I came across one sentence below, I am not able to see it... Any comment suggestion, reference is welcome. Thanks in advance.

"Let $V$ be a finite dimensional inner product space. The complexification of metric endows $\mathbb C\otimes V$ with a complex quadratic form. And this quadratic form $Q$is degenerate. That is there are nonzero element in $\omega\in\mathbb C\otimes V$ such that $Q(\omega)=0$."

Such elements are called isotropic vector. I can't see it, why these element are called isotropic vector(why this terminology).

Edit: Above sentence I found from Laszlo Lempert paper " loop space as complex manifold" page 533, Last paragraph; where he is making above statement for normal bundle of over a loop....

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Copied from wiki: "Isotropy is uniformity in all orientations; it is derived from the Greek iso (equal) and tropos (direction)." See here: en.wikipedia.org/wiki/Isotropy – Paul Feb 23 '12 at 10:09
@Paul, thanks for the comment.. but why in this context, element of complexified vector space are called isotropic.. and what will be $Q$, etc... this is my question.. – zapkm Feb 23 '12 at 10:14
Are you sure about the part "$Q$ is degenerate"? For an Euclidean space (real inner product space of finite dimension) you will get a non-degenerate quadratic form, but which (if $\dim V>1$) has isotropic vectors nonetheless. – Marc van Leeuwen Feb 23 '12 at 12:29
@MarcvanLeeuwen, I am really confuse.. I am also thinking that we will get degenerate form...Above sentence I found from Laszlo Lempert paper " loop space as complex manifold" page 533, Last paragraph; where he is making above statement for normal bundle of over a loop..... – zapkm Feb 23 '12 at 12:41
@MarcvanLeeuwen,, I could not figure out which $Q$ he is using... – zapkm Feb 23 '12 at 12:41

I don't think there is a good rationale for calling these vectors isotropic vectors. Nor is there a good reason to call the stabiliser subgroup of a point its isotropy group. The original meaning of isotropy, "being similar in different directions" as it is used in physics, would apply to a whole space or to a physical medium, but it seems to make little sense when applied to a vector or to a group. One thing that one could say about isotropic vectors is that, in contrast to other vectors, the quadratic form distinguishes them in no way from their scalar multiples, but that is more like "being the similar in the same direction". A particular perverse use of the term is to call the quadratic form itself isotropic if it has isotropic vectors (as Wikipedia claims, I haven't seen this used myself), since this ensures (except in rare cases where all vectors are isotropic) that the quadratic form is not the same in all directions (some directions are isotropic, others anisotropic), whereas an anisotropic quadratic form, like $x^2+y^2+z^2$ in $\mathbb R^3$, may very well be similar in all directions.
Let $\langle \cdot, \cdot \rangle$ denote the inner product on $V$. Suppose $\omega = 1\otimes v + i \otimes w$ then $Q(\omega) = \langle v,v \rangle - \langle w, w \rangle + 2i \langle v,w \rangle$. So such $\omega$ for which $Q(\omega) = 0$ correspond with pairs $(v, w)$ of orthogonal vectors of the same norm in $V$.