samples/texts/2274792/page_16.md

I then evaluated the resulting eigenvalues numerically by substituting the values for $b$ and $t$ over the range explored in the main text where $b > t$ (the conditions for the existence of the equilibrium). This equilibrium is not stable over this range, which corresponds to the results of the numerical simulations from the main text where the system never evolved to this equilibrium.

A.3 Stability of $A_1$ equilibrium without culture

To determine the stability of this equilibrium, I calculated the characteristic equation for the system given by Equations 2 and 3 evaluated where $x_{f10} = x_{m10} = 1$.

$$\frac{{\lambda }^9(1-\lambda {)}^3(\lambda -(1+b-2t))(2\lambda -(1+b-t))}{2}=0\backslash frac\{\backslash lambda^9(1-\backslash lambda)^3(\backslash lambda\; -\; (1+b-2t))\; (2\backslash lambda\; -\; (1+b-t))\}\{2\}\; =\; 0$$

Since three of the resulting eigenvalues are equal to one, the equilibrium is unstable or neutrally stable for all parameters. This makes intuitive sense since the fitness of all four alleles in both sexes are equivalent in the absence of the cultural trait, allowing for drift between the alleles.

A.4 Stability of $A_2$ equilibrium without culture

To determine the stability of this equilibrium, I calculated the characteristic equation for the system given by Equations 2 and 3 evaluated where $x_{f20} = x_{m20} = 1$.

$$\frac{{\lambda }^9(1-\lambda {)}^3(\lambda -(1+b-t))(2\lambda -(1+b-2t))}{2}=0\backslash frac\{\backslash lambda^9(1-\backslash lambda)^3(\backslash lambda\; -\; (1+b-t))\; (2\backslash lambda\; -\; (1+b-2t))\}\{2\}\; =\; 0$$

Since three of the resulting eigenvalues are equal to one, the equilibrium is unstable or neutrally stable for all parameters. This makes intuitive sense since the fitnesses of all four alleles in both sexes are equivalent in the absence of the cultural trait.