Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the question below, I have the following definitions and concepts in mind: The $k^{th}$ exterior power of a real vector space $V$, denoted $\Lambda^k(V)$ can be realized as the quotient of the tensor product $\bigotimes^k V$ with the subspace of $\bigotimes^k V$ generated by all elements of the form $v_1 \otimes \dots \otimes v_k$ where $v_i = v_j$ for some $i \neq j$. The equivalence class of $(v_1, \dots, v_k)$ in $\Lambda^k(V)$ is denoted by $v_1 \wedge \cdots \wedge v_k$; it can be thought of as the image of $(v_1, \dots, v_k)$ under the canonical alternating multilinear map that sends each element in $V^k$ to its equivalence class in $\Lambda^k(V)$. By the universal property of the exterior product, I understand that every alternating multilinear form defined on $V^k$ can identified with a unique linear form with domain $\Lambda^k(V)$. Finally, I know that the determinant of an endomorphism $T:V\rightarrow V$ can be defined as the unique real number $\det T$ such that $$ Tv_1 \wedge \cdots \wedge Tv_n = (\det T)v_1 \wedge \cdots \wedge v_n $$

My Question: It is a fact that if $\phi^i, \dots, \phi^k$ are linear forms on $V$ then $$ \phi^1 \wedge \cdots \wedge \phi^k(v_1, \dots, v_k) = \det[\phi^i(v_j)] $$ Can this be proved using the facts outlined above without resorting to the combinatorial definition of the determinant/wedge product?

share|improve this question
    
Could you say a word about how you're viewing this $\phi_1 \wedge \cdots \wedge \phi_k$ and its action on $k$-tuples? –  Dylan Moreland Jul 8 '12 at 1:12
    
@Dylan Well, I understand from the combinatorial perspective that the $k-$fold exterior product of forms acts like a k-linear alternating function and that, for instance, given a $p$-form $\alpha$ and a $q$-form $\beta$, $$\alpha \wedge \beta (v_1, \dots, v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_k} \epsilon(\sigma) \alpha \otimes \beta (v_{\sigma(1)}, \dots v_{\sigma(p+q)})$$ But there are many other equivalent definitions... –  ItsNotObvious Jul 8 '12 at 1:33

1 Answer 1

Please refer to my answer on my own question :) I had pretty much the same question from a homework and I eventually figured out a proof for that statement.

algebraic manipulation of differential form

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.