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I'm confused about the proof of the following claim:

Set $Z_2 = \mathbb{Z}/2\mathbb{Z} = \mathbb{F}_2$. Let $V$ be a $Z_2$-vector space of dimension $2n$ and let $e_i, f_i$ be a symplectic basis. Let $q: V \to Z_2$ be a non-degenerate quadratic form. Define $c(q) = \sum_i q(e_i) q(f_i)$. Then $c$ is independent of the choice of basis for $V$.

We can show this by showing that any $q(v) = q(x_1 e_1 + y_1 f_1 + \dots + x_n e_n + y_n f_n)$ can be written in the form $ x_1 y_1 + \dots x_n y_n + c(q) (x_n^2 + y_n^2)$.

Here's an outline of the proof(as given for example here on page 318 or here p. 98):

First, we show it for $\mathrm{dim}(V) = 2$. In dimension $2$ there are only two non-equivalent non-degenerate quadratic forms, $q_0, q_1$ and they are such that $q_0(v) = x_1 y_1$ and $q_1 (v) = x_1 y_1 + x_1^2 + y_1^2$ and $c(q_0)= 0$ and $c(q_1) = 1$.

Next we note that $q_0 + q_0$ and $q_1 + q_1$ are equivalent quadratic forms.

Then if $q$ is any quadratic form on $V$ with $\mathrm{dim}(V) = 2n$ we can either write $q = q_0 + \dots + q_0 = n q_0$ or $q = (n-1)q_0 + q_1$. The respective forms written in coordinates are $x_1 y_1 + \dots + x_n y_n$ and $x_1 y_1 + \dots + x_n y_n + x_n^2 + y_n^2$ and we have $c(nq_0 ) = 0$ and $c((n-1)q_0 + q_1) = 1$.

To conclude the proof one needs to show that $n q_0$ and $(n-1)q_0 + q_1$ are not equivalent.


Why is this last step necessary? At that time we already have that we can write any $q$ in a coordinate independent way and that there are only two non-equivalent non-degenerate quadratic forms on any symplectic vector space $V$. Thanks for your help.

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    @Matt: I think we already know that $c(nq_0)\neq c((n-1)q_0+q_1)$, so it is not an invariant, if those two forms are equivalent.2012-08-31

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