I know that Boardman spectra aren't supposed to have an on-the-nose commutative smash product, and symmetric spectra -- which look like essentially the same thing, except the spaces $X_n$ have to come with an action of the symmetric group -- do. Essentially, I'm trying to understand why adding this simple condition is the right thing to do.
If $X,Y$ are symmetric spectra, then $(X\wedge Y)_n$ is defined as the coequalizer of
$$ \alpha_X,\alpha_Y:\bigvee_{p+1+q=n}\Sigma^+_n\wedge_{\Sigma_p\times \Sigma_1\times \Sigma_q}X_p\wedge S^1\wedge Y_q\rightrightarrows \bigvee_{p+q=n}\Sigma^+_n\wedge_{\Sigma_p\times \Sigma_q}X_p\wedge Y_q$$ where $\alpha_X$ is induced by $X_p\wedge S^1\wedge Y_q\overset{\sigma_p\otimes 1}\to X_{p+1}\wedge Y_q$ (where $\sigma_p$ is the structure map) and $\alpha_Y$ is induced by $X_p\wedge S^1\wedge Y_q\overset{\text{twist}}\to X_p\wedge Y_q \wedge S^1\overset{1\otimes\sigma_q}\to X_p\wedge Y_{q+1}\to X_p\wedge Y_{1+q}$ where the last map uses the $\Sigma_{q+1}$-action.
My question is why the $\Sigma$-action is necessary: you can try to make a parallel construction, where you omit the group actions and say $(X \wedge Y)_n$ is the coequalizer of $$ \alpha_X,\alpha_Y:\bigvee_{p+1+q=n}X_p\wedge S^1\wedge Y_q\rightrightarrows \bigvee_{p+q=n}X_p\wedge Y_q$$ where $\alpha_X$ is induced by $X_p\wedge S^1\wedge Y_q\overset{\sigma_p\otimes 1}\to X_{p+1}\wedge Y_q$ and $\alpha_Y$ is induced by
- $X_p\wedge S^1\wedge Y_q\overset{1\otimes\sigma_q}\to X_p\wedge Y_{q+1}$ (where I'm assuming the existence of structure maps $S^1\wedge X_n\to X_n$ as well as $X_n\wedge S^1\to X_n$ with some compatibility properties) or just
- $X_p\wedge S^1\wedge Y_q\overset{\text{twist}}\to X_p\wedge Y_q \wedge S^1\overset{1\otimes\sigma_q}\to X_p\wedge Y_{q+1}$.
I feel like I'm sort of missing the point of this construction; can someone fill me in? Thanks!