right) = \lef... | Yes |
Example 5.3.9 (Area of a torus). Choose \( R > r > 0 \) . We obtain the torus shown in Figure 5.3.2 by taking the circle of radius \( r \) in the \( \left( {x, z}\right) \) - plane that is centered at \( x = R, z = 0 \), and rotating it around the \( z \) -axis. This surface is parametrized by\n\n\[ \gamma \left( \begi... | Then\n\n\[ \left\lbrack {\mathbf{D}\gamma \left( \mathbf{u}\right) }\right\rbrack = \left\lbrack \begin{matrix} - r\sin u\cos v & - \left( {R + r\cos u}\right) \sin v \\ - r\sin u\sin v & \left( {R + r\cos u}\right) \cos v \\ r\cos u & 0 \end{matrix}\right\rbrack \]\n\nand the surface area of the torus is given by the ... | Yes |
What is the area of the graph of the function \( {x}^{2} + {y}^{3} \) above the unit square \( Q \subset {\mathbb{R}}^{2} \) ? | Applying formula 5.2.6, we parametrize the surface by\n\n\[ \gamma \left( \begin{array}{l} x \\ y \end{array}\right) \mapsto \left( \begin{matrix} x \\ y \\ {x}^{2} + {y}^{3} \end{matrix}\right) \]\n\nand use equation 5.3.33 to compute\n\n\[ {\int }_{Q}\left| {\overset{\overrightarrow{{D}_{1}}}{\overbrace{\left\lbrack ... | Yes |
The subset of \( {\mathbb{R}}^{4} \) given by the two equations in four unknowns\n\n\[ \n{x}_{1}^{2} + {x}_{2}^{2} = {r}_{1}^{2}\;\text{ and }\;{x}_{3}^{2} + {x}_{4}^{2} = {r}_{2}^{2} \n\]\n\nis a surface. It can be parametrized by\n\n\[ \n\gamma \left( \begin{array}{l} u \\ v \end{array}\right) = \left( \begin{array}{... | and since\n\n\n\n\[ \n= \left\lbrack \begin{matrix} {r}_{1}^{2} & 0 \\ 0 & {r}_{2}^{2} \end{matrix}\right\rbrack \n\]\n\nthe area of the surface is given by\n\n\[ \n{\int }_{\left\lbrack {0,{2\pi }}\right\rbrack \tim... | Yes |
Example 5.3.12: Let us compute the area of the part of the surface of equation \( {z}_{2} = {z}_{1}^{2} \) where \( \left| {z}_{1}\right| \leq 1 \) . | Polar coordinates for \( {z}_{1} \) give a nice parametrization:\n\n\[ \gamma \left( \begin{array}{l} r \\ \theta \end{array}\right) = \left( \begin{matrix} r\cos \theta \\ r\sin \theta \\ {r}^{2}\cos {2\theta } \\ {r}^{2}\sin {2\theta } \end{matrix}\right) ,0 \leq r \leq 1,\;0 \leq \theta \leq {2\pi }.\]\n\nAgain we n... | Yes |
Example 5.3.13 (Volume of a three-dimensional manifold in \( {\mathbb{R}}^{4} \) ). Let \( U \subset {\mathbb{R}}^{3} \) be an open set, and let \( f : U \rightarrow \mathbb{R} \) be a \( {C}^{1} \) function. The graph of \( f \) is a three-dimensional manifold in \( {\mathbb{R}}^{4} \), and it comes with the natural p... | \n\[
\gamma \left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left( \begin{matrix} x \\ y \\ z \\ f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) \end{matrix}\right)
\]
\n\[
\det \left( {{\left\lbrack \mathbf{D}\gamma \left( \begin{array}{l} x \\ y \\ z \end{array}\right) \right\rbrack }^{\top }\left\lb... | Yes |
Let us compute \( {\operatorname{vol}}_{n}{S}^{n} \), where \( {S}^{n} \subset {\mathbb{R}}^{n + 1} \) is the unit sphere. | \[ {\operatorname{vol}}_{n + 1}{B}^{n + 1} = {\int }_{0}^{1}{\operatorname{vol}}_{n}{S}^{n}\left( r\right) {dr} \] \[ = {\int }_{0}^{1}{r}^{n}{\operatorname{vol}}_{n}{S}^{n}{dr} = \frac{1}{n + 1}{\operatorname{vol}}_{n}\left( {S}^{n}\right) \] | No |
Theorem 5.4.1. Let \( {D}_{r}\left( \mathbf{p}\right) \) be the set of all points \( \mathbf{q} \) in a surface \( S \subset {\mathbb{R}}^{3} \) such that there exists a curve of length \( \leq r \) in \( S \) joining \( \mathbf{p} \) to \( \mathbf{q} \) . Then \[ \operatorname{Area}\left( {{D}_{r}\left( \mathbf{p}\rig... | Proof. The proof will take about four pages. Since Theorem 5.4.1 is local, we may assume that \( \mathbf{p} = \mathbf{0} \) and that \( S \) is the graph of a smooth function defined near the origin of \( {\mathbb{R}}^{2} \), whose Taylor polynomial starts with quadratic terms, as in equation 5.4.1. These quadratic ter... | Yes |
Proposition 5.4.2. Set \( \gamma \left( \rho \right) = \left( \begin{matrix} \rho \cos \alpha \left( \rho \right) \\ \rho \sin \alpha \left( \rho \right) \end{matrix}\right) \) and \( {\delta }_{r}\left( \rho \right) = \left( \begin{matrix} \rho \cos \alpha \left( r\right) \\ \rho \sin \alpha \left( r\right) \end{matri... | Proof of Proposition 5.4.2. The proof consists of computing Taylor polynomials. We will compute, through terms in \( {r}^{3} \), the Taylor polynomials of the lengths of \( {\widetilde{\delta }}_{r}\left( \left\lbrack {0, r}\right\rbrack \right) \) and \( \widetilde{\gamma }\left( \left\lbrack {0, r}\right\rbrack \righ... | Yes |
Proposition 5.4.3. We have | Proof of Proposition 5.4.3. This is a straightforward computation:\n\n\[ \left\lbrack {\mathbf{D}g\left( \rho \right) }\right\rbrack = \left\lbrack \begin{matrix} \cos \theta & - \rho \sin \theta \\ \sin \theta & \rho \cos \theta \\ \rho \left( {a{\cos }^{2}\theta + b{\sin }^{2}\theta }\right) + o\left( \rho \right) & ... | Yes |
Corollary 5.4.5. A minimal surface has mean curvature 0. | Proof. Suppose that at some point \( \mathbf{a} \in S \) we have \( \overrightarrow{H}\left( \mathbf{a}\right) \neq \overrightarrow{\mathbf{0}} \) . The vector field \( \overrightarrow{H} \) then points on one side of the surface in some neighborhood \( V \) of a, and we can find a little \ | No |
We claim that this is a set of dimension \( \ln 3/\ln 2 \) . | At the \( n \) th stage of the construction, sum, over all the little pieces, the the whole curve, as in Figure 5.5.3. Then \( B \) consists of four copies of \( A \) . (This is true at any level, but it is easiest to see at the first level, the top graph in Figure 5.5.2.). Therefore, in any dimension \( d \), it shoul... | Yes |
Let \( {i}_{1},\ldots ,{i}_{k} \) be any \( k \) integers between 1 and \( n \) . Then \( d{x}_{{i}_{1}} \land \cdots \land d{x}_{{i}_{k}} \) is that function of \( k \) vectors \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k} \) in \( {\mathbb{R}}^{n} \) that puts these vectors side by si... | For instance,\n\n\[ \underset{\text{2-form }}{\underbrace{d{x}_{1} \land d{x}_{2}}}\left( {\left\lbrack \begin{array}{r} 1 \\ 2 \\ - 1 \\ 1 \end{array}\right\rbrack ,\left\lbrack \begin{array}{r} 3 \\ - 2 \\ 1 \\ 2 \end{array}\right\rbrack }\right) = \det \underset{\begin{matrix} \text{ 1st and 2nd rows } \\ \text{ of ... | Yes |
The function defined by \( {W}_{\overrightarrow{\mathbf{v}}}\left( \overrightarrow{\mathbf{w}}\right) = \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}} \), where both the fixed vector \( \overrightarrow{\mathbf{v}} \) and the variable \( \overrightarrow{\mathbf{w}} \) are elements of \( {\mathbb{R}}^{n} \... | The \( {a}_{i} \) are given by equation 6.1.19: \( {a}_{i} = {W}_{\overrightarrow{\mathbf{v}}}\left( {\overrightarrow{\mathbf{e}}}_{i}\right) = \overrightarrow{\mathbf{v}} \cdot {\overrightarrow{\mathbf{e}}}_{i} = {v}_{i} \) . Thus\n\n\( {W}_{\overrightarrow{\mathbf{v}}} = {v}_{1}\;d{x}_{1} + \cdots + {v}_{n}\;d{x}_{n}... | Yes |
