Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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If\n\n\[ A = \left\lbrack \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 0 \\ - 1 & 0 & 1 \end{array}\right\rbrack \text{ and }\left( {{v}^{1},{v}^{2},{v}^{3}}\right) = \left( {2,3, - 1}\right) ,\n\]\n\nfind \( \left( {{\widetilde{v}}^{1},{\widetilde{v}}^{2},{\widetilde{v}}^{3}}\right) \) . | Solution.\n\nUsing the row reduction algorithm, we find that\n\n\[ {A}^{-1} = \left\lbrack \begin{array}{rrr} - \frac{1}{2} & 1 & \frac{1}{2} \\ 1 & - 1 & - 1 \\ - \frac{1}{2} & 1 & \frac{3}{2} \end{array}\right\rbrack \]\n\nHence, from \( {\left( {2.36}\right) }_{2} \),\n\n\[ {\widetilde{v}}^{1} = \left( {-\frac{1}{2}... | Yes |
Using the change of basis defined by the matrix \( A \) in Problem 2.7, compute \( {\widetilde{T}}_{21} \) and \( {\widetilde{T}}_{2}^{\cdot 3} \), where \( \mathbf{T} \) is defined in Problem 2.6. | From \( {\left( {2.38}\right) }_{1,3} \) we have\n\n\[ \n{\widetilde{T}}_{21} = {A}_{2}^{1}\left( {{A}_{1}^{1}{T}_{11} + {A}_{1}^{2}{T}_{12} + {A}_{1}^{3}{T}_{13}}\right) \n\] \n\n\[ \n+ {A}_{2}^{2}\left( {{A}_{1}^{1}{T}_{21} + {A}_{1}^{2}{T}_{22} + {A}_{1}^{3}{T}_{23}}\right) \n\] \n\n\[ \n+ {A}_{2}^{3}\left( {{A}_{1}... | Yes |
Show that in a central force field (f parallel to \( \mathbf{x} \) )\n\n(i) \( \mathbf{x} \) lies in a plane.\n\n(ii) Kepler’s Law holds: \( \mathbf{x} \) sweeps out equal areas in equal times. | (i) If \( \mathbf{f} \) is parallel to \( \mathbf{x} \), then \( \mathbf{x} \times \mathbf{f} = \mathbf{o} \) and (3.7) implies that\n\n\[ \mathbf{x} \times \dot{\mathbf{x}} = \mathbf{c} \]\n\n(3.17)\n\na constant vector. Thus\n\n\[ \mathbf{x} \cdot \mathbf{c} = \mathbf{x} \cdot \left( {\mathbf{x} \times \dot{\mathbf{x... | Yes |
Compute the Christoffel symbols for the \( {u}^{j} \) -coordinate system defined by\n\n\[ \n{x}^{1} = {u}^{1}{u}^{2} \]\n\n\[ \n{x}^{2} = {\left( {u}^{3}\right) }^{2} \]\n\n\[ \n{x}^{3} = {\left( {u}^{1}\right) }^{2} - {\left( {u}^{2}\right) }^{2} \]\n | Solution.\n\nAs mentioned earlier, when dealing with specific coordinate systems, it often simplifies the typography to set \( x = {x}^{1}, y = {x}^{2},\ldots ,{x}_{,1}^{1} = {x}_{u},{\mathbf{g}}_{1} = {\mathbf{g}}_{u} \), etc. Thus, recalling our abc's, we have\n\n(a)\n\[ \n\mathbf{x} = {\widehat{\mathbf{x}}}^{i}\left... | Yes |
Compute the roof and physical components \( {a}^{2} \) and \( {a}^{\left( 2\right) } \) of the acceleration vector in spherical coordinates. | From (3.65), \[ {a}^{2} = {\dot{v}}^{2} + {\left( {v}^{1}\right) }^{2}{\Gamma }_{11}^{2} + 2{v}^{1}{v}^{2}{\Gamma }_{12}^{2} + 2{v}^{1}{v}^{3}{\Gamma }_{13}^{2} + {\left( {v}^{2}\right) }^{2}{\Gamma }_{22}^{2} + 2{v}^{2}{v}^{3}{\Gamma }_{23}^{2} + {\left( {v}^{3}\right) }^{2}{\Gamma }_{33}^{2}. \] From Problem 3.5 we s... | Yes |
If\n\n\\[ f = {xy} + {yz} + {zx} \\]\n\ncompute \\( \\nabla f \\) and \\( \\left| {\nabla f}\\right| \\) at \\( \\left( {{12},5, - 9}\\right) \\) . Compute the corresponding cellar components of \\( \\nabla f \\) in circular cylindrical coordinates. | ## Solution.\n\nNoting that\n\n\\[ {f}_{, x} = y + z,\\;{f}_{, y} = x + z,\\;{f}_{, z} = y + x, \\]\n\nand evaluating these partial derivatives at \\( \\left( {{12},5, - 9}\\right) \\), we find that\n\n\\[ \nabla f \sim \\left( {-4,3,{17}}\\right) ,\\;\\left| {\nabla f}\\right| = \\sqrt{{\\left( -4\\right) }^{2} + {\\l... | Yes |
Find the point(s) on the graph \( y = 1/{x}^{4} \) closest to the origin. | Let \( f = {x}^{4}y - 1 \) . Then \( \nabla f \sim \left( {4{x}^{3}y,{x}^{4}}\right) \) and (4.8) reduces to the two scalar equations\n\n\[ 4{x}^{3}y = {\lambda x}\text{ and }{x}^{4} = {\lambda y}. \]\n\n(Note that these equations constitute a nonlinear eigenvalue problem.) As \( x \neq 0 \) (why?), the first equation ... | Yes |
If \( \mathbf{v} = z{\mathbf{e}}_{x} + {xy}{\mathbf{e}}_{y} + {xyz}{\mathbf{e}}_{z} \), compute \( \mathbf{\nabla } \cdot \mathbf{v},\mathbf{\nabla } \times \mathbf{v} \), and \( \mathbf{\nabla }\mathbf{v} \) . | Solution.\n\n\( {\mathbf{v}}_{, x} = y{\mathbf{e}}_{y} + {yz}{\mathbf{e}}_{z},{\mathbf{v}}_{, y} = x{\mathbf{e}}_{y} + {xz}{\mathbf{e}}_{z},{\mathbf{v}}_{, z} = {\mathbf{e}}_{x} + {xy}{\mathbf{e}}_{z} \), and, in Cartesian coordinates,\n\n\( {\mathbf{g}}^{1} = {\mathbf{e}}_{x},{\mathbf{g}}^{2} = {\mathbf{e}}_{y},{\math... | Yes |
Compute \( \nabla \cdot \mathbf{v} \) in spherical coordinates in terms of the roof and physical components of \( \mathbf{v} \) . | From (3.57), \( J = {\rho }^{2}\sin \phi \) . Hence, with \( \mathbf{v} = {v}^{\rho }{\mathbf{g}}_{\rho } + {v}^{\phi }{\mathbf{g}}_{\phi } + {v}^{\theta }{\mathbf{g}}_{\theta } \), we have, from (4.25), \n\n\[ \mathbf{V} \cdot \mathbf{v} = \frac{1}{{\rho }^{2}\sin \phi }\left\lbrack {\frac{\partial }{\partial \rho }\l... | Yes |
Compute the acceleration of a particle using cylindrical Eulerian coordinates. Express the answer in terms of the physical components of the velocity. | With \( \mathbf{v} = {v}^{k}{\mathbf{g}}_{k} \) and \( \mathbf{a} = {a}^{k}{\mathbf{g}}_{k} \), the component form of (4.36) reads\n\n\[ \n{a}^{k} = {v}^{i}{\nabla }_{i}{v}^{k} + {v}_{, t}^{k} \n\]\n\nwhere \( {\nabla }_{i}{v}^{k} \) is given by (4.17). One can verify easily that the only non-zero Christoffel symbols i... | Yes |
Theorem 1.2 Given the short exact sequence (1.9), there exists a linear map \( S : W \rightarrow V \) such that \( T \circ S = 1 \) . We say that the exact sequence (1.9) splits. | Proof Let \( \left\{ {f}_{i}\right\} \) be a basis of \( W \) . By the surjectivity of \( T \), for each \( i \) there exists an \( {e}_{i} \in V \) such that \( T\left( {e}_{i}\right) = {f}_{i} \) . Let \( S \) be the map \( {f}_{i} \mapsto {e}_{i} \) extended by linearity. The composition of linear maps is linear, so... | Yes |
Theorem 1.3 Let the short exact sequence (1.9) be given, and let \( S \) be a section of \( T \) . Then\n\n\[ V = \ker T \oplus S\left( W\right) \]\n\nIn particular, \( \dim V = \dim \ker T + \dim W \) . | Proof Let \( v \in V \), and define \( w \mathrel{\text{:=}} S\left( {T\left( v\right) }\right) \) and \( u \mathrel{\text{:=}} v - w \) . Then \( T\left( u\right) = \) \( T\left( {v - w}\right) = T\left( v\right) - T\left( {S\left( {T\left( v\right) }\right) }\right) = 0 \), so \( v = u + w \) with \( u \in \ker T \) ... | No |
