1132: For Index Set, Sum by n-th Power of Index Set Is Sum by Quotient Set of n-th Power Index Set of Sum by Quotient Set of n-Symmetric Group

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description/proof of that for index set, sum by $n$-th power of index set is sum by quotient set of $n$-th power index set of sum by quotient set of $n$-symmetric group

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of product set.
• The reader knows a definition of quotient set.
• The reader knows a definition of $n$-symmetric group.
• The reader knows a definition of representatives set of quotient set.

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Target Context

• The reader will have a description and a proof of the proposition that for any index set, the sum by the $n$-th power of the index set is the sum by the canonical quotient set of the $n$-th power index set of the sum by the canonical quotient set of the $n$-symmetric group for each element of the quotient set of the $n$-th power index set.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$J$: $\in \{\text{ the index sets }\}$
$n$: $\in \mathbb{N}\setminus \{0\}$
$J^n$: $=\text{ the product set }$
$S_n$: $=\text{ the }n\text{ -symmetric group }$
$\sim$: $\in \{\text{ the equivalence relations on }{J}^n\}$, such that $\mathrm{\forall }j=(j_1,...,j_n),j^{\prime }=(j_1^{\prime },...,j_n^{\prime })\in J^n(j\sim j^{\prime }⟺\mathrm{\exists }\sigma \in S_n((j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_1^{\prime },...,j_n^{\prime })))$
$J^n/\sim$: $=\text{ the quotient set }$
$\overline{J^n/\sim -f}$: $=\text{ the representatives set of }{J}^n/\sim$, where $f:J^n/\sim \to J^n$, the representatives map, is chosen arbitrarily
$\{{\sim }_j|j\in \overline{J^n/\sim -f}\}$: ${\sim }_j\in \{\text{ the equivalent relations of }{S}_n\}$, such that $\mathrm{\forall }\sigma ,{\sigma }^{\prime }\in S_n(\sigma {\sim }_j{\sigma }^{\prime }⟺(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }}))$
$\{S_n/{\sim }_j|j\in \overline{J^n/\sim -f}\}\}$: $S_n/{\sim }_j\in \{\text{ the quotient sets }\}$
//

Statements:
$\sum _{j\in J^n}=\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}=\sum _{[j]\in J^n/\sim }\sum _{\sigma \in S_n/{\sim }_{f([j])}}$
//

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2: Note

What is being done is really just a common sense: $J^n$ is divided into the subsets each of which is the permutations of a combination ('combination' means any element of $J^n/\sim$).

For example, when $j=\{1,2\}$ and $n=3$, $J^n=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\}$, $J^n/\sim =\{\{(1,1,1)\},\{(1,1,2),(1,2,1),(2,1,1)\},\{(1,2,2),(2,1,2),(2,2,1)\},\{(2,2,2)\}\}$, $\overline{J^n/\sim -f}=\{(1,1,1),(1,1,2),(1,2,2),(2,2,2)\}$ as a choice, and $\sum _{j\in J^n}$ is accomplished by taking each element of $\overline{J^n/\sim -f}$, summing by all the distinct permutations of the element, and summing up the sums.

The point of this proposition is offering an exact notation of the intuitively natural procedure.

${\sim }_j$ depends on the choice of $f$ (the choice of the representative for each class).

For example, when $j=\{1,2\}$ and $n=3$, if $(1,1,2)$ is a representative, $(1,2,3)\mapsto (1,2,3)\in S_n$ and $(1,2,3)\mapsto (2,1,3)\in S_n$ will be equivalent, but if $(1,2,1)$ is a representative, $(1,2,3)\mapsto (1,2,3)\in S_n$ and $(1,2,3)\mapsto (2,1,3)\in S_n$ will not be equivalent (instead, $(1,2,3)\mapsto (1,2,3)\in S_n$ and $(1,2,3)\mapsto (3,2,1)\in S_n$ will be equivalent).

Nevertheless, $\sum _{\sigma \in S_n/{\sim }_j}$ covers the same set of permutations for the class $[j]$.

When $J$ is linearly-ordered, a natural choice for the representative of $[(j_1,...,j_n)]$ is $j_1\le ...\le j_n$ (or $j_n\le ...\le j_1$).

The motivation for this proposition is that often, the summand of $\sum _{j\in J^n}$ depends (totally or partially) on the combination $[j]$, and then, thinking $\sum _{j\in J^n}$ as $\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$ can become handy: for example, when the summand depends totally on $[j]$, $\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}=\sum _{j\in \overline{J^n/\sim -f}}|S_n/{\sim }_j|$ where $|S_n/{\sim }_j|$ denotes the cardinality of $S_n/{\sim }_j$.

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3: Proof

Whole Strategy: Step 1: see that $\sim$ is indeed an equivalence relation; Step 2: see that ${\sim }_j$ is indeed an equivalence relation; Step 3: see that $\sum _{j\in J^n}=\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}=\sum _{[j]\in J^n/\sim }\sum _{\sigma \in S_n/{\sim }_{f([j])}}$.

Step 1:

Let us see that $\sim$ is indeed an equivalence relation.

For each $j=(j_1,...,j_n)\in J^n$, $j\sim j$, because there is $\sigma =id\in S_n$ such that $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_1,...,j_n)$.

For each $j=(j_1,...,j_n),j^{\prime }=(j_1^{\prime },...,j_n^{\prime })\in J^n$ such that $j\sim j^{\prime }$, $j^{\prime }\sim j$, because while there is a $\sigma \in S_n$ such that $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_1^{\prime },...,j_n^{\prime })$, there is ${\sigma }^{-1}\in S_n$ such that $(j_{{{\sigma }^{-1}}_1}^{\prime },...,j_{{{\sigma }^{-1}}_n}^{\prime })=(j_1,...,j_n)$, which is true because while there is a ${\sigma }_l=\sigma (l)=1$, $l={\sigma }^{-1}(1)$, and $j_1=j_{{\sigma }_l}=j_l^{\prime }=j_{{\sigma }^{-1}(1)}^{\prime }$.