Theorem 6.1.10 (Dimension of \( {A}_{c}^{k}\left( {\mathbb{R}}^{n}\right) \) ). The space \( {A}_{c}^{k}\left( {\mathbb{R}}^{n}\right) \) has dimension equal to the binomial coefficient\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) = \frac{n!}{k!\left( {n - k}\right) !} \] | Proof. Just count the elements of the basis: the elementary \( k \) -forms on \( {\mathbb{R}}^{n} \) . Not for nothing is the binomial coefficient called \ | No |
Example 6.1.11 (Dimension of \( {A}_{c}^{k}\left( {\mathbb{R}}^{3}\right) \) ). The vector spaces \( {A}_{c}^{0}\left( {\mathbb{R}}^{3}\right) \) and \( {A}_{c}^{3}\left( {\mathbb{R}}^{3}\right) \) have dimension 1, and \( {A}_{c}^{1}\left( {\mathbb{R}}^{3}\right) \) and \( {A}_{c}^{2}\left( {\mathbb{R}}^{3}\right) \) ... | \[ \left( \begin{array}{l} 3 \\ 0 \end{array}\right) = \frac{3!}{0!3!} = 1\text{elementary 0 -form;}\left( \begin{array}{l} 3 \\ 1 \end{array}\right) = \frac{3!}{1!2!} = 3\text{elementary 1 -forms} \] \[ \left( \begin{array}{l} 3 \\ 2 \end{array}\right) = \frac{3!}{2!1!} = 3\text{ elementary }2\text{-forms; }\left( \be... | Yes |
Proposition 6.1.15 (Properties of the wedge product). wedge product has the following properties:\n\n1. distributivity:\n\[ \n\\varphi \\land \\left( {{\\omega }_{1} + {\\omega }_{2}}\\right) = \\varphi \\land {\\omega }_{1} + \\varphi \\land {\\omega }_{2} \n\]\n\n6.1.35\n\n2. associativity:\n\[ \n\\left( {{\\varphi }... | Note that in equation 6.1.37 the \\( \\varphi \\) and \\( \\omega \\) change positions. For example, if \\( \\varphi = d{x}_{1} \\land d{x}_{2} \\) and \\( \\omega = d{x}_{3} \\), skew commutativity says that\n\n\[ \n\\left( {d{x}_{1} \\land d{x}_{2}}\\right) \\land d{x}_{3} = {\\left( -1\\right) }^{2}d{x}_{3} \\land \... | Yes |
Example 6.1.17 (A 2-form field on \( {\mathbb{R}}^{3} \) ). The form field \( \cos \left( {xz}\right) {dx} \land {dy} \) is a 2 -form field on \( {\mathbb{R}}^{3} \) . Here we evaluate it twice, each time on the same vectors, but at different points: | \[ \cos \left( {xz}\right) {dx} \land {dy}\left( {{P}_{\left( \begin{matrix} 1 \\ 2 \\ \pi \end{matrix}\right) }\left( {\left\lbrack \begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} 2 \\ 2 \\ 3 \end{array}\right\rbrack }\right) }\right) = \left( {\cos \left( {1 \cdot \pi }\right) }\r... | Yes |
Consider a case where \( k = 1, n = 2 \), and \( \gamma \left( u\right) = \left( \begin{array}{l} R\cos u \\ R\sin u \end{array}\right) \). We will take \( U \) to be the interval \( \left\lbrack {0, a}\right\rbrack \), for some \( a > 0 \). If we integrate \( {xdy} - {ydx} \) over \( \left\lbrack {\gamma \left( U\righ... | \n\[{\int }_{\left\lbrack \gamma \left( U\right) \right\rbrack }\left( {{xdy} - {ydx}}\right) = {\int }_{\left\lbrack 0, a\right\rbrack }\left( {{xdy} - {ydx}}\right) \left( {{P}_{\left( \begin{matrix} R\cos u \\ R\sin u \end{matrix}\right) }\overset{\overrightarrow{{D}_{1}}\gamma \left( \mathbf{u}\right) }{\overbrace{... | Yes |
Example 6.2.3 (An anchored \( k \) -parallelogram). An important example of a parametrized domain is \( {P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) \), parametrized by\n\n\[ \gamma \left( \begin{matrix} {t}_{1} \\ \vdots \\ {t}_{k} \end{matrix}\right) = \m... | In this case, Definition 6.2.1 gives\n\n\[ {\int }_{\left\lbrack {P}_{\mathbf{x}}\left( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}\right) \right\rbrack }\varphi = {\int }_{0}^{1}\cdots {\int }_{0}^{1}\varphi \left( {{P}_{\gamma \left( \mathbf{t}\right) }\left( {{\overrightarrow{\mathbf{... | Yes |
Example 6.2.4 (Integrating a 2-form field over a parametrized surface in \( {\mathbb{R}}^{3} \) ). Let us integrate \( {dx} \land {dy} + {ydx} \land {dz} \) over the parametrized domain \( \left\lbrack {\gamma \left( S\right) }\right\rbrack \) where\n\n\[ \n\gamma \left( \begin{array}{l} s \\ t \end{array}\right) = \le... | Applying Definition 6.2.1, we find\n\n\[ \n{\int }_{\left\lbrack \gamma \left( S\right) \right\rbrack }{dx} \land {dy} + {ydx} \land {dz} \n\]\n\n\[ \n= {\int }_{0}^{1}{\int }_{0}^{1}\left( {{dx} \land {dy} + y\;{dx} \land {dz}}\right) \left( {{P}_{\left( \begin{matrix} s + t \\ {s}^{2} \\ {t}^{2} \end{matrix}\right) }... | Yes |
Example 6.2.5 (Different parametrizations give different results). Suppose you and your neighbor wish to integrate \( {dy} \) over the upper right quadrant of the unit circle. This can be interpreted as determining the \( y \) -component of that arc of circle; if you walk along the arc, how far will you go in the \( y ... | You choose the parametrization \( \gamma \left( t\right) = \left( \begin{matrix} \cos t \\ \sin t \end{matrix}\right) \) ; applying equation 6.2.1, you get\n\n\[{\int }_{\left\lbrack 0,\pi /2\right\rbrack }{dy}\left( {{P}_{\left( \begin{matrix} \cos t \\ \sin t \end{matrix}\right) }\left\lbrack \begin{array}{r} - \sin ... | Yes |
Example 6.3.2 (Orienting a subspace). The vector space \( {\mathbb{R}}^{n} \) has a standard orientation, but subspaces do not. For instance,\n\n\[ \underset{\left\{ {\mathbf{v}}_{1}\right\} }{\underbrace{\left\lbrack \begin{array}{r} 1 \\ - 1 \\ 0 \end{array}\right\rbrack ,\left\lbrack \begin{array}{r} 1 \\ 0 \\ - 1 \... | \[ {}^{4}\det \left\lbrack {P}_{\left\{ {\mathbf{v}}_{1}\right\} \rightarrow \left\{ {\mathbf{v}}_{2}\right\} }\right\rbrack = \det \left\lbrack \begin{array}{rr} 1 & 1 \\ - 1 & 0 \end{array}\right\rbrack = 1,\det \left\lbrack {P}_{\left\{ {\mathbf{v}}_{1}\right\} \rightarrow \left\{ {\mathbf{v}}_{3}\right\} }\right\rb... | Yes |
Example 6.3.5 (A nonorientable surface). The Möbius strip shown in Figure 6.3.4 is not orientable; it is not possible to choose a transverse vector field \( \overrightarrow{\mathbf{n}} \) varying continuously with \( \mathbf{x} \) . | If you imagine yourself walking along the surface of a Möbius strip, planting a forest of normal vectors, one at each point, all pointing \ | No |
For every \( R > 0 \) , the vector field\n\n\[ \overrightarrow{\mathbf{t}}\left( \begin{array}{l} x \\ y \end{array}\right) = \left\lbrack \begin{array}{r} - y \\ x \end{array}\right\rbrack \]\n\nis a nonvanishing vector field tangent to the circle of equation \( {x}^{2} + {y}^{2} = {R}^{2} \) , defining the counterclo... | 6.3.10\n\nis a nonvanishing vector field tangent to the circle of equation \( {x}^{2} + {y}^{2} = {R}^{2} \) , defining the counterclockwise orientation. \( \bigtriangleup \) | No |
Example 6.3.7 (Orienting a surface in \( {\mathbb{R}}^{3} \) ). Let \( S \subset {\mathbb{R}}^{3} \) be the surface of equation \( a{x}^{2} + b{y}^{2} + c{z}^{2} = {R}^{2} \), where we assume that \( {abc} \neq 0 \) (i.e., \( a, b \) , and \( c \) are all nonzero) and \( R > 0 \) . Then the radial vector field\n\n\[ \o... | Indeed, if \( f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = a{x}^{2} + b{y}^{2} + c{z}^{2} - {R}^{2} \), then\n\n\[ \left\lbrack {\mathbf{D}f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) }\right\rbrack \overbrace{\left\lbrack \begin{array}{l} x \\ y \\ z \end{array}\right\rbrack } = \left\lbrack {{... | Yes |
