<The previous article in this series | The table of contents of this series |
description/proof of that double exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary is \(0\)
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the double exterior derivative of any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary is \(0\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(t\): \(: M \to \Lambda_q (T M)\), \(\in \Omega_q (T M)\)
//
Statements:
\(d d t = 0\)
//
2: Proof
Whole Strategy: Step 1: around each \(m \in M\), take a chart, \((U_m \subseteq M, \phi_m)\), and express \(t\) as \(\sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\); Step 2: compute \(d d t\) and see that it is \(0\).
Step 1:
Let \(m \in M\) be any.
Let us take a chart, \((U_m \subseteq M, \phi_m)\).
With respect to the chart, \(t = \sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\), by the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
Step 2:
\(d t = \sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = \sum_{j_1 \lt ... \lt j_q} \partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\).
\(d d t = d (\sum_{j_1 \lt ... \lt j_q} \partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \sum_{j_1 \lt ... \lt j_q} d (\partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})\), because \(d\) is \(\mathbb{R}\)-linear.
For each fixed \((j_1, ..., j_q)\), the terms for \(l \in \{j_1, ..., j_q\}\) are \(0\), by a property mentioned in Note for the definition of wedge product of multicovectors.
For each fixed \((j_1, ..., j_q)\), the term for each \(l \notin \{j_1, ..., j_q\}\) is \(d (\partial t_{j_1, ..., j_q} / \partial x^l d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = \partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})\), because while the definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary requires \(d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\) in the increasing order, reordering it in the increasing order yields a sign, but reordering the result back in the original order eliminate the whatever sign.
For each fixed \((j_1, ..., j_q)\), the terms for \(l = m\) are \(0\), as before, and for each \((l, m)\) pair such that \(l \neq m\), there is the corresponding \((m, l)\) pair, and \(\partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) + \partial (\partial t_{j_1, ..., j_q} / \partial x^m) / \partial x^l d x^l \wedge d x^m \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}) = 0\), because while \(\partial (\partial t_{j_1, ..., j_q} / \partial x^l) / \partial x^m = \partial (\partial t_{j_1, ..., j_q} / \partial x^m) / \partial x^l\), \(d x^m \wedge d x^l \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = - d x^l \wedge d x^m \wedge d x^{j_1} \wedge ... \wedge d x^{j_q}\), by a property mentioned in Note for the definition of wedge product of multicovectors.
So, \(d d t = 0\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that exterior derivative of wedge product of \(C^\infty\) \(q_1\)-form and \(C^\infty\) \(q_2\)-form over \(C^\infty\) manifold with boundary is wedge product of exterior derivative of \(q_1\)-form and \(q_2\)-form plus \(-1\) to power \(q_1\) wedge product of \(q_1\)-form and exterior derivative of \(q_2\)-form
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the exterior derivative of the wedge product of any \(C^\infty\) \(q_1\)-form and any \(C^\infty\) \(q_2\)-form over any \(C^\infty\) manifold with boundary is the wedge product of the exterior derivative of the \(q_1\)-form and the \(q_2\)-form plus \(-1\) to power \(q_1\) the wedge product of the \(q_1\)-form and the exterior derivative of the \(q_2\)-form.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(t_1\): \(: M \to \Lambda_{q_1} (T M)\), \(\in \Omega_{q_1} (T M)\)
\(t_2\): \(: M \to \Lambda_{q_2} (T M)\), \(\in \Omega_{q_2} (T M)\)
//
Statements:
\(d (t_1 \wedge t_2) = (d t_1) \wedge t_2 + (-1)^{q_1} t_1 \wedge (d t_2)\)
//
2: Proof
Whole Strategy: Step 1: around each \(m \in M\), take a chart, \((U_m \subseteq M, \phi_m)\), and express \(t_s\) as \(\sum_{j_1 \lt ... \lt j_{q_s}} t_{s, j_1, ..., j_{q_s}} d x^{j_1} \wedge ... \wedge d x^{j_{q_s}}\); Step 2: compute \(d (t_1 \wedge t_2)\) and see that it is \((d t_1) \wedge t_2 + (-1)^{q_1} t_1 \wedge (d t_2)\).
Step 1:
Let \(m \in M\) be any.
Let us take a chart, \((U_m \subseteq M, \phi_m)\).
With respect to the chart, \(t_s = \sum_{j_1 \lt ... \lt j_{q_s}} t_{s, j_1, ..., j_{q_s}} d x^{j_1} \wedge ... \wedge d x^{j_{q_s}}\), by the proposition that the \(q\)-covectors space of any vectors space has the basis that consists of the wedge products of the increasing elements of the dual basis of the vectors space.
