Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 (Symmetries, Fields and Particles)

The matrices are rotations about the third coordinate axis. Direct multiplication givesand $g(\theta {)}^{\mathsf{T}}g(\theta )=I$. Thus they form a one-parameter subgroup of the orthogonal group $O(3)$, isomorphic to the circle group.

If $X$ belongs to the Special orthogonal Lie algebra, which is also the Lie algebra of $O(3)$, then $e^{tX}\in O(3)$ near $t=0$. Differentiatingat zero gives $X^{\mathsf{T}}+X=0$. Conversely, the matrix exponential of every real skew-symmetric matrix is orthogonal. Hence

For $A\in O(3)$ and $X^{\mathsf{T}}=-X$,so ${\mathrm{Ad}}_A$ maps $\mathfrak{o}(3)$ into itself. Moreover,It also sends the identity to the identity linear map, so $A\mapsto {\mathrm{Ad}}_A$ is the Adjoint representation of a Lie group.

For a smooth Lie-group representation $D:G\to GL(V)$, define its derived representationDifferentiation makes $dD$ linear. Applying $D$ to the group commutator curveand taking the mixed derivative at $(0,0)$ givesThus $dD:\mathfrak{g}\to \mathfrak{gl}(V)$ is a Lie algebra representation.

Differentiate ${\mathrm{Ad}}_{e^{tX}}Y=e^{tX}Y{e}^{-tX}$ at zero. This gives the Adjoint representation of a Lie algebraThe Jacobi identity is exactlyso this is a Lie-algebra representation on $\mathfrak{o}(3)$.

The Exponential map of a Lie group sends $X\in \mathfrak{o}(3)$ to $e^X\in SO(3)$ and is a local diffeomorphism at zero. Every element of $SO(3)$ is an exponential of a skew-symmetric matrix, but no determinant-$-1$ element of $O(3)$ is an exponential because $\det e^X=e^{\mathrm{tr}X}=1$.

A representation $d:\mathfrak{o}(3)\to \mathfrak{gl}(V)$ always integrates uniquely to the simply connected covering group $\mathrm{Spin}(3)\simeq SU(2)$. It descends to $SO(3)$ exactly when the nontrivial element in the kernel of $SU(2)\to SO(3)$ acts trivially. Extending it further to disconnected $O(3)$ requires an additional parity operator compatible with conjugation by a reflection. Thus the Lie-algebra representation alone need not define a representation of all of $O(3)$.

Define the ladder operators $J_{\pm }=J_1\pm i{J}_2$. The SU(2) Lie algebra relations implyThus $d(J_{\pm })v_m$ has weight $m\pm 1$ when nonzero. Starting from a maximum-weight vector $v_I$, repeated lowering gives weightsThe norm formula derived from the Casimir element,shows that lowering stops precisely at $m=-I$. Therefore $2I$ is a nonnegative integer and the irreducible representation has dimension $2I+1$.

With normalized states and $J_{-}^{†}=J_{+}$,Put $n=I-m$. Iterating from the highest-weight state givesHencewhere the factorial arguments are integers because $I\pm m\in {\mathbb{Z}}_{\ge 0}$.

The transformed generators areFor example,The other two cyclic commutators work identically, soConjugation by $C$ is therefore an automorphism of a Lie algebra.

Since $C{J}_3{C}^{-1}=-J_3$, the operator $C$ sends a weight-$m$ state to a weight-$-m$ state. AlsoFor the spin-one normalization,Write $C|1,1⟩=a|1,-1⟩$. Using $C|1,0⟩=|1,0⟩$ in the first relation gives $a=-1$, and applying $C$ to the second gives the other phase. Therefore, in the stated phase convention,A simultaneous phase redefinition of the charged pion states changes both displayed signs but not their physical interchange under charge conjugation.

A Cartan subalgebra $\mathfrak{h}$ of a complex semisimple Lie algebra is a maximal commuting subalgebra consisting of semisimple elements. A nonzero functional $\alpha \in {\mathfrak{h}}^{\ast }$ is a root when its space in the root-space decompositionis nonzero. A Cartan-Weyl basis combines a basis of $\mathfrak{h}$ with root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$.

A choice of regular hyperplane divides roots into positive and negative roots. The simple roots are the positive roots that are not sums of two positive roots; they form a basis of the real root span. The Cartan matrix isup to the equivalent transposed indexing convention.

Let $\beta$ be a positive root. If it is not simple, it is a sum of two positive roots. Repeating this decomposition terminates because the height with respect to a regular positive functional strictly decreases, and it writeswith at least one $n_i>0$. Equivalently, one may repeatedly choose a simple root ${\alpha }_i$ with $(\beta ,{\alpha }_i)>0$ and use the root string to replace $\beta$ by the root $\beta -{\alpha }_i$.

For uniqueness, the simple roots are linearly independent. Therefore two such expansions have identical coefficients. Here “positive integer coefficients” must allow zero coefficients: a simple root itself has coefficient one on its own basis vector and zero on the others.

The matrix is the Cartan matrix of type A3. The positive roots areTogether with their negatives they formso $|\Phi |=12$.

In Dynkin label coordinates, subtracting ${\alpha }_i$ subtracts row $i$ of the stated Cartan matrix. Starting from $(1,0,0)$ gives the weight chainThese are the four weights of the defining representation of $A_3\cong {\mathfrak{sl}}_4$.

The highest weight $(0,1,0)$ gives the six-dimensional second exterior power of the defining representation. Adding each unordered pair of the four defining weights from part i givesThere are no multiplicities, in agreement with the assumption in the question.

For equal-length candidate roots, the required inner products are

Set A is invalid because ${\alpha }_1\cdot {\alpha }_3=-1$ rather than zero; indeed its three vectors sum to zero and are not linearly independent. Set B has all the displayed inner products and is linearly independent, so it is valid. Set C is the standard realizationof the A3 root system and is also valid.

Requirefor every $\varphi$. Expanding and cancelling $g{\partial }_{\mu }\varphi$ givesRight multiplication by $g^{-1}$ yields the gauge-field transformation law

The first term $g{A}_{\mu }{g}^{-1}$ lies in $\mathfrak{g}$ because the adjoint action preserves the Lie algebra. For the second, fix $x$ and consider the group curve $s\mapsto g(x+s{e}_{\mu })g(x{)}^{-1}$ through the identity. Its tangent at zero is $({\partial }_{\mu }g)g^{-1}$, so this is also in $\mathfrak{g}$. Since a Lie algebra is a vector space, $A_{\mu }^{\prime }\in \mathfrak{g}$.

Direct expansion givesSince $D_{\mu }^{\prime }(g\varphi )=g{D}_{\mu }\varphi$,Therefore the non-Abelian gauge field strength transforms covariantly:

The Killing form is invariant under the adjoint action:Together with $F_{\mu \nu }^{\prime }={\mathrm{Ad}}_g{F}_{\mu \nu }$, this immediately givesso every positive integral power ${\mathcal{L}}_1$ is gauge invariant.

The adjoint covariant derivative obeysThis follows either by substituting the transformation laws or by applying the covariance of $D_{\mu }^{\prime }$ to an adjoint-valued field. A second use of invariance of the Killing form giveswhich proves gauge invariance of ${\mathcal{L}}_2$.