Consider the surface \( S \) defined by \( f\left( \mathbf{x}\right) = \sin \left( {x + {yz}}\right) = 0 \) . We saw in Example 3.1.13 that the vector field \[ \overrightarrow{\nabla }f\left( \mathbf{x}\right) \overset{\text{ def }}{ = }{\left\lbrack \mathbf{D}f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) \ri... | Thus, for \( \overrightarrow{\mathbf{v}} \in {T}_{\mathbf{x}}S \) , \[ 0 = \left\lbrack {\mathbf{D}f\left( \mathbf{x}\right) }\right\rbrack \overrightarrow{\mathbf{v}} = \overrightarrow{\nabla }f\left( \mathbf{x}\right) \cdot \overrightarrow{\mathbf{v}} \] So the vector \( \overrightarrow{\mathbf{x}} = \overrightarrow{... | Yes |
Proposition 6.3.9 (Orienting manifolds given by equations). Let \( U \subset {\mathbb{R}}^{n} \) be open, and let \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{n - k} \) be a map of class \( {C}^{1} \) such that \( \left\lbrack {\mathbf{{Df}}\left( \mathbf{x}\right) }\right\rbrack \) is surjective at all \( \mathbf{x} \i... | \[ \Omega \left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) \overset{\mathrm{{def}}}{ = }\mathrm{{sgn}}\det \left\lbrack {\overrightarrow{\nabla }{f}_{1}\left( \mathbf{x}\right) ,\ldots ,\overrightarrow{\nabla }{f}_{n - k}\left( \mathbf{x}\right) ,{\overrightarrow{\mathbf{v}}}_... | Yes |
Proposition 6.3.10 (Orientation of connected, orientable manifold). If \( M \) is a connected manifold, then either \( M \) is not orientable, or it has two orientations. If \( M \) is orientable, then specifying an orientation of \( {T}_{\mathbf{x}}M \) at one point defines the orientation at every point. | Proof. If \( M \) is orientable and \( \Omega : \mathcal{B}\left( M\right) \rightarrow \{ + 1, - 1\} \) is an orientation of \( M \), then \( - \Omega \) is also an orientation, so an orientable manifold has at least two orientations.\n\nWe must show that if \( M \) is connected and \( {\Omega }^{\prime },{\Omega }^{\p... | Yes |
Example 6.4.3 (Parametrizations of unit circle). Let \( C \) be the circle of equation \( {x}^{2} + {y}^{2} = {R}^{2}, R > 0 \) oriented by the vector field \( \overrightarrow{\mathbf{t}}\left( \begin{array}{l} x \\ y \end{array}\right) = \left\lbrack \begin{array}{r} - y \\ x \end{array}\right\rbrack \) , as in Exampl... | The parametrization \( \gamma \left( t\right) = \left( \begin{matrix} R\cos t \\ R\sin t \end{matrix}\right) \) preserves that orientation since\n\n\[ \overrightarrow{\mathbf{t}}\left( {\gamma \left( t\right) }\right) \cdot {\overrightarrow{\gamma }}^{\prime }\left( t\right) = \left\lbrack \begin{array}{r} - R\sin t \\... | Yes |
Example 6.4.4 (Orientation-preserving parametrization of surface in \( {\mathbb{C}}^{3} \) ). Consider the surface \( S \subset {\mathbb{C}}^{3} \) parametrized by \( z \mapsto \left( \begin{matrix} z \\ {z}^{2} \\ {z}^{3} \end{matrix}\right) ,\left| z\right| < 1 \). Does this parametrization preserve the standard orie... | Using Definition 6.3.1, we must compute the determinant of the change of matrix basis between the basis\n\n\[ \overrightarrow{{D}_{1}\gamma }\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} 1 \\ 0 \\ {2x} \\ {2y} \\ 3{x}^{2} - 3{y}^{2} \\ {6xy} \end{matrix}\right) ,\;\overrightarrow{{D}_{2}\gam... | Yes |
Does the parametrization\n\n\[ \gamma \left( \begin{array}{l} u \\ v \end{array}\right) = \left( \begin{matrix} \left( {R + r\cos u}\right) \cos v \\ \left( {R + r\cos u}\right) \sin v \\ r\sin u \end{matrix}\right) \]\n\npreserve that orientation? | Since\n\n\[ \overrightarrow{{D}_{1}\gamma } = \left\lbrack \begin{matrix} - r\sin u\cos v \\ - r\sin u\sin v \\ r\cos u \end{matrix}\right\rbrack ,\overrightarrow{{D}_{2}\gamma } = \left\lbrack \begin{matrix} - \left( {R + r\cos u}\right) \sin v \\ \left( {R + r\cos u}\right) \cos v \\ 0 \end{matrix}\right\rbrack ,\]\n... | Yes |
Proposition 6.4.8 (Orientation-preserving parametrizations). Let \( M \subset {\mathbb{R}}^{n} \) be a \( k \) -dimensional oriented manifold. Let \( {U}_{1},{U}_{2} \) be subsets of \( {\mathbb{R}}^{k} \), and let \( {\gamma }_{1} : {U}_{1} \rightarrow {\mathbb{R}}^{n} \) and \( {\gamma }_{2} : {U}_{2} \rightarrow {\m... | Proof. Set \( \mathbf{x} = {\gamma }_{1}\left( {\mathbf{u}}_{1}\right) = {\gamma }_{2}\left( {\mathbf{u}}_{2}\right) \) . Then (by Proposition 3.2.7), the columns \[ \left\lbrack {\mathbf{D}{\gamma }_{1}\left( {\mathbf{u}}_{1}\right) }\right\rbrack = \left\lbrack {\mathbf{D}\left( {{\gamma }_{2} \circ {\gamma }_{2}^{-1... | Yes |
Consider the subset \( X \subset \operatorname{Mat}\left( {2,3}\right) \) made of all \( 2 \times 3 \) matrices of rank 1 . This is a manifold of dimension 4 in Mat \( \left( {2,3}\right) \), as Exercise 6.4.2 asks you to show; we will see that it is nonorientable. | Indeed, suppose that \( X \) is orientable. Let\n\n\[ \n{\gamma }_{1},{\gamma }_{2} : \left( {{\mathbb{R}}^{2} - \left\{ \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \right\} }\right) \times {\mathbb{R}}^{2} \rightarrow X \n\]\n\nbe given by\n\n\[ \n{\gamma }_{1} : \left( \begin{array}{l} {a}_{1} \\ {b}_{1} \\ {c}... | No |
Theorem 6.4.10 (Integral independent of orientation-preserving parametrizations). Let \( M \subset {\mathbb{R}}^{n} \) be a \( k \) -dimensional oriented manifold, let \( {U}_{1},{U}_{2} \) be open subsets of \( {\mathbb{R}}^{k} \), and let \( {\gamma }_{1} : {U}_{1} \rightarrow {\mathbb{R}}^{n} \) and \( {\gamma }_{2}... | Proof of Theorem 6.4.10. We will prove this by trying to prove the (false) statement that the integrals are equal for any parametrizations (whether orientation preserving or reversing) and discovering where we go wrong.\n\nDefine the \ | No |
What is the integral of the 2-form \( \omega = {ydy} \land {dz} + {xdx} \land {dz} + {zdx} \land {dy} \) through the piece \( P \) of the plane defined by \( x + y + z = 1 \) where \( x, y, z \geq 0 \), and oriented by \( \overrightarrow{\mathbf{n}} = \left\lbrack \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right\rbrack \... | This surface is the graph of \( z = 1 - x - y \), so\n\n\[ \gamma \left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} x \\ y \\ 1 - x - y \end{matrix}\right) \]\n\nis a parametrization, if \( x \) and \( y \) are in the triangle \( T \subset {\mathbb{R}}^{2} \) given by \( x, y \geq 0 \), \( x + y... | Yes |
Example 6.4.14 (Integrating a 2-form field over a parametrized surface in \( {\mathbb{C}}^{3} \) ). Consider again the oriented surface \( S \subset {\mathbb{C}}^{3} \) which we saw in Example 6.4.4: the surface parametrized by \( z \mapsto \left( \begin{matrix} z \\ {z}^{2} \\ {z}^{3} \end{matrix}\right) ,\left| z\rig... | Example 6.4.14: To parametrize \( S \) using polar coordinates (equation 6.4.35) we use\n\n\[ \nz = r\cos \theta + {ir}\sin \theta \n\]\n\n(equation 0.7.10); the first two entries correspond to the real and imaginary parts of \( z \), the third and fourth to squaring the real and imaginary parts, and so on. Remember De... | Yes |
