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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.

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: unless it is dual to itself) cannot have two (linearly independent) 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.

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.

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.

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.

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.