For each $j=(j_1,...,j_n),j^{\prime }=(j_1^{\prime },...,j_n^{\prime }),j^{″}=(j_1^{″},...,j_n^{″})\in J^n$ such that $j\sim j^{\prime }$ and $j^{\prime }\sim j^{″}$, $j\sim j^{″}$, because while there are a $\sigma \in S_n$ such that $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_1^{\prime },...,j_n^{\prime })$ and a ${\sigma }^{\prime }\in S_n$ such that $(j_{{\sigma }_1^{\prime }}^{\prime },...,j_{{\sigma }_n^{\prime }}^{\prime })=(j_1^{″},...,j_n^{″})$, there is $\sigma \circ {\sigma }^{\prime }\in S_n$ such that $(j_{(\sigma \circ {\sigma }^{\prime }{)}_1},...,j_{(\sigma \circ {\sigma }^{\prime }{)}_n})=(j_1^{″},...,j_n^{″})$, which is true because while $j_1^{″}=j_{{\sigma }_1^{\prime }}^{\prime }$, ${\sigma }_1^{\prime }=l$ for an $l$ and $j_{{\sigma }_1^{\prime }}^{\prime }=j_l^{\prime }=j_{{\sigma }_l}=j_{\sigma (l)}=j_{\sigma ({\sigma }_1^{\prime })}=j_{\sigma ({\sigma }^{\prime }(1))}=j_{\sigma \circ {\sigma }^{\prime }(1)}=j_{\sigma \circ {\sigma }_1^{\prime }}$, so, $j_1^{″}=j_{\sigma \circ {\sigma }_1^{\prime }}$.

Step 2:

Let us see that ${\sim }_j$ is indeed an equivalence relation.

For each $\sigma \in S_n$, $\sigma {\sim }_j\sigma$, because $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_{{\sigma }_1},...,j_{{\sigma }_n})$.

For each $\sigma ,{\sigma }^{\prime }\in S_n$ such that $\sigma {\sim }_j{\sigma }^{\prime }$, ${\sigma }^{\prime }{\sim }_j\sigma$, because while $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }})$, $(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }})=(j_{{\sigma }_1},...,j_{{\sigma }_n})$.

For each $\sigma ,{\sigma }^{\prime },{\sigma }^{″}\in S_n$ such that $\sigma {\sim }_j{\sigma }^{\prime }$ and ${\sigma }^{\prime }{\sim }_j{\sigma }^{″}$, $\sigma {\sim }_j{\sigma }^{″}$, because while $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }})$ and $(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }})=(j_{{\sigma }_1^{″}},...,j_{{\sigma }_n^{″}})$, $(j_{{\sigma }_1},...,j_{{\sigma }_n})=(j_{{\sigma }_1^{\prime }},...,j_{{\sigma }_n^{\prime }})=(j_{{\sigma }_1^{″}},...,j_{{\sigma }_n^{″}})$.

Step 3:

Let us see that $\sum _{j\in J^n}=\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$.

There is no duplication in $\sum _{j\in J^n}$, obviously.

There is no duplication in $\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$, because while there is no duplication in $\sum _{j\in \overline{J^n/\sim -f}}$, there is no duplication in the corresponding $\{[j]\}$, and for each $[j]$, there is no duplication in $\sum _{\sigma \in S_n/{\sim }_j}$ because $\sigma \in S_n/{\sim }_j$ is determined exactly for that effect, and there is no duplication between $[j]$ and $[j^{\prime }]$ because $[j]$ and $[j^{\prime }]$ represent some different combinations and permutating any combination does not change the combination.

Each element of $\sum _{j\in J^n}$ exists in $\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$, because the element is of a combination whose representative is in $\sum _{j\in \overline{J^n/\sim -f}}$, and the element is a permutation of the combination, which is in the corresponding $\sum _{\sigma \in S_n/{\sim }_j}$.

Each element of $\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$ obviously exists in $\sum _{j\in J^n}$.

So, $\sum _{j\in J^n}=\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}$.

$\sum _{j\in \overline{J^n/\sim -f}}\sum _{\sigma \in S_n/{\sim }_j}=\sum _{[j]\in J^n/\sim }\sum _{\sigma \in S_n/{\sim }_{f([j])}}$ is just an obvious reformulation: the latter is taking the class, $[j]$, instead of its representative, $j$, but anyway, $\sum _{\sigma \in S_n/{\sim }_j}=\sum _{\sigma \in S_n/{\sim }_{f([j])}}$ because $j=f([j])$.

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References

<The previous article in this series | The table of contents of this series |

1131: Top-Covector for Vectors Space Is Determined by Result for Ordered Basis for Vectors Space

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description/proof of that top-covector for vectors space is determined by result for ordered basis for vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of top-covectors space of finite-dimensional vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that any top-covector for any vectors space is determined by the result for any ordered basis for the vectors space.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
${\mathrm{\Lambda }}_d(V:F)$: $=\text{ the top-covectors space }$
$t$: $\in {\mathrm{\Lambda }}_d(V:F)$
$t^{\prime }$: $\in {\mathrm{\Lambda }}_d(V:F)$
$(b_1,...,b_d)$: $\in \{\text{ the bases for }V\}$
//

Statements:
$t(b_1,...,b_d)=t^{\prime }(b_1,...,b_d)$
$⟹$
$t=t^{\prime }$
//

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2: Proof

Whole Strategy: Step 1: let $(v_1,...,v_d)$ be any set of vectors of $V$ and see that $t(v_1,...,v_d)=t^{\prime }(v_1,...,v_d)$.

Step 1:

Let $(v_1,...,v_d)$ be any set of vectors of $V$.

If $t(v_1,...,v_d)=t^{\prime }(v_1,...,v_d)$, $t=t^{\prime }$.

So, let us see that $t(v_1,...,v_d)=t^{\prime }(v_1,...,v_d)$.

$v_j=v_j^{l_j}{b}_{l_j}$.

$t(v_1,...,v_d)=t(v_1^{l_1}{b}_{l_1},...,v_d^{l_d}{b}_{l_d})=v_1^{l_1}...v_d^{l_d}t(b_{l_1},...,b_{l_d})$.

When $(b_{l_1},...,b_{l_d})$ is from $(b_1,...,b_d)$ by a permutation, $\sigma$, $t(b_{l_1},...,b_{l_d})=sgn\sigma t(b_1,...,b_d)=sgn\sigma t^{\prime }(b_1,...,b_d)=t^{\prime }(b_{l_1},...,b_{l_d})$.

Otherwise, $\{b_{l_1},...,b_{l_d}\}$ is not distinct, so, $t(b_{l_1},...,b_{l_d})=0=t^{\prime }(b_{l_1},...,b_{l_d})$.

