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

I wish to prove that for each alternating multilinear map $u\colon \overbrace{E^*\times\cdots\times E^*}^{p- times}\rightarrow \mathbb R$ there exist a unique linear map $f:Alt^p(E)\rightarrow \mathbb R $ such that $u(\alpha_1,\cdots,\alpha_p)=f(\alpha_1\wedge\cdots\wedge\alpha_p) \quad \forall \alpha_i\in E^*$.

I want to use only alternating multilinear maps $\textbf{without}$ using tensor product of vector spaces , $\textbf{without}$ the universal property of $k$-the exterior power of a vector space and $\textbf{without}$ $Alt^p(E^*)\simeq (Alt^p(E))^*$ and $(\Lambda^p(E))^*\simeq Alt^p(E)$ for this problem.

$\underline{\textbf{My suggestion:}}$

If someone prove that there exist an $\textbf{unique}$ $\underline{natural}$ bilinear map $B\colon Alt^p(E^*)\times Alt^p(E)\rightarrow\mathbb R$ then the problem is OK.

  • $\begingroup$ $\alpha_i\in E^*$. $\endgroup$ – bigli Jul 20 '15 at 7:37
  • $\begingroup$ Oh, ok, I get it. $\endgroup$ – geodude Jul 20 '15 at 7:39

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