10
$\begingroup$

This question is grounded firmly in numerology. It originates in an observation about some Bernoulli polynomials and the regular icosahedron. Let $F_{k+1}(n)=\sum_{i=1}^n i^k$ be the sum of the first $n$ $k$th powers of positive integers. We know that $F_k$ is a polynomial of degree $k$ that is related to Bernoulli polynomials via an affine-linear substitution for $n$. The observation here concerns $F_5$ and $F_7$. Specifically, we see $$60F_5(n)=12n^5+30n^4+20n^3-2n$$ and $$168F_7(n)=24n^7+84n^6+84n^5-28n^3+4n.$$ Notice that the coefficients of $60F_5$ coincide with the face vector of a regular icosahedron (20 triangles, 30 edges, 12 vertices). Also, the coefficients of $168F_7$ nearly coincide with the face vector of a surface related to the Klein quartic. This is a "regular" triangulation of a surface of genus 3 that has 24 vertices, 84 edges, and 56 triangles. Notice that the sum of the coefficients of $168F_7$ on the terms of degrees 5 and 3 is 56. The numbers 60 and 168 are significant because they are the orders of a couple of groups that act on these polyhedra.

Is this a coincidence? Are there similar phenomena relating higher-degree Bernoulli polynomials to regular triangulations of other surfaces? Should I lay off the moonshine?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ "a surface related to the Klein quartic", can you elaborate? From Wikipedia: "Klein quartic forms part of a "trinity" in the sense of Vladimir Arnold, which can also be described as a McKay correspondence. In this collection, the [...] groups PSL(2,5), PSL(2,7), and PSL(2,11) (orders 60, 168, 660) are analogous, corresponding to icosahedral symmetry (genus 0), the symmetries of the Klein quartic (genus 3), and the buckyball surface (genus 70).[12] These are further connected to many other exceptional phenomena, which is elaborated at "trinities"." so maybe $660F_{11}$ will be relevant? $\endgroup$
    – i9Fn
    Commented Nov 15, 2020 at 16:03
  • 1
    $\begingroup$ Maybe p=3 and p=23 would also be interesting to consider ? $\endgroup$
    – F. C.
    Commented Jan 3, 2021 at 21:16

2 Answers 2

Reset to default
4
$\begingroup$

Following up on i9Fn's suggestion, there does seem to be at least one more case along these lines: $$660F_{11}(n) = 60n^{11} + 330n^{10} + 550n^9 - 660n^7 + 660n^5 - 330n^3 + 50n.$$ This might be related to a genus 70 Riemann surface with 11 embedded buckyballs; see Martin & Singerman From biplanes to the Klein quartic and the Buckyball. Each buckyball/ buckminsterfullerene has 60 vertices, 90 edges, and 32 faces. The number 660 arises as both the order of $PSL(2,11)$ and the count of particular triangles.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ There is a "regular" triangulation of a genus 26 surface related to the group PSL(2,11). This has 60 vertices, 330 edges, and 220 triangles. It is described in this article by Ioannis Ivrissimtzis, David Singerman, and James Strudwick: arxiv.org/abs/1909.08568. Seeing $660F_{11}$ here, one again notices some coincidences in the coefficients. Here, one must add the coefficients of the terms of degree 9, 7, 5, and 3 to get the number 220 of triangles. $\endgroup$
    – David Richter
    Commented Nov 16, 2020 at 12:37
  • 1
    $\begingroup$ Another observation: The coefficients -2, 4, and 50 are the negatives of the respective Euler characteristics of these surfaces which seem to correspond to $F_5$, $F_7$ and $F_{11}$. $\endgroup$
    – David Richter
    Commented Nov 16, 2020 at 13:33
  • 1
    $\begingroup$ It would be even crazier if all of this connected to the Ramanujan partition congruences which hold (in the most thorough sense) only modulo 5, 7, and 11. $\endgroup$
    – Brian Hopkins
    Commented Nov 18, 2020 at 18:21
1
$\begingroup$

Explicitly, the polynomials are

sage: 12*(bernoulli_polynomial(x,5)-bernoulli(5))                               
12*x^5 - 30*x^4 + 20*x^3 - 2*x
sage: 24*(bernoulli_polynomial(x,7)-bernoulli(7))                               
24*x^7 - 84*x^6 + 84*x^5 - 28*x^3 + 4*x
sage: 60*(bernoulli_polynomial(x,11)-bernoulli(11))                             
60*x^11 - 330*x^10 + 550*x^9 - 660*x^7 + 660*x^5 - 330*x^3 + 50*x

where the scaling factors are the cardinalities of the groups $A_4$, $S_4$, $A_5$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ And the odd Bernoulli numbers vanish, so one can remove them from the formula. $\endgroup$
    – F. C.
    Commented Jan 3, 2021 at 20:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .