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The question The number $\pi$ and summation by $SL(2,\mathbb Z)$ considers the series $$\sum(\lVert x\rVert+\lVert y\rVert-\lVert x+y\rVert)$$ where the series runs over vectors $x,y\in{\mathbb N}^2$ such that $\det(x,y)=1$. The norm being the standard euclidean one. Its value (equal to $2$) is obtained by a telescoping argument. However, the telescopage requires the knowledge that the auxiliary series $$\sum\frac1{\lVert x\rVert\lVert y\rVert(\lVert x\rVert+\lVert y\rVert)}$$ converges. For this, we need to estimate the distribution of matrices in $\operatorname{SL}_2({\mathbb N})$, in terms of the lengths of their columns.

What is known in this direction? What is approximately the number of matrices $M=(x,y)\in \operatorname{SL}_2({\mathbb N})$ such that $\lVert x\rVert+\lVert y\rVert\le R$, as $R\rightarrow+\infty$?

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  • $\begingroup$ @ChristianRemling Yes indeed. And $M$ is the matrix whose columns are $x,y$. Thank you. $\endgroup$
    – Denis Serre
    Commented Oct 12, 2021 at 15:03
  • $\begingroup$ I am not wise enough to tell in the answers to the linked question—is the auxiliary series really $\sum \lVert x\rVert^{-2}\lVert y\rVert^{-1}$ (not $\lVert y\rVert^{-2}$)? $\endgroup$
    – LSpice
    Commented Oct 12, 2021 at 15:48
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    $\begingroup$ @LSpice Actually, it is symmetric in $x$ and $y$, homogeneous of degree $-3$. It is enough to treat the series mentionned in the question. $\endgroup$
    – Denis Serre
    Commented Oct 12, 2021 at 16:09
  • $\begingroup$ Hmm, I would understand the symmetry as a sum over $\operatorname{SL}_2(\mathbb Z)$, since if $(x, y)$ lies in $\operatorname{SL}_2(\mathbb Z)$ then so does $(y, -x)$, but why is it symmetric as a sum over $\operatorname{SL}_2(\mathbb N)$? (Also, just to check, does your $\mathbb N$ include $0$?) $\endgroup$
    – LSpice
    Commented Oct 12, 2021 at 16:32
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    $\begingroup$ This monoid is freely generated by the transvections $I+E_{12}$ and $I+E_{2,1}$ if memory recalls and so if you can estimate the word length in terms of the norms, you should be able to say something since the number of words of length at most $k$ is easy to compute. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 12, 2021 at 17:52

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Define $d_R(n)$ to be the number of ways to write $n=ab$ with $a$ and $b$ bounded by $R$. Clearly, this is bounded by the standard (unrestricted by $R$) divisor function $d(n)$. To upper-bound the desired number of matrices $M$, it suffices to bound $$\sum_{n \leq R^2} d_R(n) d_R(n+1).$$ Using only the bound $d(m) \ll m^{\varepsilon}$ shows this quantity grows at most like $R^{2+\varepsilon}$.

This can be improved slightly with more work, but maybe this is sufficient.

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