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Let $V$ be a vector space with basis $e_1, \ldots, e_n$ and $V^*$ be its dual space with dual basis $e_1^*, \ldots, e_n^*$. Let $k$ be an integer between $1$ and $n$. Why $\wedge^{n-k}V=\wedge^{k}V^*$? Thank you very much.

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One usually calls $\Lambda^k V$ an exterior power, not a wedge product. – Mariano Suárez-Alvarez Jan 23 '11 at 4:43
I've deleted the tag "algebraic geometry". – Martin Brandenburg Jan 23 '11 at 9:07
@Mariano, thank you very much. – user Jan 23 '11 at 15:15
up vote 10 down vote accepted

This is slightly false. The two are isomorphic, but not canonically so. There is a natural pairing $\Lambda^{n-k} V \times \Lambda^k V \to \Lambda^n V$ given by exterior product, but this pairing does not identify $\Lambda^{n-k} V$ with $(\Lambda^k V)^{\ast}$ until you pick an isomorphism $\Lambda^n V \simeq k$; this implies a choice of orientation, but is slightly stronger; one might say it implies a choice of "volume form." But it does not imply a choice of inner product.

The (canonical) isomorphism between $(\Lambda^k V)^{\ast}$ and $\Lambda^k V^{\ast}$ comes from the way duals commute with tensor products. It should look pretty straightforward with a specific basis.

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thank you very much. – user Jan 23 '11 at 16:32
I retract my last claim conditional on the outcome of the discussion at… . – Qiaochu Yuan Jun 8 '11 at 22:27

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