Past exam of the groups and representations course of the University of Oxford / 2019 (A15090W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The annual bundle is named “Physics Papers 2018”, but this paper's cover is dated 11 January 2019. The year here follows the cover. All four questions are solved, although only three were required.

A subgroup $N\le G$ is a normal subgroup if $gN{g}^{-1}=N$ for every $g\in G$. A group homomorphism $f:G\to \tilde{G}$ satisfies $f(gh)=f(g)f(h)$ for every $g,h\in G$. In particular $f(e)=\tilde{e}$ and $f(g^{-1})=f(g{)}^{-1}$. A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$ into the general linear group of invertible linear maps on $V$.

For the kernel of a group homomorphism, put $K=\ker f=\{g:f(g)=\tilde{e}\}$. It contains $e$, and if $k,\ell \in K$ then $f(k{\ell }^{-1})=f(k)f(\ell {)}^{-1}=\tilde{e}$, so $K$ is a subgroup. For $g\in G$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Applying this inclusion with $g^{-1}$ gives the reverse inclusion, proving that $\ker f$ is a normal subgroup.

For a normal subgroup $N$, define multiplication of left cosets byTo check independence of representatives, replace $g,h$ by $gn,hm$, with $n,m\in N$. Thenbecause normality gives $h^{-1}nh\in N$. Hence the operation is well-defined. Associativity follows from $(gh)k=g(hk)$ in $G$; the identity is $N=eN$; and the inverse of $gN$ is $g^{-1}N$. These verify the group axioms, giving the quotient group $G/N$. The map $g\mapsto gN$ is a surjective group homomorphism with kernel of a group homomorphism equal to $N$.

Nonzero scalar matrices form a subgroup $N$ of the general linear group, since products and inverses remain nonzero scalar matrices. Moreover, for every invertible $g$,Thus $N$ lies in the center of a group and is a normal subgroup. The quotient group is the complex projective linear group ${\mathrm{PGL}}_n(\mathbb{C})$, whose elements identify invertible matrices differing by a nonzero scalar.

The canonical group homomorphism $\pi :g\mapsto [g]$ obeys $[g][h]=[gh]$. Composing it with the given group representation givesConsequentlyThis is the projective representation associated with $R$: a nonzero scalar acts trivially on each line in the projective space of $V$. Its kernel of a group homomorphism consists of those $g$ for which $R(g)$ is scalar, which can be larger than the kernel of a group homomorphism of $R$ itself.

Write $A=g_1$, $B=g_2$ and $C=AB$. Direct multiplication givesThe relation $BA=-AB$ moves all occurrences of $A$ to the left of all occurrences of $B$. Removing even powers reduces every product of the generators to $\pm A^a{B}^b$ with $a,b\in \{0,1\}$. The resulting eight matrices are distinct, soExplicitly, the four pairs areSince $AB=-BA$ and $AB\ne 0$, the group is noncommutative. More precisely, $C$ has order four, $A$ has order two and $ACA=C^{-1}$, identifying $G$ with the dihedral group of a square. This fixes the convention unambiguously: it is the dihedral group of group order eight, irrespective of whether it is denoted $D_4$ or $D_8$.

Use the matrices of solution. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The group homomorphism is therefore explicitlyThese four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.

Although $A$ and $B$ anticommute as matrices, their projective classes commute:Furthermore $[A{]}^2=[B{]}^2=[I_2]$ and $[C{]}^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit group isomorphism to the Klein four-group isIndeed reordering products introduces only a sign, which disappears in the projective linear group, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker \tilde{R}=G\cap N=\{\pm I_2\}$ and the first isomorphism theorem givesThis projective image of the square symmetry group illustrates how a noncommutative group can have an Abelian projective image: its scalar commutator becomes the identity class.

Put $r=g_1=\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s=g_2$, using ${\alpha }^{\ast }={\alpha }^{-1}={\alpha }^2$. ThenEvery word can therefore be put in the form $r^k$ or $r^{k}s$, with $k=0,1,2$. These give six distinct matrices:The group isomorphism $r\mapsto (123)$, $s\mapsto (12)$ identifies this group with the symmetric group $S_3$, with permutations composed from right to left.

