Past exam of the groups and representations course of the University of Oxford / 2025 (A15090H1)

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15090h1.pdf

These solutions cover every part of the held paper. Numbered headings identify the corresponding source parts; the question statements are not reproduced.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I$. A nonzero group representation is an irreducible representation if its only invariant subspaces are $0$ and $V$. Here the representations and characters are finite-dimensional over the complex numbers.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. Three useful properties are:

In particular, ${\chi }_R(e)=\dim V$ and ${\chi }_R(g^{-1})={\chi }_R(g{)}^{\ast }$, the latter following from unitarization of a finite-group representation. The character inner product $⟨{\chi }_r,{\chi }_R⟩$ counts the multiplicity of an irreducible representation $r$ in $R$.

For any $h\in G$, left multiplication permutes the elements of the finite group. Consequently averaging over a finite group givesThus $Pv$ belongs to the fixed-point subspace of a group action $V^G$; any nonzero such vector spans a trivial representation, or singlet. The zero vector is included in the invariance statement.

Conversely $Pw=w$ for every $w\in V^G$, soThis is a projection operator onto all singlets, rather than necessarily onto a one-dimensional vector subspace. Its trace is the singlet multiplicity $|G{|}^{-1}{\sum }_g{\chi }_R(g)$.

Put $r=g_1$ and $s=g_2$. Direct matrix multiplication givesMoving every $s$ to the right shows that every word in these generators of a group has the form $r^a{s}^b$, with $a\in \{0,1,2\}$ and $b\in \{0,1\}$. The three $r^a$ are distinct diagonal matrices, and the three $r^{a}s$ are distinct off-diagonal matrices. HenceSince $srs=r^2\ne r$, this is a non-abelian group. Its group presentation identifies it with the dihedral group of an equilateral triangle, equivalently the symmetric group $S_3$.

The conjugacy classes areIndeed $s$ interchanges $r$ and $r^2$ by conjugation, while $r^{a}s{r}^{-a}=r^{2a}s$ runs through all three reflections. No reflection is conjugate to a rotation, since conjugation preserves order of a group element.

There are three inequivalent complex irreducible representations, one for each conjugacy class. The sum of squares of irreducible representation dimensions gives $d_1^2+d_2^2+d_3^2=6$. One representation is trivial, and the only possible positive integer dimensions are

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

Choose the doublet bases so that the column of barred fields $B=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $R=({\psi }_R^1,{\psi }_R^2{)}^T$ both transform by the doublet matrices in d. This choice follows the source's assignment of the barred doublet itself. Since $H$ transforms in $\epsilon$, its Yukawa coupling must multiply a fermion bilinear in $\epsilon$.

There is exactly one such bilinear in $2\otimes 2$: the Clebsch-Gordan coefficients in d give $B_1{R}_2-B_2{R}_1$. Equivalently, writing the upper block of the Yukawa matrix as $A$, invariance requiresThe first equation sets $A_{11}=A_{22}=0$, because ${\alpha }^2\ne 1$; the second sets $A_{21}=-A_{12}$.

A mixed doublet-singlet bilinear remains a doublet after multiplication by $H$, so all four mixed entries vanish. Finally ${\bar{\psi }}_L^3{\psi }_R^3$ transforms in $\epsilon \otimes 1=\epsilon$, and multiplication by $H$ makes it invariant. ThereforeThe allowed Yukawa interaction is $H\{a({\bar{\psi }}_L^1{\psi }_R^2-{\bar{\psi }}_L^2{\psi }_R^1)+b{\bar{\psi }}_L^3{\psi }_R^3\}$, together with its Hermitian conjugate when forming a real Lagrangian density.

The displayed texture depends on basis. If instead the unbarred left doublet is assigned $D$ in the same ordering as the right doublet, its barred doublet has the dual representation $D^{\ast }$ and the equivalent upper block is $a\mathrm{diag}(1,-1)$. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex Yukawa couplings.