Step 2:
\(d (t_1 \wedge t_2) = d (\sum_{j_1 \lt ... \lt j_{q_1}} t_{1, j_1, ..., j_{q_1}} d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge \sum_{l_1 \lt ... \lt l_{q_2}} t_{2, l_1, ..., l_{q_2}} d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}) = d (\sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} t_{1, j_1, ..., j_{q_1}} t_{2, l_1, ..., l_{q_2}} d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}) = \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} d (t_{1, j_1, ..., j_{q_1}} t_{2, l_1, ..., l_{q_2}} d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}) = \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} d (t_{1, j_1, ..., j_{q_1}} t_{2, l_1, ..., l_{q_2}}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}) = \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} (t_{2, l_1, ..., l_{q_2}} d t_{1, j_1, ..., j_{q_1}} + t_{1, j_1, ..., j_{q_1}} d t_{2, l_1, ..., l_{q_2}}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}\), by the proposition that the exterior derivation of \(C^\infty\) function over \(C^\infty\) manifold with boundary satisfies the Leibniz rule.
\(= \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} t_{2, l_1, ..., l_{q_2}} d t_{1, j_1, ..., j_{q_1}} \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}} + \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} t_{1, j_1, ..., j_{q_1}} d t_{2, l_1, ..., l_{q_2}} \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}} = \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} d t_{1, j_1, ..., j_{q_1}} \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge t_{2, l_1, ..., l_{q_2}} d x^{l_1} \wedge ... \wedge d x^{l_{q_2}} + (-1)^{q_1} \sum_{j_1 \lt ... \lt j_{q_1}} \sum_{l_1 \lt ... \lt l_{q_2}} t_{1, j_1, ..., j_{q_1}} d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge d t_{2, l_1, ..., l_{q_2}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}}\), because moving \(d t_{2, l_1, ..., l_{q_2}}\) there is 1st switching it with \(d x^{j_1}\), then, switching it with \(d x^{j_2}\), ..., and then, switching it with \(d x^{j_{q_1}}\).
\(= \sum_{j_1 \lt ... \lt j_{q_1}} d t_{1, j_1, ..., j_{q_1}} \wedge d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge \sum_{l_1 \lt ... \lt l_{q_2}} t_{2, l_1, ..., l_{q_2}} d x^{l_1} \wedge ... \wedge d x^{l_{q_2}} + (-1)^{q_1} \sum_{j_1 \lt ... \lt j_{q_1}} t_{1, j_1, ..., j_{q_1}} d x^{j_1} \wedge ... \wedge d x^{j_{q_1}} \wedge \sum_{l_1 \lt ... \lt l_{q_2}} d t_{2, l_1, ..., l_{q_2}} \wedge d x^{l_1} \wedge ... \wedge d x^{l_{q_2}} = (d t_1) \wedge t_2 + (-1)^{q_1} t_2 \wedge (d t_2)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of exterior derivative of \(C^\infty\) \(q\)-form over \(C^\infty\) manifold with boundary.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\( t\): \(: M \to \Lambda_q (T M)\), \(\in \Omega_q (T M)\)
\(*d t\): \(: M \to \Lambda_{q + 1} (T M)\), \(\in \Omega_{q + 1} (T M)\)
//
Conditions:
\(\forall (U \subseteq M, \phi) \in \{\text{ the charts for } M\} \text{ such that } t = \sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q} (d t = \sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q})\)
//
2: Note
Let us see that it is well-defined.
Let us see that the formula does not depend on the choice of chart.
Let \((U' \subseteq M, \phi')\) be any other chart such that \(U \cap U' \neq \emptyset\).
On \(U'\), \(t = \sum_{j_1 \lt ... \lt j_q} t'^{j_1, ..., j_q} d x'^{j_1} \wedge ... \wedge d x'^{j_q}\).
On \(U \cap U'\), \(t = \sum_{j_1 \lt ... \lt j_q} t_{j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q} = \sum_{j_1 \lt ... \lt j_q} t'^{j_1, ..., j_q} d x'^{j_1} \wedge ... \wedge d x'^{j_q}\), but \(\sum_{j_1 \lt ... \lt j_q} d t_{j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = \sum_{j_1 \lt ... \lt j_q} d t'_{j_1, ..., j_q} \wedge d x'^{j_1} \wedge ... \wedge d x'^{j_q}\), by the proposition that for any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary that is expressed as multiplication of function by wedge product of \(q\) exterior derivatives of functions in any 2 ways, the expressions with the heading functions replaced by its exterior derivatives (and the multiplications replaced by wedge products) represent the same object.