Theorem 1.4 If \( U \) is a subspace of the vector space \( V \) then \( V/U \) carries a natural vector space structure, the projection map is linear; and \( \dim \left( {V/U}\right) = \dim V - \dim U \) . | Proof Let \( v, w \in V \). Define \( \left\lbrack v\right\rbrack + \left\lbrack w\right\rbrack = \left\lbrack {v + w}\right\rbrack \). We must show that this is well defined, or independent of class representative, meaning that we would get the same answer if we chose different class representatives on the left-hand s... | Yes |
The standard example of a complex vector space (a vector space over the complex numbers) is \( {\mathbb{C}}^{n} \), the Cartesian product of \( \mathbb{C} \) with itself \( n \) times: \( \mathbb{C} \times \cdots \times \mathbb{C} \). A vector in \( {\mathbb{C}}^{n} \) is just an \( n \)-tuple \( \left( {{a}_{1},\ldots... | \[ g\left( {u, v}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}\overline{{u}_{i}}{v}_{i} \] | Yes |
Example 1.3 (The Euclidean inner product or dot product on \( {\mathbb{R}}^{n} \) ) If \( u = \) \( \left( {{u}_{1},{u}_{2},\ldots ,{u}_{n}}\right) \) and \( v = \left( {{v}_{1},{v}_{2},\ldots ,{v}_{n}}\right) \) then we may define | \[ g\left( {u, v}\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}{u}_{i}{v}_{i} \] | Yes |
Theorem 1.5 Every inner product space has an orthonormal basis. | First proof of Theorem 1.5 We use induction on \( k = \dim V \) . If \( \dim V = 1 \) then there must be at least one nonzero vector \( v \in V \) such that \( g\left( {v, v}\right) \neq 0 \) . Otherwise, for all \( v, w \in V \) ,\n\n\[ 0 = g\left( {v + w, v + w}\right) = g\left( {v, v}\right) + {2g}\left( {v, w}\righ... | Yes |
Lemma 1.6 Let \( V \) be a real vector space with basis \( \left\{ {e}_{i}\right\} \), and let \( g \) be a real symmetric bilinear form. Let \( {g}_{ij} \mathrel{\text{:=}} g\left( {{e}_{i},{e}_{j}}\right) \) and set \( \mathbf{G} = \left( {g}_{ij}\right) \) . Then \( g \) is an inner product if and only if the Grammi... | Proof Let \( v = \mathop{\sum }\limits_{i}{a}_{i}{e}_{i} \), and suppose that \( g\left( {v, w}\right) = 0 \) for all \( w \) . Then, for all \( j \) ,\n\n\[ 0 = \mathop{\sum }\limits_{i}{a}_{i}g\left( {{e}_{i},{e}_{j}}\right) = \mathop{\sum }\limits_{i}{a}_{i}{g}_{ij}. \]\n\n(1.47)\n\nIf \( \deg \left( {g}_{ij}\right)... | No |
Given a rigid body consisting of a bunch of point masses \( {m}_{\alpha } \) at positions \( {\mathbf{r}}_{\alpha } = \left( {{x}_{\alpha ,1},{x}_{\alpha ,2},{x}_{\alpha ,3}}\right) \), its inertia tensor is \[ {I}_{ij} = \mathop{\sum }\limits_{\alpha }{m}_{\alpha }\left( {{r}_{\alpha }^{2}{\delta }_{ij} - {x}_{\alpha,... | Of course, this is typical physics sloppiness: \( {I}_{ij} \) is not a tensor, it is the component of a tensor in the standard basis \( \left\{ {{e}_{1},{e}_{2},{e}_{3}}\right\} \) of \( {\mathbb{R}}^{3} \) . Moreover, the indices are in the wrong place. If we were being really pedantic we would write \[ {I}^{ij} = \ma... | No |
Example 2.2 In (2.37) we have stipulated that the components \( {a}^{{i}_{1}\ldots {i}_{p}} \) be totally antisymmetric. Suppose for the moment that we had not insisted on this. Consider what would happen in the case \( p = 2, n = 3 \) . Writing out all the components explicitly gives | \[ \frac{1}{2}\left( {{a}^{11}{e}_{1} \land {e}_{1} + {a}^{12}{e}_{1} \land {e}_{2} + {a}^{21}{e}_{2} \land {e}_{1} + {a}^{22}{e}_{2} \land {e}_{2}}\right. \]\n\[ \left. {+{a}^{13}{e}_{1} \land {e}_{3} + {a}^{31}{e}_{3} \land {e}_{1} + {a}^{23}{e}_{2} \land {e}_{3} + {a}^{32}{e}_{3} \land {e}_{2} + {a}^{33}{e}_{3} \lan... | Yes |
Theorem 2.1 Let \( V \) be \( n \) -dimensional with inner product \( g \) . Let \( \eta ,\lambda \in \mathop{\bigwedge }\limits^{p}V \) , and choose \( \sigma \in \mathop{\bigwedge }\limits^{n}V \) to satisfy \( g\left( {\sigma ,\sigma }\right) = {\left( -1\right) }^{d} \) . Then\n\n\[ \star \star \lambda = {\left( -1... | Proof Choose an orthonormal basis \( \left\{ {e}_{i}\right\} \) for \( V \) . We must have\n\n\[ \sigma = a{e}_{1} \land \cdots \land {e}_{n} \]\n\nfor some constant \( a \), and the hypothesis requires that \( a = \pm 1 \) . For definiteness we choose \( a = 1 \) . (The proof for \( a = - 1 \) is identical.)\n\nWe pro... | No |
Example 2.4 Let \( V = {\mathbb{M}}^{4} \) and let \( {e}_{0} = \left( {1,0,0,0}\right) ,{e}_{1} = \left( {0,1,0,0}\right) ,{e}_{2} = \left( {0,0,1,0}\right) \), and \( {e}_{3} = \left( {0,0,0,1}\right) \). Thus\n\n\[ \left( {g}_{ij}\right) = \left( \begin{array}{rrrr} - 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 ... | for some constants \( {a}^{{i}_{1}{i}_{2}{i}_{3}} \). Equation (2.61) gives\n\n\[ {e}_{0} \land \star {e}_{0} = - g\left( {{e}_{0},{e}_{0}}\right) \sigma = \sigma ,\]\n\n\[ {e}_{1} \land \star {e}_{0} = - g\left( {{e}_{1},{e}_{0}}\right) \sigma = 0,\]\n\n\[ {e}_{2} \land \star {e}_{0} = - g\left( {{e}_{2},{e}_{0}}\righ... | Yes |
Example 3.5 The Cartesian space \( {\mathbb{R}}^{n} \) is an orientable \( n \) -dimensional manifold. | (Surprise.) | No |
We claim that \( B \) is a two-dimensional nonorientable smooth manifold. | Let \( \left\lbrack {x, y}\right\rbrack \) denote the equivalence class of \( \left( {x, y}\right) \) in \( B \) . It’s a bit less confusing if we cover \( B \) with three patches rather than the traditional two. Let’s choose\n\n\[ \n{U}_{1} = \{ \left\lbrack {x, y}\right\rbrack : 0 < x < 2/3, y \in \mathbb{R}\} , \n\]... | Yes |
As a simple illustration of the utility of this perspective, consider again the transformation (3.6) between Cartesian and polar coordinates in the plane. By considering the effect of the gradient operator we see that the vector \( \\partial /\\partial x \) (respectively, \( \\partial /\\partial y \) ) points in the di... | From (3.6) and the chain rule we compute\n\n\[ \n\\frac{\\partial }{\\partial r} = \\frac{\\partial x}{\\partial r}\\frac{\\partial }{\\partial x} + \\frac{\\partial y}{\\partial r}\\frac{\\partial }{\\partial y} = \\cos \\theta \\frac{\\partial }{\\partial x} + \\sin \\theta \\frac{\\partial }{\\partial y}, \n\]\n\n\[... | No |
Theorem 3.3 Let \( {x}^{1},\ldots ,{x}^{n} \) be local coordinates in a neighborhood \( U \) of a point \( p \in M \) . Then \( {T}_{p}M \) is spanned by the tangent vectors \( \partial /\partial {x}^{1},\ldots ,\partial /\partial {x}^{n} \) . In particular, \( \dim {T}_{p}M = n \) . | Proof Without loss of generality we may assume that \( \left( {{x}^{1}\left( p\right) ,\ldots ,{x}^{n}\left( p\right) }\right) = \) \( \left( {0,\ldots ,0}\right) = 0 \) . Let \( {X}_{p} \) be a tangent vector at \( p \), and define \( {X}^{i} \mathrel{\text{:=}} {X}_{p}\left( {x}^{i}\right) \) . As discussed in Sectio... | Yes |
