Past exam of the groups and representations course of the University of Oxford / 2024 / 2

Use the mathematical Lie algebra convention $U(t)=I+tX+O(t^2)$. Differentiating $U^{†}U=I$ and $\det U=1$ at $t=0$ gives $X^{†}=-X$ and $\mathrm{tr}X=0$. Conversely, for such an $X$, the matrix exponential is unitary and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. ThusThe four diagonal entries are purely imaginary, and the six entries above the diagonal are arbitrary complex numbers determining the entries below it. This gives $4+2\cdot 6=16$ real parameters before imposing the one real trace constraint, so ${\dim }_{\mathbb{R}}\mathfrak{su}(4)=15$.

A Cartan subalgebra of this compact Lie algebra is a maximal commuting Lie subalgebra; chooseA simple basis is $i{H}_1,i{H}_2,i{H}_3$, whereAny matrix commuting with every diagonal element is itself diagonal, which proves maximality. Commuting skew-Hermitian matrices can also be simultaneously diagonalized, so no larger commuting Lie subalgebra is possible. ConsequentlyThe Hermitian quantum generator convention instead uses the Hermitian $H_a$ and $U=\exp (i{\theta }^a{H}_a)$. The two conventions describe the same Lie algebra.

Use the Hermitian $H_a$ from a to record real weight coordinates: $H_a{e}_j=(w_j{)}_a{e}_j$. Reading off the diagonal entries givesThese are weights in the simple-coroot coordinate system, not coordinates in an orthonormal basis. In the skew-Hermitian convention they mean $(i{H}_a)e_j=i(w_j{)}_a{e}_j$. Intrinsically, the weight on $i\mathrm{diag}(h_1,h_2,h_3,h_4)$ is the functional $h\mapsto h_j$, with ${\sum }_j{h}_j=0$.

The conjugate representation acts by $U^{\ast }$, so a diagonal transformation with entries $e^{i{h}_j}$ becomes one with entries $e^{-i{h}_j}$. Thus its four weights areAll have multiplicity one. In both representations the sum of the weights is zero, consistent with the vanishing trace of every Lie algebra generator.

Write $[r_1,r_2,\dots ]$ for a Young diagram with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the fundamental representation to another single box gives a row or a column:The corresponding symmetric tensors and antisymmetric tensors have dimensions $4\cdot 5/2=10$ and $4\cdot 3/2=6$, soFor the conjugate representation, the invariant volume tensor identifies $\bar{4}\cong {\Lambda }^{3}4$, represented by a column of three boxes. Adding one box givesA column of four boxes is the determinant representation, which is trivial on $SU(4)$. The remaining Young diagram has Dynkin labels $(1,0,1)$ and dimension fifteen. For example the hook-content formula givesEquivalently, a mixed tensor $M^i{}_j$ transforms by $M\mapsto UM{U}^{†}$ and splits into its trace and traceless matrix parts. Thereforewhere $15$ is the Adjoint representation.

The embedded $SU(3)$ acts on the first three coordinates and fixes the fourth. This immediately gives the branching rulesDistribute the tensor product of group representations over these direct sums and split $3\otimes \bar{3}=8\oplus 1$ by taking the trace:One can locate every summand using the block matrixThe traceless part of $A$ is $8$, $b$ is $3$, $c$ is $\bar{3}$, and $\mathrm{tr}A,d$ are independent trivial representations. Removing the overall trace instead gives $15\downarrow SU(3)=8\oplus 3\oplus \bar{3}\oplus 1$. The dimension check is $16=8+3+3+1+1$.

A meson has quark model flavour content $q_i{\bar{q}}^j$, so four quark flavours give sixteen flavour states inThis counts a flavour set for a specified spin and orbital assignment; it does not count every excited meson. Three quark flavours instead give the familiar nine states $3\otimes \bar{3}=8\oplus 1$.

Taking the fourth quark to be the charm quark, the first nine combinations $q_{\alpha }{\bar{q}}^{\beta }$, with $\alpha ,\beta \in \{u,d,s\}$, are the light sector. The extra seven are $q_{\alpha }\bar{c}$ in $3$, $c{\bar{q}}^{\alpha }$ in $\bar{3}$, and $c\bar{c}$ in $1$ under light-flavour $SU(3)$. For the lowest pseudoscalar mesons, the open-charm states are the three D meson/Ds meson states and their antiparticles, and the hidden-charm state is the eta c meson. The two neutral trivial representation states can mix when the flavour symmetry is broken.

The intended limitation is the accuracy of flavour symmetry: the charm quark is much heavier than the three light quarks, so a quark mass term is far from $SU(4)$ invariant. Exact flavour $SU(4)$ would relate light and charmed states with the same spin, contrary to their large observed mass splittings. However, saying that a four-flavour quark model is never considered would be too strong: flavour $SU(4)$ is used as classification bookkeeping, with much stronger symmetry breaking than light-flavour $SU(3)$.^[ref] This is a flavour special unitary group, distinct from the three-colour gauge group of Quantum chromodynamics.