So, \(d t\) does not depend on the choice of chart.
\(d\) is \(\mathbb{R}\)-linear, which means that \(d (r_1 t_1 + r_2 t_2) = r_1 d t_1 + r_2 d t_2\) for each \(r_1, r_2 \in \mathbb{R}\), because \(t_l = \sum_{j_1 \lt ... \lt j_q} t_{l, j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\), \(d (r_1 t_1 + r_2 t_2) = d (r_1 \sum_{j_1 \lt ... \lt j_q} t_{1, j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q} + r_2 \sum_{j_1 \lt ... \lt j_q} t_{2, j_1, ..., j_q} d x^{j_1} \wedge ... \wedge d x^{j_q}) = d (\sum_{j_1 \lt ... \lt j_q} (r_1 t_{1, j_1, ..., j_q} + r_2 t_{2, j_1, ..., j_q}) d x^{j_1} \wedge ... \wedge d x^{j_q}) = \sum_{j_1 \lt ... \lt j_q} d (r_1 t_{1, j_1, ..., j_q} + r_2 t_{2, j_1, ..., j_q}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = \sum_{j_1 \lt ... \lt j_q} (r_1 d t_{1, j_1, ..., j_q} + r_2 d t_{2, j_1, ..., j_q}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = r_1 \sum_{j_1 \lt ... \lt j_q} d t_{1, j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} + r_2 \sum_{j_1 \lt ... \lt j_q} d t_{2, j_1, ..., j_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = r_1 d t_1 + r_2 d t_2\).
In fact, although the definition requires \(t\) to be expressed with the increasing orders terms with \(j_1 \lt ... \lt j_q\), even when \(t = \sum_{(l_1, ... l_q)} t_{l_1, ..., l_q} d x^{l_1} \wedge ... \wedge d x^{l_q}\) with any orders terms, \(d t = \sum_{(l_1, ... l_q)} d t_{l_1, ..., l_q} \wedge d x^{l_1} \wedge ... \wedge d x^{l_q}\), because while as \(d\) is \(\mathbb{R}\)-linear, \(d t = d (\sum_{(l_1, ... l_q)} t_{l_1, ..., l_q} d x^{l_1} \wedge ... \wedge d x^{l_q}) = \sum_{(l_1, ... l_q)} d (t_{l_1, ..., l_q} d x^{l_1} \wedge ... \wedge d x^{l_q})\), as \(t_{l_1, ..., l_q} d x^{l_1} \wedge ... \wedge d x^{l_q} = sgn \sigma t_{l_1, ..., l_q} d x^{j_1} \wedge ... \wedge d x^{j_q}\) where \(j_1 \lt ... \lt j_q\) and \(\sigma\) is the permutation that permutates \((l_1, ..., l_q)\) to \((j_1, ..., j_q)\), \(d (t_{l_1, ..., l_q} d x^{l_1} \wedge ... \wedge d x^{l_q}) = d (sgn \sigma t_{l_1, ..., l_q} d x^{j_1} \wedge ... \wedge d x^{j_q}) = d (sgn \sigma t_{l_1, ..., l_q}) \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = sgn \sigma d t_{l_1, ..., l_q} \wedge d x^{j_1} \wedge ... \wedge d x^{j_q} = d t_{l_1, ..., l_q} \wedge d x^{l_1} \wedge ... \wedge d x^{l_q}\).
