1126: Wedge Product of Multicovectors|T.B.P.

<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

Topics

About: vectors space

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.

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 {the fields}

V

\in {the F vectors spaces}

* \land

: Λ_{k_{1}} (V : F) \times Λ_{k_{2}} (V : F) \to Λ_{k_{1} + k_{2}} (V : F), (t_{1}, t_{2}) \mapsto (k_{1} + k_{2})! / (k_{1}! k_{2}!) A s y m (t_{1} \otimes t_{2})

//

Conditions:
//

\land (t_{1}, t_{2})

is usually denoted as

t_{1} \land t_{2}

2: Note

Another definition defines that

t_{1} \land t_{2} = A s y m (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} \land t_{2} \in Λ_{k_{1} + k_{2}} (V : F)

, because

t_{1} \otimes t_{2} \in L (V, . . ., V : F)

A s y m (t_{1} \otimes t_{2}) \in Λ_{k_{1} + k_{2}} (V : F)

, and

(k_{1} + k_{2})! / (k_{1}! k_{2}!) A s y m (t_{1} \otimes t_{2}) \in Λ_{k_{1} + k_{2}} (V : F)

.

Let us see some properties of wedge product.

Let us see that wedge product is associative.

(t_{1} \land t_{2}) \land t_{3} = (k_{1} + k_{2})! / (k_{1}! k_{2}!) A s y m (t_{1} \otimes t_{2}) \land t_{3} = (k_{1} + k_{2} + k_{3})! / ((k_{1} + k_{2})! k_{3}!) A s y m ((k_{1} + k_{2})! / (k_{1}! k_{2}!) A s y m (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}!) A s y m (A s y m (t_{1} \otimes t_{2}) \otimes t_{3}) = (k_{1} + k_{2} + k_{3})! / (k_{1}! k_{2}! k_{3}!) A s y m (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} \land (t_{2} \land t_{3}) = t_{1} \land (k_{2} + k_{3})! / (k_{2}! k_{3}!) A s y m (t_{2} \otimes t_{3}) = (k_{1} + k_{2} + k_{3})! / (k_{1}! (k_{2} + k_{3})!) A s y m (t_{1} \otimes (k_{2} + k_{3})! / (k_{2}! k_{3}!) A s y m (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}!) A s y m (t_{1} \otimes A s y m (t_{2} \otimes t_{3})) = (k_{1} + k_{2} + k_{3})! / (k_{1}! k_{2}! k_{3}!) A s y m (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} \land t_{2}) \land t_{3} = t_{1} \land (t_{2} \land t_{3})

.

So, while

t_{1} \land . . . \land t_{n} := (. . . ((t_{1} \land t_{2}) \land t_{3}) . . . \land t_{n - 1}) \land t_{n}

, it can be associated in any way.

t_{1} \land . . . \land t_{n} = (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) A s y m (t_{1} \otimes . . . \otimes t_{n})

: to prove it inductively, it holds when

n = 2

; supposing that it holds for

n = 2, . . ., n^{'} - 1

t_{1} \land . . . \land t_{n^{'}} = (t_{1} \land . . . \land t_{n^{'} - 1}) \land t_{n^{'}} = (k_{1} + . . . + k_{n^{'} - 1})! / (k_{1}! . . . k_{n^{'} - 1}!) A s y m (t_{1} \otimes . . . \otimes t_{n^{'} - 1}) \land t_{n^{'}} = (k_{1} + . . . + k_{n^{'}})! / ((k_{1} + . . . + k_{n^{'} - 1})! k_{n^{'}}!) A s y m ((k_{1} + . . . + k_{n^{'} - 1})! / (k_{1}! . . . k_{n^{'} - 1}!) A s y m (t_{1} \otimes . . . \otimes t_{n^{'} - 1}) \otimes t_{n^{'}}) = (k_{1} + . . . + k_{n^{'}})! / ((k_{1} + . . . + k_{n^{'} - 1})! k_{n^{'}}!) (k_{1} + . . . + k_{n^{'} - 1})! / (k_{1}! . . . k_{n^{'} - 1}!) A s y m (A s y m (t_{1} \otimes . . . \otimes t_{n^{'} - 1}) \otimes t_{n^{'}}) = (k_{1} + . . . + k_{n^{'}})! / (k_{1}! . . . k_{n^{'}}!) A s y m (t_{1} \otimes . . . \otimes t_{n^{'}})

