Hyperplane arrangement $\mathcal{A}$ obtained from the essentialization of $\mathcal{A}$ From the top of page $3$ in http://www.cis.upenn.edu/~cis610/sp06stanley.pdf, it says the following 
I do not understand what is meant by "...then the arragnement $\mathcal{A}$ is obtained from $A_{W}$ by 'stretching' the hyperplane $H \cap W  \in \mathcal{A}_{W}$ orthogonally to $W$". Can someone please explain this with a very simple example? Suppose, with $V \cong \mathbb{R}^{2}$? I am having a very hard time visualizing this.
 A: I don't know if this is what you're looking for, but if $W$ is a proper subspace of a vector space $V$, and you have a symmetric, alternating, or hermitian form on $V$ (so a notion of orthogonality), and $\mathfrak h$ is a hyperplane subspace of $W$, then of course $\mathfrak h$ is no longer a hyperplane of $V$.  But $\mathfrak h \oplus W^{\perp} = \{h + w' : h \in \mathfrak h, w' \in W^{\perp} \}$ is a hyperplane of $V$, whose orthogonal complement in $V$ is equal to the orthogonal complement of $\mathfrak h$ in $W$.  This is your "stretched" version of $\mathfrak h$.
When $\mathfrak h$ is an affine hyperplane of $W$, similar reasoning applies, except now you're looking at the affine hyperplane $\mathfrak h + W^{\perp}$ of $V$ (you can't really call it a direct sum since it's not a subspace?)
As an example, take $V = \mathbb{R}^3$ with the usual inner product and let $W$ be the $xy$ plane.  Let $\mathfrak h$ be the affine hyperplane in $W$ consisting of the vertical line $x = -3$.  Then the stretched version of $\mathfrak h$ is the hyperplane in $V$ given by the same equation $x = -3$, except now this thing is a plane instead of a line.