In fact, when \(t = \sum_{j \in J} f_{j, 0} d f_{j, 1} \wedge ... \wedge d f_{j, q}\) for any index set, \(J\), and any \(C^\infty\) functions, \(\{f_{j, l} \vert j \in J, l \in \{0, ... q\}\}\), \(d t = \sum_{j \in J} d f_{j, 0} \wedge d f_{j, 1} \wedge ... \wedge d f_{j, q}\), because as \(d\) is \(\mathbb{R}\)-linear, \(d t = \sum_{j \in J} d (f_{j, 0} d f_{j, 1} \wedge ... \wedge d f_{j, q})\), but applying the proposition that the exterior derivative of the wedge product of any \(C^\infty\) \(q_1\)-form and any \(C^\infty\) \(q_2\)-form over any \(C^\infty\) manifold with boundary is the wedge product of the exterior derivative of the \(q_1\)-form and the \(q_2\)-form plus \(-1\) to power \(q_1\) the wedge product of the \(q_1\)-form and the exterior derivative of the \(q_2\)-form, \(= \sum_{j \in J} d f_{j, 0} \wedge d f_{j, 1} \wedge ... \wedge d f_{j, q} + (-1)^0 f_{j, 0} d (d f_{j, 1} \wedge ... \wedge d f_{j, q})\), but \(d (d f_{j, 1} \wedge ... \wedge d f_{j, q}) = 0\), by iteratively applying that proposition and the proposition that the double exterior derivative of any \(C^\infty\) \(q\)-form over any \(C^\infty\) manifold with boundary is \(0\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(C^\infty\) map between \(C^\infty\) manifolds with boundary, pullback of wedge product of multicovectors is wedge product of pullbacks of multicovectors
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any \(C^\infty\) map between any \(C^\infty\) manifolds with boundary, the pullback of the wedge product of any multicovectors is the wedge product of the pullbacks of the multicovectors.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\(m_1\): \(\in M_1\)
\(t\): \(\in \Lambda_q (T_{f (m_1)}M_2)\)
\(t'\): \(\in \Lambda_{q'} (T_{f (m_1)}M_2)\)
//
Statements:
\(f^*_{m_1} (t \wedge t') = (f^*_{m_1} t) \wedge (f^*_{m_1} t')\)
//
This holds also when \(q = 0\) or \(q' = 0\).
2: Note
As a corollary, \(f^*_{m_1} (t \wedge t' \wedge ... \wedge t'') = (f^*_{m_1} t) \wedge (f^*_{m_1} t') \wedge ... \wedge (f^*_{m_1} t'')\): \(f^*_{m_1} (t \wedge t' \wedge t'') = f^*_{m_1} ((t \wedge t') \wedge t'') = (f^*_{m_1} (t \wedge t')) \wedge (f^*_{m_1} t'') = ((f^*_{m_1} t) \wedge (f^*_{m_1} t')) \wedge (f^*_{m_1} t'') = (f^*_{m_1} t) \wedge (f^*_{m_1} t') \wedge (f^*_{m_1} t'')\), and so on.
3: Proof
Whole Strategy: Step 1: compute \(f^*_{m_1} (t \wedge t') (v_1, ..., v_{q + q'})\); Step 2: compute \((f^*_{m_1} t) \wedge (f^*_{m_1} t') (v_1, ..., v_{q + q'})\); Step 3: conclude the proposition.
Step 1:
Let \(q = 0\).
\(f^*_{m_1} (t \wedge t') = f^*_{m_1} (t t') = (t \circ f (m_1)) f^*_{m_1} t'\), because the pullback is linear.
Let \(q' = 0\).
\(f^*_{m_1} (t \wedge t') = f^*_{m_1} (t t') = f^*_{m_1} (t' t) = (t' \circ f (m_1)) f^*_{m_1} t\).
Let \(0 \lt q\) and \(0 \lt q'\).
Let \((v_1, ..., v_{q + q'}) \in T_{m_1}M_1 \times ... \times T_{m_1}M_1\) be any.
\(f^*_{m_1} (t \wedge t') (v_1, ..., v_{q + q'}) = t \wedge t' (d f_{m_1} v_1, ..., d f_{m_1} v_{q + q'}) = (q + q')! / (q! q'!) Asym (t \otimes t') (d f_{m_1} v_1, ..., d f_{m_1} v_{q + q'}) = (q + q')! / (q! q'!) 1 / (q + q')! \sum_{\sigma \in S_{q + q'}} sgn \sigma t \otimes t' (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma t (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_q}) t' (d f_{m_1} v_{\sigma_{q + 1}}, ..., d f_{m_1} v_{\sigma_{q + q'}})\).
Step 2:
Let \(q = 0\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') = (t \circ f (m_1)) (f^*_{m_1} t')\).
Let \(q' = 0\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') = (f^*_{m_1} t) (t' \circ f (m_1)) = (t' \circ f (m_1)) f^*_{m_1} t\).
Let \(0 \lt q\) and \(0 \lt q'\).
\((f^*_{m_1} t) \wedge (f^*_{m_1} t') (v_1, ..., v_{q + q'}) = (q + q')! / (q! q'!) Asym ((f^*_{m_1} t) \otimes (f^*_{m_1} t')) (v_1, ..., v_{q + q'}) = (q + q')! / (q! q'!) 1 / (q + q')! \sum_{\sigma \in S_{q + q'}} sgn \sigma (f^*_{m_1} t) \otimes (f^*_{m_1} t') (v_{\sigma_1}, ..., v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma (f^*_{m_1} t) (v_{\sigma_1}, ..., v_{\sigma_q}) (f^*_{m_1} t') (v_{\sigma_{q + 1}}, ..., v_{\sigma_{q + q'}}) = 1 / (q! q'!) \sum_{\sigma \in S_{q + q'}} sgn \sigma t (d f_{m_1} v_{\sigma_1}, ..., d f_{m_1} v_{\sigma_q}) t' (d f_{m_1} v_{\sigma_{q + 1}}, ..., d f_{m_1} v_{\sigma_{q + q'}})\).
