The asymmetry does not depend on the transverse wavenumbers

Setting. Let $q$ be a natural number with $q \neq 0$, so $q \ge 1$. This is the only assumption on $q$, and it makes sure the index $0$ exists in $\{0, 1, \dots, q-1\}$. Let $K, c, \beta$ be arbitrary real numbers. There are no sign or size conditions on them, so $K$, $c$ and $\beta$ may be negative or zero. Let $k = (k_0, \dots, k_{q-1})$ and $k' = (k'_0, \dots, k'_{q-1})$ be arbitrary vectors in $\mathbb{R}^q$. The only hypothesis linking them is that their $0$-th components are equal:

The other components $k_a$ and $k'_a$ for $a \ge 1$ are completely unconstrained and may differ.

Definitions used. For a vector $k \in \mathbb{R}^q$, the function $\omega$ (with parameters $q, K, c, \beta$) is

The sum runs over all indices $a = 0, \dots, q-1$, including $a = 0$. Here $\sqrt{\cdot}$ is Mathlib's Real.sqrt. It gives the usual non-negative square root when the radicand is $\ge 0$. When the radicand is negative it returns $0$, not an error or a complex number. So whenever $(\beta c \sin k_0)^2 + K + 2c\sum_a(1-\cos k_a) < 0$, which can happen because $K$ or $c$ may be negative, the formula gives $\omega(k) = \beta c \sin k_0$.

The "flip" of $k$ is the vector $\tilde k$ that equals $k$ except that its $0$-th component is negated:

This is Function.update, which overwrites exactly one coordinate. When $q = 1$, the only coordinate is $0$, so $\tilde k = (-k_0)$.

Assertion. For all such $q, K, c, \beta, k, k'$ with $k_0 = k'_0$,

In words, the difference between $\omega$ at $k$ and $\omega$ at the vector with its $0$-th component negated is the same for any two vectors that share the same $0$-th component, whatever their remaining components are. Written out,

This holds because the squared term is unchanged and the $a=0$ summand becomes $1 - \cos(-k_0)$. The statement is asserted for every real $K, c, \beta$, including the case where the radicands are negative and the square roots are truncated to $0$.

Degenerate cases. When $q = 1$, the hypothesis $k_0 = k'_0$ forces $k = k'$, so the two sides are literally the same expression. When $c = 0$, $\omega(k) = \sqrt{K}$ (truncated) for every $k$, so both sides are $0$. When $\beta = 0$ or $\sin k_0 = 0$, the linear term vanishes. The proof is left as sorry; this read-back records only what is stated.