For a curve $O(t)=I+tX+O(t^2)$ in the special orthogonal group, differentiate $O(t{)}^{T}O(t)=I$ at the identity. This gives $X^T+X=0$. Conversely, the matrix exponential of any real skew-symmetric matrix is orthogonal and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. HenceA basis is $L_{ij}=E_{ij}-E_{ji}$ for $1\le i<j\le N$, where $E_{ij}$ is a matrix unit. If $L_{ji}=-L_{ij}$ and $L_{ii}=0$, their Lie brackets areThere are $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ independent off-diagonal entries. A maximal torus rotates independently in mutually orthogonal two-planes, generated by $L_{12},L_{34},\dots$. ThusThe maximality follows by simultaneously decomposing commuting real skew-symmetric matrices into rotation two-planes, with at most one remaining fixed axis.

The unbroken group is the stabilizer subgroup of the vacuum expectation value. For a nonzero vector vacuum expectation value, an orthogonal transformation puts $⟨\varphi ⟩=v{e}_N$. A special orthogonal transformation fixing it has the block form $\mathrm{diag}(A,1)$ with $A\in \mathrm{SO}(N-1)$. ConsequentlyThe zero vacuum expectation value leaves the entire special orthogonal group unbroken.

For a real symmetric tensor write its vacuum expectation value as a real symmetric matrix $Q$. The transformation is $Q\mapsto OQ{O}^T$. The spectral theorem for real symmetric matrices gives an orthonormal basis in whichBecause $OQ{O}^T=Q$ is equivalent to $OQ=QO$, an unbroken transformation preserves each eigenspace. The complete unbroken group is thereforeIts identity component is $H^0={\prod }_i\mathrm{SO}(m_i)$. This symmetric tensor vacuum stabilizer shows why the representation alone does not determine a unique breaking pattern:

For example a rank-one tensor $Q=v{e}_N{e}_N^T$ preserves $O(N-1)$ embedded by $A\mapsto \mathrm{diag}(A,\det A)$. Its continuous unbroken group is $\mathrm{SO}(N-1)$, but it also allows reversal of the distinguished axis. Restricting to a traceless symmetric tensor imposes ${\sum }_i{m}_i{q}_i=0$ and otherwise leaves the same classification.

The complexification of a Lie algebra gives $\mathfrak{so}(2n,\mathbb{C})$ of type Dn root system and $\mathfrak{so}(2n+1,\mathbb{C})$ of type Bn root system. For $D_n$, the simple roots ${\alpha }_i=e_i-e_{i+1}$ for $1\le i<n$ and ${\alpha }_n=e_{n-1}+e_n$ give a chain $1,\dots ,n-2$ with two terminal nodes $n-1,n$ joined to $n-2$. All edges are single. For $B_n$, use ${\alpha }_i=e_i-e_{i+1}$ for $i<n$ and ${\alpha }_n=e_n$. The Bn Dynkin diagram is a chain with a double final edge and an arrow pointing to the short node $n$.

The usual family diagrams apply to $D_n$ for $n\ge 4$ and $B_n$ for $n\ge 2$. The small cases are $D_3\cong A_3$, $D_2\cong A_1\oplus A_1$, $B_1\cong A_1$; $\mathfrak{so}(2)$ is one-dimensional and Abelian.

To fix Casimir normalization for orthogonal vector representations, use Hermitian Lie algebra generators $t_{ij}=i(E_{ij}-E_{ji})/\sqrt{2}$, so the defining representation has Dynkin index $c(\mathbf{N})=1$. With long roots of squared length two, the quadratic Casimir operator has Casimir eigenvaluewhere $\rho$ is the Weyl vector and $\lambda$ in the final expression is the column of Dynkin labels. The defining fundamental representation has highest weight $(1,0,\dots ,0)$ in the stated family conventions.

For $D_n$, the row supplied in the source has sum $n-1$ and first entry one, soFor $B_n$, the supplied row has sum $n-\frac{1}{2}$, givingTaking the trace of ${\sum }_a{t}_a{t}_a=C_2(R)I_R$ gives the relation between the Dynkin index and Casimir eigenvalue, $c(R)\dim \mathfrak{g}=C_2(R)\dim R$. ThereforeThere is also a direct check: each coordinate appears in $N-1$ planes and $t_{ij}^2=(E_{ii}+E_{jj})/2$, so ${\sum }_{i<j}{t}_{ij}^2=(N-1)I/2$.