Example 6.4.15 (Integrating a 0-form over an oriented point).\n\nLet \( x \) be an oriented point and \( f \) a function (i.e., a 0-form field) defined in some neighborhood of \( \mathbf{x} \) . Then\n\n\[ \n{\int }_{+\mathbf{x}}f = + f\left( \mathbf{x}\right) \;\text{ and }\;{\int }_{-\mathbf{x}}f = - f\left( \mathbf{... | Thus\n\n\[ \n{\int }_{+\{ + 2\} }{x}^{2} = 4\text{ and }{\int }_{-\{ + 2\} }{x}^{2} = - 4.\;\bigtriangleup\n\] | Yes |
The 2-form\n\n\[ \Phi = {ydy} \land {dz} + {xdx} \land {dz} - {zdx} \land {dy} \]\n\n\n\nis the flux form of the vector field\n\n\[ \overrightarrow{F}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left\lbrack \begin{array}{r} y \\ - x \\ - z \end{array}\right\rbrack \] | Vectors are tangent to concentric cylinders around the \( z \) -axis; the \( z \) - component of the vectors is negative where \( z > 0 \) and positive where \( z < 0 \). In the \( \left( {x, y}\right) \) -plane, this vector field is simply rotation clockwise around the origin. As you would expect, \( \Phi \) returns 0... | Yes |
What is the work of \( \overrightarrow{F}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left\lbrack \begin{array}{r} y \\ - x \\ 0 \end{array}\right\rbrack \) over the helix oriented by the tangent vector equation 6.3.4 that a curve \( C \) can be oriented by \[ \operatorname{sgn}\left( {\overrightarrow{\mat... | So by equation 6.5.14 the work of the vector field \( \overrightarrow{F} \) over the helix is Equation 6.5.16: Recall that \[ {\int }_{0}^{4\pi }\overset{\overrightarrow{F}\left( {\gamma \left( t\right) }\right) }{\overbrace{\left\lbrack \begin{matrix} \sin t \\ - \cos t \\ 0 \end{matrix}\right\rbrack }} \cdot \overset... | Yes |
The flux of the vector field\n\n\[ \n\overrightarrow{F}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left\lbrack \begin{matrix} x \\ {y}^{2} \\ z \end{matrix}\right\rbrack \n\]\n\nthrough the parametrized domain \( \left( \begin{array}{l} u \\ v \end{array}\right) \mapsto \left( \begin{array}{l} {u}^{2} \\ ... | \n\[ \n{\int }_{0}^{1}{\int }_{0}^{1}\det \left\lbrack \begin{matrix} {u}^{2} & {2u} & 0 \\ {u}^{2}{v}^{2} & v & u \\ {v}^{2} & 0 & {2v} \end{matrix}\right\rbrack {dudv} = {\int }_{0}^{1}{\int }_{0}^{1}\left( {2{u}^{2}{v}^{2} - 4{u}^{3}{v}^{3} + 2{u}^{2}{v}^{2}}\right) {dudv} \n\]\n\n\[ \n= {\int }_{0}^{1}{\left\lbrack... | Yes |
The electromagnetic field, a six-component object, cannot be represented either as a function (a one-component object) or as a vector field (in \( {\mathbb{R}}^{4} \), a four-component object). | The standard way to deal with the problem is to choose coordinates \( x, y, z, t \), in particular choosing a specific space-like subspace and a specific time-like subspace, quite likely those of your laboratory. Experiments indicate the following force law: there are two time-dependent vector fields, \( \overrightarro... | No |
Proposition 6.6.4. Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \), and let \( M \subset U \) be the \( k \) -domensional manifold defined by \( \mathbf{f} = \mathbf{0} \), where \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{n - k} \) is a \( {C}^{1} \) function with \( \left\lbrack {\mathbf{{Df}}\left( \mathbf{x... | The proof of Proposition 6.6.4 is left as Exercise 6.6.11. | No |
The points \( \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \) and \( \left( \begin{array}{l} 1 \\ 1 \end{array}\right) \) are in the boundary \( {\partial }_{M}X \), but they are not smooth points: \( {V}_{1} \cap X \) cannot be defined by a single \( g \) with onto derivative; nor can \( {V}_{2} \cap X \) . They ... | Indeed, at both points we can take\n\n\[ \mathbf{g}\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{array}{l} x - {y}^{2} \\ y - {x}^{2} \end{array}\right) ,\;\text{with onto derivatives} \]\n\n\[ \left\lbrack {\mathbf{{Dg}}\left( \begin{array}{l} 0 \\ 0 \end{array}\right) }\right\rbrack = \left\lbrack... | Yes |
Example 6.6.7 (A subset with no smooth boundary). The subset pictured in Figure 6.6.5 is bounded by three copies of the Koch snowflake of Example 5.5.1, rotated by \( {120}^{ \circ } \) and \( {240}^{ \circ } \), and translated so they fit end to end. This region has no smooth boundary: | as we saw in Example 5.5.1, the length of any little piece of the boundary is always infinite, but Corollary 5.3.4 says that every point of a \( k \) -dimensional manifold has a neighborhood with finite \( k \) -dimensional volume, so by Proposition 6.6.3, the region has no smooth boundary. | Yes |
Example 6.6.10 (Piece-with-boundary). We saw in Example 6.6.6 that the shaded region shown in Figures 6.6.3 and 6.6.4 (the region set off from the rest of \( {\mathbb{R}}^{2} \) by the inequalities \( y \geq {x}^{2} \) and \( x \geq {y}^{2} \) ) is a piece-with-corners. It follows from Exercise 6.6.9 that it is a piece... | It follows from Exercise 6.6.9 that it is a piece-with-boundary. | No |
Example 6.6.13 (Another locus not a piece-with-boundary). Let \( M \subset {\mathbb{R}}^{2} \) be the \( x \) -axis and let \( X \subset M \) be the locus defined by \( g \leq 0 \), for\n\n\[ g\left( x\right) = \left\{ \begin{array}{ll} {x}^{3}\sin \frac{1}{x} & \text{ if }x \neq 0 \\ 0 & \text{ if }x = 0. \end{array}\... | This function \( g \) is \( {C}^{1} \), as required by Definition 6.6.2; this can be seen by the same computation as in Example 1.9.4, replacing the \( {x}^{2} \) by \( {x}^{3} \) .\n\nAs shown in Figure 6.6.8, the locus \( X \subset M = \mathbb{R} \) consists of disjoint intervals \( \left\lbrack {{x}_{{2i} - 1},{x}_{... | Yes |
Example 6.6.14. In constrast to the locus \( X \) of Example 6.6.13, the two-dimensional locus consisting of the area between the graph and the \( x \) -axis in Figure 6.6.8 is a piece-with-boundary of \( {\mathbb{R}}^{2} \) . The boundary is the \( x \) - axis and the graph of \( g \) ; the points where they intersect... | Both conditions of Definition 6.6.8 are met: here \( k - 1 = 1 \) and the nonsmooth points have 1-dimensional volume (i.e., length) 0, and for every \( \epsilon > 0 \) we can find an open subset \( U \subset {\mathbb{R}}^{2} \) that covers all the nonsmooth points and such that the length of \( {\partial }_{{\mathbb{R}... | Yes |
Theorem 6.6.16. If \( X \subset M \) is a \( k \) -dimensional piece-with-boundary, then \( X \) has finite \( k \) -dimensional volume. | Proof. By Corollary 5.3.4, every point has a neighborhood with finite \( k \) -dimensional volume. Since a piece-with-boundary is compact, it can be covered by finitely many such neighborhoods. | Yes |