So, $t(v_1,...,v_d)=v_1^{l_1}...v_d^{l_d}t(b_{l_1},...,b_{l_d})=v_1^{l_1}...v_d^{l_d}{t}^{\prime }(b_{l_1},...,b_{l_d})=t^{\prime }(v_1^{l_1}{b}_{l_1},...,v_d^{l_d}{b}_{l_d})=t^{\prime }(v_1,...,v_d)$.

So, $t=t^{\prime }$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1130: Top-Covectors Space of Finite-Dimensional Vectors Space

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definition of top-covectors space of finite-dimensional vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of antisymmetric tensors space with respect to field and $k$ same vectors spaces and vectors space over field.

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Target Context

• The reader will have a definition of top-covectors space of finite-dimensional vectors space.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$\ast {\mathrm{\Lambda }}_d(V:F)$:
//

Conditions:
//

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2: Note

${\mathrm{\Lambda }}_d(V:F)$ is $1$-dimensional, by the proposition that the antisymmetric tensors space with respect to any field and any $k$ same finite-dimensional vectors spaces over the field and the field has the basis that consists of the wedge products of the increasing elements of the dual basis of the same vectors space.

Each element of ${\mathrm{\Lambda }}_d(V:F)$ is called "top-covector".

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1129: Map-Related Vectors Fields Pair for C∞ Map Between C∞ Manifolds with Boundary

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of map-related vectors fields pair for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ vectors field on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.

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Target Context

• The reader will have a definition of map-related vectors fields pair for $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M_1$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_2$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$\ast (V_1,V_2)$: $V_j\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors fields over }{M}_j\}$
//

Conditions:
$\mathrm{\forall }{m}_1\in M_1(d{f}_{m_1}{V}_1(m_1)=V_2(f(m_1)))$
//

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2: Note

This definition does not claim that there is such a $V_2$ for each $V_1$.

Whether there is a $V_2$ for a $V_1$ depends on $V_1$.

For example, when $V_1=0$, $V_2=0$ will do.

When $f$ is any diffeomorphism, $V_2:m_2\to d{f}_{f^{-1}(m_2)}{V}_1(f^{-1}(m_2))$ is $C^{\mathrm{\infty }}$ (although we do not prove that fact here), so, $(V_1,V_2)$ is $f$-related.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1128: q-Covectors Space at Point on C∞ Manifold with Boundary

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of $q$-covectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of tangent vectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of antisymmetric tensors space with respect to field and $k$ same vectors spaces and vectors space over field.
• The reader knows a definition of $(p,q)$-Tensors Space at Point on $C^{\mathrm{\infty }}$ Manifold with Boundary.
• The reader knows a definition of canonical 'vectors spaces - linear morphisms' isomorphism between tensors space at point on $C^{\mathrm{\infty }}$ manifold with boundary and tensors space with respect to real numbers field and $p$ cotangent vectors spaces and $q$ tangent vectors spaces and field.

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Target Context

• The reader will have a definition of $q$-covectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$T_{m}M$: $=\text{ the tangent vectors space at }m$
$q$: $\in \mathbb{N}$
${\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R})$: $=\text{ the }q\text{ -covectors space }$
$f$: $:T_q^0(T_{m}M)\to L(T_{m}M,...,T_{m}M:\mathbb{R})$, $=\text{ the canonical 'vectors spaces - linear morphisms' isomorphism }$
$\ast {\mathrm{\Lambda }}_q(T_{m}M)$: $=f^{-1}({\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R}))$ when $q\ne 0$; $=\mathbb{R}$ when $q=0$, $\in \{\text{ the }\mathbb{R}\text{ vectors spaces }\}$
//

Conditions:
//

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2: Note

Another definition defines "${\mathrm{\Lambda }}_q(T_{m}M):={\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R})$", but that would make ${\mathrm{\Lambda }}_q(T_{m}M)$ non-subspace of $T_q^0(T_{m}M)$, because we have defined $T_q^0(T_{m}M):=T_m{M}^{\ast }\otimes ...\otimes T_m{M}^{\ast }$, not $=L(T_{m}M,...,T_{m}M:\mathbb{R})$.

$T_{m}M$ is an $\mathbb{R}$ vectors space as is shown in Note for the definition of tangent vectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary and ${\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R})$ is indeed a vectors subspace of $L(T_{m}M,...,T_{m}M:\mathbb{R})$ as is shown in Note for the definition of antisymmetric tensors space with respect to field and $k$ same vectors spaces and vectors space over field. So, $f^{-1}({\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R}))$ is a vectors subspace of $T_q^0(T_{m}M)$.

${\mathrm{\Lambda }}_q(T_{m}M)$ is canonically 'vectors spaces - linear morphisms' isomorphic to ${\mathrm{\Lambda }}_q(T_{m}M:\mathbb{R})$, and quite often, the 2 spaces are implicitly identified by the isomorphism, which is the reason why ${\mathrm{\Lambda }}_q(T_{m}M)$ is called "$q$-covectors space".

${\mathrm{\Lambda }}_0(T_{m}M)=T_0^0(T_{m}M)$.

${\mathrm{\Lambda }}_1(T_{m}M)=T_1^0(T_{m}M)$, because each element of $T_1^0(T_{m}M)$ is antisymmetric.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1127: q-Covectors Space Has Basis That Consists of Wedge Products of Increasing Elements of Dual Basis

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that $q$-covectors space has basis that consists of wedge products of increasing elements of dual basis

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 3: Proof

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Starting Context

• The reader knows a definition of antisymmetric tensors space with respect to field and $k$ same vectors spaces and vectors space over field.
• The reader knows a definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
• The reader knows a definition of wedge product of multicovectors.

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Target Context

• The reader will have a description and a proof of 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.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
${\mathrm{\Lambda }}_q(V:F)$: $=\text{ the }q\text{ -covectors space }$
$B$: $\in \{\text{ the bases for }V\}=\{b_l|1\le l\le dimV\}$
$B^{\ast }$: $=\text{ the dual basis for }B=\{b^l|1\le l\le dimV\}$
$\widetilde{B^{\ast }}$: $=\{b^{j_1}\wedge ...\wedge b^{j_q}|\mathrm{\forall }l\in \{1,...,q\}(1\le j_l\le dimV)\wedge j_1<...<j_q\}$
//

Statements:
$\widetilde{B^{\ast }}\in \{\text{ the bases for }{\mathrm{\Lambda }}_q(V:F)\}$
//

Let us call $\widetilde{B^{\ast }}$ "the standard basis with respect to $B$": it is not determined unless $B$ is specified.

