Past exam of the groups and representations course of the University of Oxford / 2023 / 2

Conjugacy classes in the symmetric group are specified by cycle type, equivalently a partition of $4$. If $m_j$ counts cycles of length $j$, the number of permutations of that cycle type is $4!/{\prod }_j(j^{m_j}{m}_j!)$: cyclic rotation of each cycle and interchange of equal-length cycles do not change the permutation. This givesThe sizes sum to $24$. Since the number of complex irreducible representations of a finite group equals its number of conjugacy classes, $S_4$ has $\boxed{5}$ inequivalent complex irreducible representations.

The trivial representation and sign of a permutation giveThey are one-dimensional representations, hence irreducible representations. Their characters, in the conjugacy class order of solution, areThese are the only one-dimensional representations. All transpositions are conjugate, so they have the same scalar image $t$ in any one-dimensional representation; since a transposition squares to the identity, $t=\pm 1$. Transpositions generate the symmetric group, fixing the whole representation once $t$ is chosen.

By the sum of squares of irreducible representation dimensions, the three remaining dimensions satisfy $d_2^2+d_3^2+d_4^2=24-1-1=22$, with each $d_i\ge 2$. No $d_i$ can be at least $4$, since $16+4+4>22$. Thus the only possible dimensions are

Use the composition convention $(\sigma \tau )(i)=\sigma (\tau (i))$. Then $R_{+}(\sigma )R_{+}(\tau )e_i=e_{\sigma (\tau (i))}=R_{+}(\sigma \tau )e_i$. Since the sign of a permutation is multiplicative, the same calculation gives $R_{-}(\sigma )R_{-}(\tau )=R_{-}(\sigma \tau )$. Both send the identity to $I$ and have inverses given by ${\sigma }^{-1}$, so both are group representations.

The trace of a permutation matrix counts fixed points, and multiplying that matrix by its sign of a permutation multiplies its trace by that sign. Thus the two characters areLet $u=e_1+e_2+e_3+e_4$ and $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. Both are invariant subspaces, and ${\mathbb{C}}^4=\mathbb{C}u\oplus W$. On $\mathbb{C}u$, $R_{+}$ is the trivial representation and $R_{-}$ is the sign representation. Subtracting their characters yieldsThe character inner product verifies that it is an irreducible representation:The squares of the entries of ${\chi }_4$ are identical, so its character inner product with itself is also one. Their values at a transposition differ, so these irreducible representations are inequivalent. Therefore

Only the character ${\chi }_2$ of the two-dimensional irreducible representation remains unknown. Write it as $(2,a,b,c,d)$. The regular representation has character zero off the identity and decomposes with multiplicities equal to the irreducible representation dimensions. Thus, for a nonidentity element $g$,For the four nonidentity conjugacy classes this gives $a=0$, $b=2$, $c=-1$, $d=0$. The complete character table is consequentlyFor example, $⟨{\chi }_2,{\chi }_2⟩=(4+12+8)/24=1$, agreeing with character orthogonality.

Use the realization $R_3=W\subset {\mathbb{C}}^4$ from solution. Put $x=({\varphi }_1,{\varphi }_2,{\varphi }_3,-{\varphi }_1-{\varphi }_2-{\varphi }_3)$, which identifies the three-dimensional scalar field with $W$. The polynomial invariants of the standard S4 representation include the symmetric power sumsbecause every permutation merely rearranges their summands. In these coordinates,Neither polynomial vanishes identically: at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(1,1,1)$ they equal $12$ and $-24$. Hence coefficients multiplying $I_2$ and $I_3$ provide nonzero invariant quadratic and cubic terms. A change of basis transports these polynomial invariants to any equivalent realization of $R_3$. For a real scalar field, both are real. For a complex scalar field, a real scalar potential may contain ${\lambda }_k{I}_k+{\lambda }_k^{\ast }{I}_k^{\ast }$; the quadratic Hermitian form ${\sum }_i|x_i{|}^2$ is another invariant.

As an independent character check, the symmetric square of a group representation has character $[\chi (g{)}^2+\chi (g^2)]/2$, and the symmetric cube of a group representation has character $[\chi (g{)}^3+3\chi (g)\chi (g^2)+2\chi (g^3)]/6$. With $\chi ={\chi }_3$ these are $(6,2,2,0,0)$ and $(10,2,-2,1,0)$. Their character inner products with ${\chi }_0$ are both one, so each degree has exactly one independent polynomial invariant.