# 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 $$Arf(q)=\frac{1}{|V|}\sum_{\alpha \in V} (-1)^{q(\alpha)}.$$ 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!

-
Could you link to the paper? – Colin McQuillan Mar 7 '13 at 15:26
In fact it's on the page 16 of the following paper arxiv.org/pdf/math-ph/0608070v2.pdf – mapping Mar 7 '13 at 20:38
Thanks - how strange! – Colin McQuillan Mar 8 '13 at 0:26

## 2 Answers

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 $$Arf(q)=\frac1{\sqrt{|V|}}\sum_{x\in V}(-1)^{q(x)}.$$ Then we get $$Arf(q)^2=\frac1{|V|}\sum_{x\in V}\sum_{y\in V}(-1)^{q(x)}(-1)^{q(y)}.$$ 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 $y=z-x=z+x$) $$Arf(q)^2=\frac1{|V|}\sum_{z\in V}\left((-1)^{q(z)}\sum_{x\in V}(-1)^{f(x,z+x)}\right).$$ 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 $$Arf(q)^2=\frac{|V|}{|V|}(-1)^{q(0)}=(-1)^0=1.$$ 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.

: 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} 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).

-

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

-
Thanks for your link, but after reading it, I found something different: By your link, the Arf invarian is just an element in $\mathbb{Z}_2$, namely $0$ or $1$, but in my paper, by the definition and the last statement, it's not true. – mapping Mar 8 '13 at 11:43
Finally I asked professor and there was a typing mistake in definition of $Arf$: It must be $\sqrt{|V|}$ instead of $|V|$. – mapping Mar 11 '13 at 21:35
@mapping: that comment should be an answer – Colin McQuillan Mar 11 '13 at 21:59
@McQuillan Yes finally I need to check that two definitions (one in the paper and one in your link) are equivalent. Do you have any idea? – mapping Mar 12 '13 at 8:55