If the Casimir operator is instead defined by $C(\lambda )=(\lambda ,\lambda +2\rho )$, every Lie algebra generator is larger by $\sqrt{2}$ and the corresponding answers are $C(\mathbf{N})=N-1$ and $c(\mathbf{N})=2$. These are the same Lie algebra in a different normalization. At the exceptional small ranks the family Dynkin labels of the vector change, but the direct matrix computation remains valid.

In the normalization $c(8)=1$ of c, the Adjoint representation of $D_4$ has highest weight ${\lambda }_{\mathrm{adj}}=(0,1,0,0)$, the highest root. The supplied Gram matrix has second row $(1,2,1,1)$, whose sum is five. HenceThe Adjoint representation has the same dimension as the Lie algebra, so its Dynkin index is also $c(\mathrm{adj})=6$. The $k$ Weyl fermions form a direct sum of $k$ vector representations; additivity of the trace gives $c(r_W)=k$.

Substitution into the given one-loop renormalization-group beta function yieldsFor positive $g$, asymptotic freedom at this order requires a negative beta function, equivalentlyIndeed, if $b_0=22-2k/3>0$, integrating $\mu dg/d\mu =-b_0{g}^3/(16{\pi }^2)$ givesso the gauge coupling decreases towards zero as the energy scale increases. At $k=33$ the one-loop coefficient vanishes and the given equation cannot decide asymptotic freedom; $k>33$ has the opposite one-loop running.

With the alternative convention $C(\lambda )={\lambda }^{T}G(\lambda +21)$ described in c, $c(8)=2$, $c(\mathrm{adj})=12$, and the same source formula reads $\beta (g)=-(44-4k/3)g^3/(16{\pi }^2)$. Rescaling the gauge coupling together with the Lie algebra generators relates the conventions; the bound $k<33$ is identical.

Adopt the Minkowski metric $\eta =\mathrm{diag}(-1,+1)$ and the Clifford algebra convention $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\eta }_{\mu \nu }I$. The source's Pauli matrices giveThey satisfy ${\gamma }_0^2=-I$, ${\gamma }_1^2=I$ and ${\gamma }_0{\gamma }_1=-{\gamma }_1{\gamma }_0=-{\sigma }_3$, which verifies the required Clifford algebra. Raising the indices gives ${\gamma }^0=-{\gamma }_0$ and ${\gamma }^1={\gamma }_1$.

Fix the sign of the rapidity $\theta$ by taking the Lorentz boostThe corresponding Spinor representation of the Lorentz group isFor example $S^{-1}{\gamma }^{0}S={\gamma }^0\cosh \theta -{\gamma }^1\sinh \theta$ and $S^{-1}{\gamma }^{1}S={\gamma }^1\cosh \theta -{\gamma }^0\sinh \theta$, establishing Lorentz covariance by $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$.

Thus the Dirac spinor transforms as ${\psi }^{\prime }(x^{\prime })=S(\theta )\psi (x)$, with infinitesimal lawThese variations compare the fields at corresponding spacetime points. At a fixed coordinate one adds the usual orbital term $-\delta x^{\mu }{\partial }_{\mu }\psi$. Reversing the rapidity convention reverses both boost signs consistently. Infinitesimal transformations describe the proper orthochronous Lorentz group.

Choose the chirality operatorA Weyl spinor is a Dirac spinor restricted to one eigenspace of $\Gamma$. The chiral projectors are $P_R=(I+\Gamma )/2$ and $P_L=(I-\Gamma )/2$. ConsequentlyThe two independent Weyl spinors each contain one complex component. Since $[S(\theta ),\Gamma ]=0$, each chirality is preserved by every connected Lorentz transformation, with ${\psi }_R\mapsto e^{\theta /2}{\psi }_R$ and ${\psi }_L\mapsto e^{-\theta /2}{\psi }_L$. A spatial reflection exchanges the two chiralities.