Step 3:
The result of Step 1 and the result of Step 2 are the same.
So, \(f^*_{m_1} (t \wedge t') = (f^*_{m_1} t) \wedge (f^*_{m_1} t')\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for unitary map from vectors space with inner product with induced topology into same vectors space, adjoint of map is unitary and double adjoint of map is map
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any unitary map from any vectors space with any inner product with the induced topology into the same vectors space, the adjoint of the map is unitary and the double adjoint of the map is the map.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any inner product, with the topology induced by the metric induced by the norm induced by the inner product
\(f\): \(: V \to V\), \(\in \{\text{ the unitary maps }\}\)
\(f^*\): \(\text{ = the adjoint of } f\)
//
Statements:
\(f^* \in \{\text{ the unitary maps }\}\)
\(\land\)
\({f^*}^* = f\)
//
2: Proof
Whole Strategy: Step 1: see that \(\langle f^* (v), f^* (v) \rangle = \langle v, v \rangle\) and conclude that \(f^*\) is unitary; Step 2: see that \({f^*}^*\) is unitary and \({f^*}^* = {f^*}^{-1} = {f^{-1}}^{-1} = f\).
Step 1:
\(f^* \circ f = f \circ f^* = id\) means that \(f^* = f^{-1}\).
\(f^* = f^{-1}\) is a linear surjection, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
\(\langle f^* (v), f^* (v) \rangle = \langle f^{-1} (v), f^{-1} (v) \rangle\).
By the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary, \(\langle f (f^{-1} (v)), f (f^{-1} (v)) \rangle = \langle f^{-1} (v), f^{-1} (v) \rangle\).
But the left hand side is \(\langle v, v \rangle\), so, \(\langle f^* (v), f^* (v) \rangle = \langle v, v \rangle\).
By the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary, \(f^*\) is unitary.
Step 2:
By Step 1, \({f^*}^*\) is unitary.
So, \({f^*}^* \circ f^* = f^* \circ {f^*}^* = id\), which means that \({f^*}^* = {f^*}^{-1}\). But as \(f^* = f^{-1}\), \({f^*}^{-1} = {f^{-1}}^{-1} = f\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for map from vectors space with inner product with induced topology into same vectors space, if map is unitary, it preserves norm and if map is linear surjection that preserves norm, it is unitary
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any map from any vectors space with any inner product with the induced topology into the same vectors space, if the map is unitary, it preserves norm and if the map is any linear surjection that preserves norm, it is unitary.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any inner product, with the topology induced by the metric induced by the norm induced by the inner product
\(f\): \(: V \to V\)
//
Statements:
(
\(f \in \{\text{ the unitary maps }\}\)
\(\implies\)
\(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\)
\(\land\)
(
\(f \in \{\text{ the linear surjections }\} \land \forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\)
\(\implies\)
\(f \in \{\text{ the unitary maps }\}\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f\) is unitary; Step 2: see that \(\langle f (v), f (v) \rangle = \langle v, v \rangle\); Step 3: suppose that \(f\) is a linear surjection and \(\langle f (v), f (v) \rangle = \langle v, v \rangle\); Step 4: see that for each \(v, v' \in V\), \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\); Step 5: see that \(f\) is unitary.
Step 1:
Let us suppose that \(f\) is unitary.
Step 2:
Let \(v \in V\) be any.
\(\langle f (v), f (v) \rangle = \langle f^* \circ f (v), v \rangle = \langle id (v), v \rangle = \langle v, v \rangle\).
Step 3:
Let us suppose that \(f\) is a linear surjection and \(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\).
Step 4:
Let us see that for each \(v, v' \in V\), \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
\(\langle f (v + v'), f (v + v') \rangle = \langle v + v', v + v' \rangle\).
The left hand side is \(\langle f (v) + f (v'), f (v) + f (v') \rangle = \langle f (v), f (v) \rangle + \langle f (v), f (v') \rangle + \langle f (v'), f (v) \rangle + \langle f (v'), f (v') \rangle\).
The right hand side is \(\langle v, v \rangle + \langle v, v' \rangle + \langle v', v \rangle + \langle v', v' \rangle\).
