In my research, I am trying to use the following construction by Benson Farb and John Franks, which proves that for all $n$, the group of $n\times n$ matrices with 1's on the diagonal, 0's above the diagonal and integer entries below embeds as a subgroup of $C^1(S^1)$.


The construction hinges on defining interval lengths with the following sums:

Let $K>0$ and $B_K : \mathbb{Z}^n \to \mathbb{R}$ be defined by \begin{align*} B_K(q_1,q_2,\ldots,q_n) &= K + \sum_{j=1}^{n} q^{2n-2j+2}_j\\ &=q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K \end{align*} and let $S_K$ be defined by $$ S_K = \sum_{(q_1,q_2,\ldots,q_n)\in\mathbb{Z}^n} \frac{1}{B_K(q_1,q_2,\ldots,q_n)}.$$ The authors off-handedly say the sum defining $S_K$ converges by the integral and comparison tests. When I first saw the sum, I was like "Pfff of course, I'll just do an inductive comparison to the harmonic series. Easy squeezy lemons."

Two days later and I have no idea how to prove convergence.

I've tried induction on $n$. Clearly converges for $n=1$, assume true for $n-1$ and write \begin{align*} S_K =\sum_{q_1\in\mathbb{Z}}\ \sum_{(q_2, q_3,\ldots,q_n)\in\mathbb{Z}^{n-1}} \frac{1}{q^{2n}_1+q^{2n-2}_2+\cdots+q^4_{n-1}+q^2_n +K}\end{align*} but I cannot figure out a way to extract a $q_1$ term to prepare for the second summation.

I've also tried to sum one $q_i$ at a time, performing the integral test at each step, i.e.

$$S_K=\sum_{(q_1, q_2,\ldots,q_{n-1})\in\mathbb{Z}^{n-1}}\ \sum_{q_n\in\mathbb{Z}} \frac{1}{A + q^2_n}$$

where $A$ is all of the other terms. Then, by the integral test $$S_K \leq \sum_{(q_1, q_2,\ldots,q_{n-1})}\frac{1+\frac{\pi}{2}\sqrt{A}}{A}$$ but this square root stirs up trouble for me because I'm forced to halve the exponents of all the terms in $A$ for my next approximation. This works for $n=2$ and $n=3$, but I can't push it further.

I'm close to foaming at the mouth, so if y'all have any input it would be greatly appreciated.

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    $\begingroup$ I think they may have meant \sum_i q_i^{2^i} or something. In your case the number of q\in \Z^n with A\leq B_K(q) < 2A is \asymp A^{(1/2 + o(1)) \log{n}} [estimate term by term and you’ll see the answer is the product of A^{1/(2i)}], and so I think the sum actually diverges when n\geq 4 or something. In the corrected case the number of such q is \asymp A^{1 - 2^{-n} (1+o(1))}, which, by virtue of having an exponent < 1, converges. (I’ve written the sum as \asymp \sum_{a\geq \log{K}} 2^{-a} * \sum_{q : 2^a\leq B_K(q) < 2^{a+1}} 1.). This, by the way, is a superbly written question. :) $\endgroup$ – alpoge Apr 13 '18 at 19:09
  • $\begingroup$ @alpoge Thank you. I thought the same thing, that they must have meant the exponents to be $2^i$. However, they directly use that same definition of $S_K$ to prove two crucial lemmas afterward (Lemma 2.8 and 2.10). They seem to really mean it. I'll see if those lemmas can be adjusted to fit your corrected definition. $\endgroup$ – P. May Apr 13 '18 at 19:28
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    $\begingroup$ Note that "which proves there exists a nilpotent subgroup of $C^1(S^1)$" is more than vague; literally this is a vacuously true statement. The actual statement is that for every $n$ the group of lower triangular $n\times n$ integral matrices with $1$ on the diagonal, embeds into $C^1(S^1)$ (and hence as a corollary, every finitely generated torsion-free nilpotent group embeds in the latter group). If the proof has a gap, this is quite embarrassing! $\endgroup$ – YCor Apr 13 '18 at 20:41
  • $\begingroup$ @Ycor You are correct. I was trying to be concise and ended up being silly. I'll make the edit. $\endgroup$ – P. May Apr 13 '18 at 21:24
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    $\begingroup$ I asked A. Navas, who told me that his student D. Jorquera noticed the error, which is repaired in sciencedirect.com/science/article/pii/S0166864112000922; he provided the further references cambridge.org/core/journals/…, link.springer.com/article/10.1007/s00208-013-0995-1, mat.usach.cl/images/Profesores/navas-papers/30-13-critic.pdf $\endgroup$ – YCor Apr 14 '18 at 17:39

The sum diverges already for $n=4$. To see this, let $L\geq K$ be a dyadic parameter, and consider the contribution of $q_1\asymp L^{1/8}$, $q_2\asymp L^{1/6}$, $q_3\asymp L^{1/4}$, $q_4\asymp L^{1/2}$. If the implied constants are sufficiently close to each other, these ranges are pairwise disjoint. However the contribution of such a range is $\asymp L^{1/8+1/6+1/4+1/2}/L>1$. Summing up over the various $L$'s, the claim follows.


The series indeed diverges for $n\ge4$. Indeed, without loss of generality $K\ge1$. For real $A\ge1$ and $p>1$ we have
\begin{equation} \int_1^\infty\frac{dq}{A+q^p}\ge\int_1^\infty\frac{dq}{(A^{1/p}+q)^p} \gg\frac1{A^{1-1/p}}. \end{equation} So, for $n\ge4$ \begin{align*} S_K&\ge\int_{[1,\infty)^n} \frac{dq_1\cdots dq_n}{K+q^{2n}_1+\cdots+q^{8}_{n-3}+q^{6}_{n-2}+q^4_{n-1}+q^2_n} \\ &\gg\,\int_{[1,\infty)^{n-1}} \frac{dq_1\cdots dq_{n-1}}{K+q^{n}_1+\cdots+q^{4}_{n-3}+q^{3}_{n-2}+q^2_{n-1}} \\ &\gg\,\int_{[1,\infty)^{n-2}} \frac{dq_1\cdots dq_{n-2}}{K+q^{n/2}_1+\cdots+q^{2}_{n-3}+q^{3/2}_{n-2}} \\ &\gg\,\int_{[1,\infty)^{n-3}} \frac{dq_1\cdots dq_{n-3}}{K+q^{n/6}_1+\cdots+q^{2/3}_{n-3}} =\infty. \end{align*}


The authors just posted a correction to the paper on arXiv, where they say:

"In an earlier version of this paper we used $2n − 2j + 2$ as the exponent in the definition of BK instead of $4n − 2j + 2.$ As a result (as was pointed out to us) the series SK might not converge. However, the only properties we use of BK are that SK converges and that the ratio of the value of BK at certain points limits to 1."


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