Theorem 3.4 Let \( {x}^{1},\ldots ,{x}^{n} \) be local coordinates in a neighborhood \( U \) of a point \( p \in M \) . Then \( {T}_{p}^{ * }M \) is spanned by the 1 -forms \( d{x}_{p}^{1},\ldots, d{x}_{p}^{n} \) . In particular, \( \dim {T}_{p}^{ * }M = n. \) | \( {\text{Proof }}^{25} \) Let \( f \) be a smooth function on \( M \) . Choose local coordinates \( {x}^{1},\ldots ,{x}^{n} \) in a neighborhood of a point \( p \in M \) and set \( x = \left( {{x}^{1},\ldots ,{x}^{n}}\right) \) as before. Without loss of generality we can choose our coordinates so that \( x\left( p\ri... | Yes |
Theorem 3.5 Let \( \Gamma \left( {TM}\right) \) (respectively, \( \Gamma \left( {{T}^{ * }M}\right) \) ) denote the space of all vector (respectively, covector) fields on \( M.{}^{29} \) Then a tensor field \( \Psi \) of type \( \left( {r, s}\right) \) is a function linear map\n\n---\n\n\[ \Psi : \underset{r\text{ time... | Proof For simplicity we will prove the theorem for the case \( r = 0 \) . The general case is similar. Suppose that \( \widetilde{\Psi } \) satisfies the conditions of the theorem. We claim that the value of \( \widetilde{\Psi } \) at \( p \) depends only on the values of \( {X}_{i}\left( p\right) ,1 \leq i \leq s \) .... | Yes |
Consider again the 2-sphere with the atlas obtained from stereographic projection, as in Example 3.6. Now we do not have global coordinates, so we do not have obvious global forms. Instead, we must define everything locally and then hope that we can patch things together consistently. Typically one does this by first d... | The obvious method turns out to work: we simply define \( {\omega }_{U} = {\omega }_{V} \) on the overlap (so that they trivially agree), reexpress \( {\omega }_{U} \) in terms of the \( u \) coordinates, and define \( {\omega }_{U} \) elsewhere on \( U \) using this expression. The reason why this works is that changi... | Yes |
Example 3.11 Let \( \omega = {Adx} + {Bdy} + {Cdz} \) be a 1-form on \( {\mathbb{R}}^{3} \). Then | \n\[ \n{d\omega } = {dA} \land {dx} + A \land {d}^{2}x \n\] \n\n\[ \n+ {dB} \land {dy} + B \land {d}^{2}y \n\] \n\n\[ \n+ {dC} \land {dz} + C \land {d}^{2}z \n\] \n\n(by (1) and (2)) \n\n\[ \n= {dA} \land {dx} + {dB} \land {dy} + {dC} \land {dz} \n\] \n\n(by (3)) \n\n\[ \n= \left( {\frac{\partial A}{\partial x}{dx} + \... | Yes |
Example 3.12 Let \( \omega = {Ady} \land {dz} + {Bdz} \land {dx} + {Cdx} \land {dy} \) be a 2 -form on \( {\mathbb{R}}^{3} \) . Then | \[ {d\omega } = \left( {\frac{\partial A}{\partial x} + \frac{\partial B}{\partial y} + \frac{\partial C}{\partial z}}\right) {dx} \land {dy} \land {dz}, \] | Yes |
Example 3.15 If \( \alpha \) and \( \beta \) are 1 -forms and \( X \) and \( Y \) are vector fields,\n\n\[ \n\left( {\alpha \land \beta }\right) \left( {X, Y}\right) = {i}_{Y}{i}_{X}\left( {\alpha \land \beta }\right) \n\] | \n\[ \n= {i}_{Y}\left( {{i}_{X}\alpha \land \beta - \alpha \land {i}_{X}\beta }\right) \n\]\n\n\[ \n= {i}_{Y}\left( {\alpha \left( X\right) \beta - \beta \left( X\right) \alpha }\right) \n\]\n\n\[ \n= \alpha \left( X\right) \beta \left( Y\right) - \beta \left( X\right) \alpha \left( Y\right) . \n\] | Yes |
Example 3.16 Let \( \alpha = {6xydx} + {2yzdy} + {x}^{2}{dz} \) be a 1 -form in \( {\mathbb{R}}^{3} \), and let \( \gamma : I \rightarrow {\mathbb{R}}^{3} \) be a parameterized curve given by \( t \mapsto \left( {t,{t}^{2},{t}^{3}}\right) \) . Then | \[ \left( {{\gamma }^{ * }\alpha }\right) \left( t\right) = 6\left( t\right) \left( {t}^{2}\right) \left( {dt}\right) + 2\left( {t}^{2}\right) \left( {t}^{3}\right) \left( {2tdt}\right) + \left( {t}^{2}\right) \left( {3{t}^{2}{dt}}\right) \] \[ = \left( {6{t}^{3} + 3{t}^{4} + 4{t}^{6}}\right) {dt} \] | Yes |
Example 3.17 Let \( \omega = {ydy} \land {dz} + {2ydz} \land {dx} + {xzdx} \land {dy} \) be a 2 -form in \( {\mathbb{R}}^{3} \) and let \( \sigma : I \times I \rightarrow {\mathbb{R}}^{3} \) be a parameterized surface given by \( \left( {u, v}\right) \rightarrow \left( {u\cos v, u\sin v,2{u}^{2}}\right) \) . Then | \[ \left( {{\sigma }^{ * }\omega }\right) \left( {u, v}\right) = \left( {u\sin v}\right) \left( {\sin {vdu} + u\cos {vdv}}\right) \land \left( {4udu}\right) \] \[ + 2\left( {u\sin v}\right) \left( {4udu}\right) \land \left( {\cos {vdu} - u\sin {vdv}}\right) \] \[ + \left( {u\cos v}\right) \left( {2{u}^{2}}\right) \left... | Yes |
Theorem 3.7 The Lie bracket of vector fields is natural with respect to pushfor-ward by a diffeomorphism. That is, for any diffeomorphism \( \varphi : M \rightarrow N \) and vector fields \( X \) and \( Y \) on \( M \) , | Proof It suffices to show that both sides of (3.87) yield the same result when acting on any smooth function \( g : N \rightarrow \mathbb{R} \) . Applying (3.86) twice gives\n\n\[ {YX}\left( {{\varphi }^{ * }g}\right) = Y\left\lbrack {{\varphi }^{ * }\left\lbrack {\left( {{\varphi }_{ * }X}\right) \left( g\right) }\rig... | Yes |
Example 4.1 (The cylinder construction) Let \( D \) be a disk in \( {\mathbb{R}}^{2} \) . Then the cylinder \( I \times D \) is homotopic to \( D \) . | Let \( f : I \times D \rightarrow D \) be given by \( \left( {s, y}\right) \mapsto y \), and let \( g : D \rightarrow I \times D \) be given by \( y \mapsto \left( {0, y}\right) \) . Then \( f \circ g = i{d}_{D} \), while \( g \circ f \sim i{d}_{I \times D} \) via the homotopy \( F\left( {t,\left( {s, y}\right) }\right... | Yes |
Theorem 4.2 Let \( M \) be a smooth manifold. Then\n\n\[ \dim {H}_{\mathrm{{dR}}}^{0}\left( M\right) = \text{the number of connected components of}M\text{.} \] | Proof No 0-form \( f \) on \( M \) is exact (there are no \( \left( {-1}\right) \) -forms!), so every closed 0- form defines its own cohomology class. But \( {df} = 0 \) if and only if \( f \) is constant on each connected component \( {M}_{i} \) of \( M \), and a basis for the vector space of constant functions on the... | Yes |
Theorem 4.3 If \( f, g : M \rightarrow N \) are homotopic maps then \( {f}^{ * } = {g}^{ * } : {H}_{\mathrm{{dR}}}^{ \bullet }\left( N\right) \rightarrow \) \( {H}_{\mathrm{{dR}}}^{ \bullet }\left( M\right) .{}^{6} In other words, homotopic maps induce the same map in cohomology. | Proof If \( f, g : M \rightarrow N \) are homotopic then by (4.4) and (4.5) there is a homotopy \( F : I \times M \rightarrow N \) such that\n\n\[ F \circ {s}_{0} = f \]\n\n\[ F \circ {s}_{1} = g \]\n\nwhere \( {s}_{0},{s}_{1} : N \rightarrow I \times M \) are the inclusion maps into the top and bottom of the cylinder ... | Yes |