Conjugation by $s$ exchanges $r,r^2$. Conjugation by $r$ cycles $s,rs,r^{2}s$, because $r(r^{k}s)r^{-1}=r^{k+2}s$. Their conjugacy classes arewith sizes $1,2,3$. There is no mixing between them: their elements have orders one, three and two, respectively.

The matrix $r$ has distinct eigenvalues $\alpha ,{\alpha }^2$. Any one-dimensional invariant subspace for the defining representation would therefore have to be one of its two coordinate eigenspaces. But $s$ exchanges those lines, so neither is invariant under the whole group. Hence $R_2$ is an irreducible representation over the complex numbers.

There are three complex irreducible representations up to equivalence of group representations, because there are three conjugacy classes. The trivial representation and the sign representation already give two inequivalent one-dimensional representations. Together with $R_2$, their dimensions satisfy the sum of squares of irreducible representation dimensions:Thus there are no further irreducible representations. Alternatively the character of $R_2$ is $(2,-1,0)$, and its character inner product with itself is $(4+2)/6=1$, independently confirming irreducibility.

The irreducible representations can all be specified on the generators:The first is the trivial representation and the second is the sign representation. To see directly that these exhaust the one-dimensional representations, their scalar images commute, so the relation $srs=r^{-1}$ forces $r=r^{-1}$. Combining $r^2=1$ with $r^3=1$ gives $r=1$, and then $s$ may be either sign.

Taking traces gives the character table, in the conjugacy class order of solution:Here $\alpha +{\alpha }^2=-1$ gives the middle entry in the last row. The weighted character inner products of distinct rows vanish, and the norm of each row is one. Together with solution, this proves completeness of the table.

Let $S=s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Entrywise complex conjugation interchanges the diagonal entries of $r$ and fixes $s$, soSince $r,s$ generate the group, these identities hold for every element. Thus $S$ is an invertible intertwiner, provingThe character of a tensor product representation is the product of the factor characters. Hence ${\chi }_{R_2\otimes R_2}=(4,1,0)$. Its character inner products with the three rows of solution give multiplicities $(1,1,1)$, soFor explicit Clebsch-Gordan decomposition of group representations bases, let $e_{+},e_{-}$ be the coordinate basis. The symmetric tensor $e_{+}\otimes e_{-}+e_{-}\otimes e_{+}$ spans $1$; the antisymmetric tensor $e_{+}\otimes e_{-}-e_{-}\otimes e_{+}$ spans $\epsilon$; and the ordered pair $e_{-}\otimes e_{-},e_{+}\otimes e_{+}$ carries $R_2$, since $r$ acts by $\mathrm{diag}(\alpha ,{\alpha }^2)$ and $s$ exchanges them. In particular $S^2{R}_2=1\oplus R_2$ and ${\Lambda }^2{R}_2=\epsilon$.

Multiplying by the sign representation preserves the character of $R_2$, because its character is zero on all three reflections. Thus $\epsilon \otimes R_2\cong R_2$, and associativity givesThereforeThe dimensions give $8=1+1+3\cdot 2$. Its two one-dimensional summands can also be seen directly: $e_{+}^{\otimes 3}+e_{-}^{\otimes 3}$ and $e_{+}^{\otimes 3}-e_{-}^{\otimes 3}$ transform as $1$ and $\epsilon$, respectively.

For $z\in U(1)$ and $U\in SU(n)$, the matrix $zU$ is unitary, since $(zU{)}^{†}(zU)=|z{|}^2{U}^{†}U=I_n$. Thus the map has its image in the unitary group $U(n)$. Scalar matrices commute with all matrices, so the direct product of groups multiplication givesAlso $f(1,I_n)=I_n$. HenceThe special unitary group condition $\det U=1$ imposes no requirement that $zU$ itself have unit determinant; its determinant is $z^n$, as expected for a general element of the unitary group.

