Arf invariant for quadratic forms I am reading a paper in which they define the Arf invariant as follows:
Let $V$ be a vector space of dimension $2n$ over $\mathbb{Z}_2$ and $f$ is a non-degenerate bilinear form $V$. We call $q$ a quadratic form on $(V,f)$ if $$q(x+y)=q(x)+q(y)+f(x,y).$$ Then we define the Arf invariant of $q$ as $$\operatorname{Arf}(q)=\frac{1}{|V|}\sum_{\alpha \in V} (-1)^{q(\alpha)}.$$ Now they state that there are exactly $2^{2n-1}+2^{n-1}$ quadratic forms on $(V,f)$ with Arf invarian $1$ and $2^{2n-1}-2^{n-1}$ quadratic forms with Arf invariant $-1$. 
I don't understand why but if the statement is true, since there are exactly $2^{2n}$ quadratic forms on $(V,f)$, so each quadratic form always has Arf invariant $1$ or $-1$. It's not true by the definition.
Some one can help me? Thanks a lot!
 A: Let's first explain why that formula always gives $\pm1$.
First let's make the observation that $f(x,x)=0$ for all $x\in V$. This is because
$$
q(x+x)=q(x)+q(x)+f(x,x),
$$ 
and here $q(x+x)=q(0)=0$ and $q(x)+q(x)=2q(x)=0$.
Let then
$$
\operatorname{Arf}(q)=\frac1{\sqrt{|V|}}\sum_{x\in V}(-1)^{q(x)}.
$$
Then we get
$$
\operatorname{Arf}(q)^2=\frac1{|V|}\sum_{x\in V}\sum_{y\in V}(-1)^{q(x)}(-1)^{q(y)}.
$$
Here we get
$$
(-1)^{q(x)}(-1)^{q(y)}=(-1)^{q(x)+q(y)}=(-1)^{q(x+y)+f(x,y)}.
$$
Let us take $z=x+y\in V$ as a new variable. As $(x,y)$ range over $V\times V$ so do $(x,z)$, so we get (solving $y=z-x=z+x$)
$$
\operatorname{Arf}(q)^2=\frac1{|V|}\sum_{z\in V}\left((-1)^{q(z)}\sum_{x\in V}(-1)^{f(x,z+x)}\right).
$$
Here in the inner sum
$$
f(x,z+x)=f(x,z)+f(x,x)=f(x,z).
$$
If the fixed value of $z$ in the inner sum is $\neq0$, then the form $f(x,z)$ takes both values,$0$ and $1$, equally often by non-degeneracy of $f$. OTOH if $z=0$ then $f(x,z)=f(x,0)=0$ for all $x\in V$. Therefore the inner sum (over $x$) is equal to zero, unless $z=0$ in which case the inner sum is equal to $|V|$. So in the end we get
$$
\operatorname{Arf}(q)^2=\frac{|V|}{|V|}(-1)^{q(0)}=(-1)^0=1.
$$
Therefore this definition of the Arf-invariant always gives you $(-1)^\epsilon,$ with $\epsilon\in\{0,1\}$. I dare guess that
the (more common?) Arf-invariant is just that exponent $\epsilon$. I need to dig a bit deeper to recall how that is gotten using a symplectic basis.

[Edit]: Assume that $a_1,a_2,\ldots,a_n$, $b_1,b_2,\ldots,b_n$ is a symplectic basis of the
quadratic space $V$. In other words we have $f(a_i,a_j)=f(b_i,b_j)=0$, $f(a_i,b_j)=\delta_{ij}$.
An arbitrary vector $v\in V$ can be written using the symplectic basis as
$$
v=\sum_{i=1}^nx_ia_i+\sum_{j=1}^ny_jb_j
$$
with coordinates $x_i,y_j\in\{0,1\}$ for all $i,j$. Repeated application of the relation between $q$ and $f$ as well as our assumptions then allow us to evaluate $q(v)$ to be equal to (remember that always $x_i^2=x_i$ and $y_j^2=y_j$, so $q(x_ia_i)=x_i^2q(a_i)=x_iq(a_i)$ et cetera)
$$
\begin{aligned}
q(v)&=q(\sum_i x_i a_i+\sum_j y_jb_j)\\
&=q(\sum_i x_i a_i)+q(\sum_j y_j b_j)+f(\sum_i x_ia_i,\sum_j y_jb_j)\\
&=\sum_i x_iq(a_i)+\sum_i y_iq(b_i)+\sum_i x_iy_i\\
&=\sum_i (x_i+q(b_i))(y_i+q(a_i))+\sum_i q(a_i)q(b_i).
\end{aligned}
$$
Write $x_i'=x_i+q(b_i)$, $y_i'=y_i+q(a_i)$. As all the coordinates $x_i$, $y_i$, $i=1,2,\ldots,n$, range over $\mathbb{F}_2$ so do $x_i',y_i'$. Therefore
$$
\begin{aligned}\operatorname{Arf}(q)&=\frac1{2^n}\sum_{x_1,x_2,\ldots,x_n\in\mathbb{F}_2}\sum_{y_1,y_2,\ldots,y_n\in\mathbb{F}_2}(-1)^{q(\sum_i x_i a_i+\sum_j y_jb_j)}\\
&=\frac1{2^n}(-1)^{\sum_iq(a_i)q(b_i)}
\sum_{x_1',x_2',\ldots,x_n'\in\mathbb{F}_2}\sum_{y_1',y_2',\ldots,y_n'\in\mathbb{F}_2}(-1)^{x_1'y_1'+x_2'y_2'+\cdots x_n'y_n'}\\
&=(-1)^{\sum_{i=1}^nq(a_i)q(b_i)}.
\end{aligned}
$$
In the last step I used the easy to prove fact that the inner sum with the usual inner product involving primed coordinates gives a total $2^n$. The proof is similar to what I did earlier with the inner sum.
Thus your definition of the Arf-invariant is, as suspected, equal to $(-1)^\epsilon$, where
$\epsilon$ is the Arf-invariant from the source given by Colin McQuillan (+1).
A: The Arf invariant is $0$ if that sum is positive, and $1$ if that sum is negative. This is the "democratic invariant" definition of the Arf invariant; see for example
http://books.google.co.uk/books?id=fqexbFMISJ8C&lpg=PA109&ots=ggLVoA30gY&dq=browder%20arf%20invariant&pg=PA111#v=onepage&q=corollary%209.5&f=false
