Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

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.


[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 $$ \begin{aligned} Arf(q)&=\frac1{2^n}\sum_{x_1,x_2,\ldots,x_n\in\mathbb{F}_2}\sum_{y_1,y_2,\ldots,y_n\in\mathbb{F}_2}(-1)^{q(\sum_i x_i a_i+\sum_j y_jb_j)}\\ &=\frac1{2^n}(-1)^{\sum_iq(a_i)q(b_i)} \sum_{x_1',x_2',\ldots,x_n'\in\mathbb{F}_2}\sum_{y_1',y_2',\ldots,y_n'\in\mathbb{F}_2}(-1)^{x_1'y_1'+x_2'y_2'+\cdots x_n'y_n'}\\ &=(-1)^{\sum_{i=1}^nq(a_i)q(b_i)}. \end{aligned} $$ 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).

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.