2025-06-16

1167: Wedge Product of 1-Covectors Is Sum of Signed Reordered Tensor Products of 1-Covectors

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description/proof of that wedge product of 1-covectors is sum of signed reordered tensor products of 1-covectors

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 the wedge product of any 1-covectors is the sum of the signed reordered tensor products of the 1-covectors.

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: { the fields }
V: { the F vectors spaces }
{t1,...,tn}: Λ1(V:F)
t1...tn: Λn(V:F)
//

Statements:
t1...tn=σSnsgnσtσ1...tσn
//


2: Note


Some cumbersomeness of the definition of wedge product is that the wedge product cannot be handled without some arguments put into it.

This proposition allows the wedge product to be handled as a combination of some tensor products.


3: Proof


Whole Strategy: see that the results of the both hand sides on any arguments combination are the same; Step 1: take any finite sequence of vectors, (v1,...,vn); Step 2: put (v1,...,vn) into t1...tn; Step 3: put (v1,...,vn) into σSnsgnσtσ1...tσn and see that the result equals that of Step 2.

Step 1:

Let us take any finite sequence of vectors, (v1,...,vn), where vjV.

As each element of Λn(V:F) is a tensor and it is uniquely determined by how it operates on the arguments, if the results of t1...tn and σSnsgnσtσ1...tσn on (v1,...,vn) are the same, t1...tn and σSnsgnσtσ1...tσn will be the same element.

Step 2:

t1...tn(v1,...,vn)=n!/(1!...1!)Asym(t1...tn)(v1,...,vn)=n!1/n!σSnsgnσt1...tn(vσ1,...,vσn)=σSnsgnσt1...tn(vσ1,...,vσn)=σSnsgnσt1(vσ1)...tn(vσn).

Step 3:

On the other hand, σSnsgnσtσ1...tσn(v1,...,vn)=σSnsgnσtσ1(v1)...tσn(vn).

tσ1,...,tσn is a permutation of t1,...,tn, so, tσ1(v1)...tσn(vn) can be reordered to t1(vλ(1))...tn(vλ(n)), where λ:{1,...,n}{1,...,n} is a function.

What is λ really like? For each j{1,...,n}, tj(vλ(j))=tσl(vl) for an l{1,...,n}. So, j=σl=σ(l) (σ is indeed a bijective map), so, l=σ1(j). λ(j)=l, so, λ(j)=l=σ1(j). So, λ=σ1.

So, tσ1(v1)...tσn(vn)=t1(vσ1(1))...tn(vσ1(n)).

So, σSnsgnσtσ1...tσn(v1,...,vn)=σSnsgnσt1(vσ1(1))...tn(vσ1(n)), but as sgnσ=sgnσ1, =σSnsgnσ1t1(vσ1(1))...tn(vσ1(n)), but as σSn=σ1Sn, =σ1Snsgnσ1t1(vσ1(1))...tn(vσ1(n)).

That equals σSnsgnσt1(vσ1)...tn(vσn), the result of Step 2.

So, t1...tn=σSnsgnσtσ1...tσn.


References


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