Let $E$ be a finite dimensional vector space over field $\mathbb R$ with $E^*$ as the dual space of $E$.($\dim E=n$)

$\Omega^P(E):=\{ \alpha\colon \overbrace{E\times\cdots\times E}^{p- times}\rightarrow \mathbb R\ \ , \alpha \text{ is alternating multilinear map}\}$ $\Omega^P(E^*):=\{ u\colon \overbrace{E^*\times\cdots\times E^*}^{p- times}\rightarrow \mathbb R\ \ , u \text{ is alternating multilinear map}\}$

I wish to prove that there exist an $\textbf{unique}$ bilinear form $B\colon \Omega^p(E^*)\times\Omega^p(E)\rightarrow\mathbb R$

such that for all $ f_i\in E^* \ , u_j\in E^{**}\simeq E \qquad i,j=1,\cdots,p$, it has the following rule:

$B(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p)=\det[f_i(u_j)]$.

$\wedge$ is wedge product between alternating maps i.e. $\alpha\wedge\beta(v_1,\cdots,v_{p+p})=\dfrac{1}{p! p!}\displaystyle\sum_{\sigma\in S_{p+p}}sign(\sigma)\alpha(v_{\sigma(1)},\cdots,v_{\sigma(p)})\beta(v_{\sigma(p+1)},\cdots,v_{\sigma(p+p)})$

How can I do this?

  • $\begingroup$ How do you show its existence? I think that's the harder part, and the proof gives uniqueness almost as a side-product. $\endgroup$ – darij grinberg Jul 14 '15 at 19:13
  • $\begingroup$ This is not a proof of existence. If you want to compute $B\left(U, F\right)$ using your definition, you have to write $U$ as a linear combination of "pure wedges" of the form $u_1 \wedge \cdots \wedge u_p$, and write $F$ likewise. Both times, there are many ways to write $U$ (resp. $F$), and you need to check that the resulting values obtained for $B\left(U, F\right)$ will be the same for all of these ways. This is not a triviality; for example, if you replaced the determinant by the trace or by the permanent, then these values would not be all equal. $\endgroup$ – darij grinberg Jul 14 '15 at 20:00
  • 1
    $\begingroup$ I don't understand what you are talking about. $\endgroup$ – darij grinberg Jul 14 '15 at 20:23

To show that a multilinear map such as $B$ is unique, it is sufficient to specify its action on a basis. Since that is how you define $B$ you are done.

Here is a proof of the proceeding statement. Suppose $B$ and $C$ are bilinear maps: $V \times W \to \mathbb{R}$ and suppose $\{v_i\},\{w_i\}$ are bases for the vectors spaces $V, W$. Then if $B(v_i, w_j) = C(v_i, w_j)$ for all $i,j$, then $B = C$.

Let $v, w$ be elements of $V, W$. Then \begin{eqnarray*} B(v, w) &=& B(\sum a_i v_i, \sum b_j w_j) \\ &=& \sum a_i \sum b_j B(v_i, w_j)\\ &=& \sum a_i \sum b_j C(v_i, w_j)\\ &=& C(\sum a_i v_i, \sum b_j w_j)\\ &=& C(v, w) \end{eqnarray*}

  • $\begingroup$ There is not assumption $B(v_i, w_j) = C(v_i, w_j)$ here. $\endgroup$ – bigli Jul 14 '15 at 19:50
  • $\begingroup$ @bigli By definition $B(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p)=C(u_1\wedge\cdots\wedge u_p,f_1\wedge\cdots\wedge f_p) = \det[f_i(u_j)]$ $\endgroup$ – muaddib Jul 14 '15 at 19:52

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