So, \(\langle f (v), f (v') \rangle + \langle f (v'), f (v) \rangle = \langle v, v' \rangle + \langle v', v \rangle\), but the left hand side is \(\langle f (v), f (v') \rangle + \overline{\langle f (v), f (v') \rangle} = 2 Re (\langle f (v), f (v') \rangle)\) and the right hand side is \(\langle v, v' \rangle + \overline{\langle v, v' \rangle} = 2 Re (\langle v, v' \rangle)\), so, \(Re (\langle f (v), f (v') \rangle) = Re (\langle v, v' \rangle)\).
When \(F = \mathbb{R}\), that already means that \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
Let us suppose that \(F = \mathbb{C}\).
\(\langle f (i v + v'), f (i v + v') \rangle = \langle i v + v', i v + v' \rangle\).
The left hand side is \(\langle i f (v) + f (v'), i f (v) + f (v') \rangle = - \langle f (v), f (v) \rangle + i \langle f (v), f (v') \rangle - i \langle f (v'), f (v) \rangle + \langle f (v'), f (v') \rangle\).
The right hand side is \(- \langle v, v \rangle + i \langle v, v' \rangle - i \langle v', v \rangle + \langle v', v' \rangle\).
So, \(i \langle f (v), f (v') \rangle - i \langle f (v'), f (v) \rangle = i \langle v, v' \rangle - i \langle v', v \rangle\), but the left hand side is \(i \langle f (v), f (v') \rangle - i \overline{\langle f (v), f (v') \rangle} = i (\langle f (v), f (v') \rangle - \overline{\langle f (v), f (v') \rangle}) = 2 i Im (\langle f (v), f (v') \rangle) i\) and the right hand side is \(i \langle v, v' \rangle - i \overline{\langle v, v' \rangle} = i (\langle v, v' \rangle - \overline{\langle v, v' \rangle}) = 2 i Im (\langle v, v' \rangle) i\), so, \(Im (\langle f (v), f (v') \rangle) = Im (\langle v, v' \rangle)\).
So, \(\langle f (v), f (v') \rangle = \langle v, v' \rangle\).
Step 5:
As \(f\) is a surjection, for each \(v'' \in V\), there is a \(v \in V\) such that \(v'' = f (v)\).
By Step 4, for each \(v' \in V\), \(\langle v'', f (v') \rangle = \langle f (v), f (v') \rangle = \langle v, v' \rangle\), which means that the domain of \(f^*\) is \(V\) and \(f^* (f (v)) = v\), according to the definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space.
As that holds for each \(v \in V\), \(f^* \circ f = id\).
For each \(v' \in V\), \(f \circ f^* (v') = f \circ f^* (f (v)) = f (v) = v'\), which means that \(f \circ f^* = id\).
\(\forall v \in V (\langle f (v), f (v) \rangle = \langle v, v \rangle)\) means that \(f\) is bounded.
So, \(f\) is a unitary map.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space
The table of contents of this article
Main Body
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of top-form over \(C^\infty\) manifold with boundary
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of top-form over \(C^\infty\) manifold with boundary.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary } \}\)
\( (T^0_d (TM), M, \pi)\): \(= \text{ the } C^\infty (0, d) \text{ -tensors bundle over } M\)
\( (\Lambda_d (TM), M, \pi)\): \(= \text{ the } C^\infty d \text{ -covectors bundle over } M\)
\(*f\): \(: M \to T^0_d (TM)\) such that \(Ran (f) \subseteq \Lambda_d (TM)\) or \(: M \to \Lambda_d (TM)\), \(\in \{\text{ the sections }\}\)
//
Conditions:
//
2: Note
In short, 'top-form' is any \(d\)-form.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for complex Euclidean vectors space with complex Euclidean norm and \(2\) vectors with same norm, there is orthogonal linear map that maps one of vectors to other whose canonical representative matrix is unitary, and when dimension is equal to or larger than \(2\), determinant can me made \(1\)
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any complex Euclidean vectors space with the complex Euclidean norm and any \(2\) vectors with any same norm, there is an orthogonal linear map that maps any one of the vectors to the other whose canonical representative matrix is unitary, and when the dimension is equal to or larger than \(2\), the determinant of the matrix can me made \(1\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{C}^d\): \(= \text{ the complex Euclidean vectors space with the complex Euclidean norm }\)
\(\{v, v'\}\): \(\subseteq \mathbb{C}^d\), such that \(\Vert v \Vert = \Vert v' \Vert\)
//
Statements:
\(\exists M \in \{\text{ the } d \times d \text{ unitary matrices } \} (v'^t = M v^t)\)
\(\land\)
(
\(2 \le d\)
\(\implies\)
\(\exists M \in \{\text{ the determinant } 1 d \times d \text{ unitary matrices } \} (v'^t = M v^t)\)
)
//
2: Note
Also the proposition that for any Euclidean vectors space with the Euclidean norm and any \(2\) vectors with any same norm, there is an orthogonal linear map that maps any one of the vectors to the other whose canonical representative matrix is orthogonal, and when the dimension is equal to or larger than \(2\), the determinant of the matrix can me made \(1\) holds.
