Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 60 / 1

For a vector $x$, its isotropy group is the stabilizer subgroupFor a subgroup $H\le G$, its fixed-point subspace of a group action isIt is a linear subspace, the intersection of the kernels of $R_G(h)-I$. The subgroup fixes a particular vector pointwise; the fixed-point subspace of a group action consists of all vectors it fixes, and need not be one-dimensional.

Suppose $f$ is sufficiently smooth and equivariant, $f(0,\mu )=0$, and the group representation on the critical space is absolutely irreducible. Its linearization is then $D_{x}f(0,\mu )=a(\mu )I$. Assume $a(0)=0$ and $a^{\prime }(0)\ne 0$. The equivariant branching lemma states that each isotropy subgroup $H$ with $\dim \mathrm{Fix}(H)=1$ has a local nontrivial branch of equilibria with that isotropy, apart from the common origin. Conjugate isotropy subgroups give symmetry-related branches, so branch types are counted up to conjugacy.

To see the scalar mechanism, write the fixed-point subspace of a group action as $\mathbb{R}e$ and restrict $f(se,\mu )=eh(s,\mu )$. Equivariance makes the fixed-point subspace of a group action invariant. Since $h(0,\mu )=0$, write $h=sk(s,\mu )$; $k(0,0)=0$ and $k_{\mu }(0,0)=a^{\prime }(0)\ne 0$. The implicit function theorem solves $k(s,\mu (s))=0$ near zero. For small nonzero $s$, the stabilizer subgroup is exactly that of $e$. A normalizer element reversing $e$ makes $h$ odd and $\mu (s)$ even; the generic branch is then a pitchfork bifurcation. The lemma ensures existence, not dynamical stability, and does not exclude further branches of smaller isotropy.

Conjugating the coordinate reflection by the rotations supplies reflections in the other two coordinates. The quarter-turns, together with those reflections, also supply coordinate transpositions. Consequently the represented group consists of all signed permutation matrices, the full symmetry group of a cube of order $2^3\cdot 3!=48$.

Let $C$ be any real or complex matrix commuting with the group representation. Commutation with the three independent coordinate reflections forces all off-diagonal entries of $C$ to vanish. Commutation with the quarter-turn about the first axis equates its second and third diagonal entries, and the quarter-turn about the second axis equates its first and third entries. ThereforeFor completeness, if the complexified group representation had a proper invariant subspace, its orthogonal complement would be invariant too, since the generating matrices are unitary. The orthogonal projection onto that subspace would be a nonscalar commuting matrix, contradicting the calculation. Thus the complexification is irreducible, proving absolute irreducibility of the real group representation.

Choose three representative vectors, with nonzero coordinate scale suppressed. Their stabilizer subgroups and fixed-point subspaces of a group action areHere ${\kappa }_x{r}_z$ exchanges the first two coordinates and ${\kappa }_y{r}_x$ exchanges the last two. On the first line the stabilizer subgroup is every signed permutation of the two zero coordinates; fixing it forces those coordinates to vanish. On the second line it exchanges the equal coordinates and independently reflects the zero coordinate, forcing $x=y,z=0$. On the third line all coordinate permutations fix the vector, forcing $x=y=z$.

The three subgroups have different orders and hence cannot be conjugate. Their group orbits contain respectively six face-axis, twelve edge-axis and eight vertex-axis directions. Each fixed-point subspace of a group action is one-dimensional, so the equivariant branching lemma guarantees three distinct symmetry types of bifurcating equilibria. The axial isotropy types of the full cube symmetry group are precisely these: a vector with several distinct nonzero absolute coordinate values has a separate freely variable coordinate block for each such value in the fixed-point subspace of a group action of its stabilizer subgroup, giving dimension greater than one.

Coordinate reflection in $x_i$ requires the $i$th component to be odd in $x_i$, and reflections in the other coordinates require it to be even in those coordinates. Thus no constant or quadratic terms are permitted. The only cubic monomials in component $i$ are $x_i^3$ and $x_i{x}_j^2$ for $j\ne i$. Coordinate permutations equate their coefficients in all components. After using the transverse linear eigenvalue as parameter, the normal form isThe remainder starts at fifth spatial order, together with parameter-dependent corrections such as $\mu |x{|}^3$. This is the cubic equivariants of the full cube symmetry group form. On a branch with $m=1,2,3$ equal nonzero absolute coordinates $s$, respectively,Hence the three leading branch amplitudes are $s^2=-\mu /a$, $-\mu /(a+b)$ and $-\mu /(a+2b)$. Their generic nonzero denominators determine on which side the branches occur; existence alone does not make them stable.