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Proof Verification

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.

Note by Ethan Robinett
2 years, 4 months ago

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