, by the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.

t_{1} \land . . . \land (r t_{j} + r^{'} t_{j}^{'}) \land . . . \land t_{n} = r t_{1} \land . . . \land t_{j} \land . . . \land t_{n} + r^{'} t_{1} \land . . . \land t_{j}^{'} \land . . . \land t_{n}

, because

t_{1} \land . . . \land (r t_{j} + r^{'} t_{j}^{'}) \land . . . \land t_{n} = (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) A s y m (t_{1} \otimes . . . \otimes (r t_{j} + r^{'} t_{j}^{'}) \otimes . . . \otimes t_{n}) = (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) A s y m (r t_{1} \otimes . . . \otimes t_{j} \otimes . . . \otimes t_{n} + r^{'} t_{1} \otimes . . . \otimes t_{j}^{'} \otimes . . . \otimes t_{n})

, by the property of tensor product of tensors,

= (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) (r A s y m (t_{1} \otimes . . . \otimes t_{j} \otimes . . . \otimes t_{n}) + r^{'} A s y m (t_{1} \otimes . . . \otimes t_{j}^{'} \otimes . . . \otimes t_{n}))

, by the proposition that any antisymmetrization-of-tensor map is linear,

= r (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) A s y m (t_{1} \otimes . . . \otimes t_{j} \otimes . . . \otimes t_{n}) + r^{'} (k_{1} + . . . + k_{n})! / (k_{1}! . . . k_{n}!) A s y m (t_{1} \otimes . . . \otimes t_{j}^{'} \otimes . . . \otimes t_{n}) = r t_{1} \land . . . \land t_{j} \land . . . \land t_{n}) + r^{'} t_{1} \land . . . \land t_{j}^{'} \land . . . \land t_{n}

.

When

(t^{1}, . . ., t^{k})

is any combination of elements of

V^{*}

and

σ \in S_{k}

is any, for each

v_{1}, . . ., v_{k} \in V

t^{σ_{1}} \land . . . \land t^{σ_{k}} (v_{σ_{1}}, . . ., v_{σ_{1}}) = t^{1} \land . . . \land 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^{σ_{1}} \land . . . \land t^{σ_{k}} = k! A s y m (t^{σ_{1}} \otimes . . . \otimes t^{σ_{k}})

and

t^{1} \land . . . \land t^{k} = k! A s y m (t^{1} \otimes . . . \otimes t^{k})

.

When

(t^{1}, . . ., t^{k})

is any combination of elements of

V^{*}

and

σ \in S_{k}

is any,

t^{σ_{1}} \land . . . \land t^{σ_{k}} = s g n σ t^{1} \land . . . \land t^{k}

, because

t^{σ_{1}} \land . . . \land t^{σ_{k}} (v_{1}, . . ., v_{k}) = t^{σ_{1}} \land . . . \land t^{σ_{k}} (v_{(σ \circ σ^{- 1})_{1}}, . . ., v_{(σ \circ σ^{- 1})_{k}}) = t^{1} \land . . . \land t^{k} (v_{σ_{1}^{- 1}}, . . ., v_{σ_{k}^{- 1}})

, by the above result,

= s g n σ^{- 1} t^{1} \land . . . \land t^{k} (v_{1}, . . ., v_{k})

, because

t^{1} \land . . . \land t^{k}

is antisymmetric, but

s g n σ^{- 1} = s g n σ

.