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3: Proof

Whole Strategy: Step 1: see that $\widetilde{B^{\ast }}$ spans ${\mathrm{\Lambda }}_q(V:F)$; Step 2: see that $\widetilde{B^{\ast }}$ is linearly independent.

Step 1:

Let us see that $\widetilde{B^{\ast }}$ spans ${\mathrm{\Lambda }}_q(V:F)$.

Let $t\in {\mathrm{\Lambda }}_q(V:F)$ be any.

As ${\mathrm{\Lambda }}_q(V:F)\subseteq L(V,...,V:F)$, $t\in L(V,...,V:F)$.

By the proposition that for any field and any $k$ finite-dimensional vectors spaces over the field, the tensors space with respect to the field and the vectors spaces and the field has the basis that consists of the tensor products of the elements of the dual bases of any bases of the vectors spaces, $L(V,...,V:F)$ has the standard basis with respect to $B$, $\{b^{j_1}\otimes ...\otimes b^{j_k}|\mathrm{\forall }l\in \{1,...,k\}(1\le j_l\le dimV)\}$.

So, $t=t_{j_1,...,j_k}{b}^{j_1}\otimes ...\otimes b^{j_k}$.

As $t$ is antisymmetric, $t=Asym(t)=Asym(t_{j_1,...,j_k}{b}^{j_1}\otimes ...\otimes b^{j_k})=t_{j_1,...,j_k}Asym(b^{j_1}\otimes ...\otimes b^{j_k})=t_{j_1,...,j_k}1/k!b^{j_1}\wedge ...\wedge b^{j_k}$, by a property of wedge product of multicovectors.

For each term such that $\{j_1,...,j_k\}$ is not distinct, the term is $0$, by a property of wedge product of multicovectors.

For each term such that $(j_1,...,j_k)$ is not increasing, there is a permutation, $\sigma \in S^k$, such that $(j_{{\sigma }_1},...,j_{{\sigma }_k})$ is increasing, and $b^{j_1}\wedge ...\wedge b^{j_k}=c{b}^{{\sigma }_1}\wedge ...\wedge b^{{\sigma }_k}$ where $c$ is $1$ or $-1$, by a property of wedge product of multicovectors.

So, $t$ is a linear combination of $\widetilde{B^{\ast }}$.

Step 2:

Let us see that $\widetilde{B^{\ast }}$ is linearly independent.

Let $c_{j_1,...,j_k}{b}^{j_1}\wedge ...\wedge b^{j_k}=0$ where $j_1<...<j_k$.

For each fixed $(j_1^{\prime },...,j_k^{\prime })$, let $c_{j_1,...,j_k}{b}^{j_1}\wedge ...\wedge b^{j_k}$ operate on $(b_{j_1^{\prime }},...,b_{j_k^{\prime }})$.

$c_{j_1,...,j_k}{b}^{j_1}\wedge ...\wedge b^{j_k}((b_{j_1^{\prime }},...,b_{j_k^{\prime }}))=0((b_{j_1^{\prime }},...,b_{j_k^{\prime }}))=0$.

But the only possibly nonzero term of the left hand side is $c_{j_1^{\prime },...,j_k^{\prime }}{b}^{j_1^{\prime }}\wedge ...\wedge b^{j_k^{\prime }}((b_{j_1^{\prime }},...,b_{j_k^{\prime }}))$, because any other term has a $b^{j_l}$ such that $j_l\notin \{j_1^{\prime },...,j_k^{\prime }\}$ and $b^{j_1}\wedge ...\wedge b^{j_k}((b_{j_1^{\prime }},...,b_{j_k^{\prime }}))=0$, because $b^{j_l}(b_{j_m}^{\prime })=0$ for each $m$.

$b^{j_1^{\prime }}\wedge ...\wedge b^{j_k^{\prime }}((b_{j_1^{\prime }},...,b_{j_k^{\prime }}))=det(b^{\prime j_l}(b_{j_m}^{\prime }))=detI=1$, by a property of wedge product of multicovectors.

So, $c_{j_1^{\prime },...,j_k^{\prime }}=0$.

So, $\widetilde{B^{\ast }}$ is linearly independent.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>

1126: Wedge Product of Multicovectors

<The previous article in this series | The table of contents of this series | The next article in this series>

definition of wedge product of multicovectors

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

---

Starting Context

• The reader knows a definition of antisymmetric tensors space with respect to field and $k$ same vectors spaces and vectors space over field.
• The reader knows a definition of tensor product of tensors.
• The reader knows a definition of antisymmetrization of tensor with respect to some arguments.

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Target Context

• The reader will have a definition of wedge product of 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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$\ast \wedge$: $:{\mathrm{\Lambda }}_{k_1}(V:F)\times {\mathrm{\Lambda }}_{k_2}(V:F)\to {\mathrm{\Lambda }}_{k_1+k_2}(V:F),(t_1,t_2)\mapsto (k_1+k_2)!/(k_1!k_2!)Asym(t_1\otimes t_2)$
//

Conditions:
//

$\wedge (t_1,t_2)$ is usually denoted as $t_1\wedge t_2$.

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2: Note

Another definition defines that $t_1\wedge t_2=Asym(t_1\otimes t_2)$.

The difference in the coefficients results in the differences in the coefficients for some properties of wedge product.

Indeed, $t_1\wedge t_2\in {\mathrm{\Lambda }}_{k_1+k_2}(V:F)$, because $t_1\otimes t_2\in L(V,...,V:F)$, $Asym(t_1\otimes t_2)\in {\mathrm{\Lambda }}_{k_1+k_2}(V:F)$, and $(k_1+k_2)!/(k_1!k_2!)Asym(t_1\otimes t_2)\in {\mathrm{\Lambda }}_{k_1+k_2}(V:F)$.

Let us see some properties of wedge product.

Let us see that wedge product is associative.

$(t_1\wedge t_2)\wedge t_3=(k_1+k_2)!/(k_1!k_2!)Asym(t_1\otimes t_2)\wedge t_3=(k_1+k_2+k_3)!/((k_1+k_2)!k_3!)Asym((k_1+k_2)!/(k_1!k_2!)Asym(t_1\otimes t_2)\otimes t_3)=(k_1+k_2+k_3)!/((k_1+k_2)!k_3!)(k_1+k_2)!/(k_1!k_2!)Asym(Asym(t_1\otimes t_2)\otimes t_3)=(k_1+k_2+k_3)!/(k_1!k_2!k_3!)Asym(t_1\otimes t_2\otimes t_3)$, by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.