First the group homomorphism is surjective. For any $W\in U(n)$, choose $z\in U(1)$ with $z^n=\det W$ and set $U=z^{-1}W$. Then $U$ is unitary and $\det U=z^{-n}\det W=1$, so $W=f(z,U)$.

Next $f(z,U)=I_n$ is equivalent to $U=z^{-1}{I}_n$, and the special unitary group condition becomes $z^n=1$. With $\zeta =e^{2\pi i/n}$, the kernel of a group homomorphism is thereforeIt is the cyclic central subgroup generated by $(\zeta ,{\zeta }^{-1}{I}_n)$; the scalar and special unitary group factors are paired with inverse phases. The first isomorphism theorem now givesExplicitly the isomorphism sends the coset $[(z,U)]$ to $zU$. The $n$ possible choices of the root $z$ differ precisely by multiplication by elements of $K$, which explains why the product before quotienting is not a unique factorization.

Use upper indices for the defining representation of SU(6) and lower indices for its dual representation. The Young tableaux below display tensor-slot labels; their shapes determine the Young diagrams. Full columns of height six are determinant representations and can be removed for SU(6).Here all indices run from one to six. The highest weight Dynkin labels follow from the differences of successive row lengths. In the same order the row lengths are $(1)$, $(1,1,1,1,1)$, $(2)$, $(1,1)$ and $(2,1,1,1,1)$.

The defining vector transforms as $v^a\mapsto U^a{}_b{v}^b$. For a unitary representation, the conjugate representation is equivalent to the dual representation, so $\overline{6}$ can be written as a covector with $w_a\mapsto w_b(U^{-1}{)}^b{}_a$. Its five-box column is equivalently a totally antisymmetric tensor $A^{a_1\cdots a_5}$, related by the invariant alternating tensor $w_a={ϵ}_{a{a}_1\cdots a_5}{A}^{a_1\cdots a_5}/5!$.

The symmetric tensor transforms as $S^{ab}\mapsto U^a{}_c{U}^b{}_d{S}^{cd}$, with $6(6+1)/2=21$ independent components. The antisymmetric tensor has the same transformation law but obeys $A^{ab}=-A^{ba}$, giving $6(6-1)/2=15$ independent components. For the Adjoint representation, $T\mapsto UT{U}^{-1}$; the trace constraint removes one component from the $36$ entries of a mixed tensor, leaving $35$. Its six-box Young diagram is obtained by converting the five-index column to a lower index and selecting the traceless mixed tensor.

Thus the requested dimensions, in the order above, are

In this part the earlier variable $n$ specializes to six. A group representation of $U(1)\times SU(6)$ factors through the quotient group $U(6)$ exactly when the central subgroup $K$ in solution acts trivially. For an irreducible SU(6) tensor representation represented by a Young diagram with $b$ boxes, the scalar matrix ${\zeta }^{-1}{I}_6$ acts as ${\zeta }^{-b}$, where $\zeta =e^{2\pi i/6}$.

The prescribed representation $\tilde{R}(z,U)=R(U)$ gives no compensating phase from the first factor. Its value on the generator of $K$ is therefore $R({\zeta }^{-1}{I}_6)={\zeta }^{-b}I$. The box counts in solution giveConsequentlyConcretely, conjugation $T\mapsto WT{W}^{-1}$ defines it for every $W\in U(6)$, and the scalar phase cancels.

The qualification “prescribed lifts” matters. The defining representation of $U(6)$ certainly exists, but its pullback is $(z,U)\mapsto zU$, not $(z,U)\mapsto U$. For instance $(\zeta ,{\zeta }^{-1}{I}_6)$ and $(1,I_6)$ represent the same quotient element, while the latter prescription assigns them different matrices in the defining representation. More generally, for an integer $q$, a lift $z^{q}R(U)$ descends precisely whenThis descent criterion for representations of U(n) explains how all the listed SU(6) representations can be extended after choosing a suitable first-factor U(1) charge, even though only one descends with that charge fixed to zero.

