Past exam of the groups and representations course of the University of Oxford / 2021 / 1

Let $T(v\otimes w)=w\otimes v$ on the tensor product $V\otimes V$. The linear map $R(g)\otimes R(g)$ commutes with $T$, so preserves its $+1$ and $-1$ eigenspaces, namely the symmetric square $S^{2}V$ and the exterior square ${\wedge }^{2}V$. Restricting the tensor product representation to these invariant subspaces defines $S^{2}R$ and ${\wedge }^{2}R$. The representation identity follows by restricting $(R(g)\otimes R(g))(R(h)\otimes R(h))=R(gh)\otimes R(gh)$.

For a finite group, $g$ has finite order $m$, so $R(g{)}^m=I$. Over the complex numbers, the polynomial $t^m-1$ has distinct roots; hence $R(g)$ is diagonalizable. Choose an eigenbasis $v_1,\dots ,v_n$ with eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$. A basis of the symmetric square consists of $v_i\otimes v_i$ and $v_i\otimes v_j+v_j\otimes v_i$ for $i<j$. A basis of the exterior square consists of $v_i\otimes v_j-v_j\otimes v_i$ for $i<j$. Their eigenvalues are respectively ${\lambda }_i^2,{\lambda }_i{\lambda }_j$ and ${\lambda }_i{\lambda }_j$. Taking traces,But $\chi (g{)}^2={\sum }_i{\lambda }_i^2+2{\sum }_{i<j}{\lambda }_i{\lambda }_j$ and $\chi (g^2)={\sum }_i{\lambda }_i^2$. Thus the characters areAt $g=e$, these also give the dimensions $n(n+1)/2$ and $n(n-1)/2$.

Write the two Pauli matrices asThey satisfy $A^2=B^2=I$, $BA=-AB$ and $C^2=-I$. Moving every $A$ to the left of every $B$ reduces each word to one of eight distinct matrices. ConsequentlyTaking $r=C$, $s=A$ gives $r^4=s^2=I$ and $srs=r^{-1}$, identifying the group with the dihedral group of a square.

The elements $\pm I$ are central. Conjugating $A$ by $B$ changes its sign, and similarly for $B$ and $C$. No conjugation mixes these pairs: modulo the central subgroup $\{\pm I\}$ the quotient group is the Klein four-group and hence Abelian. Thus the conjugacy classes are preciselyThere are five complex irreducible representations, one for each conjugacy class. In a one-dimensional representation, the images of $A,B$ are signs $s,t\in \{1,-1\}$. Since they commute, $-I$ maps to $1$, and all four choices of $s,t$ satisfy the relations. These are four inequivalent one-dimensional representations. The sum of squares of irreducible representation dimensions leaves $8-4=4$ for the fifth, so the dimensions are

Use the conjugacy class order of solution. The defining representation has characterIts character inner product with itself is $(4+4)/8=1$, proving that it is an irreducible representation. There is also a direct invariant subspace argument: $B$ has two distinct coordinate eigenspaces, and $A$ exchanges them, so no line is invariant under both.

Let $L_{st}$ denote the one-dimensional representation with $L_{st}(A)=s$ and $L_{st}(B)=t$. Then $L_{st}(-I)=1$ and $L_{st}(C)=st$. The complete character table isThe four scalar rows obey character orthogonality, and the dimensions exhaust the group order as established in solution.

The squares of $I,-I,\pm A,\pm B$ are $I$, whereas $(\pm C{)}^2=-I$. Therefore ${\chi }_{\rho }(g^2)=(2,2,2,2,-2)$. The symmetric square and exterior square formulas from solution giveComparison with the character table gives the Clebsch-Gordan decomposition of group representationsFor explicit bases, put $e_1=(1,0{)}^T,e_2=(0,1{)}^T$. The symmetric tensors $e_1\otimes e_1+e_2\otimes e_2$, $e_1\otimes e_2+e_2\otimes e_1$, and $e_1\otimes e_1-e_2\otimes e_2$ transform as $L_{++},L_{+-},L_{-+}$ respectively. The antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ transforms as $L_{--}$, also the determinant representation of $\rho$. Adding these direct sums recovers $\rho \otimes \rho ={⨁}_{s,t}{L}_{st}$.

Write a homogeneous quadratic polynomial as $p(x,y)=a{x}^2+bxy+c{y}^2$. Invariance under $B:(x,y)\mapsto (x,-y)$ forces $b=0$. Invariance under $A:(x,y)\mapsto (y,x)$ then forces $a=c$. Since $A,B$ generate the group, this proves that all homogeneous quadratic polynomial invariants areThis agrees with the single trivial representation in the symmetric square in solution; more precisely, coordinate polynomials lie in the symmetric square of the dual representation, which here is equivalent to $\rho$. If “quadratic” permits terms of smaller degree, arbitrary constants can be added. There are no invariant linear terms because $-I\in G$ negates every linear polynomial.