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I speculated in 2008 that the modified Neretin polynomials presented in A145900 of the On-line Encyclopedia of Integer Sequences, which can be summed to give a normalized Schwarzian derivative for a complex function and are related to a representation of the Virasoro algebra, all have integer coefficients. Definitions, references, and links are provided in the entry. Can anyone prove or disprove this conjecture?

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    $\begingroup$ Could you please give a self-contained statement of the conjecture? The OEIS stuff is not really readable. $\endgroup$ – darij grinberg Jan 17 '12 at 2:04
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    $\begingroup$ Is it so difficult to make this question self-contained at least to some extent? $\endgroup$ – Vladimir Dotsenko Jan 17 '12 at 10:21
  • $\begingroup$ The Question and Remark sections of Darij's answer are rewordings of briefer statements in the OEIS entry and contain no new substantive content. The Question section does conform to the lingo and formality of an algebraist; however, it would only confuse a highschool algebra student. $\endgroup$ – Tom Copeland May 6 '16 at 15:44
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Before I give the answer, let me try to formulate the question in the way I would have asked it here. Fortunately I am not bound by the OEIS requirements of brevity and ASCII, and there is LaTeX here...

Question. Let $A$ be the polynomial ring $\mathbb Z\left[c_1,c_2,c_3,...\right]$ in infinitely many commuting indeterminates $c_1$, $c_2$, $c_3$, .... Let $g$ be the formal power series $x+c_1x^2+c_2x^3+c_3x^4+... \in A\left[\left[x\right]\right]$. By considering $A=\mathbb Z\left[c_1,c_2,c_3,...\right]$ as a subring of $\mathbb Q\left[c_1,c_2,c_3,...\right]$, we can define a power series $S = \dfrac{x^2}{6} \left(\dfrac{g^{\prime\prime\prime}}{g^{\prime}} - \dfrac32 \left(\dfrac{g^{\prime\prime}}{g^{\prime}}\right)^2\right) \in \left(\mathbb Q\left[c_1,c_2,c_3,...\right]\right)\left[\left[x\right]\right]$. To prove that this $S$ actually lies in $A\left[\left[x\right]\right]$.

Remark. If we let $f=g^{\prime}$, then $S$ can also be written as $\dfrac{x^2}{6}\left(D^2\left(\ln f\right) - \dfrac12 \left(D\left( \ln f\right)\right)^2\right)$. We will not need this, however.

Answer to the question. Since the constant term of the power series $g^{\prime} \in A\left[\left[x\right]\right]$ is $1$, the power series $g^{\prime}$ has a multiplicative inverse $\dfrac{1}{g^{\prime}}$ in $A\left[\left[x\right]\right]$.

For every $k\in\mathbb N$, the power series $\dfrac{1}{k!}g^{(k)}$ lies in $A\left[\left[x\right]\right]$ (because for every $n\in\mathbb N$, the coefficient of this power series $\dfrac{1}{k!}g^{(k)}$ before $x^n$ is

$\dfrac{1}{k!}\left(n+k\right)\left(n+k-1\right)...\left(n+1\right)c_{n+k} = \dbinom{n+k}{k} c_{n+k} \in A$ (where $c_0$ denotes $1$)

). Applied to $k=2$, this yields $\dfrac{g^{\prime\prime}}{2} \in A\left[\left[x\right]\right]$. On the other hand, applied to $k=3$, it yields $\dfrac{g^{\prime\prime\prime}}{6} \in A\left[\left[x\right]\right]$.

Now,

$S = \dfrac{x^2}{6} \left(\dfrac{g^{\prime\prime\prime}}{g^{\prime}} - \dfrac32 \left(\dfrac{g^{\prime\prime}}{g^{\prime}}\right)^2\right) = x^2 \left(\dfrac{g^{\prime\prime\prime}}{6}\cdot\dfrac{1}{g^{\prime}} - \left(\dfrac{g^{\prime\prime}}{2}\right)^2\cdot\left(\dfrac{1}{g^{\prime}}\right)^2\right)$

is in $A\left[\left[x\right]\right]$ (because each of $x^2$, $\dfrac{g^{\prime\prime\prime}}{6}$, $\dfrac{1}{g^{\prime}}$ and $\dfrac{g^{\prime\prime}}{2}$ is in $A\left[\left[x\right]\right]$).

Meta-Question. The formula $S = \dfrac{x^2}{6}\left(D^2\left(\ln f\right) - \dfrac12 \left(D\left( \ln f\right)\right)^2\right)$ reminds me of $p_2 = e_1^2 - 2e_2$ (one of the formulae for power sums in terms of elementary symmetric functions). Does this mean that the $S$ is actually the $2$nd member of a series of differential operators with interesting divisibility properties?

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  • $\begingroup$ Also reminiscent of the Maurer-Cartan form. $\endgroup$ – Tom Copeland Jan 17 '12 at 16:20
  • $\begingroup$ I'm giving up on the meta-question; but if anyone finds a reasonable generalized conjecture, I'd be happy to hear! $\endgroup$ – darij grinberg Jan 17 '12 at 18:23
  • $\begingroup$ Ivey and Landsberg in Cartan for Beginners ... on pg. 21 and 22 relate the Cartan-Maurer form to the Schwarzian. $\endgroup$ – Tom Copeland Jan 17 '12 at 19:25

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