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This is a small piece of article from

<...> So, first of all, consider a vector arrangement $S=\{v_1,\ldots,v_n\}\subset V$ where V is your favorite vector space over your favorite coefficient field $\mathbb{K}$. This is just an arbitrary set of non-zero vectors (with repetition allowed), though, for simplicity, I’ll assume it spans all of V. You could also think of this as a hyperplane arrangement on $V^*$, by looking at the zero set of $v_i$ thought of as a function on $V^*$.

Now, let $W$ be the vector space of solutions to the equation $\sum_{i=1}^nw_iv_i=0$, where $w_i\in \mathbb{K}$. This is not just a vector space, but one equipped with a set distinguished functions $w_i:W\to \mathbb{K}$. That is, $S^\vee = \{w_1,\ldots, w_n\}\subset W^*$ is a new vector arrangement, which we call the Gale dual to S. <...>

Can you explain me, how functions $w_i \in W^*$ act on W.

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The article in question is… – Vladimir Sotirov Jul 12 '11 at 6:28
up vote 1 down vote accepted

One way to see this formally , is to consider $A= \mathbb{K}^n$ with an explicit basis $e_1, \ldots, e_n$, and a map $A \mapsto V $ by $e_i \mapsto v_i$. Then you can think of $W$ as the kernel of this. Dualizing the inclusion map of $W$ into $A$ get a map from $A^*$ into $W^*$. The $w_i$ are the dual vectors to $e_i$ and we identify them with their images in $W^*$.

Same thing said slightly more informally, a vector in $W$ is actually a collection of coefficients $w_i$ (this is saying a vector in $W$ is actually a special vector in $A$), and you can take one of those coefficients - that's a function on $W$ (the coordinates are functions on $A$, and we can restrict them to $W$).

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Am I right that for any $a \in W ~~~ w_i(a) = i-$th coordinate if a? – Aspirin Jul 12 '11 at 16:59
@Alexey: Yes, exactly. – Max Jul 12 '11 at 17:56

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