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For the Dedekind eta function, defined as usual by $\eta(q) = q^{\frac1{24}} \prod\limits_{n=1}^{\infty} (1-q^{n}) $, let for brevity $e_k:=\eta(q^k)$. With this notation, a blog entry of Michael Somos gives three beautiful identities for sums of $\eta$-products where all exponents are only $0$ or $1$:

$I_{60}:\qquad \ \ e_{1}e_{12}e_{15}e_{20} + e_{3}e_{4}e_{5}e_{60}=e_{2}e_{6}e_{10}e_{30} $

$I_{210}:\qquad e_{1}e_{30}e_{35}e_{42 }+ e_{3}e_{10}e_{14}e_{105} = e_{2}e_{15}e_{21}e_{70} +e_{5}e_{6}e_{7}e_{210}$

$I_{30}:\qquad\ \ e_{1}e_{3}e_{5}e_{15}+ 2e_{2}e_{6}e_{10}e_{30}=e_{1}e_{2}e_{15}e_{30}+ e_{3}e_{5}e_{6}e_{10} $

For $I_{60}$ and $I_{30}$ the structure is clear at one glance if we write the divisors of $60$ as vertices of a Cayley-like graph (here: 'union' of two cube graphs $Q_3$ with the common face $(2,6,30,10)$ ):

4                  20

    2          10

        1   5

        3  15

    6          30

12                 60

Alternatively, if we define

$a_0:=e_{1} e_{15} \qquad b_0:= e_{3} e_{5} $

$a_1:=e_{2} e_{30} \qquad b_1:= e_{6} e_{10} $

$a_2:=e_{4} e_{60} \qquad b_2:= e_{12} e_{20} $

then

$I_{60}\iff a_0b_2+b_0a_2=a_1b_1$

$I_{30}\iff a_0a_1+b_0b_1=a_0b_0+2a_1b_1$.

For $I_{210}$ the symmetry is a bit less obvious to see. We can identify the divisors of $210$ with the vertices of a tesseract graph $Q_4$ or write the factors of the four products as lines of a matrix and note $a_{i,j}a_{4-i,j}=210$ as well as the factor $3$ between the two pairs of lines:

  1  30  35  42   
  3  10 105  14   

 70  21   2  15   
210   7   6   5      

I'd suggest to call identities of this type linear eta product identities. Their linearity seems to enforce a high degree of symmetry in the way these three identities $I_n$ feature the divisors of $n$, which makes them very special among the thousands of known eta product identities. It looks like there is something deeper behind. And: Why do all products have exactly $4$ factors?

So, more precisely, for naturals $a\ge b$ let's define a linear eta product identity of type $\mathbf{(a,b)}$ as an identity $L_1+\cdots+L_a=R_1+\cdots+R_b$, where each $L_i$ and each $R_i$ is a finite product of pairwise different terms of form $\eta(q^{\lambda})$ with $\lambda\in\mathbb N$. (The products $L_i$ and $R_i$ don't need to be all different, e.g. the above $I_{30}$ is of type $(3,2)$ with $L_2=L_3$. But of course we want $\{L_i\}\cap\{R_i\}=\emptyset$, and also that the gcd of all the $\lambda$'s is $1$.)
Somos conjectures that $I_{60}$ is the only linear identity of type $(2,1)$.

Is it possible that the three above identities are only the first ones of a whole (infinite?) set of linear eta product identities, and/or that for naturals $a\ge b$, there is at most one such identity of type $(a,b)$?

share|improve this question
    
Kyoji Saito has papers on various eta product identities... I am not sure this is relevant, just a comment. –  Alexander Chervov Mar 20 '12 at 7:23
    
If we define $a_k =\eta(q^k)\,\eta(q^{PQk})$ and $b_k= \eta(q^{Pk})\,\eta(q^{Qk})$, then for $P,Q =3,5$ we have $I_{60} \iff a_1b_4+a_4b_1 = a_2b_2$. For $P,Q =5,7$, it is $I_{210} \iff a_1b_6+a_3b_2 = a_2b_3+a_6b_1$. It's so tempting to speculate that these belong to an infinite family for appropriately chosen primes $P,Q$. –  Tito Piezas III Feb 1 at 1:13
    
I tried $P,Q = 11,13$. Since $LCM(11\cdot12\cdot13) =1726$ (which has 24 divisors), I hoped to find linear relations between the 6 real numbers $a_1 b_{12},\, a_2 b_6,\, a_3 b_4,\, a_4 b_3,\, a_6 b_2,\, a_{12} b_1$. Unfortunately, Mathematica couldn't seem to find anything. Sigh. –  Tito Piezas III Feb 1 at 1:15
    
@TitoPiezasIII Your notation $a_k$ and $b_k$ is better than mine (which only handles $k$'s that are powers of 2), moreover it shows that for $I_{60}$ and $I_{210}$, all terms $a_kb_\ell$ have $k\ell=const$. I agree with you, that seems to cry for generalization, but in between I have gained the impression that in spite of the thousands of existing eta-identities, everything is finite there in terms of re-occuring patterns. –  Wolfgang Feb 1 at 14:32
    
There are some more linear ones, 14 altogether in Somos' collection, some of them with products of four, but many terms (see the 2 last ones in Somos’ level 300 file, the two of level 450 and the level 945 one), others with products of six (to wit, two pairs for each 180 & 300, one for 252 somewhat similar to $I_{60}$, and one for 240). Somos has searched in vain for linear ones with products of eight. They all have many internal symmetries, as may be expected. E.g. the pairs for 180 and 300 are perfectly "isomorphic" to each other: switch all factors 3 with factors 5. Interesting! –  Wolfgang Feb 2 at 18:50
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