Corollary 4.4 (Homotopy invariance of de Rham cohomology) If \( M \) and \( N \) are homotopic then \( {H}_{\mathrm{{dR}}}^{ \bullet }\left( M\right) \cong {H}_{\mathrm{{dR}}}^{ \bullet }\left( N\right) \) . | Proof Now, if \( M \) and \( N \) are homotopic then there exist \( f : M \rightarrow N \) and \( g : N \rightarrow \) \( M \) such that \( f \circ g \sim i{d}_{N} \) and \( g \circ f \sim i{d}_{M} \) . Hence \( {g}^{ * } \circ {f}^{ * } \) equals the identity on \( {H}_{\mathrm{{dR}}}^{ \bullet }\left( N\right) \) and... | Yes |
Example 4.3 The space \( {\mathbb{R}}^{n} \) is contractible, so it has the homotopy type of a point. It follows from Theorem 4.2 and Corollary 4.4 that | \[ {H}_{\mathrm{{dR}}}^{k}\left( {\mathbb{R}}^{n}\right) = \left\{ \begin{array}{ll} \mathbb{R}, & \text{ if }k = 0,\text{ and } \\ 0 & \text{ otherwise. } \end{array}\right. \] | "No" |
To illustrate this, consider the punctured Euclidean space \( X \mathrel{\text{:=}} {\mathbb{R}}^{n + 1} - \{ 0\} \) . The sphere \( {S}^{n} \) is a deformation retract of \( X \) via the map \( r : X \rightarrow {S}^{n} \) given by \( x \mapsto \) \( x/\parallel x\parallel \) . To see this, let \( \iota : {S}^{n} \rig... | \[ F\left( {t, x}\right) = \left( {1 - t}\right) x + {tx}/\parallel x\parallel \] say. Therefore, by Corollary 4.4, \( {r}^{ * } : {H}_{\mathrm{{dR}}}^{ \bullet }\left( {S}^{n}\right) \rightarrow {H}_{\mathrm{{dR}}}^{ \bullet }\left( X\right) \) is an isomorphism. | Yes |
Lemma 4.5 The sequence\n\n\[ 0 \rightarrow {\Omega }^{ * }\left( M\right) \overset{{\varphi }^{ * }}{ \rightarrow }{\Omega }^{ * }\left( U\right) \oplus {\Omega }^{ * }\left( V\right) \overset{{\psi }^{ * }}{ \rightarrow }{\Omega }^{ * }\left( {U \cap V}\right) \rightarrow 0, \]\n\nwhere \( {\psi }^{ * }\left( {\mu, v}... | Proof To show exactness at the first term, \( {\Omega }^{ * }\left( M\right) \), we must show that \( {\varphi }^{ * } \) is injective. But this is clear, because if a form vanishes on both \( U \) and \( V \) then it must vanish on \( U \cup V = M \), so the only thing sent to zero by \( {\varphi }^{ * } \) is the zer... | Yes |
We can use the result of this last example to prove a famous theorem in topology called the Brouwer fixed point theorem. Let \( {B}^{n} \) denote the \( n \) -ball, namely the set \( \left\{ {x \in {\mathbb{R}}^{n} : \parallel x\parallel \leq 1}\right\} \) . The Brouwer fixed point theorem asserts that any continuous m... | We first show by contradiction that there is no retraction \( r : {B}^{n + 1} \rightarrow {S}^{n} \) of the \( n \) -ball onto its boundary. Suppose that there were. Then, by pulling back everything, we would have the composition\n\n\[ \n{H}_{\mathrm{{dR}}}^{n}\left( {S}^{n}\right) \overset{{r}^{ * }}{ \rightarrow }{H}... | Yes |
Example 5.3 We have\n\n\[ \n{\partial }^{2}\left\langle {{p}_{0},{p}_{1},{p}_{2}}\right\rangle = \partial \left\{ {\left\langle {{p}_{1},{p}_{2}}\right\rangle + \left\langle {{p}_{2},{p}_{0}}\right\rangle + \left\langle {{p}_{0},{p}_{1}}\right\rangle }\right\} \n\] | \[ \n= {p}_{2} - {p}_{1} + {p}_{0} - {p}_{2} + {p}_{1} - {p}_{0} \n\]\n\n\[ \n= 0\text{.} \n\] | Yes |
Theorem 5.1 The operator \( {\partial }^{2} = 0 \), i.e., the boundary of a boundary is zero. | Proof The general proof is similar to that of Example 5.3, in that it uses pairwise cancellation. The only tricky thing is keeping track of the summations:\n\n\[ \partial \left\lbrack {\partial \left\langle {{p}_{0},\ldots ,{p}_{m + 1}}\right\rangle }\right\rbrack = \partial \left\lbrack {\mathop{\sum }\limits_{{j = 0}... | Yes |
Example 5.5 Just for kicks, let's do one more example. This time we want to compute the homology of the two-dimensional disk (2-ball) \( {D}^{2} \) . The disk is homeomorphic to the complex \( K \) consisting of a triangle together with its interior. | Proceeding as before, we find \( {\beta }_{0} = 1 \) again. But \( {\beta }_{1} \) is different, for now there is a nontrivial 2-chain, namely \( K \) itself: \( {\bar{c}}_{2} = \left\langle {{p}_{0},{p}_{1},{p}_{2}}\right\rangle \) . It follows that \( {\partial }_{2} \) has rank 1, because it maps \( {C}_{2}\left( K\... | Yes |
Theorem 6.1 For a smooth manifold \( M \) , \[ {H}_{k}^{\infty }\left( M\right) = {H}_{k}\left( M\right) \] | (For this reason, one often omits the superscript \( \infty \) in the smooth setting.) One way to prove this result is to exploit the fact that every smooth manifold admits a smooth triangulation. \( {}^{2} \) More precisely, a smoothly triangulated manifold is a triple \( \left( {M, K,\phi }\right) \), where \( M \) i... | No |
If \( \omega = {3dy} \land {dx} \land {dz} \) and if \( {\bar{s}}^{n} = \left\langle {0,{e}_{1},{e}_{2},\ldots ,{e}_{n}}\right\rangle \) is the standard simplex \( \left( \left\{ {e}_{i}\right\} \right. \) being the standard basis of \( \left. {\mathbb{R}}^{n}\right) \) then | \[ {\int }_{{\bar{s}}^{3}}\omega = - 3{\int }_{{\bar{s}}^{3}}{dxdydz} = - 3{\int }_{0}^{1}{dx}{\int }_{0}^{1 - x}{dy}{\int }_{0}^{1 - x - y}{dz} = - 1/2. \] | Yes |
Example 6.2 The change of variables theorem can also be used to recover the classical expressions for line and surface integrals. Let \( \omega = {A}_{1}d{x}^{1} + {A}_{2}d{x}^{2} + \) \( {A}_{3}d{x}^{3} \) be a 1 -form on \( {\mathbb{R}}^{3} \), and let \( \gamma \) be a curve. In our new terminology, a curve is just ... | We have\n\n\[ \n{\gamma }^{ * }\omega = {\gamma }^{ * }\left( {{A}_{1}d{x}^{1} + {A}_{2}d{x}^{2} + {A}_{3}d{x}^{3}}\right) \n\]\n\n\[ \n= \left( {{\gamma }^{ * }{A}_{1}}\right) \left( {{\gamma }^{ * }d{x}^{1}}\right) + \left( {{\gamma }^{ * }{A}_{2}}\right) \left( {{\gamma }^{ * }d{x}^{2}}\right) + \left( {{\gamma }^{ ... | Yes |
Theorem 6.4 (Stokes’ theorem - chain version) For any \( k \) -form \( \omega \) and \( \left( {k + 1}\right) \) -chain \( c \) , \[ {\int }_{c}{d\omega } = {\int }_{\partial c}\omega \] | Proof By linearity it suffices to prove that \[ {\int }_{\sigma }{d\omega } = {\int }_{\partial \sigma }\omega \] for some \( \left( {k + 1}\right) \) -simplex \( \sigma \) . If \( \sigma = \left( {{\bar{s}}^{k + 1}, U,\phi }\right) \) then this is equivalent to \[ {\int }_{{\bar{s}}^{k + 1}}d\left( {{\phi }^{ * }\omeg... | Yes |