Let \( C \) be a curve oriented by \( \Omega, P \) a piece-with-boundary of \( C \), and \( \mathbf{x} \) a smooth point in its boundary. Then \[ {\Omega }_{\mathbf{x}}^{\partial }\left( \phi \right) = {\Omega }_{\mathbf{x}}\left( {\overrightarrow{\mathbf{v}}}_{\text{out }}\right) \] where \( \phi \) is the unique basi... | If the outward-pointing vector \( {\overrightarrow{\mathbf{v}}}_{\text{out }} \) is a direct basis of \( {T}_{\mathbf{x}}C \) (pointing in the same direction as the tangent vector field orienting the curve), then \( {\Omega }_{\mathbf{x}}^{\partial } = + 1 \) ; if it is an indirect basis, then \( {\Omega }_{\mathbf{x}}... | Yes |
Example 6.6.23 (Oriented boundary of a piece-with-boundary of \( {\mathbb{R}}^{2} \) ). Let a smooth curve \( C \) be the smooth boundary of a piece-with-boundary \( S \subset {\mathbb{R}}^{2} \) . Give \( {\mathbb{R}}^{2} \) the standard orientation \( \Omega = \operatorname{sgn} \) det. Then at | Example 6.6.22: An oriented a point \( \mathbf{x} \in C \), the boundary \( C \) is oriented by curve with a piece-with-boundary\n\n\( P \) marked in bold. The boundary\n\[ \n{\Omega }^{\partial }\left( \overrightarrow{\mathbf{v}}\right) = \operatorname{sgn}\det \left( {{\overrightarrow{\mathbf{v}}}_{\text{out }},\over... | No |
Example 6.6.24 (Oriented boundary of a piece-with-boundary of an oriented surface in \( {\mathbb{R}}^{3} \) ). Let \( P \subset S \) be a piece-with-boundary of a surface \( S \) oriented by a normal vector field \( \overrightarrow{\mathbf{n}} \) ; it is shown as the shaded region in Figure 6.6.12. In this case Definit... | The shaded area is the piece-with-boundary \( P \) of the surface \( S \) . The vector \( {\overrightarrow{\mathbf{v}}}_{\text{out }} \) is tangent to \( S \) at a point in the boundary of \( S \) and points out of \( P \) . The vector \( \overrightarrow{\mathbf{v}} \) is tangent to the boundary. The vectors \( \overri... | Yes |
Suppose \( U \) is a piece-with-boundary of \( {\mathbb{R}}^{3} \) with the standard orientation by det. Then by Definition 6.6.21 its boundary \( S \) is oriented by \[ {\Omega }^{\partial }\left( {{\overrightarrow{\mathbf{v}}}_{1},{\overrightarrow{\mathbf{v}}}_{2}}\right) = \mathrm{{sgn}}\det \left( {{\overrightarrow... | Thus, if a surface \( S \) bounds a piece \( U \subset {\mathbb{R}}^{3} \), with \( {\mathbb{R}}^{3} \) given the standard orientation, the orientation of \( S \) by the outward-pointing normal is the boundary orientation of \( U \) . \( A \) basis \( \overrightarrow{\mathbf{u}},\overrightarrow{\mathbf{v}} \) for \( {T... | Yes |
Proposition 6.6.26 (Oriented boundary of oriented \( k \) -parallelogram). Let the \( k \) -parallelogram \( {P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) \) have the standard orientation (see Example 6.6.19). Then its oriented boundary is given by the follo... | \[ \mathop{\sum }\limits_{{i = 1}}^{k}{\left( -1\right) }^{i - 1}\left( {{P}_{\mathbf{x} + {\overrightarrow{\mathbf{v}}}_{i}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\widehat{\overrightarrow{\mathbf{v}}}}_{i},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) - {P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{... | Yes |
The boundary of \( {P}_{\mathbf{x}}\left( \overrightarrow{\mathbf{v}}\right) \) is | \[ \partial {P}_{\mathbf{x}}\left( \overrightarrow{\mathbf{v}}\right) = {P}_{\mathbf{x} + \overrightarrow{\mathbf{v}}} - {P}_{\mathbf{x}} \] where \( - {P}_{\mathbf{x}} \) is the point \( \mathbf{x} \) with a minus sign and \( {P}_{\mathbf{x} + \overrightarrow{\mathbf{v}}} \) is the point \( \mathbf{x} + \overrightarro... | Yes |
Example 6.6.28 (The boundary of an oriented 2-parallelogram). As shown by Figure 6.6.13, the boundary of an oriented parallelogram is | \[ \underset{\text{boundary }}{\underbrace{\partial {P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},{\overrightarrow{\mathbf{v}}}_{2}}\right) }} = \underset{\text{1st side }}{\underbrace{{P}_{\mathbf{x}}\left( {\overrightarrow{\mathbf{v}}}_{1}\right) }} + \underset{\text{2nd side }}{\underbrace{{P}_{\mathbf{x... | Yes |
Example 6.6.29 (Boundary of a cube). For the faces of a cube shown in Figure 6.6.14 we have \[ \left( {i = 1\text{ so }{\left( -1\right) }^{i - 1} = 1}\right) ;\; + \left( {\underset{\text{right side }}{\underbrace{{P}_{\mathbf{x} + {\overrightarrow{\mathbf{v}}}_{1}}\left( {{\overrightarrow{\mathbf{v}}}_{2},{\overright... | Proof of Proposition 6.6.26. The boundary of \( {P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) \) is composed of its \( {2k} \) faces (four for a parallelogram, six for a cube \( \ldots \) ), each of The vector \( {\overrightarrow{\mathbf{v}}}_{1} \) anchored... | No |
When \( \varphi \) is a 0-form field (i.e., a function), the exterior derivative is just the derivative: | \[ \mathbf{d}f\left( {{P}_{\mathbf{x}}\left( \overrightarrow{\mathbf{v}}\right) }\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}{\int }_{\partial {P}_{\mathbf{x}}\left( {h\overrightarrow{\mathbf{v}}}\right) }f = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {\mathbf{x} + h\overrightarrow{\mathbf... | Yes |
Theorem 6.7.4 (Computing the exterior derivative). Let\n\n\\[ \n\\varphi = \\mathop{\\sum }\\limits_{{1 \\leq {i}_{1} < \\cdots < {i}_{k} \\leq n}}{a}_{{i}_{1},\\ldots ,{i}_{k}}d{x}_{{i}_{1}} \\land \\cdots \\land d{x}_{{i}_{k}}\n\\]\n\nbe a \\( k \\) -form of class \\( {C}^{2} \\) on an open subset \\( U \\subset {\\m... | Part 4 of Theorem 6.7.4 is proved in Example 6.7.3. The remainder of the proof is in Appendix A22. | No |
Example 6.7.5 (Computing the exterior derivative of an elementary 2-form on \( {\mathbb{R}}^{4} \) ). The exterior derivative of \( {x}_{2}{x}_{3}\left( {d{x}_{2} \land d{x}_{4}}\right) \) is | \n\[
\mathbf{d}\left( {{x}_{2}{x}_{3}}\right) \land d{x}_{2} \land d{x}_{4} = \underset{\mathbf{d}\left( {{x}_{2}{x}_{3}}\right) }{\underbrace{\overset{0}{\overbrace{({D}_{1}\left( {{x}_{2}{x}_{3}}\right) }}d{x}_{1} + {D}_{2}\left( {{x}_{2}{x}_{3}}\right) d{x}_{2} + {D}_{3}\left( {{x}_{2}{x}_{3}}\right) d{x}_{3} + \ove... | Yes |
Compute the exterior derivative of the 2-form on \( {\mathbb{R}}^{4} \), \[ \psi = {x}_{1}{x}_{2}d{x}_{2} \land d{x}_{4} - {x}_{2}^{2}d{x}_{3} \land d{x}_{4} \] | \[ \mathbf{d}\psi = \mathbf{d}\left( {{x}_{1}{x}_{2}\;d{x}_{2} \land d{x}_{4}}\right) - \mathbf{d}\left( {{x}_{2}^{2}\;d{x}_{3} \land d{x}_{4}}\right) \] \[ = \left( {{D}_{1}\left( {{x}_{1}{x}_{2}}\right) d{x}_{1} + {D}_{2}\left( {{x}_{1}{x}_{2}}\right) d{x}_{2} + {D}_{3}\left( {{x}_{1}{x}_{2}}\right) d{x}_{3} + {D}_{4... | Yes |
Theorem 6.7.8. For any \( k \) -form \( \varphi \) of class \( {C}^{2} \) on an open subset \( U \subset {\mathbb{R}}^{n} \) , | Proof. This can just be computed. Let us see it first for 0 -forms:\n\n\[ \mathbf{{dd}}f = \mathbf{d}\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{D}_{i}{fd}{x}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\mathbf{d}\left( {{D}_{i}{fd}{x}_{i}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{n}\mathbf{d}{D}_{i}f \l... | Yes |