$t_1\wedge (t_2\wedge t_3)=t_1\wedge (k_2+k_3)!/(k_2!k_3!)Asym(t_2\otimes t_3)=(k_1+k_2+k_3)!/(k_1!(k_2+k_3)!)Asym(t_1\otimes (k_2+k_3)!/(k_2!k_3!)Asym(t_2\otimes t_3))=(k_1+k_2+k_3)!/(k_1!(k_2+k_3)!)(k_2+k_3)!/(k_2!k_3!)Asym(t_1\otimes Asym(t_2\otimes t_3))=(k_1+k_2+k_3)!/(k_1!k_2!k_3!)Asym(t_1\otimes t_2\otimes t_3)$, by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.

So, $(t_1\wedge t_2)\wedge t_3=t_1\wedge (t_2\wedge t_3)$.

So, while $t_1\wedge ...\wedge t_n:=(...((t_1\wedge t_2)\wedge t_3)...\wedge t_{n-1})\wedge t_n$, it can be associated in any way.

$t_1\wedge ...\wedge t_n=(k_1+...+k_n)!/(k_1!...k_n!)Asym(t_1\otimes ...\otimes t_n)$: to prove it inductively, it holds when $n=2$; supposing that it holds for $n=2,...,n^{\prime }-1$, $t_1\wedge ...\wedge t_{n^{\prime }}=(t_1\wedge ...\wedge t_{n^{\prime }-1})\wedge t_{n^{\prime }}=(k_1+...+k_{n^{\prime }-1})!/(k_1!...k_{n^{\prime }-1}!)Asym(t_1\otimes ...\otimes t_{n^{\prime }-1})\wedge t_{n^{\prime }}=(k_1+...+k_{n^{\prime }})!/((k_1+...+k_{n^{\prime }-1})!k_{n^{\prime }}!)Asym((k_1+...+k_{n^{\prime }-1})!/(k_1!...k_{n^{\prime }-1}!)Asym(t_1\otimes ...\otimes t_{n^{\prime }-1})\otimes t_{n^{\prime }})=(k_1+...+k_{n^{\prime }})!/((k_1+...+k_{n^{\prime }-1})!k_{n^{\prime }}!)(k_1+...+k_{n^{\prime }-1})!/(k_1!...k_{n^{\prime }-1}!)Asym(Asym(t_1\otimes ...\otimes t_{n^{\prime }-1})\otimes t_{n^{\prime }})=(k_1+...+k_{n^{\prime }})!/(k_1!...k_{n^{\prime }}!)Asym(t_1\otimes ...\otimes t_{n^{\prime }})$, by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.

$t_1\wedge ...\wedge (r{t}_j+r^{\prime }{t}_j^{\prime })\wedge ...\wedge t_n=r{t}_1\wedge ...\wedge t_j\wedge ...\wedge t_n+r^{\prime }{t}_1\wedge ...\wedge t_j^{\prime }\wedge ...\wedge t_n$, because $t_1\wedge ...\wedge (r{t}_j+r^{\prime }{t}_j^{\prime })\wedge ...\wedge t_n=(k_1+...+k_n)!/(k_1!...k_n!)Asym(t_1\otimes ...\otimes (r{t}_j+r^{\prime }{t}_j^{\prime })\otimes ...\otimes t_n)=(k_1+...+k_n)!/(k_1!...k_n!)Asym(r{t}_1\otimes ...\otimes t_j\otimes ...\otimes t_n+r^{\prime }{t}_1\otimes ...\otimes t_j^{\prime }\otimes ...\otimes t_n)$, by the property of tensor product of tensors, $=(k_1+...+k_n)!/(k_1!...k_n!)(rAsym(t_1\otimes ...\otimes t_j\otimes ...\otimes t_n)+r^{\prime }Asym(t_1\otimes ...\otimes t_j^{\prime }\otimes ...\otimes t_n))$, by the proposition that any antisymmetrization-of-tensor map is linear, $=r(k_1+...+k_n)!/(k_1!...k_n!)Asym(t_1\otimes ...\otimes t_j\otimes ...\otimes t_n)+r^{\prime }(k_1+...+k_n)!/(k_1!...k_n!)Asym(t_1\otimes ...\otimes t_j^{\prime }\otimes ...\otimes t_n)=r{t}_1\wedge ...\wedge t_j\wedge ...\wedge t_n)+r^{\prime }{t}_1\wedge ...\wedge t_j^{\prime }\wedge ...\wedge t_n$.

When $(t^1,...,t^k)$ is any combination of elements of $V^{\ast }$ and $\sigma \in S_k$ is any, for each $v_1,...,v_k\in V$, $t^{{\sigma }_1}\wedge ...\wedge t^{{\sigma }_k}(v_{{\sigma }_1},...,v_{{\sigma }_1})=t^1\wedge ...\wedge t^k(v_1,...,v_k)$, by the proposition that for any vectors space and its any covectors combination, the antisymmetrization of the tensor product of any permutation of the covectors combination operated on the same permutation of any vectors combination is the antisymmetrization of the tensor product of the covectors combination operated on the vectors combination, because $t^{{\sigma }_1}\wedge ...\wedge t^{{\sigma }_k}=k!Asym(t^{{\sigma }_1}\otimes ...\otimes t^{{\sigma }_k})$ and $t^1\wedge ...\wedge t^k=k!Asym(t^1\otimes ...\otimes t^k)$.

When $(t^1,...,t^k)$ is any combination of elements of $V^{\ast }$ and $\sigma \in S_k$ is any, $t^{{\sigma }_1}\wedge ...\wedge t^{{\sigma }_k}=sgn\sigma t^1\wedge ...\wedge t^k$, because $t^{{\sigma }_1}\wedge ...\wedge t^{{\sigma }_k}(v_1,...,v_k)=t^{{\sigma }_1}\wedge ...\wedge t^{{\sigma }_k}(v_{(\sigma \circ {\sigma }^{-1}{)}_1},...,v_{(\sigma \circ {\sigma }^{-1}{)}_k})=t^1\wedge ...\wedge t^k(v_{{\sigma }_1^{-1}},...,v_{{\sigma }_k^{-1}})$, by the above result, $=sgn{\sigma }^{-1}{t}^1\wedge ...\wedge t^k(v_1,...,v_k)$, because $t^1\wedge ...\wedge t^k$ is antisymmetric, but $sgn{\sigma }^{-1}=sgn\sigma$.