Theorem 6.5 (Stokes' theorem - manifold version) Let \( M \) be an oriented smooth \( n \) -dimensional manifold with boundary, and let \( \omega \) be an \( \left( {n - 1}\right) \) -form with compact support. Then\n\n\[{\int }_{M}{d\omega } = {\int }_{\partial M}\omega\]\n\nwhere \( \partial M \) is given the induced... | Proof Let \( \left\{ {U}_{i}\right\} \) be a locally finite covering of \( M \) by coordinate neighborhoods with subordinate partition of unity \( \left\{ {\rho }_{i}\right\} \) . Clearly \( \omega = \mathop{\sum }\limits_{i}{\rho }_{i}\omega \) . Thus, by the linearity of integration, it suffices to prove Stokes’ theo... | Yes |
Theorem 6.7 (de Rham's second theorem) Let per be a linear map that assigns a real number to every cycle. If per vanishes on all boundaries then there exists a closed form \( \omega \) whose periods are given by per: | \[{\int }_{z}\omega = \operatorname{per}\left( z\right)\] | No |
Theorem 6.8 (de Rham) Let \( M \) be a smooth manifold. Then the de Rham cohomology coincides with the smooth singular cohomology of \( M \) : | \[ {H}_{\mathrm{{dR}}}^{\ell }\left( M\right) \cong {H}^{\ell }\left( M\right) \] | Yes |
Theorem 7.1 A vector bundle \( E \) is trivial if and only if it has a global smooth basis of sections. | Proof Assume \( E \) to be trivial. Then there is a global trivialization map \( \varphi : E \rightarrow \) \( M \times Y \) which is smooth and preserves the linear structure on fibers. Pick a basis \( \left\{ {f}_{i}\right\} \) for \( Y \), and define \( {e}_{i}\left( p\right) \mathrel{\text{:=}} {\varphi }^{-1}\left... | Yes |
Example 7.2 The Möbius band \( B \) that we encountered in Example 3.7 can be viewed as a nontrivial smooth vector bundle over the circle with fiber \( \mathbb{R} \) . The coordinate charts used to define the manifold structure can be modified slightly to provide a local trivialization. Recall that \( B \) was defined ... | Consider the open cover of \( {S}^{1} \) given by\n\n\[ \n{U}_{1} = \{ \left\lbrack x\right\rbrack : 0 < x < 2/3\} \n\] \n\n\[ \n{U}_{2} = \{ \left\lbrack x\right\rbrack : 1/3 < x < 1\} \n\] \n\n\[ \n{U}_{3} = \{ \left\lbrack x\right\rbrack : 2/3 < x < 4/3\} . \n\] \n\nDefine local trivializations \( {\varphi }_{i} : {... | Yes |
The archetypical example of a vector bundle is the tangent bundle \( E = {TM} \) of a manifold \( M \), which consists of the union of all the tangent spaces \( {T}_{p}M \) together with the natural projection \( \pi : {TM} \rightarrow M \) that sends \( {T}_{p}M \mapsto p \) . A point of \( {TM} \) may be thought of a... | If \( \dim M = n,{TM} \) is a \( {2n} \) -dimensional manifold whose bundle transition functions are precisely the Jacobian matrices relating two local charts of \( M \) . To see this, first observe that we may cover \( {TM} \) by the countable collection of open sets \( {\pi }^{-1}\left( {U}_{i}\right) \), where \( \l... | Yes |
Let \( E \rightarrow M \) and \( F \rightarrow M \) be two vector bundles over \( M \) . For simplicity, write \( {E}_{p} \) to denote the fiber over \( p \) . Then the tensor product bundle \( \left( {E \otimes F}\right) \rightarrow M \) is the vector bundle over \( M \) whose fibers are just the tensor products of th... | By iterating this construction we can obtain the tensor product bundle \[ \underset{r\text{ times }}{\underbrace{{TM} \otimes \cdots \otimes {TM}}} \otimes \underset{s\text{ times }}{\underbrace{{T}^{ * }M \otimes \cdots \otimes {T}^{ * }M}} \rightarrow M, \] sections of which are just tensor fields of type \( \left( {... | No |
Theorem 8.1 Every smooth manifold admits a Riemannian metric. | Proof Let \( \left\{ {\rho }_{i}\right\} \) be a partition of unity subordinate to a locally finite open cover \( \left\{ {U}_{i}\right\} \) on \( M \) . Let \( h \) be the usual Euclidean metric on \( {\mathbb{R}}^{n} \) and define \( {g}_{i} = {\varphi }_{i}^{ * }h \) , where \( {\varphi }_{i} \) is the coordinate ma... | Yes |
Theorem 8.2 A frame field \( \left\{ {e}_{a}\right\} \) is locally holonomic (i.e., a coordinate frame field) if and only if\n\n\[ \left\lbrack {{e}_{a},{e}_{b}}\right\rbrack = 0\;\text{ for all }a\text{ and }b. \] | Proof If \( {e}_{a} = {\partial }_{a} \) for some local coordinates \( \left\{ {x}^{a}\right\} \) then\n\n\[ \left\lbrack {{e}_{a},{e}_{b}}\right\rbrack = \left\lbrack {{\partial }_{a},{\partial }_{b}}\right\rbrack = 0, \]\n\n(8.4)\n\nbecause mixed partials commute. Conversely, suppose that we have a frame field \( \le... | Yes |
In a general basis \( \\left\\{ {e}_{a}\\right\\} \) and cobasis \( \\left\\{ {\\theta }^{a}\\right\\} \) we have\n\n\\[ \n{g}_{ab} = g\\left( {{e}_{a},{e}_{b}}\\right) \n\\]\n\n(8.8)\n\nand\n\n\\[ \ng = {g}_{ab}{\\theta }^{a} \\otimes {\\theta }^{b} \n\\]\n\n(8.9) | Note that the nondegeneracy of the inner product is equivalent to the nonsingularity of the matrices whose entries are the components of the metric tensor in the various frames. The frame \( \\left\\{ {e}_{\\widehat{a}}\\right\\} \) is orthonormal if\n\n\\[ \n{g}_{\\widehat{a}\\widehat{b}} = \\pm {\\delta }_{\\widehat{... | No |
Theorem 8.3 In the neighborhood of any point of a geometric manifold \( \left( {M, g}\right) \) there exists an orthonormal frame field. | Proof Pick a point \( p \in M \) and let \( \left\{ {x}^{i}\right\} \) be local coordinates on a neighborhood \( U \) of \( p \) . Then \( \left\{ {\partial }_{i}\right\} \) is a frame field on \( U \) . If \( \left( {M, g}\right) \) is Riemannian then Gram-Schmidt applied to this frame field yields an orthonormal fram... | No |
Lemma 8.5 (Koszul) A connection satisfying (C1)-(C3), (C7), and (C8) obeys the Koszul formula,\n\n\[ \n{2g}\left( {{\nabla }_{X}Y, Z}\right) = {Xg}\left( {Y, Z}\right) + {Yg}\left( {X, Z}\right) - {Zg}\left( {X, Y}\right) \n\]\n\n\[ \n+ g\left( {\left\lbrack {X, Y}\right\rbrack, Z}\right) - g\left( {\left\lbrack {X, Z}... | Proof By metric compatibility\n\n\[ \n{Xg}\left( {Y, Z}\right) = g\left( {{\nabla }_{X}Y, Z}\right) + g\left( {Y,{\nabla }_{X}Z}\right) ,\n\]\n\n\[ \n{Yg}\left( {X, Z}\right) = g\left( {{\nabla }_{Y}X, Z}\right) + g\left( {X,{\nabla }_{Y}Z}\right) ,\n\]\n\n\[ \n{Zg}\left( {X, Y}\right) = g\left( {{\nabla }_{Z}X, Y}\rig... | Yes |
Theorem 8.6 The Christoffel symbols are determined uniquely by the metric tensor and the structure functions. | Proof Let \( X = {e}_{a}, Y = {e}_{b} \), and \( Z = {e}_{c} \) and apply the Koszul formula (8.26) to get\n\n\[ 2g\left( {{\nabla }_{{e}_{a}}{e}_{b},{e}_{c}}\right) = {e}_{a}g\left( {{e}_{b},{e}_{c}}\right) + {e}_{b}g\left( {{e}_{a},{e}_{c}}\right) - {e}_{c}g\left( {{e}_{a},{e}_{b}}\right) + g\left( {\left\lbrack {{e}... | Yes |