Theorem 6.7.9 (Exterior derivative of wedge product). If \( \varphi \) is a \( k \) -form and \( \psi \) is an \( l \) -form, then\n\n\[ \mathbf{d}\left( {\varphi \land \psi }\right) = \mathbf{d}\varphi \land \psi + {\left( -1\right) }^{k}\varphi \land \mathbf{d}\psi . \] | 6.7.11 Prove Theorem 6.7.9 concerning the exterior derivative of the wedge product:\n\na. Show it for 0 -forms:\n\n\[ \mathbf{d}\left( {fg}\right) = f\mathbf{d}g + g\mathbf{d}f \]\n\nb. Show that it is enough to prove the theorem when\n\n\[ \varphi = a\left( \mathbf{x}\right) d{x}_{{i}_{1}} \land \cdots \land d{x}_{{i}... | Yes |
Let \( \overrightarrow{F}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left\lbrack \begin{matrix} x + y \\ {x}^{2}{yz} \\ {yz} \end{matrix}\right\rbrack \) . Then | \[ \operatorname{curl}\overrightarrow{F} = \left\lbrack \begin{matrix} {D}_{2}\left( {yz}\right) - {D}_{3}\left( {{x}^{2}{yz}}\right) \\ {D}_{3}\left( {x + y}\right) - {D}_{1}\left( {yz}\right) \\ {D}_{1}\left( {{x}^{2}{yz}}\right) - {D}_{2}\left( {x + y}\right) \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} z... | No |
Theorem 6.8.3 (Exterior derivative of form fields on \( {\mathbb{R}}^{3} \) ). Let \( f \) be a function on \( {\mathbb{R}}^{3} \) and let \( \overrightarrow{F} \) be a vector field. Then\n\n1. \( \;{df} = {W}_{\overrightarrow{\nabla }f} = {W}_{\operatorname{grad}f}\; \) ( \( \mathrm{d}f \) is the work form field of \(... | Proof of Theorem 6.8.3. Exercise 6.8.4 asks you to prove part 3. The following proves parts 1 and 2:\n\n\[ {df} = {D}_{1}f\;{dx} + {D}_{2}f\;{dy} + {D}_{3}f\;{dz} = {W}_{\left\lbrack \begin{matrix} {D}_{1}f \\ {D}_{2}f \\ {D}_{3}f \end{matrix}\right\rbrack } = {W}_{\overrightarrow{\nabla }f} \]\n\n\[ \mathbf{d}{W}_{\ov... | No |
Suppose a building is located near a fault line. In Figure 6.8.3, we will suppose that the fault line is roughly north-south, which we think of as being on the \( y \) -axis, with north positive; the front of the building (the side with the door) is parallel to the fault line. During an earthquake, the lot on which the... | Since \[ \operatorname{curl}\left\lbrack \begin{matrix} 0 \\ - 1/x \\ 0 \end{matrix}\right\rbrack = \left\lbrack \begin{array}{l} {D}_{1} \\ {D}_{2} \\ {D}_{3} \end{array}\right\rbrack \times \left\lbrack \begin{matrix} 0 \\ - 1/x \\ 0 \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} 0 \\ 0 \\ 1/{x}^{2} \end{mat... | Yes |
Proposition 6.9.2 (Computing the pullback by a linear transformation). Let \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) be a linear transformation. Denote by \( {x}_{1},\ldots ,{x}_{n} \) the coordinates in \( {\mathbb{R}}^{n} \) and by \( {y}_{1},\ldots ,{y}_{m} \) the coordinates in \( {\mathbb{R}}^{m} \) ... | \[ {T}^{ * }\left( {d{y}_{{i}_{1}} \land \cdots \land d{y}_{{i}_{k}}}\right) = \mathop{\sum }\limits_{{1 \leq {j}_{1} < \cdots < {j}_{k} \leq n}}{b}_{{j}_{1},\ldots ,{j}_{k}}d{x}_{{j}_{1}} \land \cdots \land d{x}_{{j}_{k}}, \] | Yes |
Example 6.9.3 (Computing the pullback). Let \( T : {\mathbb{R}}^{4} \rightarrow {\mathbb{R}}^{3} \) be the linear transformation given by the matrix \( \left\lbrack T\right\rbrack = \left\lbrack \begin{array}{llll} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right\rbrack \) . Then | Since we are computing the pullback \( {T}^{ * }d{y}_{2} \land d{y}_{3} \), we take the second and third rows of \( T \), and then select out columns 1 and 2 for \( {b}_{1,2} \), columns 1 and 3 for \( {b}_{1,3} \) , and so on.\n\n\[ \n{T}^{ * }\left( {d{y}_{2} \land d{y}_{3}}\right) = {b}_{1,2}d{x}_{1} \land d{x}_{2} ... | Yes |
We will compute \( {\mathbf{f}}^{ * }\left( {{y}_{2}d{y}_{1} \land d{y}_{3}}\right) \) . | Certainly\n\n\[ {f}^{ * }\left( {{y}_{2}\;d{y}_{1} \land d{y}_{3}}\right) = b\;d{x}_{1} \land d{x}_{2} \]\n\nfor some function \( b : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) . The object is to compute that function:\n\n\[ b\left( \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right) = b\left( \begin{array}{l} {x}_{... | Yes |
Proposition 6.9.6 (Pullbacks by compositions). If \( X \subset {\mathbb{R}}^{n} \) , \( Y \subset {\mathbb{R}}^{m} \) and \( Z \subset {\mathbb{R}}^{p} \) are open, \( \mathbf{f} : X \rightarrow Y \) and \( \mathbf{g} : Y \rightarrow Z \) are \( {C}^{1} \) mappings, and \( \varphi \) is a \( k \) -form on \( Z \), then... | Proof. This follows from the chain rule:\n\n\[{\left( \mathbf{g} \circ \mathbf{f}\right) }^{ * }\varphi \left( {{P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) }\right) = \varphi \left( {{P}_{(\mathbf{g}\left( {\mathbf{f}\left( \mathbf{x}\right) }\right) }\lef... | Yes |
Proposition 6.9.7 (Pullback and wedge products). Let \( X \subset {\mathbb{R}}^{n} \) and \( Y \subset {\mathbb{R}}^{m} \) be open, \( \mathbf{f} : X \rightarrow Y \) a \( {C}^{1} \) mapping, and \( \varphi \) and \( \psi \) respectively a \( k \) -form and an \( l \) -form on \( Y \) . Then \[ {\mathbf{f}}^{ * }\varph... | Proof. This is one of those proofs where you write down the definitions and follow your nose. Let us spell it out when \( \mathbf{f} = T \) is linear; we leave the general case as Exercise 6.9.2. Recall that the wedge product is a certain sum over all permutations \( \sigma \) of \( \{ 1,\ldots, k + l\} \) such that \[... | No |
Theorem 6.9.8 (Exterior derivative is intrinsic). Let \( X \subset {\mathbb{R}}^{n} \) , \( Y \subset {\mathbb{R}}^{m} \) be open sets, and let \( \mathbf{f} : X \rightarrow Y \) be a \( {C}^{1} \) mapping. If \( \varphi \) is a \( k \) -form field on \( Y \), then\n\n\[ \n{\mathbf{{df}}}^{ * }\varphi = {\mathbf{f}}^{ ... | Proof. We will prove this theorem by induction on \( k \) . The case \( k = 0 \) , where \( \varphi = g \) is a function, is an application of the chain rule:\n\n\[ \n{f}^{ * }\mathbf{d}g\left( {{P}_{\mathbf{x}}\left( \overrightarrow{\mathbf{v}}\right) }\right) = \mathbf{d}g\left( {{P}_{\mathbf{f}\left( \mathbf{x}\righ... | Yes |
Theorem 6.7.9\n\n\[ d{\mathbf{f}}^{ * }\left( {\psi \land d{x}_{i}}\right) = \mathbf{d}\left( {{\mathbf{f}}^{ * }\psi \land {\mathbf{f}}^{ * }d{x}_{i}}\right) \overbrace{ = }\left( {\mathbf{d}\left( {{\mathbf{f}}^{ * }\psi }\right) }\right) \land {\mathbf{f}}^{ * }d{x}_{i} + {\left( -1\right) }^{k - 1}{\mathbf{f}}^{ * ... | \[ = \left( {\mathbf{d}\left( {{\mathbf{f}}^{ * }\psi }\right) }\right) \land {\mathbf{f}}^{ * }d{x}_{i} + {\left( -1\right) }^{k - 1}{\mathbf{f}}^{ * }\psi \land \mathbf{d}\mathbf{d}{\mathbf{f}}^{ * }{x}_{i} \]\n\n\[ = \iint \left( {\mathbf{d}\left( {{\mathbf{f}}^{ * }\psi }\right) }\right) \land {\mathbf{f}}^{ * }d{x... | Yes |
