Suppose we have the following system of differential equations:
\begin{cases} \frac{dx_{i}}{dt}=f_{i}\left(\boldsymbol{x},\boldsymbol{y}\right), & i=1,\ldots M\\ \\ \frac{dy_{j}}{dt}=g_{j}\left(\boldsymbol{x},\boldsymbol{y}\right), & j=1,\ldots N \end{cases}
where $\boldsymbol{x}=\left(x_{1},\ldots,x_{M}\right)$ and $\boldsymbol{y}=\left(y_{1},\ldots,y_{N}\right)$. We also suppose that the $M$ equations in $x_{i}$ are symmetric under exchange of the index $i$, while the $N$ equations in $y_{j}$ are symmetric under exchange of the index $j$. Moreover, we suppose that in general there is no symmetry if we exchange a particle $i$ with a particle $j$. In other words, these differential equations describe a system with two categories of particles (e.g. neutrons and protons), and within each category the particles are identical or indistinguishable.
My questions are:
- what is the symmetry group of the whole system?
- if for some reason a particle in one category starts to behave differently from all the other particles in that category (through a branching point bifurcation), can we talk about spontaneous symmetry breaking?
Many thanks in advance for any help you may provide.
Well, I don't know anything about equivariant bifurcation theory, but I can say a couple of things about this system.
As the problem is stated, the symmetry group is $S_M \times S_N$ (permutation groups of $M$ and $N$ objects) where the first factor acts on the $x$-space and the second on the $y$ one.
The spontaneous symettry breaking mechanism is when you have a symmetry in your equations, but not in your solutions, so a bifurcation point, as you described, breaks - let's say $S_N$ to $S_{N-1}$ - but it is not a spontaneous symmetry breaking, just a symmetry breaking.