Lemma 8.8 Let \( \\left\\{ {e}_{a}\\right\\} \) be a frame field on a neighborhood \( U \) of a point \( p \), and let \( \\omega \) and \( \\Omega \) be the corresponding matrices of connection and curvature forms, respectively. If \( \\Omega = 0 \) everywhere on \( U \), there exists a unique matrix \( A \) of smooth... | Proof Uniqueness is easy. If \( \\omega = {B}^{-1}{dB} \) with \( B\\left( p\\right) = I \) then\n\n\[ \nd\\left( {A{B}^{-1}}\\right) = \\left( {dA}\\right) {B}^{-1} - A{B}^{-1}{dB}{B}^{-1} = \\left( {A\\omega }\\right) {B}^{-1} - A\\left( {\\omega {B}^{-1}}\\right) = 0.\n\]\n\nHence \( A{B}^{-1} = A{B}^{-1}\\left( p\\... | Yes |
Theorem 8.9 \( \widetilde{H}\left( {\nabla ;p}\right) \) is trivial (i.e., it consists only of the identity element) if and only if the curvature vanishes everywhere. | In Appendix E we show that the holonomy around an infinitesimal coordinate loop of parameter area \( {\alpha }^{2} \) centered at a point \( p \) can be written\n\n\[ \vartheta = 1 + {\alpha }^{2}\mathcal{R} + \mathcal{O}\left( {\alpha }^{3}\right) \]\n\n(8.87)\n\nwhere \( \mathcal{R} \) is basically the Riemann curvat... | No |
Theorem 8.10 If \( \eta \in {\Omega }^{k}\left( M\right) \) and \( \lambda \in {\Omega }^{k + 1}\left( M\right) \) then\n\n\[ \left( {{d\eta },\lambda }\right) = \left( {\eta ,{\delta \lambda }}\right) \] | Proof The manifold \( M \) is compact without boundary so, by Stokes’ theorem and the properties of the exterior derivative operator,\n\n\[ 0 = {\int }_{M}d\left( {\eta \land \star \lambda }\right) = {\int }_{M}{d\eta } \land \star \lambda + {\left( -1\right) }^{k}{\int }_{M}\eta \land d \star \lambda . \]\n\nNote that... | Yes |
Theorem 8.11 (Hodge decomposition theorem) Let \( M \) be a compact, orientable, Riemannian manifold without boundary. Any \( k \)-form \( \omega \) on \( M \) can be decomposed uniquely into the sum of a closed form, a co-closed form, and a harmonic form: | Proof Uniqueness is easy. By linearity, it suffices to show that\n\n\[ {d\alpha } + {\delta \beta } + \gamma = 0 \]\n\n(8.98)\n\nimplies that \( {d\alpha } = {\delta \beta } = \gamma = 0 \) . Applying \( d \) to (8.98) (and remembering that \( \gamma \) is harmonic and therefore closed) gives\n\n\[ {d\delta \beta } = 0... | No |
Theorem 8.13 (Poincaré duality) Let \( M \) be a compact, connected, orientable, \( n \) -dimensional manifold without boundary. For every \( k \) ,\n\n\[ \n{H}_{\mathrm{{dR}}}^{n - k}\left( M\right) \cong {H}_{\mathrm{{dR}}}^{k}{\left( M\right) }^{ * }.\n\]\n\n(8.100)\n\nIn particular, \( {\beta }_{n - k} = {\beta }_{... | Proof Let \( \eta \in {\Omega }^{k}\left( M\right) \) and \( \lambda \in {\Omega }^{n - k}\left( M\right) \) be closed. There is a natural pairing between \( {H}_{\mathrm{{dR}}}^{k}\left( M\right) \) and \( {H}_{\mathrm{{dR}}}^{n - k}\left( M\right) \), given by\n\n\[ \n\left( {\left\lbrack \eta \right\rbrack ,\left\lb... | No |
Theorem 9.1 (Stack of records theorem) Let \( f : M \rightarrow N \) be a smooth map between manifolds of the same dimension, with \( M \) compact. If \( q \) is a regular value of \( f \) then \( {f}^{-1}\left( q\right) \) is a finite set of points \( \left\{ {{p}_{1},\ldots ,{p}_{r}}\right\} \) . Moreover, there exis... | Proof Let \( P \mathrel{\text{:=}} {f}^{-1}\left( q\right) \subseteq M \) . By the definition of a regular value together with the inverse function theorem, each point \( p \in P \) has an open neighborhood \( {U}_{p}^{\prime } \) with the property that \( f \) restricted to \( {U}_{p}^{\prime } \) is a diffeomorphism.... | Yes |
Theorem 9.2 Let \( f, M \), and \( N \) be as above. Then\n\n\[ \text{Deg}f = \deg f\text{.} \] | Proof We apply the stack of records theorem (and use the same notation). Assume \( q \) is a regular value of \( f \) . Let \( \omega \in {\Omega }^{n}\left( N\right) \) be an \( n \) -form whose support is contained in \( {V}_{q} \) . Define \( {f}_{i} \mathrel{\text{:=}} {\left. f\right| }_{{U}_{i}} \) . Then, as eac... | Yes |
Theorem 9.3 (The hairy ball theorem) The sphere \( {S}^{n} \) admits a nowhere-zero tangent vector field if and only if \( n \) is odd. | Proof Identify \( x \in {S}^{n} \) with the vector from the origin to \( x \) in \( {\mathbb{R}}^{n + 1} \) . Define the unit normal vector \( N = x/\parallel x\parallel \) . If \( n = {2k} - 1 \) is odd then the vector field \( X = \) \( \left( {{x}^{2}, - {x}^{1},{x}^{4}, - {x}^{3},\ldots ,{x}^{2k}, - {x}^{{2k} - 1}}... | Yes |
Theorem 9.4 (Poincaré-Hopf index theorem) Let \( M \) be a compact manifold without boundary, and let \( X \) be a vector field on \( M \) with only isolated zeros. Then\n\n\[ \mathop{\sum }\limits_{p}\operatorname{ind}\left( {X, p}\right) = \chi \left( M\right) \]\n\nwhere \( p \) ranges over all zeros of \( X \) and ... | We refer the reader to [33] or [58] for a proof. | No |
Theorem 9.5 (Gauss-Bonnet) Let \( M \) be a compact, oriented, two-dimensional Riemannian manifold without boundary. Then\n\n\[{\int }_{M}{KdA} = {2\pi \chi }\left( M\right)\] | The theorem was known to Gauss and was published by Bonnet in 1848. Although it was first proved using more elementary methods, one can prove it, as well as a generalization to \( {2n} \) -dimensional submanifolds of \( {\mathbb{R}}^{{2n} + 1} \), using some of the fancy machinery at our disposal. Very roughly, one emb... | No |
Although it was constructed using an orthonormal basis, the Pfaffian pf \( \Omega \) is globally defined so it cannot depend on the basis. This is made evident by rewriting (9.5) in terms of the Levi-Civita tensor: | \[ \text{ pf }\Omega = \frac{1}{{2}^{n}n!}{\epsilon }^{{i}_{1}{i}_{2}\ldots {i}_{{2n} - 1}{i}_{2n}}{\Omega }_{{i}_{1}{i}_{2}} \land \cdots \land {\Omega }_{{i}_{{2n} - 1}{i}_{2n}}. \] This expression, in turn, can be written in terms of the Riemann tensor using \( {\Omega }_{ab} = \frac{1}{2}{R}_{abcd}{\theta }^{c} \la... | Yes |
Theorem 1.1. The set of \( n \) non-zero vectors \( {\mathbf{x}}_{1},{\mathbf{x}}_{2},\ldots ,{\mathbf{x}}_{n} \) is linearly dependent if and only if some vector \( {\mathbf{x}}_{k}\left( {2 \leq k \leq n}\right) \) is a linear combination of the preceding ones \( {\mathbf{x}}_{i}\left( {i = 1,\ldots, k - 1}\right) \)... | Proof. If the vectors \( {\mathbf{x}}_{1},{\mathbf{x}}_{2},\ldots ,{\mathbf{x}}_{n} \) are linearly dependent, then\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{\mathbf{x}}_{i} = \mathbf{0} \]\n\nwhere not all \( {\alpha }_{i} \) are zero. Let \( {\alpha }_{k}\left( {2 \leq k \leq n}\right) \) be the last non... | No |