When furthermore

(t^{1}, . . ., t^{k})

has any duplication,

t^{1} \land . . . \land t^{k} = 0

, because supposing

t^{m} = t^{n}

, taking

σ

as the permutation that switches

m

and

n

(

s g n σ = - 1

t^{1} \land . . . t^{m} \land . . . \land t^{n} . . . \land t^{k} = - t^{1} \land . . . t^{n} \land . . . \land t^{m} . . . \land t^{k} = - t^{1} \land . . . t^{m} \land . . . \land t^{n} . . . \land t^{k}

.

When

t_{1} \in Λ_{k_{1}} (V : F)

, a

k_{1}

-covector, and

t_{2} \in Λ_{k_{2}} (V : F)

, a

k_{2}

-covector,

t_{2} \land t_{1} = (- 1)^{k_{1} k_{2}} t_{1} \land 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}} \land . . . \land b^{j_{k_{1}}}

and

t_{2} = {t_{2}}_{l_{1}, . . ., l_{k_{2}}} b^{l_{1}} \land . . . \land b^{l_{k_{2}}}

, and

t_{2} \land t_{1} = ({t_{2}}_{l_{1}, . . ., l_{k_{2}}} b^{l_{1}} \land . . . \land b^{l_{k_{2}}}) \land ({t_{1}}_{j_{1}, . . ., j_{k_{1}}} b^{j_{1}} \land . . . \land b^{j_{k_{1}}}) = {t_{1}}_{j_{1}, . . ., j_{k_{1}}} {t_{2}}_{l_{1}, . . ., l_{k_{2}}} b^{l_{1}} \land . . . \land b^{l_{k_{2}}} \land b^{j_{1}} \land . . . \land b^{j_{k_{1}}}

, but as each

b^{m}

is an element of

V^{*}

, 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 moved to in front of

b^{l_{1}}

with the

(- 1)^{k_{2}}

factor, ..., after all,

b^{l_{1}} \land . . . \land b^{l_{k_{2}}} \land b^{j_{1}} \land . . . \land b^{j_{k_{1}}}

can be changed to

b^{j_{1}} \land . . . \land b^{j_{k_{1}}} \land b^{l_{1}} \land . . . \land 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}} \land . . . \land b^{j_{k_{1}}} \land b^{l_{1}} \land . . . \land b^{l_{k_{2}}} = {- 1}^{k_{1} k_{2}} t_{1} \land t_{2}

.

When furthermore

k_{1} = k_{2} = k

and

t_{1} = t_{2} = t

, when

k

is odd,

t \land t = 0

, because

t \land t = (- 1)^{k^{2}} t \land t = - t \land t

; when

k

is even,

t \land t

is not necessarily

0

t \land t = (- 1)^{k^{2}} t \land t = t \land t

does not imply that

t \land t = 0

: for example,

k = 2

and

t = t_{1, 2} b^{1} \land b^{2} + t_{3, 4} b^{3} \land b^{4}

, then,

t \land t = (t_{1, 2} b^{1} \land b^{2} + t_{3, 4} b^{3} \land b^{4}) \land (t_{1, 2} b^{1} \land b^{2} + t_{3, 4} b^{3} \land b^{4}) = t_{1, 2} t_{3, 4} b^{1} \land b^{2} \land b^{3} \land b^{4} + t_{1, 2} t_{3, 4} b^{3} \land b^{4} \land b^{1} \land b^{2} = t_{1, 2} t_{3, 4} b^{1} \land b^{2} \land b^{3} \land b^{4} + t_{1, 2} t_{3, 4} b^{1} \land b^{2} \land b^{3} \land b^{4} = 2 t_{1, 2} t_{3, 4} b^{1} \land b^{2} \land b^{3} \land b^{4} \neq 0

; if

t = t_{1, 2} b^{1} \land b^{2}

t \land t = 0

, so, sometimes

t \land t = 0

for a

t \neq 0

References

The Bias Planet
Definitions and Propositions
School Mathematics from Higher Viewpoints
To Disentangle Confusing Terms or Discourses
Information Tables
Exploiting LibreOffice or OpenOffice
Exposing the Java Programming Language
Exposing C++
Exposing Python
Exposing C#
Exposing Gradle
Exposing Git
UNO Dispatch Commands

2025-05-25