Hey Brilliant users. Below is a proof I wrote as a response to a problem in Calc on Manifolds and I was wondering if any of you can verify its correctness so that I can at least know I am on the right track. The proof may be more detailed than is necessary, but I wrote it this way to make each step clear.

Let {\(\hat{e}_i\)} \(\subset\) \(\mathbb{R}^n\) be the usual basis for \(\mathbb{R}^n\) and let {\(\hat{\phi}_i\)} \(\subset (\mathbb{R}^n)^*\) be the dual basis. Then:

\(\displaystyle \wedge_{j=1}^k \phi_{i_j} (e_{i_1},...,e_{i_k}) = 1\)

Proof:

Clearly: \(\phi_{i_j} \in J^1(\mathbb{R}^n) \forall j \in \mathbb{N}, 1 \leq j \leq k\)

Hence:

\(\displaystyle \wedge_{j=1}^k \phi_{i_j} (e_{i_1},...,e_{i_k}) = k! Alt (\displaystyle \bigotimes_{j=1}^k \phi_{i_j})\)

It follows that:

\(\displaystyle \wedge_{j=1}^k \phi_{i_j} (e_{i_1},...,e_{i_k}) = \displaystyle \sum_{\sigma \in S_k} sgn(\sigma) \bigotimes_{j=1}^k \phi_{i_j}(e_{\sigma(i_j)})\)

Where \(S_k\) denotes the set of all permutations over the numbers \(1,2,...k\). Now:

\(\phi_{i_j}(e_{\sigma(i_j)}) = \delta_{i_j}\)

Where \(\delta_{i_j} =1\) if \(\sigma(i_j) = i_j\) and \(0\) otherwise. Then \(\exists\) only one \(\sigma' \in S_k\) such that:

\(\bigotimes_{j=1}^k \phi_{i_j}(e_{\sigma(i_j)}) \neq 0\). Then:

\(\displaystyle \wedge_{j=1}^k \phi_{i_j} (e_{i_1},...,e_{i_k}) = sgn(\sigma') \bigotimes_{j=1}^k \phi_{i_j}(e_{\sigma'(i_j)})=sgn(\sigma')\)

(Since \(\delta_{i_j} =1\) if \(\sigma(i_j) = i_j\)). Now:

\(sgn(\sigma') = (-1)^m\) where \(m \in \mathbb{N}\) is the number of transpositions of pairs of elements of {\(1,2,...k\)} required to obtain the permutation \(\sigma'\). Clearly, \(m=0\), hence we must have:

\(\displaystyle \wedge_{j=1}^k \phi_{i_j} (e_{i_1},...,e_{i_k}) = 1\)

Note: \(J^1(\mathbb{R}^n)\) denotes the set of all 1-tensors on \(\mathbb{R}^n\) or equivalently the dual space \((\mathbb{R}^n)^*\).

Any feedback is appreciated as usual.

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TopNewestCan you provide the link to the problem, I would love to see the question?

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