Theorem 1.2. All the bases of a finite-dimensional vector space \( \mathbb{V} \) contain the same number of vectors. | Proof. Let \( \mathcal{G} = \left\{ {{\mathbf{g}}_{1},{\mathbf{g}}_{2},\ldots ,{\mathbf{g}}_{n}}\right\} \) and \( \mathcal{F} = \left\{ {{\mathbf{f}}_{1},{\mathbf{f}}_{2},\ldots ,{\mathbf{f}}_{m}}\right\} \) be two arbitrary bases of \( \mathbb{V} \) with different numbers of elements, say \( m > n \) . Then, every ve... | Yes |
Theorem 1.3. Every set \( \mathcal{F} = \left\{ {{\mathbf{f}}_{1},{\mathbf{f}}_{2},\ldots ,{\mathbf{f}}_{n}}\right\} \) of linearly independent vectors in an n-dimensional vectors space \( \mathbb{V} \) forms a basis of \( \mathbb{V} \) . Every set of more than \( n \) vectors is linearly dependent. | Proof. The proof of this theorem is similar to the preceding one. Let \( \mathcal{G} = \) \( \left\{ {{\mathbf{g}}_{1},{\mathbf{g}}_{2},\ldots ,{\mathbf{g}}_{n}}\right\} \) be a basis of \( \mathbb{V} \) . Then, the vectors (1.3) are linearly dependent and non-zero. Excluding a vector \( {\mathbf{g}}_{k} \) we obtain a... | Yes |
Theorem 1.4. Every set \( \mathcal{F} = \left\{ {{\mathbf{f}}_{1},{\mathbf{f}}_{2},\ldots ,{\mathbf{f}}_{m}}\right\} \) of linearly independent vectors in an \( n \) -dimensional vector space \( \mathbb{V} \) can be extended to a basis. | Proof. If \( m = n \), then \( \mathcal{F} \) is already a basis according to Theorem 1.3. If \( m < n \), then we try to find \( n - m \) vectors \( {\mathbf{f}}_{m + 1},{\mathbf{f}}_{m + 2},\ldots ,{\mathbf{f}}_{n} \), such that all the vectors \( {\mathbf{f}}_{i} \), that is, \( {\mathbf{f}}_{1},{\mathbf{f}}_{2},\ld... | Yes |
Theorem 1.5. The representation (1.4) with respect to a given basis \( \mathcal{G} \) is unique. | Proof. Let\n\n\[ \mathbf{x} = \mathop{\sum }\limits_{{i = 1}}^{n}{x}^{i}{\mathbf{g}}_{i}\;\text{ and }\;\mathbf{x} = \mathop{\sum }\limits_{{i = 1}}^{n}{y}^{i}{\mathbf{g}}_{i} \]\n\nbe two different representations of a vector \( \mathbf{x} \), where not all scalar coefficients \( {x}^{i} \) and \( {y}^{i}\left( {i = 1... | "No" |
Theorem 1.6. To every basis in an Euclidean space \( {\mathbb{E}}^{n} \) there exists a unique dual basis. | Relation (1.19) enables to determine the dual basis. However, it can also be obtained without any orthonormal basis. Indeed, let \( {\mathbf{g}}^{i} \) be a basis dual to \( {\mathbf{g}}_{i}\left( {i = 1,2,\ldots, n}\right) \) . Then\n\n\[ \n{\mathbf{g}}^{i} = {g}^{ij}{\mathbf{g}}_{j},\;{\mathbf{g}}_{i} = {g}_{ij}{\mat... | Yes |
Theorem 1.7. Let \( \mathcal{F} = \left\{ {{\mathbf{f}}_{1},{\mathbf{f}}_{2},\ldots ,{\mathbf{f}}_{n}}\right\} \) and \( \mathcal{G} = \left\{ {{\mathbf{g}}_{1},{\mathbf{g}}_{2},\ldots ,{\mathbf{g}}_{n}}\right\} \) be two arbitrary bases of \( {\mathbb{E}}^{n} \) . Then, the tensors \( {\mathbf{f}}_{i} \otimes {\mathbf... | Proof. First, we prove that every tensor in \( {\operatorname{Lin}}^{n} \) represents a linear combination of the tensors \( {\mathbf{f}}_{i} \otimes {\mathbf{g}}_{j}\left( {i, j = 1,2,\ldots, n}\right) \) . Indeed, let \( \mathbf{A} \in {\operatorname{Lin}}^{n} \) be an arbitrary second-order tensor. Consider the foll... | Yes |
Theorem 1.8. A tensor \( \mathbf{A} \) is invertible if and only if \( \mathbf{A}\mathbf{x} = \mathbf{0} \) implies that \( x = 0 \) . | Proof. First we prove the sufficiency. To this end, we map the vector equation \( \mathbf{A}\mathbf{x} = \mathbf{0} \) by \( {\mathbf{A}}^{-1} \) . According to (1.124) it yields: \( \mathbf{0} = {\mathbf{A}}^{-1}\mathbf{A}\mathbf{x} = \mathbf{I}\mathbf{x} = \mathbf{x} \) . To prove the necessity we consider a basis \(... | Yes |
Theorem 2.1. If the transformation of the coordinates \( {y}^{i} = {\widehat{y}}^{i}\left( {{x}^{1},{x}^{2},\ldots ,{x}^{n}}\right) \) admits an inverse form \( {x}^{i} = {\widehat{x}}^{i}\left( {{y}^{1},{y}^{2},\ldots ,{y}^{n}}\right) \left( {i = 1,2,\ldots, n}\right) \) and if \( J \) and \( K \) are the Jacobians of... | \[ J = \left| \frac{\partial {y}^{i}}{\partial {x}^{k}}\right| \neq 0 \] | "No" |
Let us consider three linearly independent vectors \( {\mathbf{x}}_{i}\left( {i = 0,1,2}\right) \) specifying three points in three-dimensional space. The plane going through these points can be defined by | \[ \mathbf{r}\left( {{t}^{1},{t}^{2}}\right) = {\mathbf{x}}_{0} + {t}^{1}\left( {{\mathbf{x}}_{1} - {\mathbf{x}}_{0}}\right) + {t}^{2}\left( {{\mathbf{x}}_{2} - {\mathbf{x}}_{0}}\right) . \] | Yes |
Example 2. Cylinder. A cylinder of radius \( R \) with the axis parallel to \( {\mathbf{e}}_{3} \) is defined by\n\n\[ \mathbf{r}\left( {{t}^{1},{t}^{2}}\right) = R\cos {t}^{1}{\mathbf{e}}_{1} + R\sin {t}^{1}{\mathbf{e}}_{2} + {t}^{2}{\mathbf{e}}_{3}, \] | where \( {\mathbf{e}}_{i}\left( {i = 1,2,3}\right) \) again form an orthonormal basis in \( {\mathbb{E}}^{3} \) . With the aid of the cylindrical coordinates (2.16) we can alternatively write\n\n\[ \varphi = {t}^{1},\;z = {t}^{2},\;r = R. \] | Yes |
A sphere of radius \( R \) with the center at \( \mathbf{r} = \mathbf{0} \) is defined by\n\n\[ \mathbf{r}\left( {{t}^{1},{t}^{2}}\right) = R\sin {t}^{1}\sin {t}^{2}{\mathbf{e}}_{1} + R\cos {t}^{2}{\mathbf{e}}_{2} + R\cos {t}^{1}\sin {t}^{2}{\mathbf{e}}_{3}, \] | Using a parametric representation (see, e.g., [25])\n\n\[ {t}^{1} = {t}^{1}\left( t\right) ,\;{t}^{2} = {t}^{2}\left( t\right) \]\n\none defines a curve on the surface (3.52). The vector tangent to the curve (3.60) can be expressed by\n\n\[ {\mathbf{g}}_{t} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \frac{\partial \m... | Yes |
Theorem 4.1. Let \( \lambda \) be an eigenvalue of the tensor \( \mathbf{A} \) and let \( g\left( \mathbf{A}\right) = \) \( \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}{\mathbf{A}}^{k} \) be a polynomial of \( \mathbf{A} \) . Then \( g\left( \lambda \right) = \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}{\lambda }^{k} \) is... | Proof. Let \( \mathbf{a} \) be an eigenvector of \( \mathbf{A} \) associated with \( \lambda \) . Then, in view of (4.13)\n\n\[ g\left( \mathbf{A}\right) \mathbf{a} = \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}{\mathbf{A}}^{k}\mathbf{a} = \mathop{\sum }\limits_{{k = 0}}^{m}{a}_{k}{\lambda }^{k}\mathbf{a} = \left( {\math... | Yes |
Theorem 4.2. The eigenvectors of a second-order tensor corresponding to pairwise distinct eigenvalues are linearly independent. | Proof. Suppose that these eigenvectors are linearly dependent. Among all possible nontrivial linear relations connecting them we can choose one involving the minimal number, say \( r \), of eigenvectors \( {\mathbf{a}}_{i} \neq \mathbf{0}\left( {i = 1,2,\ldots, r}\right) \) . Obviously, \( 1 < r \leq n \) . Thus,\n\n\[... | Yes |
