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

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

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

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References

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