3: Proof
Whole Strategy: Step 1: conclude the proposition for when \(\Vert v \Vert = \Vert v' \Vert = 0\); Step 2: suppose that \(0 \lt \Vert v \Vert = \Vert v' \Vert\); Step 3: take any orthonormal bases, \((b_1 := v / \Vert v \Vert, b_2, ..., b_d)\) and \((b'_1 := v' / \Vert v' \Vert, b'_2, ..., b'_d)\), and define the linear map, \(f: \mathbb{C}^d \to \mathbb{C}^d, v = v^j b_j \mapsto v^j b'_j\); Step 4: see that \(f\) is orthogonal and that the canonical representative matrix is unitary; Step 5: suppose that \(2 \le d\); Step 6: take the vector, \(v'' := (\Vert v \Vert, 0, ..., 0)\), and a unitary matrix, \(N\), such that \(v^t = N v''^t\) and multiply the last column of \(N\) by \(\overline{det N}\) with the result unitary matrix, \(\widetilde{N}\), and take a unitary matrix, \(N'\), such that \(v'^t = N' v''^t\) and multiply the last column of \(N'\) by \(\overline{det N'}\) with the result unitary matrix, \(\widetilde{N'}\): Step 7: see that \(v'^t = \widetilde{N'} {\widetilde{N}}^{-1} v^t\) and \(M := \widetilde{N'} {\widetilde{N}}^{-1}\) will do.
Step 1:
Let us suppose that \(\Vert v \Vert = \Vert v' \Vert = 0\).
That means that \(v = v' = 0\).
Then, any unitary matrix, for example, \(I\), will do: \(v'^t = I v^t\); \(I^* = I^{-1}\); \(det I = 1\) (even when \(d \lt 2\)).
Step 2:
Let us suppose that \(0 \lt \Vert v \Vert = \Vert v' \Vert\).
Step 3:
Let \(\mathbb{C}^d\) have the complex Euclidean inner product, by which the complex Euclidean norm is induced.
Let us take any orthonormal basis, \((b_1 := v / \Vert v \Vert, b_2, ..., b_d)\) for \(\mathbb{C}^d\), which is possible, by the proposition that for any finite-dimensional vectors space, any linearly independent subset can be expanded to be a basis by adding a finite number of elements and the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
Let us take any orthonormal basis, \((b'_1 := v' / \Vert v' \Vert, b'_2, ..., b'_d)\) for \(\mathbb{C}^d\), which is possible, likewise.
Let us define the map, \(f: \mathbb{C}^d \to \mathbb{C}^d, v = v^j b_j \mapsto v^j b'_j\), which is linear (in fact, a 'vectors spaces - linear morphisms' isomorphism), by the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism: \(f\) maps \(b_j\) to \(b'_j\) and is expanding the mapping between the bases linearly.
Step 4:
\(f\) is orthogonal, because for each \(v = v^j b_j \in \mathbb{C}^d\), \(\Vert f (v) \Vert = \Vert f (v^j b_j) \Vert = \Vert v^j b'_j \Vert = \sqrt{\langle v^j b'_j, v^l b'_l \rangle} = \sqrt{v^j \overline{v^l} \langle b'_j, b'_l \rangle} = \sqrt{v^j \overline{v^l} \delta_{j, l}} = \sqrt{\langle v^j b_j, v^l b_l \rangle} = \Vert v^j b_j \Vert = \Vert v \Vert\).
So, its canonical representative matrix, \(M\), is unitary, by the proposition that for any linear map from any complex-Euclidean-normed complex Euclidean vectors space into itself, the map is orthogonal if and only if the canonical representative matrix is unitary.
As \(b'_1 = f (b_1)\), \(v' / \Vert v' \Vert = f (v / \Vert v \Vert) = 1 / \Vert v \Vert f (v) = 1 / \Vert v' \Vert f (v)\), which means that \(v' = f (v)\).
So, \(v'^t = M v^t\).
Step 5:
Let us suppose that \(2 \le d\).