Theorem 4.3. Let \( {\mathbf{b}}_{i} \) be a left and \( {\mathbf{a}}_{j} \) a right eigenvector associated with distinct eigenvalues \( {\lambda }_{i} \neq {\lambda }_{j} \) of a tensor \( \mathbf{A} \) . Then,\n\n\[ \n{\mathbf{b}}_{i} \cdot {\mathbf{a}}_{j} = 0 \n\] | Proof. With the aid of (1.73) and taking (4.11) into account we can write\n\n\[ \n{\mathbf{b}}_{i}\mathbf{A}{\mathbf{a}}_{j} = {\mathbf{b}}_{i} \cdot \left( {\mathbf{A}{\mathbf{a}}_{j}}\right) = {\mathbf{b}}_{i} \cdot \left( {{\lambda }_{j}{\mathbf{a}}_{j}}\right) = {\lambda }_{j}{\mathbf{b}}_{i} \cdot {\mathbf{a}}_{j}... | Yes |
Theorem 4.4. The eigenvalues of a symmetric second-order tensor \( \mathbf{M} \in \) \( {\operatorname{Sym}}^{n} \) are real, the eigenvectors belong to \( {\mathbb{E}}^{n} \) . | Proof. Let \( \lambda \) be an eigenvalue of \( \mathbf{M} \) and \( \mathbf{a} \) a corresponding eigenvector such that according to (4.11)\n\n\[ \mathbf{M}\mathbf{a} = \lambda \mathbf{a}. \]\n\nThe complex conjugate counterpart of this equation is\n\n\[ \overline{\mathbf{M}}\text{ 和 }\overline{\mathbf{a}} = \overline... | Yes |
Theorem 4.5. Eigenvectors of a symmetric second-order tensor corresponding to distinct eigenvalues are mutually orthogonal. | Proof. According to Theorem 4.3 right and left eigenvectors associated with distinct eigenvalues are mutually orthogonal. However, for a symmetric tensor every right eigenvector represents the left eigenvector associated with the same eigenvalue and vice versa. For this reason, right (left) eigenvectors associated with... | Yes |
Theorem 5.1. Let \( \mathcal{F} = \left\{ {{\mathbf{F}}_{1},{\mathbf{F}}_{2},\ldots ,{\mathbf{F}}_{{n}^{2}}}\right\} \) and \( \mathcal{G} = \left\{ {{\mathbf{G}}_{1},{\mathbf{G}}_{2},\ldots ,{\mathbf{G}}_{{n}^{2}}}\right\} \) be two arbitrary (not necessarily distinct) bases of \( \mathbf{L}{\text{in}}^{n} \) . Then, ... | ## Proof. See the proof of Theorem 1.6. | No |
Theorem 5.2. Let \( \mathcal{E} = \left\{ {{e}_{1},{e}_{2},\ldots ,{e}_{n}}\right\} ,\mathcal{F} = \left\{ {{\mathbf{f}}_{1},{\mathbf{f}}_{2},\ldots ,{\mathbf{f}}_{n}}\right\} ,\mathcal{G} = \left\{ {{\mathbf{g}}_{1},{\mathbf{g}}_{2},\ldots ,{\mathbf{g}}_{n}}\right\} \) and finally \( \mathcal{H} = \left\{ {{\mathbf{h}... | Proof. In view of (5.23)\n\n\[{\mathbf{e}}_{i} \otimes {\mathbf{f}}_{j} \otimes {\mathbf{g}}_{k} \otimes {\mathbf{h}}_{l} = \left( {{\mathbf{e}}_{i} \otimes {\mathbf{h}}_{l}}\right) \odot \left( {{\mathbf{f}}_{j} \otimes {\mathbf{g}}_{k}}\right) .\]\n\nAccording to Theorem 1.6 the second-order tensors \( {\mathbf{e}}_{... | Yes |
Functional basis of one skew-symmetric second-order tensor \( \mathbf{W} \in {\operatorname{Skew}}^{3} \) . | With the aid of (6.15) and (4.87) we obtain the basis consisting of only one invariant\n\n\[ \operatorname{tr}{\mathbf{W}}^{2} = - 2{\Pi }_{\mathbf{W}} = - \parallel \mathbf{W}{\parallel }^{2} \]\n\n(6.16) | No |
Example 2. Functional basis of an arbitrary second-order tensor \( \mathbf{A} \in \) \( {\operatorname{Lin}}^{3} \) . By means of (6.15) one can write the following functional basis of \( \mathbf{A} \) | \[ \operatorname{tr}\mathbf{M},\operatorname{tr}{\mathbf{M}}^{2},\operatorname{tr}{\mathbf{M}}^{3}, \]\n\[ \operatorname{tr}{\mathbf{W}}^{2},\operatorname{tr}\left( {\mathbf{M}{\mathbf{W}}^{2}}\right) ,\operatorname{tr}\left( {{\mathbf{M}}^{2}{\mathbf{W}}^{2}}\right) ,\operatorname{tr}\left( {{\mathbf{M}}^{2}{\mathbf{W... | No |
Example 3. Functional basis of two symmetric second-order tensors \( {\mathbf{M}}_{1},{\mathbf{M}}_{2} \in {\operatorname{Sym}}^{3} \). According to (6.15) the functional basis includes in this case the following ten invariants | \[ \operatorname{tr}{\mathbf{M}}_{1},\operatorname{tr}{\mathbf{M}}_{1}^{2},\operatorname{tr}{\mathbf{M}}_{1}^{3},\operatorname{tr}{\mathbf{M}}_{2},\operatorname{tr}{\mathbf{M}}_{2}^{2},\operatorname{tr}{\mathbf{M}}_{2}^{3}, \] \[ \operatorname{tr}\left( {{\mathbf{M}}_{1}{\mathbf{M}}_{2}}\right) ,\operatorname{tr}\left(... | Yes |
Derivative of the quadratic norm \( \parallel \mathbf{A}\parallel = \sqrt{\mathbf{A} : \mathbf{A}} \) : | \n\[{\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left\lbrack \left( \mathbf{A} + t\mathbf{X}\right) : \left( \mathbf{A} + t\mathbf{X}\right) \right\rbrack }^{1/2}\right| }_{t = 0}\]\n\n\[= {\left. \frac{\mathrm{d}}{\mathrm{d}t}{\left\lbrack \mathbf{A} : \mathbf{A} + 2t\mathbf{A} : \mathbf{X} + {t}^{2}\mathbf{X} : \mathbf{X}... | Yes |
Derivatives of the principal traces \( \operatorname{tr}{\mathbf{A}}^{k}\left( {k = 1,2,\ldots }\right) \) | \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}\left\lbrack \operatorname{tr}{\left( \mathbf{A} + t\mathbf{X}\right) }^{k}\right\rbrack \right| }_{t = 0} = {\left. \frac{\mathrm{d}}{\mathrm{d}t}\left\lbrack {\left( \mathbf{A} + t\mathbf{X}\right) }^{k} : \mathbf{I}\right\rbrack \right| }_{t = 0} = {\left. \frac{\mathrm{d}}{\... | Yes |
Derivatives of \( \operatorname{tr}\left( {{\mathbf{A}}^{k}\mathbf{L}}\right) \left( {k = 1,2,\ldots }\right) \) with respect to \( \mathbf{A} \) | \[ \operatorname{tr}\left( {{\mathbf{A}}^{k}\mathbf{L}}\right) ,\mathbf{A} = \mathop{\sum }\limits_{{i = 0}}^{{k - 1}}{\left( {\mathbf{A}}^{i}\mathbf{L}{\mathbf{A}}^{k - 1 - i}\right) }^{\mathrm{T}}. \] | Yes |
Derivatives of the principal invariants \( {\mathrm{I}}_{\mathbf{A}}^{\left( k\right) }(k = 1,2 \) , \( \ldots, n) \) of a second-order tensor \( \mathbf{A} \in {\operatorname{Lin}}^{n} \) | \[ {\mathrm{I}}_{\mathbf{A}}^{\left( 1\right) },\mathbf{A} = \left( {\operatorname{tr}\mathbf{A}}\right) ,\mathbf{A} = \mathbf{I} \] \[ {\mathrm{I}}_{\mathbf{A}}^{\left( 2\right) },{}_{\mathbf{A}} = \frac{1}{2}\left( {{\mathrm{I}}_{\mathbf{A}}^{\left( 1\right) }\operatorname{tr}\mathbf{A} - \operatorname{tr}{\mathbf{A}... | Yes |
First, we show that simple eigenvalues of a second-order tensor \( \mathbf{A} \) are differentiable. To this end, we consider the directional derivative (6.44) of an eigenvalue \( \lambda \) : | \[ {\left. \frac{\mathrm{d}}{\mathrm{d}t}\lambda \left( \mathbf{A} + t\mathbf{X}\right) \right| }_{t = 0}. \]\n\nHerein, \( \lambda \left( t\right) \) represents an implicit function defined through the characteristic equation\n\n\[ \det \left( {\mathbf{A} + t\mathbf{X} - \lambda \mathbf{I}}\right) = p\left( {\lambda, ... | Yes |
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