Proof of a Dedekind eta-product identity of level 8 and degree 23.

Question

Define the notation $\, u_n := \eta(n\tau) \,$ for the Dedekind $\eta$ function and $\, q := e^{2\pi i\tau}.$

The simplest identity of level $4$ and degree $24$ seems to be

$$ 0 = u_1^{16}u_4^8 + 16 u_1^8 u_4^{16} - u_2^{24}. $$

The simplest identity of level $8$ and degree $12$ seems to be

$$ 0 = u_1^4 u_2^2 u_4^2 u_8^4 + 4 u_2^4 u_8^8 - u_4^{12}. $$

What are possible proofs of this level $8$ and degree $23$ identity

$$ 0 = u_1^4 u_4^{19} + 2 u_1^2 u_2^7 u_4^{12} u_8^2 + u_2^{14} u_4^5 u_8^4 - 2 u_1^8 u_2^2 u_4^9 u_8^4 - 2 u_1^6 u_2^9 u_4^2 u_8^6 - 12 u_1^4 u_2^4 u_4^7 u_8^8 - 8 u_1^2 u_2^{11} u_8^{10}? $$

Here is the background for this identity. Given Ramanujan's theta function $\,f(a,b)\,$ define

$$ w_n := q^{(4-n)^2/16} f(-q^n, -q^{8-n}). $$

Notice that $\,w_n = w_{n+8} = -w_{-n}\,$ for all integer $n$.

Suppose we are given the identities

$$ w_2 = \frac{u_2u_8}{u_4},\; w_4 = \frac{u_4^2}{u_8},\; w_1w_3 = \frac{u_1u_8^2}{u_2},\; w_1 = u_1 u_8^3\sqrt{\frac{2u_1 u_2 u_4}{u_1^2 u_4^7 + u_2^7 u_8^2}}. $$

Then the identity

$$ 0 = -w_4^2 w_3^2 u_2 + w_4^2 w_1^2 u_2 + 4w_2^2 u_1 u_8^2 + u_1^5 $$

is equivalent to the degree $23$ identity.

By the way, the $\,w_n\,$ sequence is an elliptic divisibility sequence and also an $\,(\alpha,\beta)\,$ Somos-4 sequence where $\,\alpha=(w_2/w_1)^2,\,\beta=-w_3/w_1.$

Answer 1

Define the notation $\, u_n := \eta(n\tau) \,$ for the Dedekind $\eta$ function and $\, q := e^{2\pi i\tau} . $

The question is how to prove the seven term level $8$ and degree $23$ identity

$$ 0 = u_1^4 u_4^{19} + 2 u_1^2 u_2^7 u_4^{12} u_8^2 + u_2^{14} u_4^5 u_8^4 - 2 u_1^8 u_2^2 u_4^9 u_8^4 \\ - 2 u_1^6 u_2^9 u_4^2 u_8^6 - 12 u_1^4 u_2^4 u_4^7 u_8^8 - 8 u_1^2 u_2^{11} u_8^{10} . $$

For convenience, define the following $\eta$ products

\begin{align*} v_1 &:= u_1^4 u_4^{19}, \\ v_2 &:= u_1^2 u_2^7 u_4^{12} u_8^2, \\ v_3 &:= u_2^{14} u_4^5 u_8^4, \\ v_4 &:= u_1^8 u_2^2 u_4^9 u_8^4, \\ v_5 &:= u_1^6 u_2^9 u_4^2 u_8^6, \\ v_6 &:= u_1^4 u_2^4 u_4^7 u_8^8, \\ v_7 &:= u_1^2 u_2^{11} u_8^{10} . \end{align*}

The identity to prove can now be written as

$$ T := v_1 + 2v_2 + v_3 - 2v_4 - 2v_5 - 12v_6 - 8v_7 = 0. $$

After discovering this identity I searched for linear relations between its terms .

Thus, define the following linear combinations

$$ t_1 := v_5 + 4 v_7 - v_2, \; t_2 = v_1 + 4 v_6 - v_3, \; t_3 = v_3 + v_4 - 2 v_1 $$

and verify that $\, -T = 2 t_1 + 3 t_2 + 2 t_3.\,$

Notice the factorizations

\begin{align*} t_1 &= u_1^2 u_2^7 u_8^2(u_1^4 u_2^2 u_4^2 u_8^4 + 4 u_2^4 u_8^8 - u_4^{12}), \\ t_2 &= u_4^5(u_1^4 u_4^{14} + 4 u_1^4 u_2^4 u_4^2 u_8^8 - u_2^{14}u_8^4), \\ t_3 &= u_4^5(u_2^{14}u_8^4 + u_1^8 u_2^2 u_4^4 u_8^4 - 2 u_1^4 u_4^{14}). \end{align*}

The polynomial factor of these sums equated to zero are denoted $\texttt{t8_12_48}$, $\texttt{t8_18_60a}$, and $\texttt{t8_18_60b}$ respectively in my "Dedekind Eta Function Product Identities" collection. References to proofs for each of these is given there.