Step 6:
Note that \(det N \overline{det N} = 1\), because as \(N^* N = I\), \(det (N^* N) = det I = 1\), but \(det (N^* N) = det N^* det N = \overline{det N} det N\).
Let us take the vector, \(v'' := (\Vert v \Vert, 0, ..., 0)\).
\(\Vert v'' \Vert = \Vert v \Vert\), so, there is a unitary matrix, \(N\), such that \(v^t = N v''^t\), by Step 4.
Let us multiply the last column of \(N\) by \(\overline{det N}\) with the result matrix, \(\widetilde{N}\).
\(\widetilde{N}\) is unitary, because for \(N = \begin{pmatrix} N^1_1 & ... & N^1_d \\ ... \\ N^d_1 & ... & N^d_d \end{pmatrix}\), \(\widetilde{N} = \begin{pmatrix} N^1_1 & ... & N^1_{d - 1} & \overline{det N} N^1_d \\ ... \\ N^d_1 & ... & N^d_{d - 1} & \overline{det N} N^d_d \end{pmatrix}\), and \(\widetilde{N}^* \widetilde{N} = \begin{pmatrix} \overline{N^1_1} & ... & \overline{N^d_1} \\ ... \\ \overline{N^1_{d - 1}} & ... & \overline{N^d_{d - 1}} \\ det N \overline{N^1_d} & ... & det N \overline{N^d_d} \end{pmatrix} \begin{pmatrix} N^1_1 & ... & N^1_{d - 1} & \overline{det N} N^1_d \\ ... \\ N^d_1 & ... & N^d_{d - 1} & \overline{det N} N^d_d \end{pmatrix}\), whose \((j, l)\) component is \(\sum_{m \in \{1, ..., d\}} \overline{N^m_j} N^m_l\) for \(j, l \lt d\); is \(\sum_{m \in \{1, ..., d\}} det N \overline{N^m_j} N^m_l\) for \(j = d\) and \(l \lt d\); is \(\sum_{m \in \{1, ..., d\}} \overline{N^m_j} \overline{det N} N^m_l\) for \(j \lt d\) and \(l = d\); and is \(\sum_{m \in \{1, ..., d\}} det N \overline{N^m_j} \overline{det N} N^m_l = \sum_{m \in \{1, ..., d\}} \overline{N^m_j} N^m_l\) for \(j = l = d\), but \(\overline{det N}\) or \(det N\) appears only when \(j \neq l\), and then, \(\sum_{m \in \{1, ..., d\}} \overline{N^m_j} N^m_l = 0\), so, the component is \(0\), and when \(j = l\), \(\sum_{m \in \{1, ..., d\}} \overline{N^m_j} N^m_l = 1\), so, the component is \(1\).
\(det \widetilde{N} = \overline{det N} det N\), by a property of determinant of matrix, \(= 1\).
\(v^t = \widetilde{N} v''^t\), because as \(v'' = (\Vert v \Vert, 0, ..., 0)\), the last column of \(\widetilde{N}\) does not influence the result at all (which is true only because \(2 \le d\), so, the last column is not the 1st column).
\(\Vert v'' \Vert = \Vert v' \Vert\), so, there is a unitary matrix, \(N'\), such that \(v'^t = N' v''^t\), by Step 4.
Let us multiply the last column of \(N'\) by \(\overline{det N'}\) with the result matrix, \(\widetilde{N'}\).
\(\widetilde{N'}\) is unitary, as before.
\(det \widetilde{N'} = 1\), as before.
\(v'^t = \widetilde{N'} v''^t\), as before.
Step 7:
So, \(v'^t = \widetilde{N'} v''^t = \widetilde{N'} {\widetilde{N}}^{-1} v^t\).
Let \(M := \widetilde{N'} {\widetilde{N}}^{-1}\).
\(M\) is unitary, because \(M^* M = (\widetilde{N'} {\widetilde{N}}^{-1})^* \widetilde{N'} {\widetilde{N}}^{-1} = ({\widetilde{N}}^{-1})^* \widetilde{N'}^* \widetilde{N'} {\widetilde{N}}^{-1} = ({\widetilde{N}}^{-1})^* I {\widetilde{N}}^{-1} = ({\widetilde{N}}^{-1})^* {\widetilde{N}}^{-1} = ({\widetilde{N}}^*)^* {\widetilde{N}}^* = \widetilde{N} {\widetilde{N}}^* = I\).
\(det M = det (\widetilde{N'} {\widetilde{N}}^{-1}) = det \widetilde{N'} det {\widetilde{N}}^{-1} = 1\).
So, \(M\) satisfies the conditions for the proposition.
References
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