Let \( S \) be the square described by the inequalities \( \left| x\right| ,\left| y\right| \leq 1 \), with the standard orientation. To compute the integral \( {\int }_{C}{xdy} - {ydx} \), where \( C \) is the boundary of \( S \), with the boundary orientation, one possibility is to parametrize the four sides of the s... | \[ {\int }_{C}{xdy} - {ydx} = {\int }_{S}\mathbf{d}\left( {{xdy} - {ydx}}\right) = {\int }_{S}{2dx} \land {dy} = {\int }_{S}2\left| {dxdy}\right| = 8. \] (The square \( S \) has sidelength 2, so its area is 4.) \( \bigtriangleup \) | Yes |
Let us integrate the 2-form\n\n\\[ \n\\varphi = \\left( {x - {y}^{2} + {z}^{3}}\\right) \\left( {{dy} \\land {dz} + {dx} \\land {dz} + {dx} \\land {dy}}\\right) \n\\]\n\nover the boundary of the cube \\( {C}_{a} \\) given by \\( 0 \\leq x, y, z \\leq a \\) . | It is quite possible to do this directly, parametrizing all six faces of the cube, but Stokes's theorem simplifies things substantially.\n\nComputing the exterior derivative of \\( \\varphi \\) gives\n\n\\[ \n{d\\varphi } = {dx} \\land {dy} \\land {dz} - {2y}\\{dy} \\land {dx} \\land {dz} + 3{z}^{2}\\{dz} \\land {dx} \... | Yes |
Now let's try something similar but harder, integrating\n\n\\[ \n\\varphi = \\left( {{x}_{1} - {x}_{2}^{2} + {x}_{3}^{3} - \\cdots \\pm {x}_{n}^{n}}\\right) \\left( {\\mathop{\\sum }\\limits_{{i = 1}}^{n}d{x}_{1} \\land \\cdots \\land \\widehat{d{x}_{i}} \\land \\cdots \\land d{x}_{n}}\\right) \n\\]\n\nover the boundar... | This time, the idea of computing the integral directly is pretty awesome: parametrizing all \\( {2n} \\) faces of the cube, etc. Doing it using Stokes’s theorem is also pretty awesome, but much more manageable. We know how to compute \\( \\mathbf{d}\\varphi \\), and it comes out to\n\n\\[ \n\\mathbf{d}\\varphi = \\unde... | Yes |
Let \( S \) be an open box without its top: the union of the faces of the cube \( C \) given by \( - 1 \leq x, y, z \leq 1 \) except the top face, oriented by the outward-pointing normal. What is \( {\int }_{S}{\Phi }_{\overrightarrow{F}} \), where \( \overrightarrow{F} = \left\lbrack \begin{array}{l} x \\ y \\ z \end{... | Stokes’s theorem says that the integral of \( {\Phi }_{\overrightarrow{F}} \) over the entire boundary \( \partial C \) is the integral over \( C \) of \( \mathbf{d}{\Phi }_{\overrightarrow{F}} = {M}_{\text{div }\overrightarrow{F}} = {M}_{3} = {3dx} \land {dy} \land {dz} \), so\n\n\[ \n{\int }_{\partial C}{\Phi }_{\ove... | Yes |
Proposition 6.10.7. Let \( M \subset {\mathbb{R}}^{n} \) be a \( k \) -dimensional manifold, and let \( Y \subset M \) be a piece-with-corners. Then every \( \mathbf{x} \in Y \) is the center of a ball \( U \) in \( {\mathbb{R}}^{n} \) such that there exists a diffeomorphism \( \mathbf{F} : U \rightarrow {\mathbb{R}}^{... | Proof. The proof of Proposition 6.10.7 is illustrated by Figure 6.10.4. At any \( \mathbf{x} \in Y \), Definition 6.6.5 of a corner point and Definition 3.1.10 defining a manifold known by equations give us a neighborhood \( V \subset {\mathbb{R}}^{n} \) of \( \mathbf{x} \), and a collection of \( {C}^{1} \) functions ... | No |
Proposition 6.10.8 (Stokes’s theorem for quadrants). Let \( Z \subset {\mathbb{R}}^{k} \) be a closed quadrant; let \( W \) be a bounded open subset of \( {\mathbb{R}}^{k} \), oriented by sgn det on \( {\mathbb{R}}^{k} \). Give \( \partial \left( {W \cap Z}\right) \) the boundary orientation. Let \( \varphi \) be a \( ... | Proof. Choose \( \epsilon > 0 \). Take the dyadic decomposition \( {\mathcal{D}}_{N}\left( {\mathbb{R}}^{k}\right) \), where the sidelength of the cubes is \( h = {2}^{-N} \). By taking \( N \) sufficiently large, we can guarantee that the difference between the integral of \( \mathbf{d}\varphi \) over \( W \cap Z \) a... | Yes |
Theorem 6.10.9 (Stokes’s theorem for pieces-with-corners). Let \( M \subset {\mathbb{R}}^{n} \) be a \( k \) -dimensional manifold, and let \( Y \subset M \) be a piece-with-corners. Let \( \varphi \) be a \( \left( {k - 1}\right) \) -form defined on a neighborhood of \( Y \) . Then\n\n\[ \n{\int }_{{\partial }_{M}Y}\v... | Proof. It is enough to prove \( {\int }_{Y}\mathbf{d}{\varphi }_{i} = {\int }_{{\partial }_{M}Y}{\varphi }_{i} \), since then\n\n\[ \n{\int }_{Y}\mathbf{d}\varphi \underset{\text{eq. 6.10.37 }}{\underbrace{ = }}{\int }_{Y}\mathop{\sum }\limits_{i}^{N}\mathbf{d}{\varphi }_{i} = \mathop{\sum }\limits_{i}^{N}{\int }_{Y}\m... | Yes |
Proposition 6.10.10 (Trimming \( X \) to make a piece-with-corners). For all \( \epsilon > 0 \), there exist points \( {\mathbf{x}}_{1},\ldots ,{\mathbf{x}}_{p} \in X \) and \( {r}_{1} > 0,\ldots ,{r}_{p} > 0 \) , such that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{p}{r}_{i}^{k - 1} < \epsilon \;\text{ and }\;{X}_{\epsilo... | Although Proposition 6.10.10 is almost obvious, its proof is the hardest part of the proof of the generalized Stokes's theorem; we have relegated it to Appendix A23. | Yes |
Proposition 6.10.11. Let\n\n\[ \psi = \mathop{\sum }\limits_{{1 \leq {i}_{1} < \cdots < {i}_{k} \leq n}}{a}_{{i}_{1},\ldots ,{i}_{k}}d{x}_{{i}_{1}} \land \cdots \land d{x}_{{i}_{k}} \]\n\nbe a \( k \) -form on an open set \( U \subset {\mathbb{R}}^{n} \), and let \( M \subset U \) be a \( k \) -dimensional oriented man... | Proof. We need the following lemma from linear algebra; you are asked to prove it in Exercise 6.10.9, which is not all that easy.\n\nLemma 6.10.12. Let \( A | No |
Theorem 6.11.2 (Green’s theorem). Let \( S \) be a bounded region of \( {\mathbb{R}}^{2} \), bounded by a curve \( C \) (or several curves \( {C}_{i} \) ), carrying the boundary orientation as described in Definition 6.6.21. Let \( \overrightarrow{F} \) be a vector field defined on a neighborhood of \( S \) . Then\n\n\... | Suppose \( \overrightarrow{F} = \left( \begin{array}{l} f \\ g \end{array}\right) \) . Then Green’s theorem is traditionally written\n\n\[ \n{\int }_{S}\left( {{D}_{1}g - {D}_{2}f}\right) {dxdy} = {\int }_{C}{fdx} + {gdy}. \n\]\n\nTo see that the two versions are the same, write\n\n\[ \n{W}_{\overrightarrow{F}} = f\lef... | Yes |
What is the integral\n\n\[ \n{\int }_{\partial U}{2xydy} + {x}^{2}{dx} \n\]\n\nwhere \( U \) is the part of the disc of radius \( R \) centered at the origin where \( y \geq 0 \), with the standard orientation? | This corresponds to Green’s theorem, with \( f\left( \begin{array}{l} x \\ y \end{array}\right) = {x}^{2} \) and \( g\left( \begin{array}{l} x \\ y \end{array}\right) = {2xy} \), so that \( {D}_{1}g = {2y} \) and \( {D}_{2}f = 0 \) . Using\n\npolar coordinates (see Proposition 4.10.3) we have\n\n\[ \n{\int }_{\partial ... | Yes |
Theorem 6.11.4 (Stokes’s theorem for surfaces in \( {\mathbb{R}}^{3} \) ). Let \( S \) be an oriented surface in \( {\mathbb{R}}^{3} \), bounded by a curve \( C \) that is given the boundary orientation. Let \( \varphi \) be a 1-form field defined on a neighborhood of \( S \) . Then | \[ {\int }_{S}\mathbf{d}\varphi = {\int }_{C}\varphi \] | Yes |