The left/right names depend on the sign convention for $\Gamma$. Choosing ${\gamma }_0{\gamma }_1=-{\sigma }_3$ as the chirality operator exchanges the labels $L$ and $R$; the two invariant one-component representations are the same.

All gamma matrices in this basis are real, and the Lorentz transformation $S(\theta )$ is real. Therefore complex conjugation commutes with the connected Spinor representation of the Lorentz group. With a suitable phase convention, charge conjugation is simply ${\psi }^c={\psi }^{\ast }$. The Majorana spinor condition ${\psi }^c=\psi$ becomesThe reality constraint is preserved by Lorentz transformations because $(S{\psi }_M{)}^{\ast }=S{\psi }_M$. For fermion fields, real components mean self-conjugate Grassmann variables or self-adjoint field components, rather than ordinary commuting real numbers.

Moreover, $\Gamma ={\sigma }_3$ is real, so charge conjugation commutes with the chiral projectors. The Majorana spinor and Weyl spinor conditions can therefore be imposed simultaneously. The Majorana-Weyl spinors areThe independent real component counts areThese are component counts before imposing an equation of motion. In particular the simultaneous reality and chirality constraints are compatible in this spacetime dimension, unlike the familiar four-dimensional case.

Because the present ${\gamma }_0$ is anti-Hermitian, choose the Dirac adjoint with Hermitian intertwining matrix $A=i{\gamma }_0$:This convention makes the Dirac scalar bilinear $\bar{\psi }\psi$ Hermitian. For real $m$, a Lorentz invariant mass term isTo check infinitesimal Lorentz invariance, put $K={\sigma }_3/2$. Then $K^{†}=K$ and $K^{†}A+AK=\{K,A\}=0$. Using a,Equivalently, $\delta ({\psi }_{+}^{\ast }{\psi }_{-})=\theta {\psi }_{+}^{\ast }{\psi }_{-}/2-\theta {\psi }_{+}^{\ast }{\psi }_{-}/2=0$, and similarly for its complex conjugate. Exponentiation gives $S^{†}AS=A$ and therefore ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$.

At fixed coordinates ${\mathcal{L}}_m$ has only the orbital variation of a Lorentz scalar; the spacetime measure is invariant under the connected Lorentz group, so the integrated action is invariant. The mass term pairs the two opposite chiralities; retaining only one Weyl spinor removes this Dirac mass. An equivalent spinor phase convention gives a real multiple of ${\psi }_{+}^{\ast }{\psi }_{-}+{\psi }_{-}^{\ast }{\psi }_{+}$ instead.

For the A5 root system, take simple roots ${\alpha }_i=e_i-e_{i+1}$, $i=1,\dots ,5$, in the hyperplane ${\sum }_{i=1}^6{x}_i=0$. Their squared lengths are two, adjacent simple roots have inner product $-1$, and all other pairs are orthogonal. Hence the An Dynkin diagram is the five-node single-edge chainIts Cartan matrix isThe complex Lie algebra is ${\mathfrak{sl}}_6(\mathbb{C})$; the corresponding compact special unitary group is $SU(6)$, whose Lie algebra is its compact real form.

Let $V={\mathbb{C}}^6$ be the defining fundamental representation of $SU(6)$. Its weights are $w_i=e_i-\frac{1}{6}{\sum }_{j=1}^6{e}_j$. Pairing with the simple coroots gives Dynkin labels $(w_i{)}_a={\delta }_{ia}-{\delta }_{i,a+1}$. Thus the complete list isSuccessive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation $\overline{6}\cong V^{\ast }$ is the dual representation and has the negatives of these weights. Its highest weight is $-w_6=(0,0,0,0,1)$, and its entire weight decomposition isEvery entry has weight multiplicity one.

The two-index antisymmetric representation is ${\Lambda }^{2}V=15$, of dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ and highest weight $w_1+w_2=(0,1,0,0,0)$. A weight basis is $e_i\wedge e_j$ for $i<j$, carrying weight $w_i+w_j$. The fifteen sums, all distinct and all of weight multiplicity one, areThe exclusion of $i=j$ comes from the exterior square, where $e_i\wedge e_i=0$.

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.