Past exam of the groups and representations course of the University of Oxford / 2025 / 2

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.