When furthermore $(t^1,...,t^k)$ has any duplication, $t^1\wedge ...\wedge t^k=0$, because supposing $t^m=t^n$, taking $\sigma$ as the permutation that switches $m$ and $n$ ($sgn\sigma =-1$), $t^1\wedge ...t^m\wedge ...\wedge t^n...\wedge t^k=-t^1\wedge ...t^n\wedge ...\wedge t^m...\wedge t^k=-t^1\wedge ...t^m\wedge ...\wedge t^n...\wedge t^k$.

When $t_1\in {\mathrm{\Lambda }}_{k_1}(V:F)$, a $k_1$-covector, and $t_2\in {\mathrm{\Lambda }}_{k_2}(V:F)$, a $k_2$-covector, $t_2\wedge t_1=(-1{)}^{k_1{k}_2}{t}_1\wedge t_2$, because 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, $t_1={t_1}_{j_1,...,j_{k_1}}{b}^{j_1}\wedge ...\wedge b^{j_{k_1}}$ and $t_2={t_2}_{l_1,...,l_{k_2}}{b}^{l_1}\wedge ...\wedge b^{l_{k_2}}$, and $t_2\wedge t_1=({t_2}_{l_1,...,l_{k_2}}{b}^{l_1}\wedge ...\wedge b^{l_{k_2}})\wedge ({t_1}_{j_1,...,j_{k_1}}{b}^{j_1}\wedge ...\wedge b^{j_{k_1}})={t_1}_{j_1,...,j_{k_1}}{t_2}_{l_1,...,l_{k_2}}{b}^{l_1}\wedge ...\wedge b^{l_{k_2}}\wedge b^{j_1}\wedge ...\wedge b^{j_{k_1}}$, but as each $b^m$ is an element of $V^{\ast }$, 1st, $b^{j_1}$ can be moved to in front of $b^{l_1}$ by the $k_2$ transpositions each of which gives the $-1$ factor with the total $(-1{)}^{k_2}$ factor, then, $b^{j_2}$ can be move to in front of $b^{l_1}$ with the $(-1{)}^{k_2}$ factor, ..., after all, $b^{l_1}\wedge ...\wedge b^{l_{k_2}}\wedge b^{j_1}\wedge ...\wedge b^{j_{k_1}}$ can be changed to $b^{j_1}\wedge ...\wedge b^{j_{k_1}}\wedge b^{l_1}\wedge ...\wedge b^{l_{k_2}}$ with the ${{-1}^{k_2}}^{k_1}={-1}^{k_1{k}_2}$ factor, and $={-1}^{k_1{k}_2}{t_1}_{j_1,...,j_{k_1}}{t_2}_{l_1,...,l_{k_2}}{b}^{j_1}\wedge ...\wedge b^{j_{k_1}}\wedge b^{l_1}\wedge ...\wedge b^{l_{k_2}}={-1}^{k_1{k}_2}{t}_1\wedge t_2$.

When furthermore $k_1=k_2=k$ and $t_1=t_2=t$, when $k$ is odd, $t\wedge t=0$, because $t\wedge t=(-1{)}^{k^2}t\wedge t=-t\wedge t$; when $k$ is even, $t\wedge t$ is not necessarily $0$: $t\wedge t=(-1{)}^{k^2}t\wedge t=t\wedge t$ does not imply that $t\wedge t=0$: for example, $k=2$ and $t=t_{1,2}{b}^1\wedge b^2+t_{3,4}{b}^3\wedge b^4$, then, $t\wedge t=(t_{1,2}{b}^1\wedge b^2+t_{3,4}{b}^3\wedge b^4)\wedge (t_{1,2}{b}^1\wedge b^2+t_{3,4}{b}^3\wedge b^4)=t_{1,2}{t}_{3,4}{b}^1\wedge b^2\wedge b^3\wedge b^4+t_{1,2}{t}_{3,4}{b}^3\wedge b^4\wedge b^1\wedge b^2=t_{1,2}{t}_{3,4}{b}^1\wedge b^2\wedge b^3\wedge b^4+t_{1,2}{t}_{3,4}{b}^1\wedge b^2\wedge b^3\wedge b^4=2t_{1,2}{t}_{3,4}{b}^1\wedge b^2\wedge b^3\wedge b^4\ne 0$; if $t=t_{1,2}{b}^1\wedge b^2$, $t\wedge t=0$, so, sometimes $t\wedge t=0$ for a $t\ne 0$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1125: 2 Possible Meanings of Permutation of Sequence

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of 2 possible meanings of permutation of sequence

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Topics

About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of permutation of sequence.

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Target Context

• The reader will have a description and a proof of the 2 possible meanings of permutation of sequence.

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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:
$\mathbb{N}$:
$S$: $\subseteq \mathbb{N}$
$f$: $\in \{\text{ the sequences from }S\}$
$\sigma$: $:S\to S$, $\in \{\text{ the bijections }\}$
$\sigma (f)$: $=f\circ \sigma$
$g$: $=f\circ {\sigma }^{-1}$
//

Statements:
$\sigma (f)$ is the permutation of $f$ by $\sigma$ by a definition of permutation of sequence, but some people may understand $g$ by "permutation of $f$ by $\sigma$"
//

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2: Proof

Whole Strategy: Step 1: see what $\sigma (f)$ means; Step 2: see what $g$ means.

Step 1:

Let us see what $\sigma (f):=f\circ \sigma$ means.

For each $s\in S$, $\sigma (f)(s)=f(\sigma (s))$, which means that $\sigma (f{)}_j=f(\sigma (s_j))=f({\sigma }_j)$ where $s_j\in S$ is the $j$-th element of $S$, which means that the sequence is $f({\sigma }_1),f({\sigma }_2),...$.

Note that when we regard the permutation as moving an item to a position, the permutation is moving the ${\sigma }_j$ item to the $j$-th position.

For example, when $S=\{1,2,3\}$, $f=4,5,6$, and $\sigma :(1,2,3)\mapsto (3,1,2)$, $\sigma (f)=f(\sigma (1)),f(\sigma (2)),f(\sigma (3))=6,4,5$, which is moving the $\sigma (1)=3$ item to the $1$-st position, the $\sigma (2)=1$ item to the $2$-nd position, and the $\sigma (3)=2$ item to the $3$-rd position.

