Suppose we have a $N$-dimensional vector space $V$, equipped with a scalar product $\langle\cdot{ , }\cdot\rangle$ and let $\{e_1,\dots,e_N\}$ be a orthonormal basis.
I've seen that there are different ways to define $\bigwedge^k V$, the exterior $k$-power of $V$, but the most operative one (for my interests) is:
$$\bigwedge ^k V=\text{Span}\{e_{i_1}\wedge \dots \wedge e_{i_{k}}|i_1<\dots<i_k,\, 1\leq k\leq N \},$$ where we can think to $\wedge$ as a anticommutative symbol of multiplication. We can define a scalar product on each $\bigwedge^k V$, requiring that
$$\langle { e_{i_1}\wedge \dots \wedge e_{i_{k}}, } e_{i_1}\wedge \dots \wedge e_{i_{k}} \rangle=1, \\ \langle { e_{i_1}\wedge \dots \wedge e_{i_{k}}, } e_{j_1}\wedge \dots \wedge e_{j_{k}} \rangle=0 \text { if } e_{i_1}\wedge \dots \wedge e_{i_{k}}\neq \pm e_{j_1}\wedge \dots \wedge e_{j_{k}}. $$
Now, $\bigwedge^{N-1}V$ has the same dimension as $V$, its basis is given by $$\{e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N |1\leq i \leq N \}.$$ To each vector $e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N$ it is associated the vector $e_i\in V.$ My professor has said that this kind of correspondence is "equivalent" to say that $e_i$ is orthogonal to $e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N$, but I don't understand in which sense they are orthogonal. Can we identify ${e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N}^\perp$ with $e_i$? If so, how?
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? $\endgroup$ – Asaf Karagila♦ Oct 19 '15 at 10:21