# Fixed point: a consequence of symmetry?

I'm studying a dynamical system with $\mathbf{D}_{3}$ symmetry (the symmetry group of an equilateral triangle), which is given by:

\begin{align*} d\mathbf{x}_{0}/dt &= \mathbf{f}(\mathbf{x}_{2}, \mathbf{x}_{0}, \mathbf{x}_{1}) \\ d\mathbf{x}_{1}/dt &= \mathbf{f}(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}) \\ d\mathbf{x}_{2}/dt &= \mathbf{f}(\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{0}) \end{align*},

where $\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2} \in \mathbb{R}^{k}$ and $\mathbf{f}(\mathbf{u}, \mathbf{v}, \mathbf{w}) = \mathbf{f}(\mathbf{w}, \mathbf{v}, \mathbf{u})$.

Is a fixed point at $(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}) = (\mathbf{0}, \mathbf{0}, \mathbf{0})$ guaranteed by the symmetry in the problem?

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Hardly: if (0,0,0) is a fixed point for some f, it is not anymore for the (quite symmetric) system based on f+1. –  Did Aug 13 '11 at 0:59
Make $f$ a constant vector and there are no fixed points... –  anon Aug 13 '11 at 1:15
$f+1$ doesn't make sense, since $\mathbf f$ is a vector. –  joriki Aug 13 '11 at 4:29
@joriki, right. Use $f$ plus your favorite nonzero vector. (I read your comment by accident, please use the @ thing.) –  Did Aug 13 '11 at 9:49

No, a fixed point would only be guaranteed if there were no non-zero vectors $\mathbf f$ invariant under the symmetry group. In that case, all non-zero vectors $\mathbf f(\mathbf 0,\mathbf 0,\mathbf 0)$ would break the symmetry, which would guarantee $\mathbf f(\mathbf 0,\mathbf 0,\mathbf 0)=0$ and thus a fixed point at $(\mathbf 0,\mathbf 0,\mathbf 0)$. However, since you only have $D_3$ symmetry, you can choose $\mathbf f$ to be a constant vector in any direction orthogonal to the triangle; then the symmetry operations will leave it invariant, and the flow will lead away from the origin.
P.S.: I'm wondering, since $D_3$ is isomorphic to $S_3$, whether you don't actually mean spatial symmetry of an equilateral triangle but permutation symmetry of the coordinates. In that case, there's no reason to expect a fixed point at $(\mathbf 0,\mathbf 0,\mathbf 0)$, and as was pointed out in the comments $\mathbf f$ could for instance be any constant without breaking the symmetry.
Very helpful, thank you. I was trying to express that $\mathbf{D}_{3}$ acts on $(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}) \in \mathbb{R}^{3k}$ according to a permutation $\rho(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}) = (\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{0})$ and a "flip" $\kappa(\mathbf{x}_{0}, \mathbf{x}_{1}, \mathbf{x}_{2}) = (\mathbf{x}_{0}, \mathbf{x}_{2}, \mathbf{x}_{1})$. –  G. Brown Aug 13 '11 at 14:52