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Looking at the collection of Eta Function Product Identities by Michael Somos, it seems like generally those identities come in pairs: let's call two eta product identities $\sum\limits_{i=1}^r a_iP_i=0$ and $\sum\limits_{i=1}^r b_iQ_i=0$ dual if the eta products $Q_i$ can be re-labeled and the factors of each product re-arranged such that for all $i$, we can transform $P_i$ into $Q_i$ by replacing each argument $q^k$ or $k\tau$ of an eta function ($k\in\mathbb N$) by $q^{n/k}$ or $(n/k)\tau$. Note that the scalar coefficients $a_i$ and $b_i$ need not be the same, but they should be non-zero of course.

Example: The first two identities on the n=14 page, both of degree $12$, are

q14_12_44 = +1*u1^7*u2*u7^3*u14 +7*q*u1^3*u2*u7^7*u14 -1*u2^8*u7^4 -7*q^3*u1^4*u14^8 ; 
q14_12_64 = +1*u1^8*u14^4 +7*u2^4*u7^8 -56*q^2*u1*u2^3*u7*u14^7 -8*u1*u2^7*u7*u14^3 ;

re-arranged and written in a more humanly-readable form, putting $s_k=\eta(k\tau)$, we have

q14_12_44$\iff\qquad s_1^7s_2^{\ }s_7^3s_{14}^{\ } +7s_1^3s_2^{\ }s_7^7s_{14}^{\ } -s_2^8s_7^4 -7s_1^4s_{14}^8=0$

q14_12_64$\iff -56s_{14}^7s_7^{\ }s_2^3s_1^{\ }-8s_{14}^3s_7s_2^7s_1^{\ } +7s_7^8s_2^4+s_{14}^4s_1^8=0$

so both identities are duals of each other.

It should be easy to see that an eta product identity (edit: which is not dual to itself) cannot have two (linearly independant) duals. Is that really easy to see?

An eta product identity can also be self-dual, e.g. the known linear identities mentioned in my recent MO thread or the one equivalent to the well-known theta identity $\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)$, which becomes, when written in terms of $\eta$'s, $$s_2^{24}=s_1^{16}s_4^8 +16s_4^{16}s_1^8 .$$

This concept of duality seems so logical that I wonder: Has there already been any research on that? Has each eta product identity a dual counterpart, unless it is self-dual ?

In the collection mentioned above, for $n=22$ there is only one identity listed, and that one is not self dual, it is even highly asymmetrical. Its dual should have the label x22_19_124. Maybe that one is missing just because the computer search had to stop at some point.

Given that self-dual identities have a higher degree of symmetry: can they always be expressed in terms of Ramanujan's psi, phi, and/or chi functions?

I would think the answer is yes, as there are even some not self-dual ones that can be expressed in that way.

Edit concerning the first question: Even this may be trickier than expected. For example, there are two self-dual identities, labeled q6_14_36a and q6_14_36b. They are:

$s_1^9s_3^{\ }s_6^4 +\ \ 6s_1^4s_2^{\ }s_6^9 +2s_2^9s_3^4s_6^{\ } -3s_1^{\ }s_2^4s_3^9=0$

$s_1^9s_3^{\ }s_6^4 +12s_1^4s_2^{\ }s_6^9 -4s_2^9s_3^4s_6^{\ } +3s_1^{\ }s_2^4s_3^9=0$.

They contain exactly the same $\eta$-products and so can also be considered as dual to each other (and to linear combinations of them). Moreover, by linear combinations we may eliminate any one of the four terms, resulting in four 3-term identities with coefficients

 0  1 -1  1
-1  0 -8  9
 1  8  0 -1
-1 -9  1  0. 

Note that each of those has exactly one dual counterpart again, as they come in two dual pairs.
So the above conjecture of a unique dual doesn't necessarily hold if the given identity is already self-dual.

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2 Answers 2

This looks like the Fricke involution to me. Given any positive integer $M$, define $W_M$ to be the operator given by slashing with the matrix $$\left(\begin{array}{cc} 0 & -1 \newline M & 0 \end{array}\right),$$ or, equivalently, if $f(z)$ is weight $k$ modular, by $f(z)\mid_k W_M = (-iz\sqrt{M})^{-k}f\left(\frac{-1}{Mz}\right)$. This is the weight $k$, level $M$ Fricke involution. It preserves $M_k(\Gamma_0(M))$, albeit changing the character.

Now, using the transformation properties of $\eta(z)$, in particular that $\eta(-1/z)=(-iz)^{1/2}\eta(z)$, it's possible to see that hitting any $\eta$-product with the appropriate Fricke involution takes it to another $\eta$-product, perhaps after scaling. Thus, hitting an $\eta$-product identity with the Fricke involution produces another identity, which is moreover dual to the original. At least the first example you give has this property, i.e., it follows from the Fricke involution.

As to the question of unique duals, you've already seen that things get wonky if you allow for self-dual identities. But even if you require the identities to not be self-dual, things are still bad (although perhaps unsatisfyingly so). Let $I_1$ and $I_2$ be two linearly independent, not dual to each other, and not self-dual identities. Let $I_1^\prime$ and $I_2^\prime$ denote their duals, respectively (if you want, just take them to be images under the Fricke involution). Consider now the identity $I=I_1+I_2$. By your definition of dual, any of the identities $I^\prime=aI_1^\prime+bI_2^\prime$ will be dual to $I$, and so the dual-space is at least two-dimensional. Modifying this in the obvious way suggests that the dimension of the dual-space can be arbitrarily large, with the requisite $I_1, \dots, I_n$ being of the form $I_1^aI_2^b$ varying over $a+b=n-1$, say.

I think the right way to phrase the uniqueness question is by asking whether there is a set of primitive $\eta$-product identities from which all other identities can be obtained, with the property that every dual of an identity is generated by Fricke involutions of the constituent primitive elements of the original identity. In particular, if an identity is primitive, is its dual-space one-dimensional? I don't know the answer to this question.

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I would guess that this is just due to the fact that taking the dual preserves the space (i.e. if P_1 and P_2 are in the same space, the duals Q_1 and Q_2 will be as well, or at least some same space, maybe the level goes up a bit, I can't remember) of the eta quotient, and that this space is finite dimensional.

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I suppose that's about the question of unique duals? It's not that simple. Even once we have established that a dual always exists, a simple "infinite descend" argument may not necessarily work. See my edit above. –  Wolfgang Mar 25 '12 at 21:31

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