1. Dec 17, 2016

### binbagsss

$\theta(\tau, A) = \sum\limits_{\vec{x}\in Z^{m}} e^{\pi i A[x] \tau }$

$=\sum\limits^{\infty}_{n=0} r_{A}(n)q^{n}$,

where $r_{A} = No. [ \vec{x} \in Z^{m} ; A[\vec{x}] =n]$

where $A[x]= x^t A x$, is the associated quadratic from to the matrix $A$, where here $A$ is positive definite, of rank $m$ and even. (and I think symmetric?)

So I thought that this meant to solve the quadratic $A[x]= \vec{x^t} A \vec{x} = n$, for each $n$, and the representation number is then given by the number of solutions to this?, subject to $\vec{x} \in Z^{m}$ ,

What is $Z^{m}$ here please? ( z the integer symbol)

Many thanks

2. Dec 17, 2016

### Staff: Mentor

$\vec{x} \in Z^{m}$ is simply $\vec{x} = (x_1, \ldots , x_n)^\tau \in \underbrace{\mathbb{Z} \times \ldots \times \mathbb{Z}}_{n-\ times}$.

3. Dec 17, 2016

### binbagsss

no not quite, i needed to check this before i can post my full question, and that my interpretation of what the representation number is correct? (otherwise the question im about to post may not make sense)

4. Dec 18, 2016

### binbagsss

Okay so on the attachment of extract from my book, I'm not understanding the comment '$Q_{1}(x,y)$ and $Q_{2}(x,y)$ yeild the same series since they represent the same integers.'

So as I said above my interpretation of how to compute the $r(n)$ was to :

set $2 Q(x,y) = A(x,y) = n$ , for each $n$ in turn and count the number of solutions to this for each $n$.

So looking at $Q_{0}(x,y)$, should find $2(x^{2}+xy+6y^2)=0$ has one solution (i.e $(x,y)=0$) , $2(x^{2}+xy+6y^2)=1$ should find 2 solutions and $2(x^{2}+xy+6y^2)=2,3$ has no solutions for $x \in Z^m$
Is my understanding correct here?

So then looking at $Q_1 (x,y)$ and $Q_{2} (x,y)$ which differ only on the sign of the $xy$ term, I don't see how it is obvious that these will have the same number of solutions for $Q(x,y) = n$ for each $n$?

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5. Dec 18, 2016

### Staff: Mentor

If you replace $x$ by $-x$ (or equivalently $y$ by $-y\,$) you get the same number of pairs $(x,y) \in \mathbb{Z}^2$ with $Q_i(x,y)=n$. Different pairs though, but equally many.

6. Dec 29, 2016

### binbagsss

oh right thanks,
how is this obvious? e.g how do you know you won't end up getting complex solutions for the corresponding sign change, ta

7. Dec 29, 2016

### Staff: Mentor

$$\#\{(x,y)\in \mathbb{Z}^2\,\vert \,Q_1(x,y)=0\}=\#\{(-x,y)\in \mathbb{Z}^2\,\vert \,Q_1(-x,y)=0\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_1(-x,y)=0\, \wedge \, w=-x\,\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\, \wedge \, w=-x\,\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\, \wedge \, w=x\,\}=\#\{(x,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\}$$ because we consider all pairs in $\mathbb{Z}^2$, so the sign doesn't make any difference in the total number of solutions, only in the way we write, resp. notate them: $\#\{(x,y)\in \mathbb{Z}^2\,\vert \,x=1 \wedge y=2\}=\#\{(x,y)\in \mathbb{Z}^2\,\vert \,x=-1 \wedge y=2\}$.