Step 2:

The reason why we have described this proposition is that Step 1 may not be what some people understand as "permutation of $f$ by $\sigma$".

They may understand it as moving the $j$-th item to the ${\sigma }_j$ position.

Let us see that that is $g$ not $\sigma (f)$.

Let us denote the permutation that moves the $j$-th item to the ${\sigma }_j$ position as $h$.

$f(s_j)=h({\sigma }_j)=h(\sigma (s_j))$.

Taking $t:=\sigma (s_j)$, $s_j={\sigma }^{-1}(t)$, so, $h(t)=f(s_j)=f({\sigma }^{-1}(t))=f\circ {\sigma }^{-1}(t)$.

So, $h=f\circ {\sigma }^{-1}=g$ not $=f\circ \sigma :=\sigma (f)$.

For example, when $S=\{1,2,3\}$, $f=4,5,6$, and $\sigma :(1,2,3)\mapsto (3,1,2)$, which is the same with the above example, ${\sigma }^{-1}:(1,2,3)\mapsto (2,3,1)$ and $g=f({\sigma }^{-1}(1)),f({\sigma }^{-1}(2)),f({\sigma }^{-1}(3))=5,6,4$, which is moving the $1$-st item to the $3$-rd position, the $2$-nd item to the $1$-st position, and the $3$-rd item to the $2$-nd position.

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3: Note

This is not about which of the 2 is correct, but about the necessity of clarifying what is meant by 'permutation of $f$ by $\sigma$' and being consistent.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1124: For Vectors Space and Covectors Combination, Antisymmetrization of Tensor Product of Permutated Covectors Combination Operated on Same-Permutated Vectors Combination Is Antisymmetrization of Tensor Product of Covectors Combination Operated on Vectors Combination

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for vectors space and covectors combination, antisymmetrization of tensor product of permutated covectors combination operated on same-permutated vectors combination is antisymmetrization of tensor product of covectors combination operated on vectors combination

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Topics

About: vectors space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of tensor product of tensors.
• The reader knows a definition of antisymmetrization of tensor with respect to some arguments.

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Target Context

• The reader will have a description and a proof of the proposition that for any vectors space and its any covectors combination, the antisymmetrization of the tensor product of any permutation of the covectors combination operated on the same permutation of any vectors combination is the antisymmetrization of the tensor product of the covectors combination operated on the vectors combination.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$(t^1,...,t^k)$: $\in \{\text{ the combinations of elements of }{V}^{\ast }\}$
$\sigma$: $\in S_k$
//

Statements:
$\mathrm{\forall }{v}_1,...,v_k\in V(Asym(t^{{\sigma }_1}\otimes ...\otimes t^{{\sigma }_k})(v_{{\sigma }_1},...,v_{{\sigma }_1})=Asym(t^1\otimes ...\otimes t^k)(v_1,...,v_k))$
//

---

2: Proof

Whole Strategy: Step 1: expand $Asym(t^{{\sigma }_1}\otimes ...\otimes t^{{\sigma }_k})(v_{{\sigma }_1},...,v_{{\sigma }_k})$ and see that it equals $Asym(t^1\otimes ...\otimes t^k)(v_1,...,v_k)$.

Step 1:

$Asym(t^{{\sigma }_1}\otimes ...\otimes t^{{\sigma }_k})(v_{{\sigma }_1},...,v_{{\sigma }_k})=1/k!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{t}^{{\sigma }_1}\otimes ...\otimes t^{{\sigma }_k}(v_{{\sigma }^{\prime }({\sigma }_1)},...,v_{{\sigma }^{\prime }({\sigma }_k)})$, by the definition of antisymmetrization of tensor with respect to some arguments.

Note that each of $\sigma$ and ${\sigma }^{\prime }$ is a bijection, $:\{1,...,k\}\to \{1,...,k\}$, and ${\sigma }_j$ is nothing but $\sigma (j)$.

$=1/k!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{t}^{{\sigma }_1}(v_{{\sigma }^{\prime }({\sigma }_1)})...t^{{\sigma }_k}(v_{{\sigma }^{\prime }({\sigma }_k)})$.

$(t^{{\sigma }_1},...,t^{{\sigma }_k})$ can be permutated to $(t^1,...,t^k)$ while the argument of $t^j$ is $v_{{\sigma }_j^{\prime }}$, because $t^{{\sigma }_l}=t^j$ means that $j={\sigma }_l$ and the argument of $t^{{\sigma }_l}$ is $v_{{\sigma }^{\prime }({\sigma }_l)}=v_{{\sigma }^{\prime }(j)}=v_{{\sigma }_j^{\prime }}$.

So, $=1/k!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{t}^1(v_{{\sigma }_1^{\prime }})...t^k(v_{{\sigma }_k^{\prime }})=1/k!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{t}^1\otimes ...\otimes t^k(v_{{\sigma }_1^{\prime }},...,v_{{\sigma }_k^{\prime }})=Asym(t^1\otimes ...\otimes t^k)(v_1,...,v_k)$, by the definition of antisymmetrization of tensor with respect to some arguments.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1123: Associativity for 3 Items Allows Any Association

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that associativity for 3 items allows any association

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Topics

About: structure

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

---

Starting Context

• The reader knows a definition of structure.

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Target Context

• The reader will have a description and a proof of the proposition that for any structure, the associativity for any 3 items allows any association.

---

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:
$S$: $\in \{\text{ the structures }\}$
$\ast$: $\in \{\text{ the operations of }S\}$
//

Statements:
$\mathrm{\forall }{s}_1,s_2,s_3\in S((s_1\ast s_2)\ast s_3=s_1\ast (s_2\ast s_3))$
$⟹$
$s_1\ast ...\ast s_n:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{n-1})\ast s_n$ can be associated in any way
//

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2: Note

Associativity is generally defined with respect to 3 items, which is being understood to allow any associativity. Let us confirm that that is indeed the case.

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3: Proof

Whole Strategy: Step 1: for each $1\le j\le n-1$, associate $s_j$ and $s_{j+1}$ 1st; Step 2: conclude the proposition.