The first coordinate is fixed by the embedded SU(5), while the last five transform in its defining representation. Therefore $6{\downarrow }_{SU(5)}=1\oplus 5$, and taking the dual representation gives $\overline{6}{\downarrow }_{SU(5)}=1\oplus \overline{5}$.

For the symmetric square, the fixed-fixed component is one trivial representation, the fixed-five mixed components give $5$, and the symmetric five-five block is $15$. For the exterior square, the fixed-fixed component vanishes, the mixed components give $5$, and the antisymmetric five-five block is $10$. Finally restrict $35=6\otimes \overline{6}-1$, using $5\otimes \overline{5}=24\oplus 1$.

The five SU6 tensor branchings to SU5 are thusThe symbols $15$ on the left and right refer to different groups: the left-hand one is the SU(6) exterior square, whereas the right-hand one is the SU(5) symmetric square. The dimension checks are $6=5+1$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

Explicitly, in the Adjoint representation the lower-right traceless five-by-five block gives $24$; the lower-left column and upper-right row give $5$ and $\overline{5}$; and the diagonal direction $\mathrm{diag}(-5,1,1,1,1,1)$ gives the remaining trivial representation.

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore

Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

Let $r=V(0,1)$ be the four-dimensional spinor representation. The tensor product representation splits into its symmetric square and exterior square, with dimensions ten and six. If $v$ is a highest-weight vector of $r$, then $v\otimes v$ is a highest-weight vector of weight $(0,2)$ in the symmetric square. The irreducible representation with that highest weight has dimension ten by solution, exhausting the symmetric square:For the exterior square, add each pair of distinct weights from solution. The resulting multiset of weights isIts maximal weight $(1,0)$ is the highest weight of a five-dimensional irreducible representation. It accounts for the four nonzero weights and one zero weight, leaving a single zero weight. By complete reducibility of semisimple Lie algebra representations, that remaining one-dimensional summand is the trivial representation. Thus ${\Lambda }^{2}r\cong 5\oplus 1$, and the spinor tensor square for so5 isThe dimension check is $16=1+5+10$. The trivial representation lies in the exterior square, reflecting the invariant nondegenerate alternating bilinear form of this pseudoreal representation. Although $r$ itself does not descend to $SO(5)$, its tensor square does: the central element $-1\in \mathrm{Spin}(5)$ acts as $(-I_4)\otimes (-I_4)=I_{16}$.

Use the basis matrices $M_{ab}=E_{ab}-E_{ba}$ from solution. Define the Pauli matrices and the two-by-two identity byThe following five Hermitian four-by-four matrices, written as Kronecker products, realize a Euclidean Clifford algebra:Define the Lie algebra representation on ${\mathbb{C}}^4$ byThese are anti-Hermitian matrices, appropriate to the real compact Lie algebra. For complete explicitness, all ten representation matrices areEach table entry is the image under $\rho$ of the basis matrix immediately to its left. Since the two-by-two factors are specified above, these Kronecker products specify every entry of each four-by-four matrix.

To verify the representation law, the Clifford algebra relations implyApply this identity to $\rho (M_{cd})=\frac{1}{4}[{\gamma }_c,{\gamma }_d]$ and use the commutator product rule. The result isexactly the Lie bracket relation in solution. Thus $\rho$ is a Lie algebra representation.

It remains to identify its highest weight, rather than just its dimension. The commuting Hermitian Cartan subalgebra operators areTheir simultaneous eigenvalues are $(x,y)=(\pm \frac{1}{2},\pm \frac{1}{2})$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four weights of $V(0,1)$ in solution. A state with $(x,y)=(\frac{1}{2},\frac{1}{2})$ has highest weight $(0,1)$, so this Clifford realization of the so5 spin representation is the requested $r$.

Finally, $\rho (M_{12}{)}^2=-I_4/4$, soThis explicitly exhibits the distinction between the spinor representation of $\mathfrak{so}(5)$, or of $\mathrm{Spin}(5)$, and a single-valued group representation of $SO(5)$.