Theorem 6.11.6 (The divergence theorem). Let \( X \) be a bounded domain in \( {\mathbb{R}}^{3} \) with the standard orientation of space, and let its boundary \( \partial X \) be a union of surfaces \( {S}_{i} \), each oriented by the outward normal. Let \( \varphi \) be a 2-form field defined on a neighborhood of \( ... | Again, let’s make this look a bit more classical. Write \( \varphi = {\Phi }_{\overrightarrow{F}} \), so that \( \mathbf{d}\varphi = \mathbf{d}{\Phi }_{\overrightarrow{F}} = {M}_{\text{div }\overrightarrow{F}} \), and let \( \overrightarrow{N} \) be the unit outward-pointing vector field on the \( {S}_{i} \) ; then equ... | Yes |
Example 6.11.7 (Divergence theorem). Let \( Q \) be the unit cube. What is the flux of the vector field \( \left\lbrack \begin{array}{r} {x}^{2}y \\ - {2yz} \\ {x}^{3}{y}^{2} \end{array}\right\rbrack \) through the boundary of \( Q \) if \( Q \) carries the standard orientation of \( {\mathbb{R}}^{3} \) and the boundar... | Since \( \mathbf{d}{\Phi }_{\overrightarrow{F}} = {M}_{\text{div }\overrightarrow{F}} \), the divergence theorem asserts that\n\n\[ \n{\int }_{\partial Q}{\Phi }_{\left\lbrack \begin{matrix} {x}^{2}y \\ - {2yz} \\ {x}^{3}{y}^{2} \end{matrix}\right\rbrack } = {\int }_{Q}{M}_{\operatorname{div}\left\lbrack \begin{matrix}... | Yes |
Consider an important example from physics: the magnetic field due to a constant current in an infinite straight wire. We computed in equation 6.12.46 the magnetic field due to a constant current \( I \) in a wire along the \( x \) -axis, going in the positive \( x \) direction, and found\n\n\[ \overrightarrow{\mathbf{... | We encountered this vector field (up to a constant multiple) in Example 6.7.7. You were asked there to show that \( \mathbf{d}{W}_{\overrightarrow{\mathbf{B}}} = {\Phi }_{\operatorname{curl}\overrightarrow{\mathbf{B}}} = 0 \) ; here we do it in the language of vector calculus:\n\n\[ \operatorname{curl}\overrightarrow{\... | Yes |
Proposition 6.13.2. If \( U \subset {\mathbb{R}}^{n} \) is open and \( \varphi \in {A}^{k}\left( U\right) ,\psi \in {A}^{k - 1}\left( U\right) \) satisfy \( \varphi = \mathbf{d}\psi \), then for every compact oriented \( k \) -manifold \( M \subset U \) we have \( {\int }_{M}\varphi = 0 \) . | Proof. The boundary of \( M \) is empty. Thus\n\n\[ \n{\int }_{M}\varphi = {\int }_{M}\mathbf{d}\psi = {\int }_{\partial M}\psi = 0 \n\] | Yes |
Proposition 6.13.8 (Boundary orientation of a cone). The oriented boundary of a cone over a \( k \) -parallelogram is given by\n\n\[ \n\partial C{P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k}}\right) = + \underset{\begin{matrix} \text{ base of cone } \\ \text{ over }... | Proof of Proposition 6.13.8. The cone over an oriented \( k \) -parallelogram carries the orientation given by the parametrization \( \gamma \) defined in Definition 6.13.7. Denote that orientation by \( \Omega \) . Then (Definition 6.4.2) the partial derivatives of \( \gamma \) ,\n\n\[ \n\underset{\overrightarrow{{D}_... | No |
Lemma 6.13.10. The limit given in equation 6.13.18 exists, and defines a \( \left( {k - 1}\right) \) -form on \( U \) . | Proof. It is enough to prove the result when\n\n\[ \varphi = {fd}{x}_{{i}_{1}} \land \cdots \land d{x}_{{i}_{k}}\text{, where}f\text{is a}{C}^{1}\text{function.} \]\n\n6.13.19\n\nSince the cone \( C{P}_{\mathbf{x}}\left( {{\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{k - 1}}\right) \) comes p... | No |
We will compute \( \mathbf{c}\varphi \) when \( \varphi = {x}_{3}d{x}_{1} \land d{x}_{3} \) . Since \( \varphi \) is a 2-form on \( {\mathbb{R}}^{3},\mathbf{c}\varphi \) is a 1-form on \( {\mathbb{R}}^{3} \) and can be written\n\n\[ \mathbf{c}\varphi = {fd}{x}_{1} + {gd}{x}_{2} + {hd}{x}_{3} \] | To find the coefficients \( f, g \), and \( h \), we evaluate \( \mathbf{c}\varphi \) on the standard basis vectors (see Theorem 6.1.8). To get \( f \), we evaluate \( \mathbf{c}\varphi \) on \( {P}_{\mathbf{x}}\left( {\overrightarrow{\mathbf{e}}}_{1}\right) \) :\n\n\[ f\left( \mathbf{x}\right) = \mathbf{c}\left( \unde... | No |
Let \( \varphi = {x}_{3}d{x}_{1} \land d{x}_{3} \). Note that \( \mathbf{d}\left( {{x}_{3}d{x}_{1} \land d{x}_{3}}\right) = 0 \). Therefore, by Theorem 6.13.12, we should have\n\n\[ \underset{\mathbf{d}\mathbf{c}\varphi }{\underbrace{\mathbf{d}\left( {\mathbf{c}{x}_{3}d{x}_{1} \land d{x}_{3}}\right) }} = \underset{\var... | \[ \mathbf{d}\left( {\mathbf{c}{x}_{3}d{x}_{1} \land d{x}_{3}}\right) \overset{\text{eq. 6.13.28 }}{\overbrace{ = }}\mathbf{d}\left( {-\frac{{x}_{3}^{2}}{3}d{x}_{1} + \frac{{x}_{1}{x}_{3}}{3}d{x}_{3}}\right) \]\n\n\[ = - \frac{2{x}_{3}}{3}d{x}_{3} \land d{x}_{1} + \frac{{x}_{3}}{3}d{x}_{1} \land d{x}_{3} = {x}_{3}d{x}_... | Yes |
Theorem 1.8.3 (Chain rule). Let \( U \subset {\mathbb{R}}^{n}, V \subset {\mathbb{R}}^{m} \) be open sets, let \( \mathbf{g} : U \rightarrow V \) and \( \mathbf{f} : V \rightarrow {\mathbb{R}}^{p} \) be mappings, and let a be a point of \( U \) . If \( \mathbf{g} \) is differentiable at \( \mathbf{a} \) and \( \mathbf{... | Proof. We will define two \ | No |
Theorem 2.8.13 (Kantorovich’s theorem). Let \( {\mathbf{a}}_{0} \) be a point in \( {\mathbb{R}}^{n} \) , \( U \) an open neighborhood of \( {\mathbf{a}}_{0} \) in \( {\mathbb{R}}^{n} \), and \( \mathbf{f} : U \rightarrow {\mathbb{R}}^{n} \) a differentiable mapping, with its derivative \( \left\lbrack {\mathbf{{Df}}\l... | Proof. The proof is fairly involved, so we will first outline our approach. We prove existence by showing the following four facts:\n\n1. \( \left\lbrack {\mathbf{{Df}}\left( {\mathbf{a}}_{1}\right) }\right\rbrack \) is invertible, allowing us to define \( {\overrightarrow{\mathbf{h}}}_{1} = - {\left\lbrack \mathbf{{Df... | Yes |
Lemma 2.9.5. If the conditions of Theorem 2.9.4 are satisfied, then\n\n\\[ \n\\left| {\\overrightarrow{\\mathbf{h}}}_{i + 1}\\right| \\leq c{\\left| {\\overrightarrow{\\mathbf{h}}}_{i}\\right| }^{2}\\;\\text{ for all }i \n\\] | Proof. Look back at Lemma A5.4 (rewritten for \\( {\\mathbf{a}}_{i} \\) ):\n\n\\[ \n\\left| {\\mathbf{f}\\left( {\\mathbf{a}}_{i}\\right) }\\right| \\leq \\frac{M}{2}{\\left| {\\overrightarrow{\\mathbf{h}}}_{i - 1}\\right| }^{2} \n\\]\n\nA6.2\n\nThe definition \\( {\\overrightarrow{\\mathbf{h}}}_{i} = - {\\left\\lbrack... | Yes |
Theorem 2.10.7 (The inverse function theorem). Let \( W \subset {\mathbb{R}}^{m} \) be an open neighborhood of \( {\mathbf{x}}_{0} \), and let \( \mathbf{f} : W \rightarrow {\mathbb{R}}^{m} \) be a continuously differentiable function. Set \( {\mathbf{y}}_{0} = \mathbf{f}\left( {\mathbf{x}}_{0}\right) \). If the deriva... | To quantify this statement, we will specify the radius \( R \) of a ball \( V \) centered at \( {\mathbf{y}}_{0} \), in which the inverse function is defined. First simplify notation by setting \( L = \left\lbrack {\mathbf{{Df}}\left( {\mathbf{x}}_{0}\right) }\right\rbrack \). Now find \( R > 0 \) satisfying the follow... | Yes |
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