Step 1:

$s_1\ast ...\ast s_n:=(...(((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Letting $a:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})$, it is $(...((a\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Applying the associativity for 3 items to $(a\ast s_j)\ast s_{j+1}$, $=(...(a\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n=(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n$.

So, any $s_j$ and $s_{j+1}$ can be associated.

Step 2:

Letting $b:=s_j\ast s_{j+1}$, it is $(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast b)\ast ...\ast s_{n-1})\ast s_n$.

By Step 1, its any neighboring 2 items can be associated.

That is what it means by $s_1\ast ...\ast s_n$ "can be associated in any way".

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

1122: For Map, Cardinality of Range Is Equal to or Smaller Than Cardinality of Domain

<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, cardinality of range is equal to or smaller than cardinality of domain

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Topics

About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of map.
• The reader knows a definition of cardinality of set.
• The reader knows a definition of relation.

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Target Context

• The reader will have a description and a proof of the proposition that for any map, the cardinality of the range is equal to or smaller than the cardinality of the domain.

---

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:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$f$: $:S_1\to S_2$
//

Statements:
$Card(f(S_1))\le Card(S_1)$
//

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2: Proof

Whole Strategy: Step 1: think of the relation, $R:=\{(s_2,s_1)\in S_2\times S_1|s_2=f(s_1)\}$; Step 2: apply the axiom of choice to have a function, $F\subseteq R$, such that $Dom(F)=Dom(R)$.

Step 1:

Let us think of the relation, $R:=\{(s_2,s_1)\in S_2\times S_1|s_2=f(s_1)\}$.

The domain of $R$ is $Dom(R)=f(S_1)$.

$R$ is not necessarily any function, because for an $s_2$, there may be some multiple $s_1$ s.

Step 2:

But by the axiom of choice, there is a function, $F\subseteq R$, such that $Dom(F)=Dom(R)$.

$F$ is a map from $Dom(F)$ into $S_1$.

$F$ is injective, because for any $s_2,s_2^{\prime }\in Dom(F)$ such that $s_2\ne s_2^{\prime }$, $F(s_2)\ne F(s_2^{\prime })$, because if $F(s_2)=F(s_2^{\prime })$, $s_2=f(F(s_2))=f(F(s_2^{\prime }))=s_2^{\prime }$, a contradiction.

So, $Card(f(S_1))=Card(Dom(R))=Card(Dom(F))\le Card(S_1)$.

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References

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1121: For C∞ Manifold with Boundary, Interior Point Has r′-r-Open-Balls Charts Pair and Boundary Point Has r′-r-Open-Half-Balls Charts Pair

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary, interior point has $r^{\prime }$-$r$-open-balls charts pair and boundary point has $r^{\prime }$-$r$-open-half-balls charts pair

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$r^{\prime }$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$
$r$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$ such that $r<r^{\prime }$
//

Statements:
(
$m\in \{\text{ the interior points of }M\}$
$⟹$
$\mathrm{\exists }((B_{m,r^{\prime }}\subseteq M,{\varphi }_m),(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}}))\in \{\text{ the }{r}^{\prime }\text{ - }r\text{ -open-balls charts pairs around }m\text{ on }M\}$
)
$\wedge$
(
$m\in \{\text{ the boundary points of }M\}$
$⟹$
$\mathrm{\exists }((H_{m,r^{\prime }}\subseteq M,{\varphi }_m),(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}}))\in \{\text{ the }{r}^{\prime }\text{ - }r\text{ -open-half-balls charts pairs around }m\text{ on }M\}$
)
//

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2: Proof

Whole Strategy: Step 1: suppose that $m$ is any interior point, and take any $r^{\prime }$-open-ball chart around $m$, $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$; Step 2: take $B_{{\varphi }_m(m),r}\subset B_{{\varphi }_m(m),r^{\prime }}$ and define $B_{m,r}:={\varphi }_m^{-1}(B_{{\varphi }_m(m),r})$; Step 3: see that $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is an $r$-open-ball chart around $m$; Step 4: suppose that $m$ is any boundary point, and take any $r^{\prime }$-open-half-ball chart around $m$, $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$; Step 5: take $H_{{\varphi }_m(m),r}\subset H_{{\varphi }_m(m),r^{\prime }}$ and define $H_{m,r}:={\varphi }_m^{-1}(H_{{\varphi }_m(m),r})$; Step 6: see that $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is an $r$-open-half-ball chart around $m$.

Step 1:

Let us suppose that $m$ is any interior point.

Let us take any $r^{\prime }$-open-ball chart around $m$, $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$, which is possible, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

Step 2:

Let us take $B_{{\varphi }_m(m),r}\subset B_{{\varphi }_m(m),r^{\prime }}$.

Let us define $B_{m,r}:={\varphi }_m^{-1}(B_{{\varphi }_m(m),r})\subset B_{m,r^{\prime }}$.

$B_{m,r}$ is an open neighborhood of $m$ on $B_{m,r}$ and on $M$.

Step 3:

$(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is a chart, because ${\varphi }_m{|}_{B_{m,r}}:B_{m,r}\to B_{{\varphi }_m(m),r}$ is a homeomorphism and $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is $C^{\mathrm{\infty }}$ compatible with larger $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$.

So, $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is an $r$-open-ball chart around $m$.

So, $((B_{m,r^{\prime }}\subseteq M,{\varphi }_m),(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}}))$ is an $r^{\prime }$-$r$-open-balls charts pair around $m$.

Step 4:

Let us suppose that $m$ is any boundary point.

Let us take any $r^{\prime }$-open-half-ball chart around $m$, $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$, which is possible, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

Step 5:

Let us take $H_{{\varphi }_m(m),r}\subset H_{{\varphi }_m(m),r^{\prime }}$.

Let us define $H_{m,r}:={\varphi }_m^{-1}(H_{{\varphi }_m(m),r})\subset H_{m,r^{\prime }}$.

$H_{m,r}$ is an open neighborhood of $m$ on $H_{m,r}$ and on $M$.

Step 6:

$(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is a chart, because ${\varphi }_m{|}_{H_{m,r}}:H_{m,r}\to H_{{\varphi }_m(m),r}$ is a homeomorphism and $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is $C^{\mathrm{\infty }}$ compatible with larger $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$.

So, $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is an $r$-open-half-ball chart around $m$.

So, $((H_{m,r^{\prime }}\subseteq M,{\varphi }_m),(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}}))$ is an $r^{\prime }$-$r$-open-half-balls charts pair around $m$.

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References

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