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
- The reader knows a definition of wedge product of multicovectors.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(\{t_1, ..., t_n\}\): \(\subseteq \Lambda_1 (V: F)\)
\(t_1 \wedge ... \wedge t_n\): \(\in \Lambda_n (V: F)\)
//
Statements:
\(t_1 \wedge ... \wedge t_n = \sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n}\)
//
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, \((v_1, ..., v_n)\); Step 2: put \((v_1, ..., v_n)\) into \(t_1 \wedge ... \wedge t_n\); Step 3: put \((v_1, ..., v_n)\) into \(\sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n}\) and see that the result equals that of Step 2.
Step 1:
Let us take any finite sequence of vectors, \((v_1, ..., v_n)\), where \(v_j \in V\).
As each element of \(\Lambda_n (V: F)\) is a tensor and it is uniquely determined by how it operates on the arguments, if the results of \(t_1 \wedge ... \wedge t_n\) and \(\sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n}\) on \((v_1, ..., v_n)\) are the same, \(t_1 \wedge ... \wedge t_n\) and \(\sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n}\) will be the same element.
Step 2:
\(t_1 \wedge ... \wedge t_n (v_1, ..., v_n) = n! / (1! ... 1!) Asym (t_1 \otimes ... \otimes t_n) (v_1, ..., v_n) = n! 1 / n! \sum_{\sigma \in S_n} sgn \sigma t_1 \otimes ... \otimes t_n (v_{\sigma_1}, ..., v_{\sigma_n}) = \sum_{\sigma \in S_n} sgn \sigma t_1 \otimes ... \otimes t_n (v_{\sigma_1}, ..., v_{\sigma_n}) = \sum_{\sigma \in S_n} sgn \sigma t_1 (v_{\sigma_1}) ... t_n (v_{\sigma_n})\).
Step 3:
On the other hand, \(\sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n} (v_1, ..., v_n) = \sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} (v_1) ... t_{\sigma_n} (v_n)\).
\(t_{\sigma_1}, ..., t_{\sigma_n}\) is a permutation of \(t_1, ..., t_n\), so, \(t_{\sigma_1} (v_1) ... t_{\sigma_n} (v_n)\) can be reordered to \(t_1 (v_{\lambda (1)}) ... t_n (v_{\lambda (n)})\), where \(\lambda: \{1, ..., n\} \to \{1, ..., n\}\) is a function.
What is \(\lambda\) really like? For each \(j \in \{1, ..., n\}\), \(t_j (v_{\lambda (j)}) = t_{\sigma_l} (v_l)\) for an \(l \in \{1, ..., n\}\). So, \(j = \sigma_l = \sigma (l)\) (\(\sigma\) is indeed a bijective map), so, \(l = \sigma^{-1} (j)\). \(\lambda (j) = l\), so, \(\lambda (j) = l = \sigma^{-1} (j)\). So, \(\lambda = \sigma^{-1}\).
So, \(t_{\sigma_1} (v_1) ... t_{\sigma_n} (v_n) = t_1 (v_{\sigma^{-1} (1)}) ... t_n (v_{\sigma^{-1} (n)})\).
So, \(\sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n} (v_1, ..., v_n) = \sum_{\sigma \in S_n} sgn \sigma t_1 (v_{\sigma^{-1} (1)}) ... t_n (v_{\sigma^{-1} (n)})\), but as \(sgn \sigma = sgn \sigma^{-1}\), \(= \sum_{\sigma \in S_n} sgn \sigma^{-1} t_1 (v_{\sigma^{-1} (1)}) ... t_n (v_{\sigma^{-1} (n)})\), but as \(\sum_{\sigma \in S_n} = \sum_{\sigma^{-1} \in S_n}\), \(= \sum_{\sigma^{-1} \in S_n} sgn \sigma^{-1} t_1 (v_{\sigma^{-1} (1)}) ... t_n (v_{\sigma^{-1} (n)})\).
That equals \(\sum_{\sigma \in S_n} sgn \sigma t_1 (v_{\sigma_1}) ... t_n (v_{\sigma_n})\), the result of Step 2.
So, \(t_1 \wedge ... \wedge t_n = \sum_{\sigma \in S_n} sgn \sigma t_{\sigma_1} \otimes ... \otimes t_{\sigma_n}\).
